Every explanation I've seen so far about voltage dividers, only consider the two main resistors (that do the "dividing"), and pretty much ignore what happens after Vout.

The circuits in the explanations look more of less like this:

Vanilla Voltage Divider

In this example the voltage at Vout should be 2.5V.

This really gives the impression, that whatever goes on for the rest of the circuit, Vout is at 2.5V.

But what happens in practice, if I get something connected to Vout? E.g, I connect some components, and their equivalent resistance is 100Ω:

Voltage Divider with load

Since R2 and R4 are in parallel, an equivalent resistor would be of 50Ω. Now, if that's correct, then the voltage divider really operates between a 100Ω and a 50Ω resistor. I.e. the voltage at Vout would be 1.67V and not 2.5V, as I originally wanted it to be.

Is this really how it works?

How should one design the part that connect to Vout to avoid messing up the divider?

  • \$\begingroup\$ As a side note, the loaded voltage divider is sometimes done on purpose to approximate an audio-taper potentiometer using a linear one. The pot is connected as a variable voltage divider as usual, then a loading resistor is added between the wiper and ground. Ignoring impedance variations, this combination comes acceptably close to audio-taper in a lot of applications. (much better than linear, anyway) S-curves can be made with two loading resistors that form their own divider in parallel with the pot. \$\endgroup\$
    – AaronD
    Commented Feb 11, 2015 at 16:11
  • \$\begingroup\$ A voltage follower opamp after the divider is one way. \$\endgroup\$
    – rioraxe
    Commented Feb 12, 2015 at 6:04

4 Answers 4


How should one design the part that connect to Vout to avoid messing up the divider?

A simple voltage follower:


simulate this circuit – Schematic created using CircuitLab

Q1 conducts Vin to load while Vload < (Vdivide - VbeQ1)
Edit: As Andy_aka pointed out, Vload <= Vdivide - VbeQ1 (~0.7V)

  • 1
    \$\begingroup\$ What about the 0.7 volt biasing drop across base and emitter - this will produce a significant error. A 2.5 volt input will look more like 1.8 volts on the emitter. \$\endgroup\$
    – Andy aka
    Commented Feb 11, 2015 at 8:34
  • \$\begingroup\$ Good point; will need to set Vdivide accordingly. \$\endgroup\$
    – Jon
    Commented Feb 11, 2015 at 8:39
  • \$\begingroup\$ Am I thinking in the wrong direction, or is this still load dependent as the base current will flow through the top resistor but not through the bottom one creating an additional error? -> I did some research (my basics are all gone) and well, the input resistance is approximately \$\beta_0\$ * Rload, so it's not as bad as I thought, but might be considered as well. \$\endgroup\$
    – Arsenal
    Commented Feb 11, 2015 at 9:49
  • \$\begingroup\$ The input resistance at the base node is (hie + hfe*Rload) with hfe=β and hie=r,be. \$\endgroup\$
    – LvW
    Commented Feb 11, 2015 at 9:58

Yes this is how it works. The load needs to have very high impedance so that it doesn't affect the voltage division. An OP buffer could be used between R2 and R4.

Perhaps using a zener diode as a voltage divider could be of use here. This picture is from the site http://www.electronics-tutorials.ws/diode/diode_7.html which explains this further. Zener diode as a voltage divider

Depending on which zener diode you're using the voltage, Vz, will (under the correct circumstances) be the same independent on the load, Rl, connected to the circuit.


Yes, that is really how it works - the load does appear as a resistor in parallel with the bottom voltage divider resistor.

Voltage dividers should only be used to provide a reference voltage to a high-impedance load.

They are very inefficient when used to provide a reduced voltage to a low-impedance load since the currrent through the divider must be ten times the current delivered to the load to hold the load voltage within reasonable limits.


You are quite correct and, when taking your load into account, the top resistor R1 would need to reduce to 50 ohms to accommodate the extra load you have or, use a voltage reference chip that provides a stable 2.5 volts up to several milli amps being drawn from its output.

Alternatively use an op-amp to buffer the output of the divider - this is basically how a voltage reference works.

The same applies to capacitive dividers when trying to create an AC voltage ratio that is precise. It gets more complex when resistor loads are added to capacitive dividers though.


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