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.