# Is voltage divider output affected by load resistance?

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!

## 3 Answers

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.

• A few days ago, I remember seeing a circuit using the voltage from the divider to 3 difference IC's as reference. (dont remember where I saw it< but it definitely did this). Wouldn't the voltage across each reference be different? – Sherby Jan 18 '14 at 18:06
• @Sherby When you wire the middle point of a resistive divider to three IC pins then all three IC pins may influence the voltage of the node based on the load they introduce but in the end they are all connected to the same node and share the same voltage, it's not possible for any of them to have a different voltage that the others. – alexan_e Jan 18 '14 at 18:31

doesn't this parallel circuit reduce the total resistance across the load resistor causing the output voltage to be less than the desired 24v?

Indeed it does. Perhaps the most illuminating way to see this is to form the Thevenin equivalent of the voltage divider (without the added parallel load).

The actual circuit: simulate this circuit – Schematic created using CircuitLab

The Thevenin equivalent circuit is simulate this circuit

By Thevenin's Theorem, the voltage across an added load resistor will be identical for both circuits.

It's easy to see in the equivalent circuit that an added load $R_L$ forms a voltage divider with the equivalent resistance and so, the voltage across $R_L$ must be less than 24V:

$$V_L = 24V \frac{R_L}{50k\Omega + R_L} \lt 24V$$

Yes, loading the output of a voltage divider will lower the output voltage.

One way to look at this is to characterize the voltage source produced by the voltage divider. It so happens that any ideal voltage source with some resistive network after it can be modeled as a single voltage source with a single resistor in series. This is called the Thevenin equivalent. It is often useful to simplify more complicated voltage sources into their Thevenin equivalent.

In your case, you have: Which as you say will produce 24 V when left open. The Thevenin equivalent of what you have is: Note that from the point of view of any circuit connected to Vout, these two sources can't be distinguised. That's what Thevenin equivalence is all about.

Now it should be obvious what happens as the source is loaded. It will be 24 V with 0 current. Any current drawn will also flow thru R1, which causes a voltage drop accross R1, which subtracts from the 24 V source voltage as seen by the output. This relationship is linear. We already have the point with 0 current. Another point that gives us some intuition is the 0 voltage case. If you short Vout, all the voltage will be accross R1. By Ohms law, the current thru R1 is 24 V / 50 kΩ = 480 µA. Therefore as the current drawn from Vout varies from 0 to 480 µA, the Vout voltage vary linearly from 24 V to 0.

As one more example, if you draw 100 µA from Vout, then R1 will drop 50 kΩ x 100 µA = 5 V, so Vout will be 24 V - 5 V = 19 V in that case.