When upgrading the hardware on a computer, a lot of pre-planning is required before doing the actual physical upgrading task itself.

First, you have to know why you’re upgrading in order to know what particular components to upgrade. – For example; if your graphics are insubstantial; that is to say the screen appears to stick when you scroll and scroll in a chunky manner for instance, or you’re unable to get a proper frame-rate on a particular game – yet you know that your processor and RAM are well above the recommended system specs suggested for playing the game, then you’ll be looking to upgrade the graphics card., or to fit a graphics card that will take over from the motherboard’s onboard graphics capabilities.

There are several main components inside a computer which you can upgrade: -

Motherboard

Graphics card

Hard-drive(s)

Power supply unit (PSU)

Random Access Memory (RAM)

Processor (CPU)

One or two of the above may have to be upgraded when other main components are upgraded in order to maintain component compatibility. – For example; if you were upgrading the motherboard, unless the new motherboard takes the same processor and/or same processor socket designation as the previous one, then a new, compatible processor is called for. The same goes with the RAM: Unless the previous RAM sticks will fit into the slots and have the same slot designation as the RAM slots on the new motherboard, then new RAM is also called for. Sometimes a new power supply unit may also be called for too.

When upgrading a motherboard; which is about the biggest upgrade task you can do on a computer, you’ll probably want to change the processor and RAM for something faster that performs better anyway: There doesn’t seem much point in upgrading a motherboard just to have it bogged down by the CPU and RAM.

A well-documented way to increase a computer’s performance is to increase the amount of RAM. This can be accomplished by adding sticks, and/or by replacing existing sticks with new sticks with more capacity on them.

Replacing the processor with a faster and better-specced model of the same socket designation may also be a performance aid in some cases. (See this article, for instance.) Ideally, though, a processor upgrade should normally always be a part of a motherboard upgrade in my opinion.

Hard-drives are another thing you might upgrade: Either adding one or more, or replacing one or more. – You might just want more room to store data; and therefore might, for instance, change a 500GB HD for a 1TB HD, or even just add a 1TB HD as an extra storage drive. Perhaps your old hard-drive is making unsavoury noises, or is experiencing a recent glut of bad sectors, and you feel that it’s time to replace it.

Power supply units wear out and require replacing at times too. Alternatively you might require more wattage for the new graphics card that you’ve just installed, and therefore the upgrade is merely a matter of course.

I once described a computer as an adult Lego set; with wires, electricity, and data added to make it a little more difficult. Once you master how it fits together and at least the basics of how it works, you’ll find that it’s fairly easy to do your own upgrades, and save yourself time and a lot of money in the process. – It really isn’t as difficult as you may imagine: All you need is some basic electronics knowledge, general software knowledge, a static-free work area, take proper precautions, a small toolkit, a steady hand and relatively good hand-eye-coordination, and you’re away: You can upgrade your own computer(s), upgrade other people’s too for a fee, build new computers for yourself and others – The world’s your oyster.

There’s not a lot of money in building & upgrading computers currently, unless you do it en masse; but it’s a useful talent to have all the same, even if only for your own purposes: Whenever your PC stops working you can just usually fix it on the spot with a minimum of effort and cost, compared to calling a geek in to umm and arr, take it away for a few days, and charge you a small fortune when they return it in a working condition. Also, whenever you feel that you could do with a performance-increase in whatever respect, you can just plan whatever upgrade you need, buy the parts, and do it whenever it’s convenient to do so, all at a cost decided to the penny by you – in that you yourself are in total control over your spending, you decide what components you’ll use, what quality of components to utilise, everything.

Have you ever built, repaired, or upgraded a computer? Share your experiences.

At times you may find that you have a supply of DC (Direct Current) electricity that is too high a voltage for the circuit that you’re intending to power: While changing the value of resistors in the circuit in line with Ohm’s Law can be the answer to this problem; there is a simpler answer which, especially in the case of a large circuit requiring the supply of power, will cost less and use a lot less components: -

In a situation where you have, for instance, a 12 volt DC supply, and the circuit you intend to power has an optimal working voltage of 5 volts DC, you can add a little circuit in between the 12 volt supply rails from the power supply unit and the circuit’s + and – rails.

The active component in this little circuit, consisting of an integrated-circuit and two capacitors, is a voltage-regulator IC. This device is available in a number of different builds for the purpose of supplying different currents and voltages: from a few volts at only 10mA, (Milliamperes) right up to several tens of volts at 20 or more amperes.

The circuit diagram above shows this very simple circuit: Capacitor, C1, is a 10 microfarad electrolytic capacitor with a suitable working voltage-rating for the DC input voltage applied across the input terminals on the left of the circuit: In our example above this is 12 volts. – So C1 should, in that case, have a working voltage of between 15 and 25 volts, to be on the safe side. Capacitor C2 is any 0.1 microfarad (100 nanofarad.) capacitor, connected across the output terminals of the circuit.

The voltage regulator IC itself should be of the correct voltage output and should be able to handle the necessary current load flowing through it: Therefore if, in this case, the circuit you want to supply with 5 volts DC draws 450 milliamperes of current at peak load; you would be able to get away with a 5 volt, 500mA regulator.

- That’s how it’s done; ‘simple as that: There is one main limiting factor though; and that is that the input voltage must always be at least 3 volts higher than the required output voltage. This is the case with any regulator IC. – Therefore if you wanted the output voltage to be 10 volts, you’d fit a 10 volt regulator and supply it with at least 13 volts.

The circuit doesn’t work the other way round: So you can’t input 5 volts to the output and expect 12 volts to appear at the input. Conversely, there is no such device that will increase a DC voltage above the input voltage.

(– Don’t get me wrong here: Such devices do exist; but they work by sacrificing input current for the sake of output voltage. (I mean that, with such a type of device, which is a totally different and much more complex device than the voltage regulator that this article is about; you could input, say, 10 volts at 1 amp and get 12 volts at 100mA at the output… But that’s another article’s worth maybe.))

Mathematically; the function can be expressed as:

Vout = Vout(IC1) = Vin + =/> 3V

In plain English: The output voltage is equal to the designated output voltage of IC1, which is equal to the input voltage plus 3 or more volts.

If a certain amount of current is required at the output, then two conditions must be met: As I said earlier; IC1 must be rated to be able to handle that amount of current, and the required amount of current must also be present at the input, plus a few tens of milliamperes more, as the device itself will use some current in its operation.

It’s a simple little circuit which, provided that the current required at the output isn’t huge (By huge I mean 5A or more.), can be built in minutes on a 5 tag piece of tagstrip with a missed-tag between the poles of the capacitors, or by utilising 3 connector-blocks if you can’t get hold of a soldering-iron at the time. The total cost would be less than £2.00UKP including components, tagstrip, and solder used. If you want to encase it in a case of its own then that’ll add to the cost. – Maybe you could fit it inside the case of the circuit that you’d like to power through it?

‘Comments invited.

Although this is a bit of a strange subject; it is nevertheless something which I’ve been mulling over in my mind for a number of years.

When a computer is booted up after a significant period of time since it was last used; it and all of its parts are , at that infinitessimately  (Did I spell that right?) minute moment in time, operating at room temperature: -

So yes; in essence a computer does need to “warm up”: The result of electricity flowing through its parts, causing the millions of microscopic transistors to switch at incredibly high-frequencies is heat. – That heat has to go somewhere, and it does: It flows through the individual component’s casing and most of it is caught by any type of heatsink/cooler attached to the component and radiated into the air inside the case, to be extracted from said case by a case fan.

-The question is: How long does it take for a computer to warm up?

By asking that question, I am, in reality, asking “How long does it take for the active and inactive sub-components; by which I mean the individual transistors within an integrated circuit or “chip”, as well as the passive components such as resistors, capacitors, and inductors, to reach their optimal operating temperature? What I’m not asking is “How long does it take for the entire computer or the major components of a computer to get as hot as they normally do.”

You see; by the time everything is hot inside a computer, it means the heat from the operations of the individual transistors and the electricity passing through them has permeated the casing of the component of which they are a sub-component. For instance; in a situation where the upper surface of an in-use soldered-in component consisting of thousands of transistors in a package; such as the SATA controller chip, gets hot to the touch, the actual working components of that chip have become hot and reached their thermal maximum comparatively long before the heat permeated through the chip’s casing.

- So the working transistors become hot long before the heat is actually noticeable by the observer. What about the passive components? ‘Much the same: Although they don’t actually actively switch the electricity, they nevertheless do have electricity flowing into and/or through them; which produces heat.

While it is true that the cumulative heat from an individual component plus any effects from nearby components would cause the temperature of an individual component to rise slightly beyond the thermal effect caused by the operations within; the overall effect is a uniform distribution of heat due to the design of the motherboard, and thus we can basically discount this for the purposes of this article.

Why write this article?

When I was at college, my tutor was around retirement age, and although a fully-qualified and experienced electronics engineer, he grew up in the days of valves. (Vacuum-tubes.) He used to allow his computer to “warm up” before using it. He said that it takes a few minutes to reach its working temperature. I didn’t dispute this at the time; nevertheless I do now to some extent.

Valves are a totally different commodity to transistors; even though their operation is similar; at least to some degree: Whereas the flow of current between collector and emitter, or source and drain, in a transistor, is regulated by a voltage on the base or gate; the flow of current between anode and cathode in a valve is regulated by a grid, or series of grids. There ends the similarity.

Valves need to be hot, and for all their sub-components to reach a thermal equilibrium within their glass envelope before they can work optimally, partially due to their massive size in comparison with that of a transistor. (This is achieved by what is known as a heater: It’s a heating element powered by a separate, usually AC, voltage, which is wound inside the cathode at the centre of the valve itself, and it literally heats the cathode to a point where any surplus electrons applied to it fly off into the vacuum contained within the valve’s glass envelope, being regulated by the grid on the way, and hitting the anode. To achieve this effect the voltage potential between cathode and anode must be in the hundreds of volts.) Transistors, on the other hand, work straight away. Heat is not needed; but is nevertheless produced by their operation.

- So valves; especially in the case of audio valves, need to warm up before they reach their full potential. if you ever have listened to or you ever get the chance to listen to the sound output from a valve amplifier, you’ll notice that the fidelity of the sound changes with time: The more time goes by the less the change, as the valve(s) gets nearer maximum operating temperature and it/their electrical characteristics change as a result.

The same is true, to a very limited extent, with transistors; but it’s hardly noticeable as it happens so fast: Mainly within a fraction of a second, and tails off within the first couple of seconds. Why? Because a transistor is minute compared to a valve, and therefore it heats up and reaches maximum operating temperature much faster. Also it needs no assistance from a separate heater to begin working: It just works, and any change in its electrical characteristics  due to heat occurs almost instantaneously.

To summarise…

In answer to the original question, then, I would say that a computer does need to warm up; but only for a couple of seconds at most. By the time the BIOS has kicked in, all the active components (transistors and diodes mainly.) have reached their optimum thermal state. By the time the BIOS has finished its initial work, after switching on, and the operating system’s ready to start loading up, even the passive components (Resistors, capacitors, and inductors mainly.) have likewise reached their potential operating temperature.

What do you think? Do you agree or disagree? Have you anything to add? Please do comment below. – I’m very much open to debate on this subject.

Earlier, in the article How to Change Alternating Current into Direct Current and Supply DC Electronic Circuitry With Power, I wrote about supplying power to electronic equipment by transforming AC household mains electricity into a low voltage stabilised DC supply with which to run the circuitry incorporated in the device. In that article I mentioned the transformer – which transforms the high-voltage AC into low-voltage AC current, the rectifier – which changes AC electricity into a crude type of DC current, and the smoothing capacitors/inductors plus the voltage regulator chips – which stabilise and regulate the waveform before the final product is outputted to the electronic circuitry which it powers.

What I touched upon there was a single-supply, with a negative and a positive output: Straight DC, no frills. Sometimes; however, more than one supply of power is needed in order to power more than a single circuit, and/or to maintain separation between two parts of a single circuit; such as in a stereo amplifier, for instance: if both channels were supplied by exactly the same power feed then there would be a loss of separation and significant bleedover between the two channels. To avoid this a dual balanced supply is used.

A dual balanced supply is, basically, a mirrored supply: Think of the zero-volt or ground line as the mirror. Since the mirror is reflecting +V and 0V, then the reflection is the negative of that; supplying 0V and –V. To make this a little clearer, let’s say that +V = +12V and –V = –12V: Across the extremities, then, from +12V to –12V, exists a potential difference of 24 volts. – Across +12V and 0V exists a potential difference of only 12V, as well as across 0V and –12V. There are in effect two 12 volt supplies OR a single 24-volt supply; depending upon whichever way you choose to look at it.

The simplest type of dual balanced supply can be made from two batteries connected in series. The 0V tap is run from the join between the two batteries: -

Each battery supplies six volts; therefore the potential across +6V and –6V will be 12V. Between +6V and 0v, as well as, separately, between 0V and –6V, the potential will be only 6V respectively.

Moving on to bigger things; a dual balanced supply (Below) can be extracted from normal household mains electricity too. – Using the principles I mentioned in the other article, but mirrored, like I was saying above: -

Notice the two regulated supplies are series-connected, just as with the batteries.

In the diagram below you’ll see an example of how a dual balanced supply powers a stereo amplifier; maintaining channel separation, which I talked about above.

Well I’ve not gone into gargantuan detail by any means: Nevertheless I hope that this article makes the issue a little clearer as to what a dual balanced supply is, as well as its usage.

A capacitor is an electronic component, of which the primary purpose is to store an electrical charge electrostatically between two plates. Since a charged capacitor will block the passage of DC, (Direct Current.) capacitors are sometimes used as decoupling components where only the flow of AC (Alternating Current.) electricity is desired.

The many different types of capacitor are named by the type of insulation, known as the dielectric, used between their plates. This insulating material has a large bearing upon the electrical characteristics of the individual type of capacitor.

Electrolytic and tantalum capacitors are types of capacitor known as ‘polarised capacitors’. This means that they have to be connected into a circuit observing the correct polarity. Failure to do so may result in anything from the component’s eventual failure to its explosion.

A capacitor’s capacitance is measured in Farads, named after the discoverer of electrical capacitance; Michael Faraday:

Microfarad (uF, MFD) = 0.000001 of a Farad or 1 x 10 to the minus six of a Farad.

Nanofarad (nF)           = 0.000000001 of a Farad or 1 x 10 to the minus nine of a Farad.

Picofarad (pF)            = 0.000000000001 of a Farad or 1 x 10 to the minus twelve of a Farad.

- For example: 1000 pF = 1 nF = 0.001 uF.

If an individual capacitor has a voltage-rating printed on it then this is the maximum in-circuit voltage that it can operate at. Anything greater than that and the capacitor’s dielectric could rupture and become breached by an arc, causing the capacitor to explode in a worst-case scenario.

This article is unconnected to the 1980s group Frankie Goes to Hollywood.

A lot of people go out and throw their money away on alcohol, which ends up as part of the contents of the local sewer system rather quickly, on a Friday night, particularly in the UK; but I prefer to stay in and expand my mind: No, not with some new-fangled narcotic substance. – Rather; with study and experimentation in the style of geeky-relaxation:

This particular Friday night in July it was a bit damp outside anyway with somewhat lower than average night-time temperatures for the time of year, so unless I had something amazing planned at some external venue, which I didn’t, I was definitely going to stay indoors. There was some good TV on that evening too, so I’d spend a couple hours watching that and then relax in front of and away from the computer.

I was starting to get a little vacant-minded eventually, and I started to doodle. Now when most people doodle they draw physical things or swirls or patterns or shapes or something similar. When I start to doodle then I usually start drawing an electronic circuit diagram. – Honestly I kid you not. – It’s usually something fairly simple like a Hartley oscillator or a single transistor emitter-follower output stage; but very occasionally I become fully alert while I’m doing it, realise what I’m drawing, and suddenly an idea pops into my head from which I develop something else or it takes me onto another level mentally.

This Friday night was one such event: I’d been contemplating the Darlington transistor in a kind of semi-conscious state, and went on to remark to myself inside my head on the surprising number of hits I’d had on my article regarding a Darlington-pair amplifier circuit. Still in a dreamlike state I put that thought on hold and went on to imagine ways to mix a timebase signal with a direct current to produce an alternating current using a matched pair of bipolar power transistors. – That’s when I realised that I was doodling again; and I’d started to draw a matched pair of bipolar Darlington transistors configured as a high-gain audio amplifier.

I recoiled a little with a start: That was something I’d never thought of before, despite the concept staring me in the face. I thought it might be worth taking further while the idea was fresh in my mind. I started consciously working further on what had been my doodle: I added extra decoupling to the ground points, controlled variable simultaneous negative feedback across both Darlington pairs, 2 sets of potential dividers for biasing the Darlington bases separately…

After faffing about for a while and drawing a circuit diagram with so many corrections it was barely legible, I transcribed the circuit to a fresh diagram in order that it would be legible to anyone else… Then I decided to blog it.

So – fresh out of my mind, totally unrevised and untested, I present to you my idea for a single-channel monaural audio amplifier with gain controlled by means of negative feedback utilising a ganged potentiometer.

I think it’ll work, but I have no idea how well. It’s one of these ideas I draw up that I never actually build, and it remains a theoretical triumph of unstarted construction in my head to times unlimited. Here it is anyway: -

There seems to be an error in the diagram: It appears that I’ve drawn D2 the wrong way round.

