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.

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.

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.

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.

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!

Advertisment : Word Press Link Cloak: Easily disguise and protect your affiliate links to increase your revenue. Click Here!

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.

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.

Advertisment:

Buy “WordPress on Crack” – Build your own WordPress plugins:
Click Here!

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.

Advertisment : Word Press Link Cloak: Easily disguise and protect your affiliate links to increase your revenue. Click Here!