Voltage divider - measure resistance of load

Thread Starter


Joined Oct 24, 2011
Hi everyone

I have build a voltage divider from to resistors R1 and R2 (as shown below). I calculated R1 and R2 to let me have Vout=3V. Then I attached a load and soon realized that Vout changes under load.

Then I thought of modelling the whole thing as a voltage devider with R2 and the RL1 in parallel. After reading a bit this turns out to be the thing to do. But what value to use for RL1 (and maybe RL2, please see discussion below) ?

I've tried to abstract my circuit as much as possible, if it's too abstract and I've thrown away features, please let me know and I'll draw it in full detail.

Basically RL1 is PCB from a wireless doorbell receiver and the wires 1 and 2 are there to supply this pcb with 3V as was normally supplied to the doorbell receiver by 2x 1.5V batteries.

Wire 3 and 4 are two wires (ground and positive) which originally led to a LED in the doorbell (supplies ~3V to the diode). Now the ground is connected to ground instead of a pin on the LED and positive is connected to the base on a NPN-transistor in a 555 timer circuit.

Wires 5 and 6 supplies the 555 circuits (one monostable and one astable) with 9V and is hence connected to Vcc = 9V and ground.

I simply do not know where to measure the resistive value of RL1 (and/or RL2) so I can perform my voltage devider calculation.

Should I connect my ohm-meter to

  1. Wire 1 and ground, and let everything be connected.
  2. Wire 1 (disconnected from the voltage divider) and ground, and let everything else be connected
  3. Wire 1 and wire 2 with wire 4 connected to RL2 (555 timer circuit)
  4. etc
And what should be powered when I perform the measurement?

Thank you very much in advance



Joined Feb 19, 2009
Is there a reason you do not want to use a voltage regulator?

Resistor dividers are only marginally good for known loads, with the voltage lowering every time the 555 switches/changes state. A regulator for $1 gives a steady voltage source for up to 1A of current.


Joined Oct 23, 2010
It usually requires a more sophisticated measuring technique to determine the input resistance of an active device than by simply using an ohm meter. And circuit loading is a fact of life in electronics that must be accounted for by one method or another. And it is also important to determine how important of a factor it is for the reliable operation of your circuit. In other words 'do you really care' if the loading of RL1 on your divider occurs and how much is tolerable. Only you can answer that.

A way to determine the resistive value of RL1 is to calculate it from an indirect measurement. By measuring how much the voltage across RL1 decreases from a no load condition (RL1 disconnected) to a load condition (RL1 connected), you can calculate the effective input resistance of RL1.

Just a thought.


Joined Jul 17, 2007
You can use an LM317T or LM317L to get 3v out. Use a 120 Ohm resistor from out to adj, and 170 ohms from adj to gnd. 150+20=170; there are many other combinations that will get you close.

Be sure to use a cap on the input and output [eta] to ground. 0.1uF/100nF will work very well. If you don't use the caps, the regulator may randomly oscillate at high frequency, and that will be hard to diagnose without an oscilloscope.
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Thread Starter


Joined Oct 24, 2011
A voltage regulator seems to be what I need. I am a physicist, not a electronics engineer, so I didn't think about such a component / even knew it existed :)

wmodavis: your method for determining the RL1 value seems like a good approximation. And regarding your "question" if it was important for the circuit operation or not, it turns out not to only matter whether the V_out is in the range 1.8-3.2 V. So maybe I can estimate the RL1 load good enough as to get V_out to go between 1.8V and 3.0V.

SgtWookie: Thanks for suggestion a voltage regulator.