Very basic question. Assume a voltage divider where both divider resistors are equal value, i.e. 100K. If 48v is the applied voltage, it's easy to see that we'll measure 24v across the "load resistor". However, when we add a load in parallel with the load resistor, doesn't this parallel circuit reduce the total resistance across the load resistor causing the output voltage to be less than the desired 24v? Thanks for enlightening a non-engineer!

# Electronic – Is voltage divider output affected by load resistance

voltage divider

#### Related Solutions

By your values, I have to assume that you are measuring off of the 10kOhm resistor on the bottom. I would hazard a guess that your 10kOhm resistor is a little high and the 100kOhm resistor is a little low. This would mean if your 10kOhm resistor is taking more of the voltage in the voltage division than expected.

There are a few problems at work here.

The first is the resistor value error. Ohm's Law: V=I*R. So if there is 5% error in R, then for a constant I, there will be a %5 error in the expected value of V. The %error from resistor tolerances are the maximum +/- error. So in reality, there could be a resistor with +5% error and other with -5% error. So there is a much larger possible range in output voltages for a constant current.

- This can be eliminated by either hard-coding the real resistances in software or adding a reference voltage that is more accurate than the resistor or A/D quantization error. This reference voltage would be used to get the real resistance values as the A/D sees them.

A/Ds can have an offset error. This really has to be calibrated out. Some A/Ds have this built in.

A/Ds also have quantization error. So if a voltage falls between two consecutive quantization levels, it will have to be rounded to one of those two, introducing error. There are ways of increasing the number of bits for an A/D by oversampling and averaging blocks of samples into one sample.

There are other errors that could be at work, but those three are the main ones I have run into. If the A/D is fairly linear, then measuring two accurate voltages with the A/D circuitry would let you build an affine linear equation to correct the data. This builds off of the problems in 1 and 2.

A/Ds can appear linear, but end up being very nonlinear at a particular range. I have heard of some implementations using a look-up table to correct the nonlinear behavior. But that is getting a little beyond your current problems.

**Edit**:
One more item. It is a good idea to buffer your analog signals to the A/D. It does a few things, like add another device to protect the microcontroller, and limit any possible transient sampling behavior from the A/D go into the analog signal.

What do I take into consideration when designing the circuit?

Think it through.

- The load voltage will be highest when the current is lowest
- The load voltage will be lowest when the current is highest.
- You're given the high and low load current specification as well as the high and low load voltage specification.

So, you have two points on the load line. This will allow you calculate the equivalent resistance (Thevenin resistance) of the voltage divider as well as the voltage division factor \$\frac{R_2}{R_1 + R_2}\$.

Can you take it from here?

I edited the question to remove the current fluctuation part. I was just warned to keep it in mind, it wasn't part of the actual question.

Since there are two resistor values to pick, you need two independent specifications. Certainly, one of those specifications is that the voltage must be within 20% of 6V when the load current is 500mA.

If you form the Thevenin equivalent circuit of the 15V source and voltage divider resistors, the voltage across the load resistor is seen to be given by

$$15V \frac{R_2}{R_1 + R_2} - I_L\cdot R_1||R2 $$

So, you need another specification. One approach would be to limit the open-circuit voltage to be 20% higher than the nominal voltage.

$$15V \frac{R_2}{R_1 + R_2} = 7.2V$$

Now you have two equations and two unknowns:

$$\frac{R_2}{R_1 + R_2} = \frac{7.2}{15} $$

$$R_1||R_2 = \frac{7.2 - 6}{0.5A}$$

This is just an example of choosing an additional reasonable constraint in order to uniquely specify the resistor values. You might reasonably choose a different additional constraint.

## Best Answer

If you put a load (= a resistor) in parallel with the 'lower' resistor of your voltage divider the output voltage will indeed be lower. Did anything lead you to believe it would be otherwise?

Let's say your load resistor is 100K too. It is in parallel with the 100K resistor of the divider, so together they are effectively a 50K resistor. The total circuit now consist of a 100K resistor and a 50K resistor, so the output voltage will be 1/3 of the input.