I was thinking of building myself a Kelvin-Varley divider as described for example in this article about a mini metrology lab in the March 1996 issue of Electronics Now by Conrad R. Hoffman.

The normal layout is well described in the following illustration on Wikipedia. One stage for every decade you want to divide, with decreasing resistor values for every new stage. The total resistance for the new stage must equal to twice the resistance of a single resistor on the previous stage.

Kelvin-Varley example from Wikipedia:

Kelvin-Varley example from Wikipedia

However, because you can't easily get for example 400 Ohm resistors, and because of resistor tolerances and the importance of matching the total resistance, it seems customary to select a value that's slightly higher, then connect a shunting resistor and a trim-pot in parallel with the whole stage, to get the resistance down to whatever is necessary for the previous stage.

What would be wrong with simply using the same value of resistors for every stage, and using a shunt with a trim-pot to get the combined resistance down to the required range again:


simulate this circuit – Schematic created using CircuitLab

There are at least three benefits here:

  1. The tedious job of hand-selecting 11 closely matching resistors for each value is reduced to sorting a lot of the same value into bins.
  2. No need to buy hundreds each of odd value resistors when you only need 11 of each. The parallel and serial shunt is from E24 and does not have to be matched with everything since they are trimmed out.
  3. It's more modular. You can construct any number of these and chain. (Of course, the accuracy is still limited by the first segment.)

Perhaps there is some hidden drawback with this scheme. Will it be more sensitive to temperature? More noise?

  • \$\begingroup\$ I don't see any problem with your approach. As you noted, the crucial point is that "The total resistance for the new stage must equal to twice the resistance of a single resistor on the previous stage:" Your trimming achieves this. \$\endgroup\$ Commented Dec 20, 2018 at 7:39
  • 1
    \$\begingroup\$ It's a KeLvin-Varley divider by the way. \$\endgroup\$
    – Bart
    Commented Jul 8, 2021 at 7:56
  • \$\begingroup\$ @Bart Yep, I've been meaning to fix the title for years, but didn't want to bump it until I was back to thinking about the project again. :) \$\endgroup\$
    – pipe
    Commented Jul 9, 2021 at 20:09

2 Answers 2


I've not seen this type before, but here is a 3 minute answer.

The arrangement uses 2P10T switches to select the voltage drop across 2 of 1011 R's in a string. The resistance of each stage decreases by a factor of 5:1. Thus 20k is shunted by 10 11 x 2k = 22k and 2k and so on.

If you change the R ratio from 5 to 1 then the the spreading between dual poles might need to increase to 10 then you need a 2P20T rotary switch for each and twice as many resistors

With this design the tolerances are not equal. The sensitivity to mismatch error is reduced by the number in the string=10. The requirement to be 1/5th the previous Digit value becomes LESS critical by 5 so you can choose parts for each digit 0.1% 0.5% 2.5% and get 0.1% accuracy

Commercial examples of Kevin-Varley instruments enter image description hereenter image description here

  • \$\begingroup\$ Here is my model that proves the design. Your suggestions do not work. tinyurl.com/ybozn2jx It's a clever design. \$\endgroup\$ Commented Dec 20, 2018 at 0:26

I know this is old, but since there is no accepted answer, I'll give it a shot anyhow.

First, as you correctly say, the accuracy is limited by the first decade(s). So the reward of selecting resistors is largest in the first couple of decades. Conrad also doesn't select resistors for the last couple of decades but uses 1% resistors out of the box. He does use shunts to trim each decade, though. So if you had more sets of matched resistors, you would always use them in the first decade when building multiple units.

Second, and that is maybe the more important aspect, consider the errors when building a decade out of a shunt R_S and a ladder R_L: R = R_S * R_L /(R_S + R_L). The error is then dR = (R_L / (R_S + R_L))^2 dR_S + (R_S / (R_S + R_L))^2 dR_L. So the accuracy/stability of your decade will be dominated by the accuracy/stability of the shunt dR_S if R_L >> R_S and by the accuracy/stability of the ladder dR_L if R_S >> R_L. Hence, if you use large resistor in the ladder, e.g. 10k when 2k are called for as in your example, you need a small shunt and that will dominate the errors of the entire decade. This is not what you want, because the resistance of the ladder, in particular its temperature coefficient, are more stable since it's made of ten or eleven resistors and you get the benefit of averaging. With the shunt, you depend on fewer parts and your stability will suffer - in particular, the potentiometer is hard to stabilise. So the better choice seems to be a large shunt for trimming and a ladder that is close to the desired values from the start.

Third, and here I'm speculating, KVDs are made commercially from wire wound precision resistors due to their temperature stability - they simply are cheaper for lower resistance values as you need less wire.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.