If you’re qualified in electronics please feel free to criticise, critique, comment, other words starting with C; even build it and/or improve on the design if you like: ‘Your choice. (I deliberately left the circuit diagram small enough so that you could hopefully get it all in a single browser window in FireFox at a resolution of 1024 x 768 px.)

I didn’t choose any component values other than those of the 10 nanofarad capacitors across the base and emitter of Q1 and Q3: Including them like this does actually increase audio frequency response at bass frequencies. I heard about it somewhere ages ago and have actually tried it to prove that it works: It does; to a limited extent.

Having blogged that I’m now going to get a coffee and do something else. I’ll decide exactly what as I drink the coffee.

Tatty-bye for now.

In April 2009 I wrote an article on the basics of changing alternating current (AC) into direct current (DC) called “How to Change Alternating Current into Direct Current and Supply DC Electronic Circuitry With Power. . This kind of transformation (Known as rectification.) is extremely common; and is the first thing that happens (Although in much more detail with far more complicated circuits than presented therein.) to the AC Mains supply in some form or another as soon as it enters most modern electronic equipment.

Less common, though just as useful, is the inverter circuit: This circuit (The very basics of which are shown here in this article.) changes straight DC electricity (From a rechargeable lead-acid battery usually. – Because it requires quite a bit of power.) into AC electricity; similar to the AC mains power that runs your computer and pretty much everything else electrical in your house. This type of circuitry appears in units such as a UPS or Uninterruptible Power Source that you may have connected onto your computer’s power lead.

The main components of an inverter are:

1) A DC current source. – Usually a rechargeable lead-acid battery.

2) A timebase: This is a piece of circuitry known as an oscillator. It produces a very specific waveform that is an engineering pattern for the final output’s AC waveform.

3) A mixer:  (see further down.)This is the stage where up to the full-force of the DC current source’s power is mixed with the waveform produced by the timebase to produce a very-precisely-tuned low-voltage sine wave, capable of driving the transformer unit, which is the last stage, to produce a pseudo-AC mains output.

4) A transformer. (see above, 3).)

The oscillator drives the mixer, a piece of circuitry regulated by purpose-built voltage regulators that mirror the control input’s actions precisely; outputting the required sinusoidal waveform. In my representation of the mixer circuit I’ve used a pair of power-NPN/PNP transistors wired as a complimentary pair to achieve this. That is a very basic circuit which will be given to a huge distortion factor on the output. In reality, heavy-duty power-CMOS voltage and current regulator chips using power-MOSFET technologies are commonly used to give a perfect stabilised output. – I’ve used the power-NPN/PNP-transistor-pair in the example-circuit for the sake of simplicity.

The AC output from this circuit is then applied to the coils of a step-up transformer which increases the voltage to the AC mains voltage.

Some of the cheaper inverters, although able to supply the power-input needs of a computer and peripherals, are nevertheless unable to supply the perfect AC waveform required to run a fluorescent or low-energy lamp. – For example, of my 2 UPS units, only one; the more expensive of the two, is able to run a Thorn 2D sidelight and a PL-lamp desk light properly. The cheaper of the two UPS’s just causes the lights to flicker dimly. This is probably due to the quality of components used as well as the type of circuitry present in the unit.

I don’t intend to go into this further and subsequently go off-topic. I hope the above gives you some idea of how DC electricity is changed into its AC counterpart.

An electrical  resistor is an electronic component that offers resistance to electrical current. Electrical resistance is measured in ohms, symbolised by the Greek letter omega, and named after physicist Georg Simon Ohm who discovered electrical resistance hundreds of years ago. There is metric terminology involved with the term’s usage, such as 1000 ohms = 1 kilohm, 1000,000 ohms = 1 megohm, and also 1/1000th of an ohm = 1 milliohm, et al.

The measurement of a single ohm was devised such that it is the amount of resistance that drops a single volt at a single amp. (See the first part of Basic Ohms’ Law/Electronic Circuit Design – Crash Course to make things a bit clearer.)

In recent times the resistor has been miniaturised down to a tiny component that is even hard to pick up with tweezers. (The amount of electrical current that the resistor can handle as a result has also been vastly reduced; although in applications and gadgetry where such tiny resistors are used, the wattages flowing through them are indeed very tiny, sometimes amounting to only a fraction of a milliwatt. (1/1000th of a watt.)

You may have seen resistors on your motherboard even and not known they were there because they’re so tiny. There are usually quite a lot of resistors on a motherboard.

The “standard” sized resistors, built to handle 1/16th, 1/8th, 1/4, 1/2, and 1 watt, which are more clearly recognisable to the untrained eye of someone that is familiar with the sight of resistors, are all still available. You might see 1 or 3 of these on your motherboard also. There are definitely a number of them inside your power-supply unit, as well as probably a few higher rated resistors capable of handling up to 10 watts perhaps.

*Note: If you go and examine your motherboard after reading this; make sure you do so while the computer’s shut down and the power is off. Also I advise you to look but don’t touch; even if you do wear an electrically-earthed anti-static wristband. (See Protect Your Parts  and  Static is Your Enemy.)

So what exactly is a resistor? How is it made? What’s it made of?

In a word; carbon. – Carbon is the principle ingredient. Carbon, in its raw state, is a rather poor electrical conductor. The more it’s compressed the better it conducts electricity. Industry has managed to compress carbon to a point where it has virtually zero electrical resistance. – Then they powder it.

The carbon that has almost no electrical resistance is powdered into an extremely fine powder of highly pure carbon, with virtually no impurities of any kind in it. Impurities, in the form of a fine clay powder, are then added to it (Clay is non-conductive.) in pre-determined amounts in order to give it electrical resistance. The final electrical resistance in ohms of a given quantity of the carbon/clay mix will depend upon two factors. 

Those being: -

1) The ratio of carbon to clay in the “powder” mix. (The amount of impurities present.)

2) The amount of compression applied to the “powder” mix.

With the right combination of these two factors, precision resistors of any resistance from a fraction of an ohm right up to around 10 megohms can be made.

In recent times, different chemicals have been used which increase the tolerance, (i.e. the variation of the component’s actual resistance with the heat generated by itself and surrounding components, as well as drift by changing chemical composition with time, et al.) considerably right down to a factor of a tiny fraction of a percentage point. Nevertheless the basic idea that I have outlined to you is still in essence the method used to this day. The process happens on a level where much smaller components are manufactured on an extremely miniature level; thus using up far less resources but being of a highly precision and exact nature using the latest machinery to mass-produce millions of tiny resistors only 4mm long in some cases.

Then we come to the subject of microelectronics; including building resistors into chips: At the microelectronic level a different method of producing electrical resistance is used, which involves incorporating highly-doped MOSFETs connected as an always-open-circuit device at the point in the circuit where the resistance is required. The extra doping of the semiconductor material provides the required electrical resistance, rather than the “doping” of carbon powder.

Going back to the resistor as a discrete component: Once the desired resistance has been achieved, the resistive chemicals are sealed inside a container of some kind, (Usually roughly cylindrical.) or on a tiny non-conductive slab of material, with an electrical connection fitted at each end of the device. The resistance value is then marked on the component’s outer casing, or a thin non-conductive protective layer sprayed on to it and the relevant markings applied to that, usually by means of a resistor colour-code, standard to the electronics industry. After final testing the resistors are shipped to industry and retailers in their millions.

The Resistor Colour Code

Some high-wattage resistors are made by winding coils of resistive wire on a non-conductive heat-resistant tube. This avoids damage to the resistor’s internal structure due to the heat produced by the resistance to the high wattages flowing through the component.

You’ll learn something new every day here; that’s why it’s always a good idea to read this blog often. Why not subscribe to the RSS feed and mailing list, and be notified by RSS and email when new content is produced?

Something you’ll be aware of if you follow Star Trek is that the Borg say that resistance is futile. Clearly on the evidence of the above it’s quite the opposite!

I look forward to your comments.

For this article I semi-plagiarised part of an earlier piece that I wrote with regard to fitting an AMD socket AM2 processor; the reason being that the procedure is almost exactly the same, save to say that you’re installing an AMD socket AM2+ processor in a socket AM2+ socket.

(Note that I plagiarised my own work; which is acceptable: Plagiarising someone else’s work is totally unacceptable in the blogosphere, and is something that I would not do. (A few bloggers actually do this!) I may at times quote from someone else’s work, in which case I put the piece in quotation marks and give the creator credit before or afterwards. (As in the case of the quotations from Wikipedia, below.) – At the same time being fully prepared to remove their work should they request me to do so.)

AMD socket AM2+ is the successor of socket AM2, and is extremely similar save to say that it has more pin connections.

From Wikipedia: “Socket AM2+ is a mid-migration from Socket AM2 to Socket AM3 and is fully compatible with Socket AM2, so that processors designed for Socket AM2 work on Socket AM2+ motherboards and vice versa.”

From the same Wikipedia article: “AMD confirmed that AM2 processors will work on AM2+ motherboards and AM2+ processors will work on AM2 motherboards. However, the operation of AM2+ processors on AM2 motherboards will be limited to the specifications of Socket AM2 (1 GHz HyperTransport 2.0, and one power plane for both cores and the IMC). AM2 processors do not benefit from the faster HyperTransport 3.0 and separate power planes on AM2+ motherboards.

Many manufacturers have yet to (and may choose not to) release BIOS updates that would enable this compatibility. Others have simply stated that their AM2 motherboards are not compatible with AM2+ processors. ^[1]

AMD confirmed that there is an upgrade path from Socket AM2+ to AM3:

• AM3 processors work on AM2+ motherboards
• AM2+ processors do not work on AM3 motherboards”

(See here for the full Wikipedia article on socket AM2+.)

It’s not exactly brain-surgery or rocket-science to fit a socket AM2+ processor; but you’ll need to prepare yourself beforehand. Here’s what you’ll need: –

• Screwdriver; if you need to remove or loosen screws to remove the side-panel of your computer’s case.

• New processor at the ready. Leave it inside its packaging until the very last moment.

• Thermal paste. You’ll need this to paste onto the surface of the new processor after you’ve fitted it into the processor socket and before you fit the cooler onto it. If your replacement processor came boxed with a new cooler then chances are that the paste is pre-applied to the cooler. The amount pre-applied should be adequate (Some computer-builders would disagree with me here; but I’ve always found it to be so.). If on the other hand you’ve bought an OEM or a secondhand processor then you’ll need to apply thermal paste as above.

• An anti-static wristband connected to electrical earth. If you don’t have one you might be able to use a crocodile-lead and connect a piece of metal jewellery or a watch that you’re wearing to the computer’s case, provided that the computer’s case remains earthed.

• Finally a positive calm attitude: I personally find it a good idea to meditate for a short while before any computer-work in order to achieve a level psyche so that I don’t panic or do anything wrong during the operation through lack of concentration. Forget the gas bill, forget your boy or girlfriend for the time being. Ensure that you won’t be disturbed for a while, adopt a levelled and intellectual thinking-pattern, and we’re off.

Ensure that the switch on the computer’s power-supply unit (Usually situated at the back of the computer.) is off. If it is off but the power-lead remains connected then the case is earthed. If you disconnect it then the case is not earthed or only partially earthed. (The theory of “touching the case to earth yourself” may or may not apply in this instance; hence it’s always better to use an anti-static wristband just to be on the safe side.) Open the computer’s case by removing the side-panel.

On one of the sides of the cooler unit which you’ll be fitting atop the processor, you will see a lever connected to a clip on the cooler heatsink. Pull this lever to the upward position. You will see a pair of wires (Maybe three(?)), (Probably red and black (and white(?)).) leading from the cooler’s fan. These connect via the a small plug to a corresponding a socket on the motherboard. (Read your motherboard’s manual beforehand to be sure you understand where everything is. (RTFM)) If you don’t have the corresponding manual for your motherboard then either download it online, or ask the manufacturer for a copy. (They may or may not charge for it.)

On one side of the processor socket you will see a lever. Lift this lever into the upward position.

Carefully remove the socket AM2+ processor from its packaging without touching any of the pins. Insert it straight into the waiting empty socket, ensuring that the marked corners of both the processor and the socket are lined up.

Once again check that the marked corners of both the processor and the socket are lined up. Also ensure that your anti-static wristband or otherwise is connecting your body to electrical earth. Apply slight pressure to the processor’s surface and push the lever downwards into the position that you originally saw it in when you removed the cooler from the old processor. If anything resists your attempts DO NOT FORCE IT. Recheck that you are doing the operation correctly and try again.

Gently and carefully, trying not to cause any scratches, clean the upward-facing surface of the processor. (Do not wipe the markings off. – You don’t need to clean it that well!) and, if using a secondhand cooler unit, clean the underside of the cooler that will end up on top of the processor, Apply a very thin layer of thermal paste to the surface of the top of the processor and to the smooth underside of the cooler heatsink. The edge of a credit card will assist you to keep it thin. You may notice that it seems slightly thicker in some places than others: This is normal. The paste is filling the gaps that would otherwise be left free of contact between processor and cooler.

If you bought a new, boxed processor complete with a new cooler; remove the plastic cover to expose the  pre-applied thermal-paste.

Place the cooler on top of the processor and slide around a little to ensure that it’s located in exactly the right place. Locate the two clips positioned opposite one another on the sides of the cooler heatsink with the lugs on the socket surround. Push the lever above one of the clips into position; flush with the motherboard’s surface.

Connect the cooler’s fan by plugging the plug into the appropriate socket on the motherboard.

The operation as regards hardware is now complete, and the processor is now fitted. – It wasn’t that difficult was it?

If you have any comments to make then please do so below.

Ignoring the obvious jokes about AC/DC; including bipolar DC currents, bistable multivibrators, and all other possible smutty electronics terminology innuendos; in this article we’ll be taking a look at how the high-voltage Alternating Current (AC) electricity that comes from the power plug and is transported to your computer becomes the low-voltage Direct Current (DC) electricity that runs your computer.

I use your computer in the above paragraph as it’s the most obvious choice; considering the fact that you’re probably reading this article on your computer; and if not have probably printed it off from a printer attached to a computer. In fact it’s not just computers that have to change the AC mains voltage into a usable working DC voltage. Most if not all mains-powered electronic devices have to do this: Even to a certain extent CRT monitors and televisions, although these also utilise the high-voltage AC current of mains electricity in addition to stepping the voltage both up and down, as well as converting AC into DC in some of their internal circuitry. For the purpose of this article I’ll be using the example of a computer (PC) power supply unless otherwise indicated.

In comes the power lead carrying between 217 and 254 volts of AC electricity in the UK (Depending upon the time of day and the geographical location in the UK.), or around 110 volts AC in the USA. The power supply’s job is to convert that voltage into three separate DC voltages; 12 volts, 5 volts, and 3.3 volts. Due to the high wattage requirements of some of the circuitry in the computer; these supplies; particularly the 12 and 3.3 volt supplies, have to also be able to supply very large currents, measured in Amperes. (Amps.)

The relationship of electrical current (Amps) to electrical power (Watts) is defined in Joule’s Law as P = IV. (Power in watts is equal to the sum of amperes in amps multiplied by voltage in volts.) Therefore if you were to have a power supply supplying 12 volts at 12 amps; the available wattage would be 12 volts X 12 amps = 144 watts.

(This equation can also be reversed to show the inverse of this: -

I = P/V (Amperage = power in watts divided by voltage in volts.) and V = P/I Voltage = power in watts divided by current in amps.))

For the purpose of this article, we’ll ignore the large currents to the greatest possible extent; and rather we’ll concentrate on the basics of changing high-voltages into low-voltages, and AC into DC.

There are 4 main component blocks in a (single-output) power supply; those being: -

1) Transformer

A transformer is a single electrical component consisting of two or more coils of wire formed around a core of varying density depending upon the type of transformer. It works by the electromagnetic field induced in the primary coil  or input coil by an AC electric current affecting the secondary or output coil and causing a proportional electric current to flow within that coil. The ratio of the two or more AC currents in question is dependant upon the construction of the transformer itself.

Recently; in electronic equipment that requires very low electrical current; the transformer has been replaced by a high-wattage resistor/ AC potential divider circuit. This has the effect of dropping the voltage by using electrical resistances rather than the electromagnetic induction principles of a transformer. Since resistors are [usually] smaller and lighter than transformers, as well as being cheaper; this type of voltage-dropping circuit is commonly used wherever possible these days.

Its advantages are reduced cost and reduced weight. Its disadvantages are that it can only output a small current: commonly considerably less than 1 amp, also that the load on the AC mains input of the circuit is always constant and unchanging; whether or not the circuit’s output is being used to power anything.

2) Rectifier

A [bridge} rectifier typically consists of four diodes connected in a certain configuration end-to-end. (see diagram.)

The action of the bridge is to use the component diodes ([Rectifier] Diodes will only allow electricity to flow one way through them dependant upon their connected polarity.) to sort the component parts of the AC wave-cycle into positive and negative; therefore changing the oscillating AC waveform into a crude type of DC current.

3) Smoothing

The crude “DC current” output of the rectifier stage of the circuit isn’t anything like pure DC electricity: It’s very unstable and resembles its former state to some extent.

A large-value capacitor placed across its path helps to iron out the remaining inconsistencies and reduce the inherent instability somewhat.

