# 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!

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.