I've tried to find a circuit for a coarse and fine adjustment (two potentiometers) voltage divider, but I don't understand it and/or they don't have a linear response.

Problem: I want to have an adjustable voltage from 0 - 5V using two potentiometers, one for coarse adjustment and the other for fine (10mV if possible) adjustment.

From the datasheets I've looked at (e.g. this) they don't seem to specify the resolution of increments possible of the pot.

Here are three circuits I currently have:

enter image description here

The third circuit's fine adjustment decreases as the coarse adjustment is set higher, so I don't think this is a good idea (unless a logarithm pot is used... no idea how those work yet).

Since the first and second are very similar, I'll consider the first one.

I assumed a 5 degree resolution out of 300 degrees, since I could not find any information regarding this.

This gives me:

  • 0.83kOhm / adjustment with the 50K pot, and a 166mV resolution
  • 0.167kOhm / adjustment with the 10K pot

The equation I obtain is:

$$ V_{out} = \frac{R_{course} + R_{fine}}{50 + R_{fine}} V_{in} $$

Plotting this in matlab for 0V course adjustment, I get the following curve:

enter image description here

At the lower end of the pot, there is a resolution of 33mV and at the higher end of the pot there is a resolution of 24.7mV.

For my application, this is adequate. However I'm unsure if there is a better (and linear) approach to a fine and course adjustment.


This is better..


simulate this circuit – Schematic created using CircuitLab

Advantages are:

  • Low sensitivity to pot tolerance and tempco (you can use precision resistors for R2/R3)
  • Quite linear and almost constant fine adjustment range in mV
  • Quite constant (+/-0.5%) and predictable output impedance (minimum 9.09K maximum 9.195)
  • Low sensitivity to CRV (contact resistance variation) of pots (1% CRV in R1 results in 0.05% variation).

This circuit draws 20mA or so from the 5V rail. If that's an issue you can increase R4 10:1, increase both R4 and R1 by another 10:1 at the expense of a bit of performance or scale all the value at the expense of output impedance.

Your circuit #1 has an output impedance of 0 ohms to 27.5K, depending on the pot settings.

Fine and coarse only takes you so far, you could also consider a switched voltage divider for the "coarse" adjustment. Expecting the "coarse" adjust to stay stable within 0.2% may be too much to ask unless it's a very nice potentiometer.

Note that your conductive plastic pot does not specify a temperature coefficient at all- that's because conductive plastic pots are generally horrible- maybe +/-1000ppm/°C typically, so using them as a rheostat rather than a voltage divider is not such a great idea. You've got that reduced by 5:1 by the ratios of the pots, but it's still pretty bad. The circuit I presented would typically be about 5x better with decent resistors for R2/R3 because the pots are used purely as voltage dividers.

Edit: as a good approximation for R4 << R3 and R1 << R2 (you can do the exact math in Matlab taking the pot resistances into account if you like), the output voltage is:

\$ V_{OUT} = 5.0 (\frac {\alpha \cdot 9.09K}{10K} + \frac {\beta \cdot 9.09K}{100K}) \$

Where 0\$\le \alpha\le 1\$ is the position of R1 and 0\$\le \beta \le 1\$ is the position of R4

So the range of R1 is 4.545V and the range of R4 is 0.4545V. If you center both pots you get 2.500V. If you can set R4 to 1% of full scale (reasonable), that's 4.5mV resolution.

  • \$\begingroup\$ I'm having trouble understanding how your circuit works - how does R2/R3 make one pot high res/low range and vice versa? \$\endgroup\$ – tgun926 Dec 19 '14 at 3:08
  • \$\begingroup\$ @tgun926, I can't answer for Spehro, but I'm seeing it as a current source feeding inverting input of TIA. \$\endgroup\$ – George Herold Dec 19 '14 at 3:09
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    \$\begingroup\$ @tgun926, imagine each potentiometer as a voltage divider between 5V and 0V. The other two resistors are a voltage divider on the output of those two dividers, taking a weighted average between the two of them. goo.gl/HRqm5G The trade-off is that the two 500Ohm resistors are small, so there will be some current flowing there. If you make the 500Ohm pots bigger than less current will flow but then you need to take the resistance of the pots into the math of the voltage divider and it will decrease sensitivity of the "fine" knob in the middle range. goo.gl/HRqm5G \$\endgroup\$ – Eyal Dec 30 '15 at 8:09
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    \$\begingroup\$ @ElectronSurf 10K in parallel with 100K is about 9.09K, I ignored the pot resistances which contribute 0 to 125 ohms each, depending on wiper positions. Taking those into account it could be as high as 9.19K. The effective resistance of each pot is (500*a)||((1-a)*500) where a is the pot electrical rotation from 0 to 1, ignoring wiper resistance. Minimum is 0 and maximum is 250||250 = 125 ohms. \$\endgroup\$ – Spehro Pefhany Apr 26 '20 at 14:23
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    \$\begingroup\$ @ElectronSurf Maybe. It really depends on the pot construction etc. and how many uV you are thinking and for how many seconds. See the answer below for some of those considerations. \$\endgroup\$ – Spehro Pefhany Apr 26 '20 at 14:46