4) Regulator

No matter how much smoothing is applied to the output of the rectifier by capacitor(s), it can never be transformed into a totally stable DC current by this method alone. A voltage-regulator IC (Integrated Circuit) is placed in-circuit at this stage to finally stabilise any residual waveform-ripple and set the exact output voltage before the current can finally be outputted to run an electronic circuit.

- So very basically; that’s how it’s done. There’s more; a huge amount more, to be learned. – But these are the very basic basics of it.

I assume that either you’ve checked your power supply’s output in one of or both of the two ways I’ve suggested earlier. Those being:

a) While the PSU’s outputs are unconnected, and also while the PSU is running under the artificial load I suggested.

OR you preferred to leave the PSU fitted and tested it: -

b) While the PSU is connected to the computer’s components in a normal way.

The question “What do I do if my power supply voltage drops Under Load?” is a bit of a multi-pronged question. I’ll try in this article to cover a number of those points in the hope that you can come to a conclusion based upon the information and guidance herein.

I’ll start by saying that, in the case of any supply of power, the voltage will always drop to some extent under sufficient load. To clarify that point somewhat, I’ll draw your attention back to Ohm’s Law and the relationship between wattage (Power), current (Amperage), voltage, and resistance (The load):

If the wattage or power (P) = the current (I) squared multiplied by the load resistance (R) in a DC circuit; then it follows that R divided by I squared = P.

If we give I the fixed value of 10 amps in this example, then we can see that as R decreases, so does P. : Translated into English, this shows that as the load increases and therefore the parallel load resistance decreases, so the amount of unused power available from the supply also decreases. Why does this occur? Because as the load increases so the resistance decreases, therefore more of the available electrical power in watts is used by the load and transformed into other kinds of energy.

We know from Ohm’s Law that V = IR (Voltage = current multiplied by resistance.). Therefore, if we use the fixed figure of 10 amps as the value for I again, we note that as the value of R (The resistance) increases then so does the value of V (The voltage.). Inversely as the (load) resistance drops as extra load is added to a DC circuit, so the voltage drops proportionally.

The higher the amount of current (I) available in the circuit; the less the decrease in R (Increasing the load.) affects the decrease of V. (Drop in voltage.)

We can prove that by increasing the value of I to 20 amps: Do the math again and you’ll see that this is true.

Conclusion: The greater the electrical load on a given supply; the greater the voltage drop on said supply.

This is always true to a given extent; no matter how much current is available.

Bearing this in mind, then; how much voltage drop is too much?

In a perfect world; any voltage drop is too much. – But as we’ve seen above; there can never be zero voltage drop unless there is zero load. As we know; if there is zero load then there is zero computer, so the point is mute. The objective of the exercise then is to minimise the voltage drop to the greatest extent possible.

In the examples above we saw that the more current we had at our disposal, the smaller the voltage drop under any given load: In Joule’s Law; I x V = P (Current multiplied by voltage = power (wattage).). We know that the voltage outputs of a computer power supply are fixed ay 12v, 5v, and 3.3v. therefore the available current (Amperage) is dependant upon the available power (Wattage): The more wattage the supply can output the more current is available; therefore the less the voltage drop will be under any given load.

Conclusion: To minimise any voltage drop under any given load conditions; make more wattage (power) accessible.

In other words; if the voltage on your supply rails drops substantially under the load that your computer’s components put on those supply rails, then your power supply isn’t outputting the required wattage.

( It is possible that a malfunctioning piece of hardware is drawing far too much current from the supply; but this is unlikely: A malfunctioning piece of hardware that draws such high currents would get very hot very quickly and would probably begin to cook, causing noticeable odur and smoke. It’s far more likely that the power supply itself is at fault if nothing else is frying itself. )

In the face of that; you have two choices. Either: -

1) Keep your existing power supply unit and decrease the load on it. (- By not running so many components: In other words removing something from the circuit, such as a hard-drive, graphics card, whatever.)

OR

2) Fit a new power supply capable of delivering the required wattage.

At this point you may say:

“I’ve checked everything and worked out that the load on whatever rails is X amps; that’s Y watts at Z volts. My power supply’s output rating on whatever rails is greater than that: Therefore something’s wrong.”

Correct. Something is indeed wrong. There is an inconsistency somewhere. The problem now is to discover exactly what that inconsistency is.

It could be three main things: -

1) The equations are wrong.

2) Your calculations have arrived at an incorrect figures.

3) The PSU ratings are wrong.

Firstly there is no chance that the equations are wrong. The equations used form the basis for modern electronics theory and have done so for more than 100 years.They are an integral part of basic electronic calculus and have been proven to be correct time and time again.

Secondly, check that your calculations are correct and that you have indeed arrived at the correct figures.

If there is no fallacy in your calculations then there is now only one possibility left – That being that the PSU ratings are wrong. How can this be?

Some manufacturers; especially the manufacturers of cheapo power supplies, overrate their product. This isn’t actually lies though. It’s an overstatement:

A PSU rated at 500 watts is indeed capable of supplying 500 watts; but not necessarily 500 watts of continuous power. In tests done by a leading UK computer magazine in the last 2 years, a number of 500 watt PSUs failed when fully loaded. One, the second cheapest under test, even detonated! There is an old saying that you get what you pay for. Sure; a $15 PSU built in China and rated at 500 watts will supply a computer that needs 300 watts to run efficiently without (m)any problems. It’s when you add a powerful graphics card that drains an extra 100 watts, plus a RAID array = another 30 watts, extra optical disk =  +10 watts, and upgrade to a more powerful processor = +35 watts that things start going askew: BSODs occur, the graphics card doesn’t perform properly, the power supply starts making strange noises…

- That’s only 475 watts total, and the PSU is rated at 500 watts, so theoretically it should work; but it doesn’t. – Because the PSU can provide 500 watts peak, – but not 500 watts continuous. – In fact probably only 400 watts or less continuous. At peaks where the power requirement is at its greatest there may be a voltage drop of over a volt causing memory failures and processor outages.

Maybe it’s not that serious in your case? Maybe you’re on the edge of the precipice; but are still noticing a voltage drop on measurement under load, even though the PSU is still holding up, the CPU’s not having outages, and the RAM is just managing on the reduced voltage?

That’s a good thing inasmuch as your computer is still working. – But it’s so close to the wire that it’ll only take a feather to tilt the balance. – So don’t expect your computer to carry on without problems much longer in those circumstances. All you need is for the PSU to get too hot, the CPU to use a few watts more, the graphics card to draw that extra piece of polygonal shading, and…

Get a better quality and / or higher-rated power supply unit: The sooner the better. Once it’s fitted, measure the voltage drop. There will still be a voltage-drop; but it’ll be smaller and your computer will be happier as a result.

Do you have any experience(s) in this kind of thing? Are you getting a comparatively large voltage drop? Is your computer underperforming due to this? Have you recently fitted a better PSU and noticed a difference? I’d be interested to hear your story in brief. Please comment.

This post is edited from an article I originally wrote in 2007, and is included herein as in a way also a re-edit of the post “Static is Your Enemy“, based upon the same source material.

Every now and again I see and / or hear, horrific things: Things like pictures of perfectly good motherboards being placed face-up on nylon carpets to photograph in the hope of selling them. Things like RAM-sticks being wrapped in bubble-wrap and popped into a plastic shopping bag. The sight of trainee-technicians combing their hair while handling processors, or a workshop junior without a care in the world trying to bend processor pins back into place with all-metal tweezers, while at the same time brushing the dust they picked up from inside the spares cupboard off their polyester garments. – Harmless activities to the layman perhaps; but fatal or potentially hazardous to the electronics. The reason – Static electricity:

In the case of all CMOS computer circuitry; anti-static precautions are a necessity at all times when handling, packing, and storing any item of computer equipment or componentry. This is even more important, in some ways, than avoiding exposure to damp and high temperatures, as damp can always be dried out before fitting and use. 9 times out of 10 there’s no lasting damage from a bit of damp: Well nothing that drying out won’t put right anyway; provided that nobody attempts to use electronic components while they’re damp.

The Technical Bit.

Computer components such as RAM sticks, processors, motherboards, graphics cards…you name it, either consist to a large extent of a combination of discrete transistors and integrated circuits or "chips", containing in some cases millions of transistors, or are themselves “chips”, as in the case of a CPU or processor. These transistors are in most cases, other than some power-controller transistors, of the MOSFET variety. : In a very basic terms and on a nanoscopic level, these consist of a microscopic layer of doped semiconductor material laid and adhered to a micro-thin silicon wafer: A tiny gate-electrode is insulated from the semiconductor material by an incredibly-thin and fragile insulating material alongside the semiconductor.

Under normal operation; the gate, being totally electrically insulated from everything else except the lead connecting to it, regulates the flow of electrons through the semiconductor material between the drain and the source connections at either end of the semiconductor material; which is how the transistor works. Due to the fragile nature of the insulating layer between the gate and the semiconductor, however, it doesn’t take much energy, in the form of electrical current, to break down the insulation between the gate and the semiconductor creating a low-current potential divider with the gate as the centre connection, i.e connected to the semiconductor through the break in the insulation, thus ruining the action and function of that individual transistor. Static electricity can build up on virtually every surface, even the human body in some cases, to a potential of thousands, sometimes millions of volts, and at currents greater than the insulated gate of a MOSFET is capable of withstanding. When these charges are applied to any type of MOSFET circuitry, usually without the culprit realising that they are even present, the obvious occurs; the transistor(s) break down due to the insulating layer depleting or being arced through due to the current of the static-charge, and in an instant the device is rendered inoperative, dead, ruined, broken, kaput, finito, had it, shagged, destroyed, fried…

Some people might at this point be of the opinion that with millions of transistors it wouldn’t hurt if one or two don’t work: After all people don’t die if a few cells in their body die, or they injure themselves slightly.

There are a different set of conditions affecting either though. The human body and the electronic circuit are totally different in many respects: The human body can replace dead cells in days, and can bypass the function of dead cells until new ones are grown by its automatic-repair process, at least to a certain extent. With an electronic circuit if a transistor dies then the function in the pathway of a particular electron flow is rendered inoperative and the device malfunctions, in some cases triggering a chain-reaction in which many other transistors along or connected to that path also die; and when a transistor’s dead there’s no magical resurrection, no afterlife or reincarnation.

Destruction of the component can happen naturally with the age of the component; causing it to break down with usage; but it’s relatively rare in modern electronics. A static charge, however, can "fry" a device; literally causing a micro-detonation of the active components within the device, rendering it totally useless for anything other than melting down and making furniture out of.

How to Avoid Damage

Bearing the above in mind, what needs to be done to prevent this involuntary micro-vandalism from happening?

Stringent anti-static precautions should be adhered to at all times, rigorously, when handling CMOS and MOSFET circuitry. – Which encompasses nearly all circuitry within a computer, except for some parts of the Power Supply Unit.

Anti static precautions are aimed at preventing static electrical charge from reaching a device, or, in many cases, from building up in the first place. – But it’s not quite as simple as it sounds: People seem to think that since polythene is a good electrical insulator, then it’ll protect components from being exposed to static electricity.

WRONG. Polythene is one of static electricity’s favourite places to lurk. A charge builds up easily on any polythene surface by means of friction with another material. Nylon also is an excellent static-capacitor, as is your carpet, your leather sofa, and most of your clothes.

“What if I pack electronics naked?” You ask.

Your naked body is a conductor of static electricity, from the carpet, your sofa, your dress, straight into the nearest transistor, and you needn’t necessarily feel a shock either. – To avoid damaging circuitry you need to keep all possibility of static discharge well away from it.

Ground Thyself

An anti-static wristband should be worn at all times when handling semiconductors or semiconductor-based circuitry; CMOS, MOSFET, whatever. Go into any electronics lab, even at college, and you’ll see everyone wearing one on their wrist. This is providing any static charge that comes into contact with their body with a path to electrical earth; the quickest path to destination, which is the path all electricity will take in all cases.

All handling of electronic circuitry and components should be within a static-proof environment. If a static electrical charge contacts the component just once then it’s fried. – End of story. Always always always pack electronics in an anti-static bag. NEVER pack them in polystyrene or polythene. Always wear an anti-static wristband when touching a circuit-board or "card"; preferably on the wrist attached to the hand with which you are holding it. NEVER allow the component to come into contact with the carpet or any manmade fibre: Even if it did happen to be made by a woman. (Avoid carpets in the packing/handling environment if at all possible.) a polythene, polycarbonate, polystyrene, poly methyl methacrylate, poly-whatever surface, and don’t allow it into a strong electromagnetic or electrostatic field. (Microwave, electrical transformer, close to a television screen, etc.) Avoid touching any exposed metal part or component on the board/card if at all possible.

Electronics are fragile. – That’s one of the reasons they’re shielded and kept away from contact with anything else wherever possible. There’s a general rule these days that newer the part the more static-sensitive it is; notwithstanding any extra built-in durability and protection.

Take care and always use proper anti-static precautions; otherwise it could cost you a fortune in parts.

In the first article in this series, we looked at measuring the voltages while the PSU had no load attached to it. In this article we’re going to give it a number of loads to run. We’ll put a total load of around 364 Watts on the unit. If the unit is rated below that figure then it’ll be fully loaded. If the unit is rated above that figure then it’ll still be fairly heavily loaded.

We’ll load the unit using five small circuits which I’ve set out in circuit-diagrammatical form below. The smaller circuit, consisting of a bulb, a resistor, and a light-emitting diode, needs to be built four times, the larger one, consisting of six bulbs, seven light-emitting diodes, and seven resistors, only once.

Why bulbs, resistors, and LEDs? The resistors are in series with the LEDs to prevent them from overloading. The LEDs are to indicate that the 5V lines are all working; also that the PSU is lighting the Power OK LED as it should. We’re not loading the 5V lines much because it’s ideally the 3V3 and the 12V lines that we need to test under load; because they’re the ones through which the main wattage is drawn from a modern PSU by modern computers.

Which leaves the bulbs: It’s the bulbs that provide the heavy loading. Incandescent tungsten-filament bulbs are notorious these days for using too much wattage and producing too little light and too much heat. Were intending to use those negative properties of them to “waste” the PSUs output. Well be buying products from the automotive industry too, which might help it on its way to recovery; as the bulbs we’re using are car headlamp and tail light bulbs. They are made to work on a voltage of 12 Volts; which means they could be bright and very hot when in operation. – Especially so with the 80 Watt bulbs in the 4 small circuits that test the 12V lines. When you are building the circuits please bear this in mind, as it is possible that looking directly at them for more than a second could result in at least temporary blindness. Also touching them during or shortly after operation could result in severe burns, and if they come into contact with flammable objects or materials they could start a fire. Be warned, and design appropriate safety parameters into the housing for the bulbs when you design the test units built from the circuit-diagrams provided.

Ok so let’s take a look at the first circuit diagram: -

This circuit is built around the P1 motherboard connector socket. The pin-out numbers used correspond to the P1 plug and not the socket, so please bear that in mind. 4 of the 24-Watt car-bulbs load the 4 x 3V3 lines to almost 7.273 Amps each, total almost 29.1 Amps; that’s 96 Watts at 3V3. – The same as the average power-drain from a processor. (These particular bulbs aren’t expected to get particularly bright; but nevertheless could get rather hot, so beware.) The other 2 bulbs provide load of 2 Amps on each of the 12V lines: Total 48 watts. (These two bulbs WILL get both bright and hot. Bear this in mind when designing the circuit layout and component housing.)

The next circuit is fairly simple, and is built around a 4-pin Molex socket. You will need to build four identical of these circuits. Please do bear in mind that the 80 Watt bulb is burning 6 and 2/3rds of an amp: That’s a lot of current, a fair amount of power, and a lot of energy. Energy doesn’t vanish. – Rather it radiates and it changes form> – In this case into heat and light. The bulb will be bright and very hot. (You might even be able to light a cigarette from it given time?) Be careful to protect eyes from it, and beware of the fire risk posed.

Note once again that the pin-out numbers correspond to the peripheral power “Molex” plug and not the socket.

These four bulbs; one on each circuit, will load the 12V lines to another 320 Watts in total; making a grand total of 320 + 96 + 48 Watts load = 364 watts; plus a couple of Watts loading by the LEDs with their series resistors.

What should happen?

When you build and connect the circuits to a known-working PSU; the following should happen:

All the LEDs should light up. If they do it shows that all the 5-volt lines are working and that the PSU is in better health than not. The 80W bulbs will produce a lot of light and heat. All but 2 of the 24W bulbs will be rather dim; but could still get quite hot.

What do these circuits accomplish?

They put a reasonable load on the PSU, so that you can test its performance under near-normal working-conditions.

You could check the output voltages again under load to see if there is any significant voltage-drop under such conditions.

Another thing to bear in mind: -

It is well documented that when some of the cheap and nasty PSUs are fully loaded to specification they can fail or even explode. Therefore if you happen to be testing out £10 worth of cheapo Chinese unknown-label 250 Watt PSU for example; expect it to explode under the full-load that these circuits will give it. If it doesn’t and it still works afterwards then you were lucky. A good-quality 250 Watt PSU will strain and maybe complain in some way, but will easily survive a minute or so of full load.

Yes the above is a bit Heath-Robinson I admit. – But if you’re on a budget you really don’t want to shell out for the proper professional test-loads that can cost anything up to a small fortune. This is the scrub-round-it approach to the matter. – And it works; so why knock it? (Just keep an eye on; or should I say keep an eye off, those hot bright bulbs.)

