I recently made a voltage divider that uses a 10 ohm (top) and ~90 ohm (bottom) to divide a source voltage of 5v to 4.5v. The resistor values I used was somewhat arbitrary, in that I just picked a random top resistor value based on the resistors I had and calculated the bottom one.

My question: is there any advantage to use higher value resistor pairs over lower value resistors? Would a 100 top and 900 bottom have been better, for example? Or will the circuit work the same regardless of the resistance as long as the ratio remains constant?


Theoretically, any correct ratio works but of importance is not loading the 5V source too much that it sags a little i.e. don't choose resistor values that are so low that what you measure changes due to the introduction of the two resistors.

Next is consideration of what is measuring the output from the potential divider - you don't want the resistors to be so high in value that the impedance of the instrument or circuit changes the ratio. For instance an oscilloscope might have an input resistance of 1Mohm - using a 100k top resistor and a 900k bottom resistor will not be good. Try to aim for 0.1% (just a figure that is reasonable to me) so that the bottom resistor would be about 900 ohms and the top resistor 100 ohms.

Some DVMs are 10Mohm input resistance so a divider formed from 1k and 9k would be OK.

  • \$\begingroup\$ Just to clarify, if the load I'm providing the voltage to is.. say 1Kohm, the optimal values of resistors would be 0.1% of that and therefore 100 ohm and 900 ohm? Or am I misunderstanding? \$\endgroup\$ – MichaelK Jul 27 '14 at 14:34
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    \$\begingroup\$ A load (attached to the pot div output) of only 1kohm means really small value resistors of 0.1 ohm and 0.9 ohm. A load that is 1Mohm would mean 100 ohm and 900 ohm to avoid too much of a loading error (the 1M is in parallel with the 900 ohms) \$\endgroup\$ – Andy aka Jul 27 '14 at 15:08
  • \$\begingroup\$ My bad, I may have forgotten how to use percentages for a minute there... /facepalm. Thanks for the answer! \$\endgroup\$ – MichaelK Jul 27 '14 at 18:50

It's a simple question which can be resolved through Thevenin's Theorem. It all depends upon what is your load (Amps) going to be on 4.5V effective new source and how much voltage drop after loading you can afford. Draw an equivalent circuit using thevenin's theorem which represents a voltage source of 4.5V and a series resistor which is a parallel combination of the two resistors you use. You would need to use the resistors in same ratio as used by you (1:9) and the drop you can afford in the equivalent series resistor. Ideally speaking you need to know your load requirement at 4.5V and calculate its effective load resistance. (Say you need 10 mA at 4.5V then equivalent load resistance is 4.5/10e-3 = 450 ohms). Calculate the parallel combination of the lower resistor (90 ohms in your example) and make an equivalent circuit to calculate the top resistor. Of course you need to keep the loading of 5V source in mind which should feed more into your load and less into the bottom resistor.


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