What not to do...

Conversation I recently had:
ME: "so whats the problem?"
Student: "OK, so I'm running 5V to my logic circuit using a potential divider and......"
ME: "....pardon?"
Student: "yeh...can you do it that way?"
ME: "errr no...not really"
Student: "oh, ok.....why not?"

<Chris Proceeds to whiteboard and beckons student over.>

Why not to use a potential divider as a power supply

What is a potential Divider?

A typical PD circuit

A potential divider or voltage divider, produces an output voltage (Vout) which is fraction or ratio of the input voltage (Vin). This doesn't have to be 2 resistors, it could be as many as you like (look up ADC's and R-2R ladder networks), or even have capacitor or inductive elements, but for now, We'll just look at simple resistors.

Further descirption available on Wikipedia

Some Maths

Assuming that there is no load at Vout (i.e. no current flowing to Vout), the Current (I) flowing in this circuit can be calcualted using ohms law as the total resistance of the circuit divided by the input voltage:

I = Vin / (R1 + R2)     -eqn 1.

So we can work out the voltage drop across R2, which is Vout:

Vout = I x R2     -eqn 2.

Substituting eqn 1 into eqn 2 gives:

Vout = Vin R2 / R1+R2     -eqn 3.

This is the equation to work out the output voltage of the potential divider. Here is an online calculator to do the leg work for you.

So why can't I use it as a voltage reg?

Take Kirchoff's Current law, the sum of the currents flowing into a point equals the sum of the currents flowing out. So if you hook a load to Vout and draw current you can derive the total current as:

I = I1 + I2     -eqn 4.

Case study

So say you want 5V as Vout and you input voltage is 12V. This means you can use a 7k for R1 and a 5k for R2 (not E-series, but it makes the maths work) - that's ok. But what happens when you specify a current for I1?

For this I will use SIMetrix (a Free circuit simulator, worth a look, definitely read the manual and tutorials) to simulate what happens rather than explain:

the PD circuit (Ignore X1 its for the POP calculations)

Probe1 = Vout, IPROBE1 = Total circuit Current, IPROBE2 = Current in R2. I1 is a constant current source, this simulates the circuit that would be connected.

I1 is set to 0.5mA - a fractional amount of current, yet when I run the simulation:

PROBE1 is the red line, IPROBE1 is the blue and IPROBE2 is green

Though you can't see it, the IPROBE2 reports the current through R2 is about 700uA and the total current in the circuit is about1.2mA, but also notice how the voltage Vout is now only just about 3.54V!

If I increase I1 to 1mA, it gets worse for your "regulated" supply!

Now Vout is only about 2V! So why is this? Well your pulling more current though R1, which gives rise to a greater voltage drop across it. Given that:

Vout = Vin - VR1

Then Vout will get smaller the greater VR1 is!

I did try ramping the current I1 up and did get some silly results. 

To conclude

Don't use a potential divider as a voltage regulator - it just doesn't work! Leave their applications to reducing large voltage to a smaller voltage for measurement and setting ratioed voltages & references.