Now that we’ve loaded it we can do the same measurements that we did in Part 1. The two readings will be slightly different. The difference corresponds to the voltage-drop under load. This amount will depend upon the power-rating of the PSU as well as the quality of the unit.

Ideally the voltage-drop should be less than 5% of the voltage-rating of that rail. If it’s any more then the unit is either wearing out or wasn’t particularly good quality to begin with. If it’s over 10% then replace the unit.

  I have, in my arsenal, a power-supply tester: All I have to do is connect the 20 or 24-pin connector to the tester; and if all the green lights light up then the PSU is supposedly working. Is this right? Yes and no:

The power to the motherboard, not including the dedicated processor supply, is working if all the green lights light up. What it’s not testing is the power supplied to the other connectors; such as the dedicated CPU supply, (The ATX12V or the EPS12V connector.) the 4-pin Peripheral connectors, (AKA Molex plugs) the 4-pin Berg connector(s), the Serial ATA power connector(s), or any others. That renders my tester fairly useless; as if I’m not getting power to the PCIe graphics card; my tester could still read that there’s no problem.

- So I need a way of testing each individual output, individually, to see if the unit is working as it should.

A voltmeter seems a good place to start. – And where’s a good place to find one of those? On a multitester or multimeter no doubt.

OK so I have the (ATX) power-supply that I want to test on my test-bench in front of me, all wired up to the mains power and switched on. – But there’s absolutely no response – or possibly only the 5V rail is powered on. Why?

On an ATX power supply there’s a wire which goes to the motherboard which allows certain pieces of hardware, including the “Power on” button on the front of the case to bring the power-supply out of standby. Some PSUs go to +5V standby, where only the 5V rail remains powered up, others switch everything off. (Some of the much older PSUs would go to +12V standby; but not any recent models built within the last 5 years.)

That wire is the green wire on the P1 connector. (The 20 / 24-pin connector that plugs in to the motherboard.) To fire up the PSU it needs to be grounded. Fortunately the wire either side of it is a ground wire; so all you need to do is short out either way on the connector, (Short green and black.) using a short piece of wire or a paper-clip or something, and voila: Power-up has been achieved.

If it doesn’t happen; connect a DC voltmeter across green and black: Positive to green and negative to black. You should read 5V. If you don’t get a reading, or you read anything less than 2V, the PSU is faulty. Bin it.

*Power supplies can be fixed. I fixed one myself by cannibalising spare parts from 2 identical dead units. The drawbacks are:

1) It’s dangerous: there are extremely high voltages present in the circuitry when the power’s on.

2) You need to know what you’re doing; which can take up to 4 years training to achieve.

3) It’s very time-consuming: Even if you know what you’re doing; it can take hours to trace and diagnose multiple faults and repair them.

It’s easier and usually more cost-effective to bin the faulty unit and buy a new one.

Now, having powered up; check the voltages of all the connectors.

The pinouts are as follows:

(Image above from Wikipedia.org)

4-pin Peripheral

4-pin Berg

6-PIN AUX

ATX 4-pin Molex 12V P4:

Serial ATA

- And that’s how to check the voltages.

The above proceedure doesn’t test the PSU under load. In the followup article I’ll show you how to test the PSU’s voltages under full load; including a circuit that’ll fully load any PSU up to 350W

In a new technological development using principles based upon a derivation of the principles of power distribution which were originally conjectured in the early 20th century by Nikola Tesla; Korean company Zalman Tech Co have succeeded in the invention and prototype building of the world’s first wireless computer power supply unit.

The unit, which is the same size as the normal computer power supply unit, is revolutionary in that it allows the distribution of power throughout the computer without requiring the mass of wires that normally protrude from the back of such a unit, and are routed throughout the case’s interior to the various components. This can result in a disruption of the airflow in the case; especially in the upper regions where the cables are densest, causing heat to build up in certain upper areas, thus reducing the life of some components considerably.

The internal layout of a working computer’s case can now be much less cluttered by the profusion of power leads which can cause the inside of a computer to look untidy in addition to restricting airflow.

Based upon an original idea of wireless power distribution envisaged by Nikola Tesla, combining his polyphasic power distribution techniques with a major development known as short-burst DC induction; believed to be based upon the common AC induction principle but using a half-wave directed DC semi-square-wave pulse created by using the electromagnetic induction of electro-motive force in a circuit which they described as a kind of digital-inductor. The pulse is then somehow directed at a number of tiny receiving antenna affixed to individual components and power receptors on the motherboard.

According to Project Manager and Chief Development Engineer Hung Baiderbals; the new technology does have a little way to go yet before it is released into the wild, as the power-efficiency of the EMF-transducer-coil as yet leaves a lot of room for improvement, as does the amount of energy lost in both the wave separation and partial rectification at the point of origin in the unit itself, and also that energy radiated as heat by the receptors. Nevertheless; the cumulative effect of the EMF pulses at a frequency of some 200MHz, combined with the re-absorption of unused distributed energy by a device which Zalman are seemingly unwilling to comment further on, other than that it is capable of re-absorbing up to 86% of radiated energy that would otherwise be wasted, all in all give the device an overall efficiency of 91% according to the figures claimed by the development team.

Even when pressed, Baiderbals wouldn’t give a definite date for when the unit was expected to reach full development, which would mean the start of marketing it. All he would say when hassled for an answer is that we can expect to see it in production within the next year.

Although Zalman revealed sketchy details about some of the working principles of some parts of the unit, they wouldn’t be drawn any further as to the way in which it works; saying that at this stage of its development they wanted to remain tight-lipped about the technology.

The new concepts being devised in the field of technology lately are quite amazing. This idea goes to show that the ideas and dreams of our forefathers and the original pioneers of the technologies that we today take for granted do indeed have substance; and that, given progress in the light of scientific advancement, ways of utilising such techniques which once seemed destined to only be used in science-fiction stories are today becoming a reality.

It’s a bit difficult to know what to ask for in the form of comments on this subject. In the light of that I’ll just invite you to comment if you would like to do so.

If you happen to have any further information or insider information that you’re willing to share then I invite you to do so below.

This article is the first in a yearly series of April Fool’s messages here on kkomp.com.

Above: An abortion of a mess of untidy wiring.

A number of years ago a friend gave me an old computer that had “just died” in part-exchange for one of mine (Not my own build.) that I was selling. She said that her husband had been tinkering with it and had upgraded the graphics card. (He works at IBM; but in the office department, not the practical computer maintenance department.) It worked after he had been fiddling with it – Until it had been moved to a different location; after which it had been totally dead from then onwards.

On opening it up the reason became apparent: All the unconnected power leads from the PSU were hanging around loose and swinging around as it was moved. It appeared that one of the the power-carrying contacts of one of the loose Molex connectors, which had become somehow extended out from the plastic sheathing, had touched the inside of the case.

A Quick Lesson in AC Power

The electricity supplied to your computer is in the form of AC (Alternating Current) electricity, at between 110 and 260 volts, depending upon where in the world you are and what time of day it is*. The job of the power supply unit in your computer is to convert this AC voltage into a very accurate number of lower DC (Direct Current) voltages with which it supplies the various components of your computer.

AC electricity is made up of a waveform which alternates in electrical polarity between negative and positive. A complete alternation; where the waveform swings from zero-volts to positive, through zero volts again to negative, and returns to zero volts again, (Also known as a 360-degree phase.) is referred to in electrical terminology as a cycle. The number of cycles per second are measured in hertz, and are the frequency of the AC waveform.

In the UK the AC mains electricity supply is approximately 220 to 260 volts (Depending upon the time of day and the area of the UK.) at 50 Hertz (Hz). (It’s actually over 400 volts before it flows through a transformer on every home’s distribution-board, which reduces the voltage. (The unmanned electrical substations you see dotted around the UK reduce the voltage to around 450 volts from several thousand volts.)) Due to the dual-polarity of AC power, it is commonly conveyed at a number of different (high) voltages using a single cable or wire, right up as far as the domestic power unit attached to each house in the UK. At that point another wire is introduced; that being the neutral (blue) wire. It is connected to a well insulated electrical earthing point. (Not to the same one as the safety electrical earth (Green/yellow) wire.) When electricity is used by an appliance in the UK the power flows from the live wire, which is indirectly connected to the supply via the transformer I mentioned, through the appliance, and to earth via the neutral wire. The safety electrical earth; which is totally unconnected to and completely separate from the neutral earth, is only there to prevent people getting electrified if the live wire somehow becomes connected to the metal case of a faulty appliance. In modern homes there are resettable circuit-breakers fitted to the power-distribution board that will cut the power to a whole section of the home’s wiring if such a fault is detected.

The computer’s case is connected to the safety electrical earth in the UK. Although it’s not the same as the neutral connection it works in the same way; carrying electricity safely to earth and away from metal objects. Although the DC power rails of the power supply unit in a computer seem totally unrelated to the AC mains electricity in your home; that AC mains electricity is nevertheless the source of that DC power that your computer uses. (The power is stepped down in voltage using transformers inside t6he computer’s PSU, changed from AC into DC current by a number of devices called rectifiers, stabilised and regulated by various high-wattage integrated circuits and other regulation devices, and supplied to your computer’s components via the supply rails from the PSU. Allowing one of these supply rails to touch the case, therefore connecting it to electrical earth, completes a circuit, and electricity flows through that circuit to earth. Due to the nature of the build of the PSU combined with the nature of the circuit created by doing so, the power supply unit overloads and stops working. (I could explain it in detail; but neither time nor space allow currently.)

Amateur Upgrade

That’s just one scenario; the one that had happened in the case of this computer, where my friend’s husband had left the power connectors lying loose around the inside of the case after he’d changed the graphics card: Moving the computer to another location had caused a protruding connector on one of the Molex plugs that should have been insulated and inside its plastic sheathing to touch the outside of the case; therefore overloading the PSU and the computer stopped working as a consequence. (In this case I just changed the PSU and the machine worked fine.)

It could have been worse of course: The connector could have touched a metal case or a connector of something on the motherboard; rendering the motherboard useless as well as the PSU itself.

To be fair my friend’s husband wasn’t an engineer or technician: A colleague at the office had told him how to change a graphics card after learning it himself from somewhere, and he’d done as instructed. It worked; but an unforeseen snag caused a problem. He had no idea of how a computer works, and no ideas as to electricity: He followed instructions, correct instructions, and fitted a new graphics card. The power connectors were tied up and tidied; but were in the way of what he needed to do; so he moved them. He just didn’t bother to tidy them up again afterwards.

Nightmare

The sight of loose power connectors (Molex or SATA) lying around inside a computer case make any engineer or technician’s hair stand on end on sight. All unused power connections should be secured by a twist-wire or cable-tie, attached to (But obviously not connected to) the chassis or to other wiring, and kept away from anything metal. Ideally they should be bunched and woven into bundle with the used power leads and secured well away from all surfaces. Anything else is just tempting negative providence and asking for Murphy’s Law to come into play.

Although it can be difficult to make the power leads secure and also look nice at the same time; that’s the constructor’s way of doing it. Practice makes perfect. Power leads/Molex & SATA power connectors (Used and unused.) should be easy to find and/or to trace back to or from the PSU itself by any repair technician or engineer at any time. They should also be safely secured, separate from other wiring, and restricted from movement to as great an extent as is possible. Wherever possible they should be kept away from and not cross the path of other leads/ data transmission leads. (Although this is often unavoidable in many cases, at least to some extent.)

Keep it tidy, keep it traceable, keep it safe.

This post is aimed at the hobbyist electronics constructor; but others might find it useful also.

Some graphics cards have a TV output that you can connect to the UHF aerial socket of a TV set using a co-axial lead with a co-axial TV aerial plug on each end. There’s a problem with this method, in that the picture is terrible.

The graphics card takes the binary H-sync + V-sync  signal, and passes them through a UHF modulator: Result = the synchronisation pulses aren’t properly attenuated into the composite sync pulses that the TV requires, and the resultant picture becomes distorted as a result; therefore it looks a mess.

There is, however, a relatively simple circuit which uses no more than a quad-exclusive-OR gate TTL chip and a couple of transistors, plus 2 capacitors and 9 resistors; which does the job better and will interface a VGA output of a graphics card to the SCART input of a television set.

If you have the necessary constructional abilities including soft-soldering (on veroboard), then this might be a fun project to build in your spare time.

The circuit basically combines and attenuates the H-sync and V-sync outputs from the graphics-card into the composite sync pulses required by a TV set. The RGB signals are left unchanged. Unlike the TV output of a graphics card; these signals haven’t been frequency-shifted to UHF in order that they can be seen at the UHF aerial input of the TV set. This in itself removes a huge chunk of the distortion that can result from such a frequency change – Only for the TV set to change the UHF to IF and split up the signal into its constituent parts…

You’ll find all the details; circuit diagram, component layout, the lot, by clicking this link. Enjoy.

If you’ve never done this before; Please read the entire article before you begin any operations.

OK. The first thing you might ask is:

“Why would I want to upgrade a socket AM2 processor?”

AMD’s Socket AM2 has been around a while; and in some cases an AM2 motherboard capable of supporting a dual-core processor has been fitted with a single-core processor; an Athlon 64 for example. You may feel the need to upgrade to dual-core but not want to go to all the expense of a total rebuild.

If your motherboard has a single-core socket AM2 processor attached then it might very well be fairly easy to upgrade the CPU without doing very much else. An AMD Athlon 64 x 2 processor is a good direct replacement for a single-core Athlon 64. In many cases it’s also a good replacement for a socket AM2 Sempron too. (Please bear in mind that the Sempron processor is commonly a 32-bit processor. A 64-bit version is available and was used on some occasions. Check that your motherboard is capable of supporting 64-bit architecture, in addition to the checks below, before upgrading a 32-bit Sempron to a 64-bit Athlon.)

The first thing I would suggest before doing any upgrade of any kind is that you back up all of your data: Operating system as well as other files. – In essence ghost your disk(s) and have a backup ready on a USB device or another computer.

Why? The reason for making a ghost backup is that if you’re upgrading from a single-core to a dual-core processor, you’ll need to perform a maintenance-reinstall of your operating system after you’ve physically upgraded your processor. If the maintenance reinstall goes tits-up then you’ll have the original backup to reinstall. (A maintenance-reinstall is completely different to a full reinstall. The only data you’ll lose, if you’re using Windows, is probably the Microsoft patches and upgrades. – But you can always download these again anyway.)

So you have backed up all of your data: In effect ghosted the contents of your hard-disk(s).

You now need to ensure that your motherboard is capable of supporting dual-core processor technology. Go to your motherboard’s manufacturer’s website and look up data on your particular model and revision of motherboard. If yes then we can proceed. If no then that total rebuild that you were trying to avoid appears to be the only option if you want to go multi-core.

Select the exact AMD Athlon 64 x 2 processor that you want to use.

A word of caution here: Some motherboards will only support a processor operating frequency of up to 2.2 GHz. This may be particularly true in the case of older boards. (pre-2007) that are fitted with a single-core socket AM2 Athlon 64 or Sempron. If this is the case then you may find that you have trouble getting hold of a new Athlon 64 x 2 with the required operating frequency, and have to purchase a secondhand Windsor – cored Athlon 64 x 2; 1.8, 2.0, or 2.2 GHz, or a Brisbane-cored 2.1 GHz model from eBay or Craigslist.

*Added 28th April 2009: A further word of caution, and something that I omitted to add to this original article: -

I started writing this with the intention of writing an e-book; but I simply don’t have the time or the patience to write the reams and reams of text to justify a full coverage of the subject: Therefore I’ve condensed some of what I wrote down into a rather large post. Some of the text isn’t quite as I’d like it to be; but overall it conveys the right message.

This article is written with a view to teaching a few of the basics of electronic circuit design using a number of equations from or based upon Ohms’ Law. It is aimed at the beginner-level student. Whilst knowledge of component function is assumed throughout, links to articles from other sources as well as kkomp.com have been included throughout in order that the reader can study and read up on the function of individual electronic components for the purpose of being able to better follow the basic tuition herein.

Electronics is such a huge subject that it is impossible to cover every aspect in the required detail in a single article. Maybe a 1.5 terabyte hard drive would be large enough to store the knowledge of the average engineer, if it were zipped and otherwise compressed.

We begin by looking at the basics of Ohms’ Law and go on to design of a very basic DC inverting amplifier stage using three resistors and a transistor. I have attempted to keep the material as light as is possible, given such an intense learning curve packed into such a small space.

The main onus is left up to the reader, as near the end of the article I leave the reader with a conundrum to solve. The problem can be solved using only resistors, transistors, and a diode or maybe two. In solving the problem the reader will construct and solve many equations using Ohms’ Law; therefore putting into practice all that they have learned herein whilst at the same time developing their skills further.

It’s short and concise. There’s a lot of study and knowledge packed into it, and I hope I’ve done the subject justice.

Foreword

Electronics is a vast and complicated field. There’s so much to learn, and that learning curve never stops. No matter how much you know; there’s always more to be known as new discoveries are constantly being made.

For instance; forty years ago, nobody would have thought that the recently-invented transistor would be at the centre of technological advancement. It would have been thought of a a crazy notion that over sixty million transistors could be compacted into a device with the volume of a standard matchbox that is the central processing unit of a powerful personal computer.

