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.