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The data

I'm trying to understand a sensor circuit I've come across, which I've simplified to the following:

schematic

simulate this circuit – Schematic created using CircuitLab

In short, it's a potentiometer segment in parallel with a a resistor of lower value. The output voltage at Vwiper is fed into an automotive ECU. I can't measure the potentiometer resistance without modifying the circuit, but for the sake of the example, I've given Rpot a value of 1K.

  • I understand that the parallel resistor (Rpar) will limit the potentiometer's resistance range to the value of the parallel resistor.
  • I've also read that the curve of resistance seen at Vwiper vs. potentiometer wiper position should be non-linear.

Update: after the feedback in answers, I've further simplified the circuit and modelled the potentiometer as two resistors the value of which depends on the wiper position:

enter image description here

The voltage increases linearly with position in the simulation results, which seems to be confirmed by my measurements on the physical circuit.

I have also measured Rwiper vs position on the real circuit, which does not seem to be linear. It's a tiny potentiometer strip, so measurement is not exactly accurate:

Rwiper resistance curve

Notice how the output resistance starts at about 65 Ω, peaks to ca. 250 Ω and ends up decreasing to the Rpar value (30 Ω). I assume it starts at 65 Ω because the wiper rest position does not go all the way down to 0 Ω.

For more background (skip if not needed), the sensor has a continuous resistor track, divided in 7 segments. Each segment of the track has a parallel resistor as the one shown, each with a decreasing value in logarithmic progression. I've lumped together the rest of the stacked segments into R4 for simplicity.

The output, taken between Vwiper and GND, is a logarithmic curve. I believe the purpose of the segments and the parallel resistors is to do piecewise approximation to obtain this voltage curve.

The question

I can measure, and plot the output voltage, but I somehow cannot grasp how the resistor network actually works.

In other words, given this simplified circuit, I would like to understand and if possible, plot, how the output resistance between Vwiper and Vinit varies as a function of the wiper travel.

It's been many years since I've done any circuit analysis, so any pointers would be helpful on the question:

  • What is the equation that determines the output resistance seen from Vwiper?
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  • \$\begingroup\$ I probably shouldn't comment because I'm not completely following the question, but have you tried to replace the circuit with the Thevenin equivalent? en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem I also question why you want to know potentiometer output resistance. It seems like you would only be concerned with the output voltage since it is probably just being read by a very high impedance input. \$\endgroup\$ – Adam Jun 11 '18 at 14:50
  • \$\begingroup\$ @Adam Thanks. As mentioned in the question, I wanted to understand how the resistance varied (which is now clear from the accepted answer). I'm aware that the important magnitude is the voltage. As per the Thevenin equivalent circuit, I tried it and failed: when the voltage sources are shortened, there is a short across Rpar. I didn't quite know what do do with it: if Rpar were shortened, it wouldn't appear in the final resistance equation, so I got stuck there. \$\endgroup\$ – John M Jun 11 '18 at 15:48
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The equation where Rlow is the potentiometer low side resistance

schematic

simulate this circuit – Schematic created using CircuitLab

R = Rlow || (Rpot-Rlow +Rpar) = Rlow * (Rpot-Rlow +Rpar)/(Rlow + Rpot-Rlow +Rpar) = Rlow * (Rpot-Rlow +Rpar)/(Rpot+Rpar)

You can find the pot value from the maximum value that is ~260 ohm

R has a maximum at the Rlow = Rpot-Rlow +Rpar (both pats are equal) and Rlow = 2 * Rmax = 520 ohm

From here we find out that Rpot = 2*Rlow - Rpar = 1010 ohm , your guess was close.

For a linear potentiometer the R graph (% Rlow from Rpot) looks like this:

enter image description here

As you see it doesn't look like your graph, that means the pot is non linear.