In this article I’m not intending to describe in great detail the various functions of individual electronic components. Where this may be necessary I’ve linked to articles containing further information on this, should the reader require it.

This article is mainly about calculation used in simple circuit design, brought about by the use of basic calculus, in part using Ohms’ Law. The publication is intended to be the first in an ongoing series of books covering the basic principles of electronics and electronic circuit design. A basic knowledge of component identification and function is assumed in the reader. If this happens to not be the case then the information provided at the destination of the links incorporated within the text should be a sufficient source of knowledge.

The article that you’re now reading does not endeavour to go into digital electronics at this stage, and concentrates rather on basic analogue DC circuit principles, which should be learned as a forerunner to the discovery of digital circuitry.

Electronics is a very intense subject; and one could devote one’s entire lifetime to the furtherance of knowledge in this field. However with the rate of new discoveries now being made it is extremely unlikely that one could ever learn everything there is to know about the subject in even a very long lifetime.

This article is intended for the beginner class on levels 2 and 3. However, by utilising the links provided I feel even an absolute beginner on level 1 would be able to keep up, with much study, and maybe even progress a degree in doing so.

One thing that has been the very basis of all electronic advancement, from the days of valves up until the present day, is a set of equations known as Ohm’s Law. In this article we’ll be taking a look at Ohm’s Law and showing how it is applicable to every aspect of electricity and electronics.

Georg Simon Ohm was born in Germany on 16th March 1789, and lived until 6 July 1854. He became a physicist, and during his career determined that there is a direct proportionality between the voltage applied across a conductor and the resultant electric current flowing in the circuit. Further experimentation meant that eventually Ohm was able to define the relationships of voltage, current, and electrical resistance.

_._

In this rather large article I’m going to be using circuit diagrams. For those not familiar with circuit diagrams I would suggest that you take a look at this link and/or here to familiarise yourself with some of the symbols used. You will notice that I don’t always stick to the usual format in a number of cases when I’m drawing my own circuit diagrams freehand or other than on the computer itself: For instance; when I’m drawing a resistor I use a diagonal zigzag line rather than a rectangular box. Also when drawing a transistor symbol I usually omit the circle around the device.

This is for a number of purposes; the main ones being speed and neatness: If you’ve ever tried to draw a perfect circle without using a pair of compasses or a jar lid, you’ll know just how difficult it is. The symbol inside the circle is the same and unique whichever way round one draws it. The reason for the circle is to indicate that the device is a discreet device, meaning a single device in a package; as opposed to part of a multi-transistor chip or an integrated circuit. For this article we’ll just use the symbol without the circle where I’ve drawn the circuit diagrams myself: It’s a transistor and that’s it.

Standard Transistor Symbol     Transistor Symbol Used Herein

(The link shows only the symbols of a bipolar NPN and a PNP transistor, and also a phototransistor. There are many other types of transistor; such as the FET or Field Effect Transistor in its various different guises. (Which, incidentally, was not named after the author; Sharron Field. (Sadly.))

I commonly use a zigzag line as the symbol for a resistor; this was once the standard symbol for a resistor. It was abandoned for the sake of clarity because it looks too similar to the symbol for an inductor , the symbol of which has curves where the old resistor symbol has angles.

= This is how I draw a resistor

Whereas this is the modern standard symbol:

I personally use the old zigzag line symbol because it’s vastly easier to draw and takes less than 1/4 of the time. If I try to draw a rectangular box I end up wishing I hadn’t.

You’ll notice that the circuit diagrams that I’ve included were drawn with a pen or pencil on paper and scanned in: That’s the way things are currently. I don’t at the moment have either the software to draw exclusively on the computer nor the time and patience to learn how to use it. The situation may be different in the future; but right now that’s the way things stand.

The Basic Triangle

So let’s look at the most basic bit of Ohm’s Law first; that being the relationship of Voltage to current to resistance in a DC (Direct Current) environment: -

The relationship can be expressed in an easy-to-remember format thus:

V

I        R

Where V is Voltage in Volts,

I is electric current in Amperes,

and R is DC electrical resistance in Ohms

From this simple illustration we can draw the following equations: -

V / I = R

V / R = I

I x R = V

If we were to substitute the figure 1 for all of the variables we would notice that the equations are all true and equal in their most basic form: If a single Ampere flowed through a resistance of a single ohm at a voltage of a single volt it would be the point of correlation between the three measurements.

If, as happens in nearly all cases in a practical working environment, we were to increase or decrease the value to a number less than or greater than one for any or all of these variables, then that correlation vanishes; yet the equations still hold together.

Let’s look at an example on the next page: -

A current of 2 Amperes, or amps for short, is flowing through a resistance of 2 Ohms.

In this case Ohm’s law tells us that the voltage present at the point where the current exits the 2 Ohm resistor is 4 Volts; as 2 amps x 2 Ohms = 4 Volts.

Another example: –

An unspecified current is flowing through a resistance of 10 Ohms. The voltage at the point where the current exits the resistor is 5 Volts.

Ohms’ Law reveals that the unspecified current must be 5 Volts / 10 Ohms = 1/2 amp.

A third example: -

A current of 0.1 amps, or 100 milliamps, is outputting a resistor at a Voltage of 0.3 Volts, or 300 millivolts.

Ohms’ Law informs us that the resistor’s value in Ohms is 0.3 Volts / 0.1 amps = 3 Ohms.

Yes it really is that simple; at least at this stage in the proceedings.

Power

The next denomination we introduce into the mix is electrical power, represented by the letter P; and which is measured in Watts.

Here we introduce another law, that being Joule’s Law, which is named after the British physicist James Joule.

Joules’ Law has 2 main equations for giving the relation of power, or wattage, to the integers that we’ve already introduced in Ohms’ Law: The following equations describe this relationship: -

P = I x V

2 ,

P = V  / IP = V squared / square root of I

and

2

P = I  R        (P = I squared x R)

(Please excuse the error in writing the equations as mathematical formulae: The text has cocked up beyond repair. Please read the words rather than the badly-printed equations.))

Let’s look at some examples of this: -

1) A lamp draws 1 amp of current at a voltage of 6 Volts. Joules’ Law combined with Ohms’ Law tells us that the lamp is burning 1 amp x 6 Volts = 6 Watts.

2) A DC circuit draws 2A of current, and has an overall resistance of 12 Ohms. Joules’ Law tells us that (2 x 2) amps of current x 12 Ohms = 48 Watts.

In Circuit

So that’s the very simple bit out of the way and dealt with. let’s now take a look at connecting resistances in parallel and also in series, as well as working out the total resistance: -

There are different equations for calculating parallel and series resistances. Let’s first take a look at series resistances:

In the example above we have a circuit diagram of 2 resistances, R1 and R2, in series. To calculate the total resistance of the series pair we simply add up the sum of the values of the two resistors thus: –

Rt = R1 + R2

That was easy.

When calculating the resistance of 2 resistors in parallel, however, things are slightly more complicated. The equation for calculating the total resistance of 2 resistors in parallel is:

Rt = (R1 x R2) / (R1 + R2)

Let’s look at an example of this: -

In the diagram above we have a 2,200 Ohm (2.2 kilohms) resistor connected in parallel with a 1,100 Ohm (1.1 kilohms) resistor. The total resistance is given by

Rt = (1,100 x 2,200) / (1,100 + 2,200)

Rt = 2,420,000 / 3300

Rt = 733.33 Ohms (0.73333 kilohms)

Here’s a reminder of the resistor colour code and how to read the resistance value of the component. (This code also applies to some capacitors too.) : -

Introducing Semiconductors

In this article I’m not going to be covering any other “passive” components, such as capacitors and inductors. – I’ll save that for you to learn elsewhere. Right now I’d like to move on to what are termed “active components”, or semiconductors.

All the many types of transistor are classed as semiconductors, as are a range of components called diodes. There are also semiconductor components called thyristors which are used for DC power control, also triacs which are used in AC power control circuitry. High-current versions of these are probably utilised in the power supply of your computer, along with capacitors – large and small, resistors, diodes, power transistors, and inductors. Here we are starting to go beyond the scope of this article, however.

Herein I’d rather stick, for now, with just resistors, diodes, and a single basic type of transistor known as a bipolar transistor.

Very briefly; the diode in its raw form is a semiconductor device that only allows electricity to flow only one way through it. Click the hyperlink at the word “diode” above and discover more about it.

Standard Transistor Symbol     Transistor Symbol Used Herein

Yes you have seen these symbols before. They appear in the Foreword.  I thought it prudent to place them here also to serve as a reminder of the point on circuit diagrammatic terminology touched upon therein, as well as to provide the circuit diagram symbol for a bipolar transistor. – No it’s not an electronic device with a mental condition. The name derives from its construction. See the link above for more information.

The bipolar transistor comes in 2 ‘flavours’; those being NPN and PNP:

ollector                                                       mitter

NPN                                                                PNP

The meaning of these terms is described in detail in the Wikipedia article linked to above. This article isn’t written with an intention of dealing with the construction and function of electronic components. Foreknowledge in this area is assumed. Links to locations which detail this are provided for those who need to know, however.

For the examples in this publication we’ll be using the NPN transistor.

Throughput

You will appreciate that every device has its limitations; therefore although there are expensive hi-current devices available that can handle several amps of power, most low-power, and small signal bipolar transistors can only deal with a fraction of an amp passing through them without burning out. With this fact in mind we have to ensure that the current supplied to the individual transistor will not overload it. This is accomplished by a resistor connected between the collector and the + supply rail (VS). This resistor is commonly referred to as the “collector load resistor”. the amount of current allowed by this resistor is calculated by means of Ohms’ Law:

I=V/R

The main amount of current flowing through the device passes from collector to emitter. A smaller current is also required to be applied to the base connection, usually about 0.1 times or 10% (maximum) of the larger current.

More Terminology

In electronics terminology we refer to the current flowing between collector and emitter as Ice0, and the current flowing between the diode junction of the emitter and base as Ieb0.

Similarly with respect to voltage, the terms Vce0 and Veb0 are used respectively.

The terms Vb, Vc, and Ve, refer to the voltage present at the transistor’s base, collector , and emitter respectively. Similarly the terms Ib, Ic, and Ie, refer to the current present likewise.

V+ usually refers to the supply voltage, otherwise referred to as VS or Vss.

Biasing the Base

A bipolar transistor requires a voltage of 0.7 volts present at its base before it will allow any current to pass between collector and emitter. This is known as the “transconductance threshold” It is for this reason, particularly where the device is used under small signal conditions such as audio amplification that the base needs to be biased with a tiny current in proportion to the input signal, to a voltage of just under 0.7 volts.

To achieve this, a pair of resistors connected in series across the supply rails is normally used as a potential divider. The resistances of each resistor are selected such that the voltage at the centre-tap to which the base is connected is just below 0.7 volts. In addition to this the resistances of the resistors are kept as high as is reasonably possible to ensure as little current as possible, and consequently as little wattage as possible, is wasted; as a potential divider will continue to burn the same amount of wattage whether or not an output is drawn from its centre point, due to it effectively being a resistance connected across the supply rails.

In the example above we use a 10 kilohm resistor as R1 and a 1.1 kilohm resistor as R2. The supply voltage, VS, is 7 volts.

To calculate the voltage at the centre tap between the two resistors, to which the transistor’s base is connected, therefore the base voltage (Vb), we use the following equation:

Vb = VS X (R2 / R1 + R2)

Therefore in this example: -

Vb = 7 X (1100 / (10000 + 1100))

Vb = 7 X (1100 / 11100)

Vb = 7 X (11 / 111)

Vb = 7 X 0.099099

Vb = 0.6693693V

- Which puts the transistor right at the edge of the threshold of transconductance. A voltage of over 31 millivolts will flip the device over into transconductance and a proportionally equivalent current will flow between collector and emitter.

Beta

No this doesn’t refer to a test-version of a new computer program: The beta of a transistor is the quantity giving the amplification factor of that transistor. There are two ways of looking at this; in-circuit and out-of-circuit.

Out-of-circuit, as a standalone unused component, a given type of transistor has a maximum beta rating that it can be run at in-circuit. This can vary from around 20 or less for some power-transistors, to up to 1500 or more for some hi-gain amplifier transistors.

Setting the beta of a transistor in-circuit is another part of circuit design.

The in-circuit beta of a given transistor can be calculated by the proportion of Ib when Vb is above the transconductance threshold to the amount of current represented as Ice0. (Unless the transistor is connected in-circuit as a voltage amplifier rather than a current amplifier; in which case the beta is calculated by replacing the term Ib with Vb and Ice0 with Vce0. That is beyond the scope of this book, so we’ll stick to the current amplifier model for now.)

For example; let’s assume that we have a transistor connected in circuit with a base voltage of 0.75 volts (Vb = 0V75), therefore biasing it into transconductance. The base current is set at 1 milliamp (1mA). The supply voltage (VS) is 10 volts, and the collector load resistor is 100 ohms:

Ic (collector current) = V / R

Ic = 10 / 100

Ic = 0.1A (100mA)

The in-circuit beta of that transistor can then be given as:

b = Ic / Ib

b = 100 / 1

b = 100

Provided that this doesn’t exceed the transistor’s out-of-circuit beta rating it’s perfectly safe to run the transistor at this beta and expect its amplification factor to be 100 X.

(In most cases, though, such a large amplification factor in a single-transistor amplifier stage would give rise to signal distortion; especially in high-frequency

AC amplifiers. For DC amplifiers such as we’re dealing with here, though, this beta rating is OK and won’t cause any distortion as there’s effectively nothing to distort in this example.)

Let’s sum up and take a look at an example of what we’re trying to achieve here:

In the circuit above we’re using the potential divider we mentioned earlier:

R1 = 10K and R2 = 1K1

That’s great with a supply voltage of 7V as it biases the base just below the transconductance threshold as we saw earlier.

- But we haven’t yet worked out Ib in this case. How do we do that? well the total current flowing in the potential divider will be:

VS / (R1 + R2)

7 / 11100 in other words; which equates to 0.00063A, or 63 microamps. That’s pretty low but it’s OK. If we want to run the transistor at a beta of 100 then we’ll need to make the collector load resistor allow 63 X 100 microamps to flow as Ice0.

So we want to arrive at a scenario where Ice0 = 6.3 mA.

We know just how to do that using Ohms’ Law: -

If Ice0 = 6.3 mA and V=7 volts, then V / I = R:

7 / 0.0063 = 1111.1111 ohms

- Is the value of resistor that we’re looking for. We look in the spares box and find that the nearest value of resistor that we have is 1100 ohms (1K1). only 11.1111 ohms out; which will make very little difference except that the beta will be a fraction over 100. That’s good enough. – So we choose 1K1 as the value for the collector load resistor.

Another Stage?

That’s it then: We’ve designed a DC inverting amplifier with a beta of 100 (+/-1%) using a single transistor and 3 resistors.

For clarity here’s a components list: –

Transistor:

Q1: BC108C (I chose this one as its quite ideal for the purpose.)

Resistors:

R1: 1K1 1/8Watt

R2: 10K 1/8 watt

R3: 1K1 1/8 watt

If we were to apply a current of 1mV to the base, then the collector current (Ic) would drop by 100 mV. That’s a very basic medium-high gain inverting amplification stage we’ve just designed. Give yourself a pat on the back – That’s quite an achievement if you started reading this article without much, if any, idea of circuit design.

What’s meant by an inverting amplification stage? Well basically the input is the opposite of the output: When the input voltage is zero the output voltage is equal to the supply rail voltage, and whatever voltage is applied to the input, the output drops by a factor proportional to the amplifier’s beta.

For instance; if a DC voltage of 41mV was applied to the base of the transistor in this circuit, then the output would drop by 1.0 volts; from 7v to 6v. That means that if 101mV (0.101V) was applied to the input at the transistor’s base, the output at the transistor’s collector would drop from 7v to zero. – that’s a pretty sensitive circuit we’ve designed there. – But we want to design a non-inverting amplifier; one where if we apply 101mV to the input then the output rises from zero volts to 7 volts.

Why does it behave like this?

When there is no input, the transistor if switched off and current flows unopposed through R3 to the output. (Remember; a resistor gives resistance to current, not voltage; so although the collector current is regulated by R3, the voltage remains unchanged.)  – Therefore the output is at 7v. As the input voltage rises and the transistor begins to switch on and allow current to flow through it to ground, the voltage at its collector falls proportionally.

We could take resistor R3 out of the collector circuit and connect it between the transistor’s emitter and ground, taking the output from the emitter. That would work fine. – Then as the transistor begins to switch on the voltage at its emitter would rise from zero volts proportionally; but R3 as an emitter-load resistor would never allow the output voltage to rise as far as the 7 volts we require. Remember the transistor’s 0v7 transconductance threshold? That would affect the output so that it would never be able to rise above 6v3. What we need is some more circuitry added to what we’ve designed so far. Let’s get designing:

We can modify our existing circuit by adding an output stage to it: -

We have a condition at the output of our device we just designed where the output is at 7v with no input. The output drops by 0.1v with every millivolt above 31 mV applied to the input. Let’s ignore the 31mV for the time being, for the sake of simplicity. – But that idea of taking the output from the emitter can be used. First we’ll redesign the circuit: -

We’re now taking the output from the emitter. This type of circuit is called an “emitter-follower” for seemingly obvious reasons. We now design a second stage for this circuit to correct the error; or should I say YOU now design it.