+1 for Spehro Pefhany. That is a very elegant circuit. As for how it works, this is how I see it:


simulate this circuit – Schematic created using CircuitLab

The asymmetry of the voltage divider (asymmetrical because R3 > R2) makes one of the pots coarse, and the other fine. Because R2 < R3, the output voltage will be mostly a function of V1, with V4 able to make fine adjustments.

The caveat here is of course that the output impedance of the pots changes with the wiper position, so the application of Thévenin's theorem in the first step is really only correct when the pots are at their midpoints -- as the pot is moved to either extreme the output impedance approaches 0Ω. However, since R2 and R3 are much bigger than either pot, this variability is relatively insignificant, both in terms of nonlinearity, and variation in output impedance of the circuit overall.

  • \$\begingroup\$ Regarding the Thevenin equivalent at the midpoint - that is two 250 ohm resistors in series (per pot), wouldn't the Rth be half that, i.e. 125ohm? Or are you not allowed to consider the pot on it's own? \$\endgroup\$ – tgun926 Dec 19 '14 at 5:51
  • \$\begingroup\$ @tgun926 You are right...brain fart. I'll edit. \$\endgroup\$ – Phil Frost Dec 19 '14 at 5:52
  • \$\begingroup\$ @tgun926 Just to be clear, the 125 ohms for each pot comes from one pot with wiper at midpoint looking like two 250 ohm resistors in parallel. They look in parallel, to the load, because the power supply (a voltage source) is replaced by a short for the purposes of calculating the Thevenin equivalent. The details are of limited practical consequence though, since the much higher impedance of the fixed resistors renders the pots relatively insignificant, which is the elegance of this circuit. \$\endgroup\$ – Phil Frost Dec 19 '14 at 6:48
  • \$\begingroup\$ Yeah I understood that, cheers \$\endgroup\$ – tgun926 Dec 19 '14 at 7:00

You have the right approach, and your numbers are probably good to within a factor of 5 or so. For a plastic element pot, 1% resolution seems reasonable, although it depends on details of construction. For the pots you link to, the problem is that the length of the arm from the shaft to the element contact is quite small, and the bearing as cheap as possible, so there may be some slop in exactly where the element contact occurs. This probably shows up as increased hysteresis (the resistance at x degrees when turning clockwise is different from the resistance when turning counter-clockwise).

Note that resolution is worst for wirewound pots, since the contact skips along the outside of a long helix of wire, so you get a stairstep effect with fixed step size.

There are basically 3 approaches to getting better resolution from a pot. First, go to a smoother element, with smaller internal grain size. Conductive plastic is best, and the pots you link to use this. Second, make the pot larger. This allows finer control of exactly where the contact meets the element, although it also requires more precision in the bearing and design of the wiper arm to keep it from flexing as it moves. Finally, you can go to multi-turn pots, 10-turn units being the norm, although I've run across 5-turn and 20-turn models. In this approach, the resistive element forms an n-turn spiral, and the contact arm displaces vertically along the shaft axis as necessary. With a longer resistive element, more precise placement of the wiper is possible, and thus better resolution.

As for your analysis, it's right on. The amount of non-linearity is directly related to the ratio of the two resistances. A larger ratio gives better linearity (although this cuts down on the range of the fine adjustment, and requires the coarse adjustment to be more precise).

Finally, if you demand ultimate (and probably unreasonable) linearity, you don't gang the pots at all. You connect their ends in parallel, and feed each wiper to an amplifier with a different gain, then sum the two results in a final amplifier.


Glad I stumbled across this answer. Thanks to Spehro Pefhanys' Answer, got me thinking and calculating to a more general approach I would like to share.


simulate this circuit – Schematic created using CircuitLab

\$ m \$ denotes a scaling ratio, which in the case presented is 10:1, \$ m=10 \$

\$ Z_{out_{MAX}} = \frac{m}{m+1}\cdot \left(R_s +\frac{R_p}{4}\right)\$

\$ Z_{out_{min}} = \frac{m}{m+1} \cdot R_s\$

\$ Z_{out_{MAX}} \$ is reached When both potentiometer wipers are in the center position \$ \alpha = \beta = 0.5 \$

\$ Z_{out_{min}} \$ is reach when both potentiometers are at either extremes.