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  • \$\begingroup\$ Excellent, this is exactly what I was looking for! I now managed to plot the same output as your graph using Rwiper = R2eqn*(R1eqn+Rpar)/(R_pot+R_par); where R1eqn and R2eqn are your Rpot-Rlow and Rlow, but defined by the equivalent functions on the initial post's equation. And indeed, my potentiometer is not linear. Each segment is wedge-like-shaped, with increasing height as the wiper travels towards its end. One last thing I cannot quite grasp yet is how the voltage is linear, though. But that might be the subjecto of another question, thanks! \$\endgroup\$ – John M Jun 11 '18 at 13:18
  • \$\begingroup\$ @JohnM It's not clear to me what happens with V5 source when measuring R , It's completely disconnected or just turned off? \$\endgroup\$ – Dorian Jun 11 '18 at 13:31
  • \$\begingroup\$ I disconnected the voltage source when I measured R. \$\endgroup\$ – John M Jun 11 '18 at 13:45
  • \$\begingroup\$ Then it's really strange. It might have something to do with the ECU input impedance? Very strange. I thought some residual current is altering the readings. \$\endgroup\$ – Dorian Jun 11 '18 at 13:55
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    \$\begingroup\$ No worries, my main question is answered. I might formulate this as a separate question and take it from there. Perhaps I did not express myself too well. When measuring resistance to produce the graph, the sensor was standalone (i.e. neither connected to a voltage source nor to the ECU). It just struck me that when, as a separate test or with the simulation (see graph on original question) the voltage output was linear, given that the resistance graph was not. But I feel I'm probably missing some very basic principle. \$\endgroup\$ – John M Jun 11 '18 at 15:42
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If you measure the resistance between Vwiper and GND then R4 and voltage source do not matter and can be removed for clarity. If you measure from Vinit then R2 also irrelevant.

What you left with is two resistors in parallel, with wiper measuring from 0 at the bottom to Rpot||Rpar at the top.

Now, to understand the resistance in the middle I suggest you replace Rpar with a wire, since it's resistance is very small comparing to Rpot. All this will do is make wiper resistance 0 at both ends, and your plot will be symmetrical.

Now, as you move wiper along, it will go further and further from the end, increasing resistance to (Rpot/2)||(Rpot/2) in the middle. Then it will be coming closer to the other end, eventually reducing resistance back to 0. BTW, this gives you an easy way to measure actual Rpot value.

If you return Rpar back into the picture you will see that all it does is raising one side of the curve, skewing it a little.

I hope this gives you enough pointers to come up with an equation.

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  • \$\begingroup\$ Thanks a lot. Your and Olin Lathrop's answers did get me further. I updated the original question with a further simplified circuit. Still struggling with calculating the output resistance at Vwiper from the now 3 resistors, though. I'm not quite sure I followed how replacing Rpar with a wire will make resistance 0 at both ends, and not 0 at one end and Rpot at the other. \$\endgroup\$ – John M Jun 11 '18 at 12:38
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Don't let the fact that there is a potentiometer in the circuit confuse you. In any one setting, you can replace the potentiometer with two fixed resistors. Once you have done that, you have a simple 4-resistor network to solve.

Solve this at as many pot settings as you like to get a graph of voltage as a function of pot position.

With a little extra math defining the two resistors you are modeling the pot as, you can write down the closed form equation of output voltage as a function of pot position.

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  • \$\begingroup\$ Thanks, let me see if I can update the circuit diagram with the two fixed resistors replacement. However, regarding the graph of voltage, I'm not looking to plot that, but rather the resistance as a function of pot position. \$\endgroup\$ – John M Jun 11 '18 at 12:09
  • \$\begingroup\$ @John: Either way, breaking the pot into two resistors gets you there, whether solving for the output voltage or output resistance. \$\endgroup\$ – Olin Lathrop Jun 11 '18 at 12:19
  • \$\begingroup\$ That's unfortunately the part I'm struggling with though (apologies if the answer is obvious, I've been out of touch with circuit analysis for many years). In any case, I've updated the original question with the potentiometer breakout into two resistors, I'm hoping this and the circuit simplification is correct. \$\endgroup\$ – John M Jun 11 '18 at 12:33

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