“But I’m no circuit designer!”

You know enough now to solve the problem.

‘Your Turn

It’s tricky, but it can be done using only what you’ve already learned herein and by clicking on the links provided. You can use as many resistors and transistors as you wish, but remember, in the interests of cost efficiency you need to keep the number of components used as low as you can. If you manage to solve the problem using 64 transistors and 184 resistors then well done for solving it; but that’s far too many components. Keep the component-count low but keep trying.

I ask two further things: The first is that you don’t modify the original emitter-follower circuit in any way. You can connect to it at any point you choose; however you must take the output from the emitter and you cannot change either the existing circuit configuration or the component values. You also cannot change the supply voltage.

Good luck. You can refer to any electronics teaching media that you wish to use. However – here’s the second thing I ask of you – you cannot ask an electronics engineer or technician to solve the problem for you. This is your project. A qualified engineer will have no problem with it; but a qualified engineer doesn’t need to learn how to do it. Hopefully by the time you’ve solved it you’ll have learned how to do everything I’ve shown you off by heart and with ease.

All the information you need is written above; but you can use whatever other media you wish. If you want to learn then this is a worthy project. if not then I hope you’ve found what you’ve learned edifying.

If you happen to be a bit unsure of component function then click the links provided again and study the material. Several months’ basic electronics tuition has been crammed into this book to this point. It would be unrealistic to expect anyone to grasp it all in one reading; even if they did click every link and study the information there in full.

After – word

So you’ve decided you want to be an electronics engineer? Good choice. I’m not the one to teach you though: I’m only qualified as a technician. I am qualified to teach you the basics, though; and that’s a start if nothing else.

I’m trying to limit what I teach herein to what I’m qualified to teach. What I know is more than I’m trained to know. Whilst I’m not up to engineer’s status in knowledge, I do have perhaps a bit more know-how than the average technician. Had I qualified at a higher level I could teach more and feel comfortable in doing so.

The engineer’s course is 4 years long. I studied the equivalent of 2 years (‘Just over a years’ intensive training.) for my technician’s qualification. (City & Guilds 300, 301.)

I’ve deliberately not tried to make this aritcle “pretty” or to give it extra appeal. What you see is what you get. Electronics is a cold hard emotionless science: There’s a lot of maths involved; on a much higher level than this article has delved. (Bode plots and Nyquist diagrams included.) What you see is the beginnings of elementary calculus and an opportunity to dip your toe into using Ohms’ Law for real. If this breif and basic look at electronics has whetted your appetite for more then you’re probably a natural to at least a certain extent. I suggest you glean as much of the elementary basics as you can from this source; following which you continue your studies both online and offline.

Keep your eye on http://kkomp.com for any electronics titbits that I throw out to my readers. Take a home study course, night school, even go for it and take an electronics engineer’s degree if you like. This article only covers a few of the basics: I’ve barely touched on capacitors and inductors, no more than mentioned diodes and some other components, and I’ve only once or twice mentioned digital electronics. – With its logic gates, pulse-triggered flip-flops, Schmitt triggers…

The material herein has barely scratched the surface of analogue electronics. There’s so much to learn; and you’ve hardly begun.

If you’re intent on learning more, or even becoming qualified in electronics, then I wish you the very best of luck. If you found this heavy going and decided that the subject’s not for you then thank you for reading. At least you now have some idea of a subject that you don’t want to pursue any further. I hope you gained some enlightenment from your reading.

Whatever you choose to do; I hope you get the very best from it.

There are several types: The main two, in layman’s terms, are a light dependant resistor – LDR  – the resistance of which increases with the amount/type of light falling on it; and conversely one the resistance of which decreases with the amount of light falling upon it.

One of the main uses for these devices is in switching on street lighting. Rumour has it that a timer mechanism was originally used for the purpose, before the LDR’s invention. That’s maybe a sound idea if the light happens to be situated at or near the Equator; where the seasonal variations between night and day are minimal. However when you start getting as far away from the Equator, even as “close” as Southern England – Where night can come as early as 3:50PM in the depths of winter; yet as late as 10:40PM in the heights of summer: The timer would require resetting at least weekly. I honestly can’t envisage electricians running round the country resetting street lamp timers every week. Having said that though; there were people who went around the major English cities in the evenings and mornings lighting and extinguishing gas street lamps in the Victorian era.

Thankfully technology has moved on a bit since then.

See the Wikipedia article hyperlinked above from the letters LDR for a full description of the component.

You may notice that, on the circuit below, I use a non-standard symbol for an LDR which is completely different from the standard symbol. The one that I use (A zig-zag line with two arrows pointing towards it.) is a shorthand circuit-diagrammatical representation. It’s the old symbol for a resistor with the two arrows indicating that its value is dependant upon the amount/type of light falling upon it. Similarly with the fixed-resistors in the circuit; a zig-zagged line without the arrows. The symbol used as standard for a resistor is a rectangular box with leads either end. i find this too much hassle and too time-consuming to draw; therefore I resort to my shorthand: I understand it even if not all other people do.

So how does the device work in-circuit? The circuit-diagram below shows a very basic circuit incorporating 2 fixed resistors, a transistor, and a light-emitting diode to display the output. The LDR is an inverse-effect type. That is to say its resistance decreases as more light is shone on its surface.

The LDR , along with R1, acts as a potential-divider, biasing the base of Q1. As the light shining upon the LDR gets brighter, so its resistance drops, and thus the voltage at Q1’s base drops via R3. (See “Ohms Law and the Potential Divider”. Also see “Base Voltage”) When the base voltage drops below 0.7 Volts the transistor switches off and the LED goes out.

The function of the circuit can be reversed by replacing R1 with the LDR and vice-versa: In that case as the light shining upon the LDR increases, so its resistance drops and the voltage at Q1’s base rises via R3. When the base voltage rises above 0.7 Volts the transistor switches on and the LED lights up.

This circuit does work; in fact I memorised it from my early self-tuition in electronics, as well as from college. (Where I gained C&G 300, 301: Analogue and digital electronics certificates. ( I also retook a Maths exam as I had a cold on the day of my original exam twenty-something years ago and didn’t do as well as I’d have liked. – I passed; but my grade wasn’t as good as I’d hoped for, and I knew that I could get a better result.))

You might like to experiment with different values of resistor for R1, although a kilohm is probably the lowest value you should use in this case. You could also try replacing R1 with a 1K resistor and a 10K linear potentiometer connected in series. I’ll leave you to experiment.

There are a few and there will be more posts on the subject of basic practical electronics in this blog. if you’re interested in the subject then do look further into the content. It’s not all listed at time of writing so Google is your friend; use it. (Or Windows Live Search, Ask, Yahoo… whatever: I’m not biased or sponsored by Google. I prefer to use Google myself as I find the GUI simple to use and the listings useful. Your opinions may vary.)

If you build the circuit and experiment with it please do tell me your results. I’ll be interested to know.

Computer components are a heck of a lot cheaper, as well as better, than they used to be. All the same it’s still a noticeable debit to your capital if you buy the components and assemble one yourself. With this in mind the last thing you need is to find that, on power-up, one or more of the components is dead; causing your machine to function in a limited fashion, improperly, or not function at all.

Whilst it is true that in rare cases a unit does escape the manufacturer’s quality control and finds its way to market in a faulty or dead condition, the most likely cause of a faulty or dead component on initial power-up is exposure to static electricity. This is especially true with regard to components such a processors ( CPU ), RAM, and motherboard, including any and all components supplied fitted to the motherboard, such as BIOS, chipset, etc.

How Do These Components Become Exposed to Static Electricity?

Static electricity is a crafty customer. It builds up everywhere, it’s high=voltage, and the only way to avoid a charge being present in most objects, commonly, is to discharge it.

You’ll notice that the more sensitive components of a computer are supplied packed in a special anti-static type of plastic and/or foam. This is to prevent exposure to static electricity in transit. They have been manufactured in a static-free environment, and are most probably working components that are faultless.

Where they become exposed to static is after delivery to destination; in this case your workroom or workshop. In short; if these components are taken out of their protective anti-static packaging and allowed to touch anything, including you yourself, then they risk exposure to static electricity.

How do I Ensure That The Components Are Not Damaged By Static Electricity?

Firstly, the environment in which the components are unpacked is important: As far as it is possible to do so, ensure all conductive surfaces and objects are connected to electrical earth.  (A water-pipe can be a good earth connection.)

Carpets aren’t a good idea to have present in the unpacking environment: As a person walks across a carpet a static charge builds up in their body.  Also ensure that any surface that the components are placed upon after unpacking is grounded and static-free. Only remove the components from their anti-static packaging when it is absolutely necessary to do so. Install the component immediately it has been unpacked: The less time you take the less chance the component has of exposure to static.

More importantly than any of the above, though, earth yourself: An anti-static wristband connected to electrical earth will remove static charge from your body. These wristbands can be purchased rather inexpensively from a wide range of suppliers. If you don’t have one of these but need to build now; connect a piece of metal jewellery or a metal wristwatch to electrical earth, or wrap a piece of bare wire round your wrist and connect it to earth.

How do The Components Become Damaged by Static Electricity? What Happens to Them?

Briefly: Most computer components consist, at least partly, of transistors; usually millions of them. Each microscopic transistor has three connections; namely drain, source, and gate.

The gate is insulated from the drain and source by an extremely thin insulator; only a few atoms thick. The transistor works by the electrostatic (Not static electricity: ‘Completely different term; although it is often used (Wrongly in my opinion.) in place of static electricity.) field on the gate electrode regulating the transfer of electrons between source and drain. A static electrical charge applied to this is too powerful for the device and breaks down the insulating layer between the gate and the rest of the transistor: Therefore the transistor is unable to function.

- So do be careful and always ensure that you make your best efforts to avoid exposure to static electricity in the computer-building environment. Static electricity is the computer-builders’ enemy, and not a good idea to have around when dealing with sensitive components.

Analogue amplification stages incorporating just a single active device, such as a transistor or op-amp, – as in a DC amplifier -  usually allow amplification of either the negative-going or the positive-going part of an AC waveform alone; basically cutting out half of the cycle and therefore causing immense distortion. There are two ways of overcoming this: Those being amplifying both halves of the waveform separately and recombining them at the output; perhaps at a decoupling transformer, (Too much copper wire, inductance, and consequential weight for my liking.) or by using two interconnected active devices to amplify both halves of the waveform at once, before passing the full AC waveform on to the next stage.

The latter method can be achieved with a push-pull amplifier.

Pictured below is a circuit diagram of a very basic push-pull amplification stage. (Please excuse the freehand drawing.)

The input signal; an alternating waveform, is fed via capacitors 1 & 2 to the bases of Q1 and Q2. The functions of C1 & 2 individually and collectively are two-fold: The first is that they individually shield the base connections of their respective transistors from any stray DC voltages from a previous stage, also they separate any DC potentials present at the base junctions of Q1 & 2; therefore preventing unintentional and accidental biasing of one another.

Resistors 3 & 4 act as a potential divider biasing the base of Q1 to 0V7. Resistors 1 & 2 have the a similar effect effect on the base of Q2: However since Q2 is a PNP transistor, the values of resistor used in the case of the R3,4 pair are reversed; therefore giving the base of Q2 a negative bias with respect to that of Q1.

Preset potentiometers PR1 & 2 set the potential of the transistor-pair with respect to the supply rails; and consequently the swing-maximum of the output-waveform. Resistors R5 & 6 are a precaution to avoid the peak output level colliding with the supply voltage and therefore causing distortion.

Capacitor C3 provides AC decoupling for the transistor pair.

Note that the emitters of the transistors are connected together and the output taken from that connection. This allows the inclusion of the traditional collector load resistance in both cases.

A waveform appearing at the input flows through both C1 and C2 to the base of the individual transistors. If the waveform is on the positive-going half of the cycle it lowers the conduction of Q2 and raises the conduction of Q1. If the waveform is on the negative-going half of the cycle the reverse occurs: Hence the output polarity mirrors the input polarity to whatever degree of amplification is involved.

A Little Background Information:

Prior to the advent of digital electronics, this type of circuit configuration was widely used in analogue receiving and amplification devices throughout the 1950s, 60s, and to some extent even the 70s. Back in the 1950s before the transistor became widely used in electronic circuitry, they would use a pair of triode or pentode  thermionic valves in the place of the transistor pair. As technology developed the manufacturers developed smaller valves with two triode or pentode sections for the purpose, screened from one another by a metal electrode.

(Example: ECC82 (European Nomenclature), or equivalent 12AU7 (American nomenclature.) AF double-triode with a 6.3 Volt heater supply. The European equivalent with a higher heater voltage was the UCC82 which required a 32 Volt heater supply.)

Valves were also manufactured with a pre-amplification or oscillator stage included, usually a triode; along with a main amplification device, usually a pentode. (Example ECL85 (6.3V heater supply), UCL85(32 V heater supply), and PCL85 (17.5 Volt heater supply, commonly used in the audio output stages of televisions. – Right up until around 1973.))

During the late 1960s/early 70s, valves began to be excluded from the designs of electronic devices, in preference for the transistor; which was lighter, lasted a lot longer, required less voltage to function, and didn’t require an internal heater powered from an external source to make it work.

(There was a period at the very end of the 1960s which lasted a little way into the 70s where equipment manufacturers would produce valve/transistor hybrids, especially in the case of televisions. These exhibited a few benefits over valve-only technology; such as they took less time to warm up before they started working, and the amount of mains-hum distortion was reduced to a large extent.)

There: A free history-lesson along with the main subject. There’s value-for-money; even though there was no charge in the first place.

Personally Speaking:

Just in case you’re wondering, I do just about remember those old days mentioned; especially the latter valve/transistor technology. I was just getting into electronics in those days. – And yes I was rather young to be messing about inside televisions et al. I practiced hobby electronics until fairly recently, when I got qualifications in the subject after a crash-refresher course at college. I became interested in computers in the late 1970s, around the time the Commodore Pet was released to market. Upon leaving school I followed the arts for a while, at the same time as running a small hobby-enterprise in analogue electronics, until I got back into both computers and digital electronics in the late 1990s.

(Oh yes. – Just in case you were wondering; I do remember the large B9D-base line-output pentode valve used in some televisions right into the 1980s; although I can’t remember the alphanumeric designation offhand. It started with a “P”, but that’s pretty obvious. – Most television valves did, other than maybe the HF triode-pentodes in the UHF/VHF tuners of some 1960s models. – Apart from some of the B8A valves of the early 1960s with the metallic base. – Now that is going back a bit too far.)

Ah I just remembered: PL504.  

. . . . . . . . .  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

I was just proof-reading after writing this lot and I remembered; I need to hyperlink. There is so much I should hyperlink. If this article takes longer than expected to produce then that’s part of the reason why.

Oh wow; this’ll be fun!

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I wouldn’t describe building a PC as "easy"; but it’s not as difficult as one might imagine. Unlike constructing an electronic circuit, such as an amplifier, for example; there’s nothing extremely fiddly, such as soldering or quality engineering to worry about: That’s all been taken care of already by the component manufacturers.

It’s like putting a jigsaw together: Every piece fits in a certain configuration as a part of the whole. The pieces are already made, so you don’t have to make them yourself: you only need to fit them together in the correct fashion.

*At this point I’ll state that this article isn’t a comprehensive how-to: It’s just some notes from my recent rebuild experience.*

You may have heard that I recently had a computer die on me. I’d built it from scrap parts as a replacement for another one that went funny earlier. I have no idea exactly what caused the fault that killed it. It blue-screened and then just died a second later. Following on from that when I tried to restart it the BIOS couldn’t find the processor; so I assumed that the chipset had fried: ‘New motherboard required if this was the case.

There’s the old construction on the left. (Excuse the picture quality.) I’d already started taking it apart at that point, so it does look rather untidy. I’ve just rebuilt this machine; and I’m actually writing this article on it.

I stripped it down and started again; therefore I in essence built the machine from scratch. While doing so I took pics of a number of stages and of some of the parts, with a view to blogging the event. This blog has suffered from a lack of posts due to this project and other work, so I  thought it a good idea to use this project as subject matter.

First things first; a motherboard:

I purchased a fairly cheap Gigabyte motherboard for this project: It cost me about £38 at the time. I’d decided to use a socket AM2 AMD Athlon 64 x 2, 2.2GHz processor, as in my other working machine, for this one.

Some people have a low-opinion of AMD chips. Myself, I’ve always found them to be reliable and sturdy. Also they’re cheaper and the motherboards that run them cost less too. Since this was a rebuild that I didn’t want to spend too much on I was quite happy with my choice.

Of course I’d need a CPU cooler too, which consists of a heatsink and fan in order to prevent the processor from overheating. I had this one in stock and was going to use it. However I found that the original cooler was a better one, and surprisingly that it fitted onto a socket AM2 fitting perfectly; therefore having cleaned it up I used it instead.

I also invested in a new hard-drive. I could have used the old one; there was nothing wrong with it. – But I added the old one to my other machine and started this build with a brand new disk.

All-in-all the motherboard, processor, and hard-drive, cost me £108 Inc. VAT at 15%.

So to construction; and the pic on the right shows the case with the new motherboard fitted.

Always remember before starting out; earth thyself: Static electricity builds up in your body and on your clothes, and it kills computer components. personally I always wear an earthed wrist-strap when building computers, just to take any static safely to earth rather than letting it flow through the components I’m using and killing them.

