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W.r.t above circuit fragment, a 100k linear potentiometer is connected to pins 1,3,4 of J12. Internally my code in the MCU reads the 10-bit ADC from pin P1.3. I am expecting linear ADC values (this is what i need) whereas i find a completely non-linear precipitous voltage drop followed by a long tail when the pot is twisted from high to low values. I am not quite clear how to explain this but feel that R12 (10 ohms) and R17 (10k ohms) have something to do with this. On using a 5k potentiometer i get a almost linear like voltage change.

I would like to understand how this part of the circuit works and how i can calculate the input voltages from the pot (and thence the expected ADC values) for different values of the pot like 100k, 10k, 5k and 1k.

Appreciate your explanations.

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  • \$\begingroup\$ Pots have three terminals and it's important how they connect so show that. Also your circuit is terribly drawn with text running over nets and nets making 4 point joins. \$\endgroup\$ – Andy aka Jun 24 '17 at 16:44
  • \$\begingroup\$ Just to add to @Andyaka 's comment, part of that circuit is completely unnecessary. (In fact, perhaps MOST of it is unnecessary for your question.) You'd do us all a favor if you simplified it to the necessary parts. I think we can assume that the two ends of the pot are connected to ground and +3 V, and that the wiper is connected to pin 3 of J12. You could throw a lot away there. \$\endgroup\$ – jonk Jun 24 '17 at 16:47
  • \$\begingroup\$ Are you saying that this is an existing board or is it a new circuit you're intending to make? In other words, how much freedom do you have to change it all? As others say, it could be much simpler. \$\endgroup\$ – TonyM Jun 24 '17 at 17:14
  • \$\begingroup\$ To clarify some points; the above is part of the circuit diagram for an existing board. I don't have the freedom to change anything for now (maybe next revision). Hence my reason to show a larger snippet of the circuit than is necessary for this question. Please ignore everything other than the top right if that helps. I don't want to redraw it in case i make any mistakes (noob in Electronics hardware). The pot's lug-1 is connected to J12-Pin1 ground, lug-3 is connected to J12-Pin4 3.3v and lug-2 (the wiper) is connected to J12-Pin3 and through that to P1.3 on the MCU. \$\endgroup\$ – RamanathanR Jun 24 '17 at 17:36
  • \$\begingroup\$ Note that J11 is open i.e. no jumpers on it. \$\endgroup\$ – RamanathanR Jun 24 '17 at 17:38
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Ignoring the capacitors, your circuit is basically this:

schematic

simulate this circuit – Schematic created using CircuitLab

This schematic is just a transcription from your schematic, but eliminating the nearby capacitors and the rest of the schematic that isn't really needed to understand why you are getting your results.

From the above, the equation is pretty simple (where \$0\le \%\le 1\$):

$$V=3\:\textrm{V}\cdot\frac{R_{17}\cdot \%}{R_{17}+R_{12}+R_X\cdot \%\cdot\left(1-\%\right)}$$

Assuming a linear taper (not an audio taper) potentiometer and if \$R_X=100\:\textrm{k}\Omega\$ then the curve for \$P\:1.3\$ is seen on the left side below. On the other hand, if \$R_X=5\:\textrm{k}\Omega\$ then the curve for \$P\:1.3\$ is now shown on the right side below:

enter image description here enter image description here

The reason for these behaviors can be seen by examining the denominator term, \$R_X\cdot \%\cdot\left(1-\%\right)\$, in the above equation. This term represents the Thevenin equivalent resistance of the potentiometer-divider. Note that this term's contribution cannot exceed 25% of the potentiometer's value, regardless of the value of the % rotation. As the denominator also includes \$R_{17}=10\:\textrm{k}\Omega\$, the potentiometer-divider term in the divisor will be insignificant if its maximum value it is much lower than \$R_{17}\$ and it will be significant if its maximum value approaches or is greater than \$R_{17}\$.

If the potentiometer is \$1\:\textrm{k}\Omega\$, then its maximum Thevenin value is \$250\:\Omega\$ and this is very much smaller than \$R_{17}\$. So the potentiometer's setting doesn't significantly impact the divisor. However, the potentiometer's setting does affect the numerator in a linear fashion. So the overall behavior is linear, as you wanted.

If the potentiometer is \$100\:\textrm{k}\Omega\$, then its maximum Thevenin value is \$25\:\textrm{k}\Omega\$ and this is actually quite significant and larger than \$R_{17}\$. So the potentiometer's setting now does very much impact the divisor. As this affect follows a parabola shape, the divisor's effect is quite non-linear in the result. The potentiometer's setting does affect the numerator in a linear fashion, but this is overwhelmed by the non-linear, parabolic shape of the divisor. So the overall behavior is not at all linear.

I really don't know why you cannot work this out using Thevenin and/or pretty basic resistor divider math. Try your hand at it. It's pretty easy algebra (and a few terms nicely cancel out, too.)

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  • \$\begingroup\$ Thank you very much! I am not quite clear on the circuit and hence couldn't figure it out. What is the purpose of R12 and R17 ? I would have imagined that the potentiometer's output would be directly fed to the pin. If i understand your circuit correctly there are two voltage dividers here; a) The voltage output of the potentiometer wiper itself b) That over R12 is further divided with R17. Why? \$\endgroup\$ – RamanathanR Jun 25 '17 at 5:44
  • \$\begingroup\$ @RamanathanR I could guess that the designer felt that the 10k and the low pass PI filter might help protect the microcontroller input from static charge and other vagaries that may occur at a connector. Also, that they never expected to see a 100k pot there. But ultimately you are asking me to get into the head of someone else. I would have done it differently. \$\endgroup\$ – jonk Jun 25 '17 at 5:55
  • \$\begingroup\$ I see your point. I was looking for insights into some of these "design patterns" used by experienced electronics engineers. Being a software guy self-studying the hardware stuff, it is often frustrating seeing something being done but not understanding why/how. Hence my posting here. Thanks for your help. I think i will remove R17 and just use a linear 1K pot to keep everything simple. \$\endgroup\$ – RamanathanR Jun 25 '17 at 6:52
  • \$\begingroup\$ @RamanathanR If you plan to use a 1k pot, I'd leave the 10k in place. The peak Thevenin of your pot will be 250 ohms. That will completely dominate the situation and should be good enough. \$\endgroup\$ – jonk Jun 25 '17 at 6:56
  • \$\begingroup\$ Right, my plan is to not touch the board but do some testing using 10k, 5k, 1k pots and look at the linearity of the voltage curve. If that satisfies the requirements, i am good. I will only modify the hardware (however trivial) only after understanding it well :-) \$\endgroup\$ – RamanathanR Jun 25 '17 at 10:29

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