Interesting to note is that in this configuration Impedance Variance, Spread is determined solely by the Potentiometers \$ \Delta_{Z_{out}} = Z_{out_{MAX}} -Z_{out_{min}} = \frac{m}{m+1}\cdot \frac{R_p}{4}\$

If you consider \$ Z_{load} >>> Z_{out} \$ then: \$ \frac{V_{out}}{V_{in}}=\frac{m}{m+1}\cdot\left(\alpha +\frac{\beta}{m}\right)\$

\$ Z_{in} \approx R_{p1} // R_{p2} \$ , where Rp1 and Rp2 are the potentiometers represented at Rp and m·Rp in the diagram.

The circuit input impedance is relatively constant, only slightly altered with different wiper positions or even different loads.

Minor \$ \Delta_{Z_{out}} \$, Impedance variance improvements:

As can be demonstrated the fine/coarse ratio is defined by \$ m , R_{s2} = m \cdot R_{s1} \$, The impedance swing is defined solely by the potentiometers \$ R_{p2} = m \cdot R_{p1} \$

The formulas presented scale the potentiometers with the ratio \$ m \$ although they need not be. As presented initially by Spehro, they can be of the same "value". Not scaling the values increases input load but can slightly improve the impedance variance. By how much can be approximated as follows.

Let \$ f(x) = \Delta_{Z_{out}} = \frac{x}{x+1}\cdot \frac{R_p}{4}\$

\$ f'(x) = \frac{R_p}{4\left(1+x\right)^2} \$

evaluate both \$ f(m) \$ and \$ f'(m) \$ we can define a linear function :

\$ g(k) = k\cdot f'(m) + b \$

where b is found by solving \$ g(m) = f(m) \$. Now we will have a linear function \$ g(k)\$ which will approximate the impedance variance given a factor k between the potentiometers \$ R_{p2} = k\cdot R_{p1}\$ while maintaining the factor \$ m \$ for the coarse/fine ratio.

For the example provided by Spehro, \$ m=10, R_p = 0.5 k\Omega \$

\$ g(k) = \frac{k}{968} + \frac{100}{968} \$

the improvement from using two \$ 500 \Omega \$ pots, \$ g(1) \approx 104 \Omega \$ instead of a \$ 500 \Omega \$ and \$ 5k \Omega \$ pot, \$ g(10) \approx 114 \Omega \$ is an impedance swing improvement of \$ \approx 10\Omega \$

Actually, if you are willing to have an input impedance of approximately \$ \approx 250\Omega \$ you can achieve a tighter impedance swing by using \$ 250 \Omega \$ and \$ 2k5 \Omega \$ pots which would reduce the impedance variation down to \$ \Delta_{Z_{out}} \approx 57\Omega \$

Some formulas for same layout but with resistors and pots that are not bound by a ratio

The output impedance can be calculated as follows:

\$ Z_{out} = \left(R_{p1}+R_{s1}\right) //\left(R_{p2}+R_{s2}\right)= \frac{\left(R_{p1}+R_{s1}\right)\cdot\left(R_{p2}+R_{s2}\right)}{R_{p1}+R_{s1}+R_{p2}+R_{s2}}\$

Where : \$ R_{p1} = R_{p1_{Total}}\cdot(1-\alpha)\alpha\$ , being \$ \alpha \$ the wiper position \$ \{0..1\} \$

\$ Z_{out_{MAX}} = \frac{\left(R_{p1T}+4R_{s1}\right)\left(R_{p2T}+4R_{s2}\right)}{4\left(R_{p1T}+4R_{s1}+R_{p2T}+4R_{s2}\right)}\$ When both potentiometer wipers are in the center position \$ \alpha = 0.5 \$

\$ Z_{out_{min}} = \frac{R_{s1}R_{s2}}{R_{s1}+R_{s2}} \$

If you consider \$ Z_{load} >>> Z_{out} \$ then: \$ \frac{V_{out}}{V_{in}}=\frac{\alpha R_{s2}+\beta R_{s1}}{R_{s1}+R_{s2}}\$

Just thought I may share my exploration and generalization of the answer.


I've done it with two pots in series, each wired as variable R's (one end to wiper), with an opamp on the output, sometimes the variable R is in a gain stage. (But I like Spehro's circuit! another advantage he forgot to mention, ~constant input impedance.)


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