After this point I got a buzz,  and I just ploughed on ahead with construction while not bothering to take any more pics of it.

In short, though, it was just a matter from here of fitting the PSU, connecting the appropriate power leads to the motherboard, inserting the processor into the socket on the motherboard, pulling the little lever while pressing down on it to seal it in the socket, smearing some heat-conducting grease on the top of it, fitting and aligning the cooler, and pulling down the lever on it to tighten it to the surface after clipping the clips onto the processor surround.

Installing the RAM: I inserted 2 x 1GB 667MHz DDR2 sticks into the memory slots and pushed down until they clicked into place.

Following that I connected up the front panel to the appropriate pins. I had problems with the sound jacks on the front as the connections didn’t match with the new motherboard at all. In the end I left the two front sound jacks unconnected, and only connected the 2 front USB ports, the HD activity LED, and the power-indicator LED, to the appropriate pins.

I popped the new HDD into a drive-bay, screwed it in, and connected it up to a SATA power lead from the PSU and to the motherboard’s SATA controller via a SATA connector lead. The same with the DVD-RW drive. (I used the existing DVD-RW as there was no point getting a new one. – Same with the existing floppy-drive.

"Floppy-drive! Why bother with a floppy?" You ask.

I like floppy-drives. I find them useful. I also still like CRT monitors and Outlook Express too. That’s just me: I don’t expect anyone to do similar if they don’t want to.)

So having put the thing together it was time for the initial power-up: Fingers crossed. Bingo: POST. I did take a pic of it, but it was so crappy I deleted it.

After a few minor adjustments to the BIOS, it’s time to install Windows XP:

Pop the XP pro CD into the DVD-RW drive… Let’s get the HDD formatted: NTFS – A decent file-system.

Install Windows…

- Et voila mes amis.

That wasn’t exactly the hardest thing on earth to do; although the construction is the easy, quick, and interesting part for me: It’s the 12 or so hours afterwards installing, optimising, and configuring, all the software that really gets my goat: That’s one reason I don’t do upgrades as a rule for customers: Even after spending 12 hours on it; they still moan about something: That’s why I just build the comp and install and optimise Windows and the motherboard drivers after a new build only, professionally. People can add their own software afterwards and screw up the operating system any way they like once the comp is delivered and paid for.

So that’s the rebuild; and it is a rather excellent job although I do say so myself.

Chipmaker, Advanced Micro Devices, has launched its first 45nm ‘Shanghai’ Opteron chips for servers and workstations.

It also has a forthcoming range of desktop processors built on 45nm technology codenamed "Deneb". Both of these ranges are constructed using a process called "immersion lithography". AMD claim that this fabrication technique will lead to ‘dramatic performance and performance-per-watt gains.’

The new Opterons will have an increased clock speed due to this fabrication process; rising from 2.3 GHz with the current Barcelona-cored Opterons, to 2.7 GHz with the Shanghai-cored chips. The current Phenom range, which run at up to 2.6GHz, may also benefit from this upward-clocking in their next incarnation.

The new Shanghai-cores also benefit from increased cache-size, as well as from HyperTransport 3.0, which increases bandwidth considerably. Let’s hope they vastly outperform the Intel competition as well as the previous/current Phenom CPUs, or AMD is going to have a hard time on its hands and will probably end up cutting retail prices to offer a cheaper though lesser alternative to the Intel  developments.

AMD also plan to introduce a new six-cored range of chips called "Istanbul" sometime next year (2009). As for Deneb; AMD will probably be launching them before the end of 2008.

The Question is can AMD ever get ahead of Intel again? Can they even catch up; and if so is this their chance? What’s your opinion?

In this article I want to demonstrate and to talk about a simple potential divider.

What is a potential divider? It’s a device or a number of devices that divide a voltage potential.

In the first (1) of the diagrams below, we see a pair of resistors R1 and 2, dividing the voltage potential between the + rail and zero volts. This could also equally be accomplished with a single resistor or any number of resistors. I’ve used two resistors in the diagram so that there is a centre-tap where they connect together. (A).

The value chosen for the resistors in this circuit will affect the voltage at point A. These same values will also affect the amount of current flowing through the series-resistor pair; and therefore also will limit the current available at point A.

Let’s put some meat on the bones and give an example:

First let’s decide on a voltage for the + rail. 10 volts sounds a nice round figure.

Now let’s select some values for our resistors. How about we make both R1 and R2 a value of ten ohms? Let’s do just that.

So between the 10V rail and 0V (Zero volts) we have 2 X 10 ohm resistors connected in series; which gives us a total resistance of 20 ohms. How much current will flow through the resistor pair? To answer that we use Ohm’s Law:

Ohms Law says that I(current) = V(voltage) divided by R(resistance). Therefore 10 volts divided by 20 ohms = 1/2 amp, or 500 milliamperes (mA). This means that under these circumstances, if you were to connect an ammeter between point A and 0V (ground), it would give a reading of a half an ampere.(Amp.).

If you’re building the circuit you’d need to account for this: Resistors are available in a number of different wattages. In this circuit we need to know what wattage resistors 1 and 2 should be. If we use components rated at too low a wattage then they’ll get too hot rather quickly and burn out. We need to know the wattage that is used by the circuit:

Once again we turn to Ohm’s Law: Ohm’s Law describes wattage with the variable P, for Power – which is what wattage is; power. Ohm’s Law says there are two ways of calculating the wattage in a circuit:

The first of these is P = Isquared(R) ; Power = current squared multiplied by resistance.

We know that we have 1/2 amp of current, and we know that we have a total of 20 ohms of resistance: Therefore the power used  in the resistor-pair circuit is ( 1/2 x 1/2 ) =1/4 (0.25) x 20 = 2.5 watts.

We could also do this calculation the other way: Ohm’s Law says that I x V = P; current multiplied by voltage = wattage:

At point A we know that we have 1/2 amp, but we don’t know what the voltage is at point A:

What would be the voltage at point A? To calculate this we use the following equation:-

V = Vx(R2 / R1 + R2)

That means; voltage (The voltage at point A.) is equal to the voltage of the + rail, multiplied by the solution of the equation where the value of R2, in ohms, is divided by the value of R1 + the value of R2, both in ohms.

Since we know the value of all the variables in the equation, we can rewrite it thus:-

V = 10 x (10/10 + 10)

V = 10 x (10/20)

V = 10 x 1/2

V = 5 volts

Therefore we now have a voltage of five volts for point A.

Using Ohm’s Law we can say that 1/2 amp x 5 volts = 2 1/2 watts, or 2.5 watts: ‘Same answer.

When we select the physical 10 ohm resistors to build the circuit then we need to bear in mind that they need to be rated at a minimum of 2.5 watts. If we use a pair rated at exactly 2.5 watts they’ll be running at their limit; so we want to use a rating somewhere above that; let’s say 5 watts, bearing in mind that resistance decreases with heat, and we want our resistors to stay at as near 10 ohms each as is possible, so that we know what’s going on.

Having done so we can build the circuit by connecting two 10 ohm 5 watt resistors in series and connecting either end across a 10 volt supply. We know that the current used by the circuit is 0.5 amps; therefore we’ll need a power supply capable of delivering that amperage.

Basically, by building this potential divider, we’ve built a very primitive voltage regulator: We know that if we supply this circuit with exactly 10 volts at the correct amperage, we’ll get exactly 5 volts from point A. We also know that we can draw up to 0.5 amps of current from that point also.

The problem with this voltage regulator is that it’s too primitive: Whether we draw 0.5 amps of current or not, this circuit will always use 2.5 watts of power from the supply, even if we leave point A unconnected. – That’s going beyond the scope of this article though.

On a final note, let’s recap on what we’ve accomplished:

We designed a potential divider out of 2 resistors. Using Ohm’s Law we calculated the current flowing through those resistors in circuit, and we calculated the power that they would drain from the supply. We used that figure to help us choose our components, and we calculated the voltage at point A.

Although it might not appear at face-value to be so; we’ve actually just learned a very important part of analogue electronic circuit design: Once again, however, we’re going outside the scope of this article if we were to dwell any further on this.

Looking at (2) in the above diagram, I’ve added a device called a "potentiometer" (‘Obvious reasons?) or variable resistor, between the two resistors. (You’ll find potentiometers in a lot of places; even though you may not have realised it: For instance; when you turn up the volume on your stereo sound machine, if you turn a knob then you’re actually adjusting a potentiometer. The same goes for a TV volume control, brightness control, maybe even the bass and treble controls, or the graphic equaliser (Might be sliding potentiometers?) on your computer’s speaker amplifier?) Using a potentiometer this way you can adjust the output voltage at point A. – Another article perhaps?

There was a typo in this article; which has now been corrected.

I hope you find what you’ve learned both enlightening as well as useful; at least perhaps in the future if not right now.

A transistor is an electronic component. It is defined as an "active" electronic component because it is able to perform more than one function at more than a single level.

The most basic of transistors is the bipolar transistor. It has three connections; those being the base, collector, and emitter. A circuit-diagrammatical representation of the bipolar transistor is shown in Fig.1 below. The collector is the connection at the top, the base is the one in the middle, and the emitter is the one at the bottom with the arrow on it. In addition to the transistor’s amplification factor, the base and emitter act as a diode. (See Fig.2 (i) and (ii).)

The direction of the arrow indicates whether the type of bipolar transistor is NPN or PNP. (Which stands for Negative Positive Negative or Positive Negative Positive.) The difference amounts to the way that the transistor is connected in a circuit with regard to the DC polarity. This polarity is caused by the transistor’s substrate layers being doped with a P-type and an N-type substrate. (See table of links.)

Fig.3 shows a PNP transistor connected into a basic circuit. Fig 4 shows an NPN transistor connected into an equivalent circuit. The resistors in the circuit limit the current flowing through the device and set the device’s voltage potential point with respect to the supply rails. The capacitor drawn in with dotted lines is a decoupling capacitor which, along with R3, decouples the collector (PNP) or the emitter (NPN) to ground; limiting distortion in the output and/or compensating for any residual ripple present in the supply rails – depending upon its value combined with that of R3 giving a certain AC reactance. (A subject beyond the scope of this article.)

Resistor, R1, is connected as a DC current-limiting resistor in both cases, to the base of the transistor; limiting the base current which in normal operation should not rise above approximately 1/10th of the current flowing between collector and emitter. (As low as 1/100th is the preferred quiescent value for maximum amplification in most high-gain devices.) The differential between the two sets the transistor’s working amplification factor or beta. This is limited by the actual electrical characteristics of the chosen device itself.

This article cannot hope to go into the full details and various functions of the bipolar transistor under all conditions, and even the AC amplification operations of said device are far too in-depth to discuss in the space allocated.

For further information on this device please visit links in the table of links below.

Table of Links:

In electronics a resistor is a component that offers resistance to current. The main DC function of a resistor in an electronic circuit is as a current limiter; such as in the collector circuit of a single-NPN transistor stage, ensuring that the transistor doesn’t pass more current than its rated value causing it to overheat and burn out. It can also be used in conjunction with a capacitor to give a form of AC resistance known as reactance, the value of which is dependant upon the values of the chosen components combined with the AC frequency passing through the resistor/capacitor pair – However this article is concerned with calculating the DC resistance of resistors only.

Single resistors are available in a number of predetermined values. The unit of DC resistance is the Ohm; symbolised by the Greek letter omega. The value of the resistance of an individual resistor in Ohms can be read by means of the coloured stripes on the body of the component:-

Black = 0

Brown = 1

Red = 2

Orange = 3

Yellow = 4

Green = 5

Blue = 6

Violet = 7

Grey = 8

White = 9

Starting from the first and usually the largest stripe marked on the body of the component; the colour of this stripe indicates the digit furthest left. The next stripe indicates the next digit, the third stripe indicates multiplier to the power of 10 (Or put more simply; the number of zeros after the first two figures.): Thus a resistor marked red, brown, red would indicate that its resistance is 21 x 10 to the power of 2 – or 21 x 100 = 2100 Ohms; 2.1 kilohms. (2K1)

The last stripe(s); usually silver or gold, but maybe another colour, indicate the component’s tolerance or degree of accuracy at staying true to its marked value.

When 2 resistors are connected in a series circuit their total DC resistance can be ascertained by the equation

RT = R1 + R2

Therefore if we connect two 2100 Ohm resistors in series they have a total resistance of 2100 + 2100 = 4200 Ohms, or 4K2.

That’s pretty simple eh? But what if we want to calculate the resistance of two resistors in parallel? The equation becomes slightly more complicated. The total resistance of two resistances in parallel is calculated by the equation:-

RT = R1 x R2 / R1 + R2

Therefore a 2100 Ohm resistor in parallel with a 4200 Ohm resistor gives a total resistance of (2100 x 4200 = 8,820,000) divided by (2100 + 4200 = 6300) = 1400 Ohms; 1.4 kilohms. (1K4)

First I’d just like to confirm that this blog is not dead or dying: The reason that there haven’t been quite as many posts appearing lately as there used to be are computer problems the last two weekends and illness the weekend before.

A capacitor is, in basic terms, an passive electronic component that stores electrical energy between the tiny gap between two electrodes. They have many different uses; and it’s rare to find a piece of electronic equipment without a capacitor or capacitors in it.

One piece of electronic equipment where a lot of capacitors are used is your computer, where there are quite a few. You’ll find them on any board, and in particular inside the power supply unit (PSU). There’s a certain type of capacitor called an electrolytic capacitor that uses an electrolyte between the electrodes or plates in order to increase its capacitance. The circuit-board mounted electrolytic looks like a tiny bean-can of varying size and height on a circuit-board.

When an electrolytic capacitor is made with good-quality materials it has a fairly good mechanical rigidity despite its high potential for electrical leakage. When a capacitor is made on the cheap, however, problems occur, and its working life is limited. Unfortunately there’s no way to tell how long an electrolytic capacitor will last if you’re not familiar with the manufacturer. The electrical characteristics of any new capacitor will be as marked on the tin, in all but a very few cases. It’s only with use that things start to go pear-shaped.

Since around 1998, some motherboard and PSU manufacturers have been using Chinese-manufactured ultra-low-cost electrolytic capacitors in the construction of their products in order to keep costs down. This has led to something that is known these days as "capacitor plague":-

Capacitor plague occurs after around a year or more of in-circuit usage of a low-quality electrolytic capacitor. Because the chemical formula for the electrolyte used in the capacitor is basically the cheapest formulation that the manufacturer could get away with and have their product still function; eventually, with use, hydrogen gas bubbles build up in it between the capacitor’s plates. This increases the internal pressure within the sealed can. Eventually the pressure builds to a point and the electrolyte/gas bubble mix escapes any way that it can: This could be from the base of the component; but more usually from the top, where usually a weakness is deliberately built into the electrolytic capacitor’s can so that any venting electrolyte can escape from the top. The main reason why it’s better that it escapes from the top is so that a visual inspection of the faulty circuit easily identifies a leaking electrolytic.

(Of course things haven’t always been this way. Perhaps you might remember the old 1940/50s television sets, where due to lower capacitor technologies at the time, huge capacitors, literally the size of a bean-can or larger, were used in places without any place for the expanding electrolyte to go. Older readers perhaps may recall their television making a huge bang; blowing the back off and leaving a cloud of acrid smoke. That was just a capacitor reaching the end of its life – Usually rendering the whole unit inoperable and beyond repair.)

Many motherboards and cheap power supplies manufactured between 1998 and 2007 incorporate these cheap and nasty electrolytic capacitors, and many fail rather quickly. Commonly there’s no massive bang, although it has been known even in computers. More than likely the faulty cheapo capacitor either bulges, leaks, or both.

If a motherboard or other device suddenly begins playing up; take a look at it and check the capacitors: If the can on any given capacitor doesn’t look perfectly-formed, or if you see a rust-coloured mark on it, or both – usually at the top – then you can assume fairly correctly and with a good degree of accuracy that your device has capacitor plague.

There is a cure; but it could be rather long-winded and not worth the hassle, in replacing all the faulty and potentially-faulty electrolytic capacitors on a given board. This is usually ultra-geek work and requires a good and steady hand at soldering, plus some knowledge of electronics and electronic circuit construction in practical application. I’m trained to that level myself, in fact beyond that to military-grade soldering standards, but in my opinion it’s much easier to buy a new board than attempt the fiddly and not-always-successful task of replacing capacitors.

It’s a shame that this happens and has been allowed to continue unpenalised for a long time; but it does, it has, and there’s not-a-lot can be done about it. These days many manufacturers, Gigabyte in particular I notice, are manufacturing their products using non-electrolytic capacitors – Thus avoiding the chances of capacitor-plague. However some are still using electrolytic capacitors in the board’s component population, and, who knows, they could be made in China and cost ten-a-penny?

What do you think? Should companies continue to cut costs and quality too just for the purpose of making their products slightly cheaper? – After all it’s the customers who suffer, not the protagonists.

On September 12th 2008 the integrated circuit or IC became 50 years old. “So what..?” You ask:

Well the integrated circuit has come a long way in only 50 years. When you think that the first IC, measuring about 3 X 21 mm odd, consisted of a single transistor and a few other components combined to make a Hartley oscillator which produced a sine-wave trace on an oscilloscope; then compare that with the processor in your computer which combines millions of transistors into complex circuitry; all built on a small piece of silicon and surrounded by a metal envelope, and you’ll see what I mean.

The first integrated circuit was demonstrated by Jack Kilby at Texas Instruments on 12 September 1958.

The integrated circuit itself consists of a germanium strip, which you can see in the middle of the glass slide. It measured 7/16in by 1/16in.

In 1958 Texas Instruments was researching a new idea called the ‘micromodule,’ in which the components of a circuit all had the same size and shape, but still didn’t address the problems concerning high numbers of transistors.

In July 1958, Kilby confined himself to his lab alone where he came up with the idea of fabricating all of a circuit’s components with a single block of the same material.

Kilby went on to develop the first handheld electronic calculator at Texas Instruments in 1967, and collected a prestigious set of awards, including the Nobel Prize in physics, the National Medal of Science and the National Medal of Technology.

Had the integrated circuit not been invented; our computers today would be absolutely massive boxes containing complex hand-soldered transistor circuitry, pouring out heat, using more natural resources, and costing a fortune. That dedicated pioneering scientist 50 years ago brought us out of the electronic dark-ages and greatly assisted technological evolution.

To the left is the circuit-diagram symbol for an operational amplifier or op-amp; no doubt you may be familiar with it if you do any work with analogue electronic circuitry.

The standard operational amplifier has 2 inputs; non-inverting (+), and inverting (-). A voltage applied to the non-inverting input only will appear on the output as a swing from 0V to Vs. An equal voltage applied to the inverting input will appear as a drop from 0V to ground. Anything subsequently applied to the inverting input will appear as a reverse action on the output to an extent dependant upon the gain of the component* (See “Negative Feedback”)

The actual circuit for a standard operational amplifier such as the 741 is in fact quite complex, and can contain anything upwards of 24 transistors; configured as current-mirrors with a differential input stage, class A gain stage, output bias circuitry, and a final output stage.

The basic circuit in this article, however, consists of just a single criss-cross multipolar current-mirror with a push-pull output stage: -

I’ve purposely omitted the resistors in this circuit diagram  for the purpose of clarity. If you want to build the circuit use Ohm’s Law to calculate the value of the 4 resistors which should be connected between the collectors of Q1 & Q5 and rail, plus between the emitters of Q2 & Q7 and rail. Remember that the currents present should only be tiny; around 1mA or less. If you require more current then the output should drive another push-pull output stage with a larger load on it.

The diodes can be LEDs if preferred; which will then allow for a larger current to flow through Q5 to 8.

This is a deviation from the normal circuit; taking advantage of the reversed polarisation of NPN and PNP transistors to make a simple inverting/non-inverting comparator as the input stage, feeding a pair of push-pull amplifiers as the output stage. The diodes are present as current-sinks; which is why I suggest using LEDs in preference to rectifier diodes.

This circuit is very basic and won’t perform anywhere near as efficiently as, say, a 741 DIL op-amp package. It’s been drawn to show the basic comparator principle only of an operational amplifier in reversing the polarity of an input to the inverting input whilst retaining the polarity of an input to the non-inverting input.

Unlike a standard op-amp; connecting the inverting and non-inverting inputs together will not result in the circuit becoming an inverter: Rather, with this basic circuit, the output would then become a zero-volt reference – the output push-pull amplifier(s) acting as a potential divider – regardless of the identical polarity applied to both inputs.

For more information see http://en.wikipedia.org/wiki/Operational_amplifier

Connecting two transistors as a Darlington pair by connecting the emitter of the first transistor directly to the base of the second transistor multiplies the beta of the first transistor by the beta of the second transistor to give an extremely high-gain device.

Imagine, for instance; two NPN devices, each with a maximum gain of 50, connected in such a way, giving a device with a maximum gain of 2500: This would be useful for boosting the output of a piezo-microphone for example; notorious for its low output.

In the basic circuit-diagram of a Darlington amplifier below; the input is DC decoupled by C1, resistors R1 and 2 form a potential divider biasing the base of Q1 at exactly 0.7 Volts, R3 is the load resistor on the collector of Q1 – which drives the base of Q2; R4 restricting its collector load, and the R4/C2 combination decoupling its emitter to ground.

Since the base of Q1 is at 0.7 Volts, both Q1 and 2 will be in an “always on” state, and sensitive to any tiny ripple passing through the input capacitor C1.

Bearing in mind that the input ripple will probably be of only a few microamperes; the R1,2 pair should be selected with as high a resistance as possible – Within the megohm range to limit the current already present on the base of Q1 to a fraction of a microampere if at all possible.

Q1 should be chosen such that its base current need only be negligible for it to respond. With a beta of 50 the resistance of R3 should be within the range of around -2 to -40 times that of R1, so as not to drive the transistor into saturation.

Again, having a beta of 50; Q2 should be run ideally at between 2 and 40 gain.

Suggested component values to run the circuit at a voltage of 1.5 Volts are as follows:-

R1: 2M2, R2:(1M with 250K Lin. Preset in series.) R3: 100K, R4: 22K, R5: 110R

C1: 1uF 10V Elect., C2: 100uF 10V Elect., C3: 10uF 10V Elect.

Q1: BC107B, Q2: BC109C

Note: I haven’t built this circuit myself; and it’s been drawn up for demonstration purposes only: It’s very basic and wouldn’t give brilliant sound quality anyway, but should nevertheless “work”.

Today we’ll look at a very basic analogue audio-amplifier circuit and examine how the principle of negative feedback can reduce distortion:

Capacitor, C1, decouples the input to potential divider, R1&2, keeping the base of Q1 at 0.7 Volts – Therefore any AC ripple will be picked up instantly by Q1 and amplified. C2 is the first example of negative feedback: With a value of a nanofarad, it feeds higher frequencies from the collector of Q1 back to its base, thus giving a better response to bass and mid-range audio-frequencies than treble notes. Q1′s Iceo is set by the combined resistance of R3 and R4, and decoupling is provided by electrolytic capacitor C4 to ground. (C4/R4′s reactance should be taken into account when deciding the component values.)

The first stage is decoupled from the following stage by C3, and resistors R6 and R5 act as a potential-divider to set Q2′s base at 0.7 Volts. The output is taken from Q2′s collector via C6.

Let’s assume that the beta for each individual transistor is 100, and that we’ve chosen components that will allow each transistor to operate with a working beta of 50: Therefore a tiny signal of 10mV p-p applied to the input will give a massive output of 1 Volt p-p. – That’s quite some amplification; and not only will the clean signal be amplified 100 times but so will any distortion on the input waveform. Second to that, if we’re amplifying an audio signal with variable amplitude on the input; there are going to be occurrences where this circuit; particularly Q2, is driven to saturation – A 2.1 Volt p-p signal from a preamplifier at the input, for example, producing an output of 20 Volts p-p with a flattened crest where Q2 saturates – Thus producing distortion.

To compensate for the above unwanted distortion we introduce negative feedback:

With the introduction of PR1 and C7 we have allowed a regulated path back to the input from the output, DC decoupled by C7: Therefore the higher the resistance of PR1 the less the entire two-stage circuit is bypassed; therefore the more current flows through Q1 and 2 and is amplified.

This is a similar function to that of R2 in the basic DC inverting operational amplifier circuit below:

Again we note that should the voltage on the inverting input at the virtual-earth junction of R1/2 reach a certain level with respect to the supply voltage it will cause IC1 to saturate. R2 gives negative feedback between input and output, therefore limiting this effect. 

In very basic terms then, the function of negative feedback is to reduce overall gain, thereby limiting distortion.

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When designing a single-transistor-amplifier stage; be it an AF, RF, IF, whatever project, one must always take into account the gain of the transistor in question and design for it accordingly. It’s best not to attempt to utilise the full amount of available gain as the transistor will probably saturate and cause distortion. In fact a tri-stage amplifier with negative-feedback, (The subject of a later article maybe.) with only half the available gain of each transistor used will produce a much better and less distorted output than a single-transistor stage utilising all of the available gain.

For now, however, I’m going to concentrate on just a single-transistor stage, and on correctly biasing the transistor’s base proportional to its base/collector.

Let’s assume that our subject transistor has a voltage-gain of 20, and a Vb(max) of 3.3V. Let’s place it in circuit:

Ignoring the type and strength of input signal as far as this example is concerned, we intend to bias the base of the transistor from the supply rails using a straightforward potential divider (R1 & 2). We know that the base voltage must not exceed 3.3V, and we’re running the circuit from a 6 Volt rail. We use the equation V=Vin (+Vcc in this case, using a NPN transistor.) X R2 / (R1 X R2). In the example above, (Which is an example only rather than a functional circuit.) I’ve taken a guess and used a 1K negative resistor (R2) and a 2.7K resistor (R1) to the rail. This gives a working base voltage of 2.223 Volts. In most cases we’d tweak that voltage, by altering the resistance values, to as near 0.7 Volts as possible. (The transistor’s transconductance threshold, assuming that we are using a silicon rather than a germanium transistor.) For this example we’ll leave it at 2.223 Volts.

Note the use of resistances in the kilohm range: This is for the purpose of limiting the base current. Although we haven’t specified a current for the input signal we’re assuming that it’s in the several-tens-of-milliamperes range. When dealing with even smaller input signal currents use higher resistance to limit the biasing current further so as not to interfere with the input signal.

We can work out the base-bias current precisely using Ohm’s Law: I=V/R; 6 volts divided by (R1 + R2 = 3700Ohms) = 1.62 milliamperes: We’d therefore be looking for a gain of about 10, driving the transistor at half-available-gain, so it would be fair to say to use a total DC resistance of 100 Ohms in the emitter circuit, and a total DC resistance of 270 Ohms in the collector circuit, ignoring any reactance from any decoupling capacitor (Such as C3 in the second circuit.) in the emitter circuit.

Looking at that second circuit we note that there are DC decoupling capacitors (C1 & C2) on the input and output of the stage: these are included to prevent any DC component bleeding back into the stage before, as well as bleeding off from the collector circuit into the following stage. Note that I’ve sketched in a decoupling capacitor into the base circuit using a broken line: This may or may not be a good idea depending upon many factors; and that’s well beyond the scope of this post.

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Today I  took the wrapping off of and opened up the September edition of Custom PC http://www.custompc.co.uk/ : I’d received it in the post some days, maybe even a week ago; but I hadn’t had time to study it or even look at the cover yet. It flipped open in the “Hands On” section; and there was this article on how to make a PC monitor from an analogue meter.

I have reproduced page 1 of their article. Not intending to break their copyright I have deliberately made it hard to read; yet clear enough to be able to see their circuit and get some idea what the page looks like. I’m afraid if you want to see the article properly you might just have to purchase the magazine, unless they have featured it on their website.

Let’s face it though; there’s probably not a lot of genuinely old-ish analogue meters around; and a decent one can cost anything from a few £s upwards. What’s more an analogue meter is so hard to read if it’s moving at all.

My brain engaged in design mode at this point: Why use a meter at all? Yes it’s maybe nostalgia and “geekracy”, but why not instead have an LED bar-graph? Just for the purpose of being ultra-creative why not make it from individual LEDs rather than a prebuilt bar-graph?

OK easy enough; but first that circuit – Theirs is so basic it sucks:

Their circuit has a signal-diode connected cross-path without anywhere to dump the current passing through it except into the diode’s semiconductor itself: That’s a recipie for disaster if the input signal has a large amount of gain: The signal-diode’s semiconductor material will overload and burn out: In their circuit that diode is connected as a half-wave short circuit.

My remedy is to add a 100 nanofarad capacitor in series with the diode on one side to block any DC component of the signal, and a LED in parallel with the capacitor to indicate any DC component’s presence as well as to act as a partial sink. The greatest current sinkage, though, comes from series pair resistors 6 and 7, which I’ve rated at 1/4 watt just to be on the safe side: Those should sink any remaining AC component of the signal and not get too warm.

Resistor R7 is connected as a current sink and limiter for the entire circuit.

The main deviation from the original design comes where I’ve replaced the analogue meter with 4 resistors and 4 LEDs. It is possible here to have as many series resistor/LED pairs here as you like to a certain extent; bearing in mind the current drain this will cause with respect to the current gain of the input signal.

I really don’t want to get too technical here or I might end up drawing bode plots and nyquist diagrams; whatever.

This isn’t a final circuit, I’ve not built it, and it would be an idea for you, if you’re an electronics hobbyist, to play around with it, adding more LED/resistor series pairs whilst experimenting with the values of individual resistor – Even going into the specialist series with fractions of an Ohm values if you like. Also experiment with the value of R7 and see what effect a higher resistance has on the overall performance of the circuit. An 8 LED bargraph would be excellent and in my opinion would look much more geeky than an analogue meter.

Having looked at this again after publishing I think I might have actually negated the original function of the diode to some extent by removing the DC component from its circuit. Removing or short-circuiting C1/LED1 will reintroduce this DC component back into the equation; in which case now that we have resistors 7 and 8 in circuit they will help sink the DC as well as the AC waveform component, but with the result of a little more heat. (Negligible.) In an ideal world it would be an idea to use two inverting buffers incorporating an op-amp such as a 741; one on each input terminal. These could be powered from the 5V USB supply – But now I run the risk of getting overly technical if I were to go any further; and I see why they used the circuit they did: Not so much built to last as built to work, on a KISS principle: Keep It Simple Stupid. The full redesign required to make this piece of kit perfect as I’d like it to be would need a lot of components, and would further complicate what started out as a simple circuit.

So yes build the circuit and leave out R6 and C1/LED1. It’ll work; but if the input gain goes really high it could end up damaging D1 as I said. Also there’s the added factor that D1 won’t pass a current if the voltage-gain on the input drops really low. (Below Approx. 0.35 volts, which IIRC is the threshold voltage of that particular diode.

OK here’s a better solution: Disonnect D1 and connect it in series with the 10 microfarad capacitor. cathode to its + terminal. Connect another capacitor’s + terminal directly to the common anode of the LED bar display and via an identical series-connected diode with its anode to the – terminal to the in-circuit side, (Bottom) of R7: Now we have two properly polarised capacitors storing the full wavelength.

I’ll alter my circuit diagram for the purpose of clarity. I’ve crossed out the bit of circuit we no longer need and I’ve drawn in the second series diode/capacitor, connected as I said:-

Revised Circuit Diagram

Ok; you get my drift? We added another diode and a second 10uF capacitor; each stores half of the incoming waveform and is kept polarised by its series-diode in both cases: I should have thought of that first time round. There’s no stress whatsoever on either diode now, and the need for a cross-path diode has been eliminated. Consequently this circuit will last much longer because none of the components are stressed in any way.

Forget all previous designs; build this: Experiment with different values for all the resistors, and enjoy the display.

Imagine a computer…

Kustom Komputa will make it a reality. 

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I once tried a 10 watt resistor in series with the exhaust fan – I forget the value in Ohms that I used – which seemed to do the trick: The only problem being that a 10W resistor is fairly large, (I think the one I used was a wirewound ceramic out of the EHT circuit of an old TV.) not really cost-friendly if bought new, and can also generate more heat too depending upon its resistance.

I’ve also tried two fans in series but that was virtually a non-starter.

An idea I’ve just thought up this second, is to have a separate transformer with a 15 Volt or higher: 15 to 20 Volts should do the trick, secondary winding – It only need provide 500mA of current Min. – and use the AC half-wave potential to drive 2 fans at a lesser speed thus:

Connect a rectifier diode (I suggest a 1N4002) in parallel with small 10 nF capacitor across each fan opposing the polarity of the fan’s motor. e.g. connect the anode of the diode to the fan’s negative connection and vice-versa in each case.

Now connect both fans, including the two components that you’ve just connected in parallel with the fans as described above, in series so that the two negative leads from the fans are connected together/the two diodes’ anodes are effectively connected to one another via the fans’ series connection.

Connect both positive leads from the series-connected fans with the extra components added across the secondary winding of the transformer; it doesn’t matter which way round.

You should see that on power-up that both fans spin slowly and quietly. 15 Volts may not be quite enough voltage depending upon the type of fan used; so I would use an 18V secondary if possible.

How it works:

What you have flowing directly out of the transformer’s secondary winding is AC or Alternating Current. Alternating current changes polarity a given number of times per second depending upon its frequency – In this case 50 or 60 Hz.

When the polarity is one way round current flows through one of the diodes thus bypassing the fan that the diode is connected across. However since current can’t flow through the diode in series with it, due to it being against its own polarity, it flows through the other fan and causes it to start to turn. When the polarity is the other way round on the following AC half-cycle the opposite happens: The diode that was conducting no longer conducts, causing the electricity to flow through the fan; however the diode across the other fan that was previously not conducting and bypassing that fan is now able to do so, thus bypassing the fan that was originally active. The capacitors are just for interference-limiting, decoupling, and smoothing in a small way. Their reactance in this circuit is so miniscule as to be discounted and no effective alternative path is provided.

So basically the circuit that has just been constructed is sharing the entire available power between the two fans; thus causing each to run at half-power, effectively sharing the 18V between them at roughly 9V each.

As I say I’ve only just thought this up just before I wrote all this, so there are no guarantees: If you’d like to build the circuit and power up then please do: I’d be interested to hear your experiences.

If it works well and you like it then please feel free to describe my idea as FANtastic. (No fanny comments please.)

The italics above are to indicate a piece of text that has been altered: There was an editing error which left the text not making sense. This has now been corrected. Apologies.