# How to replace a potentiometer with a variable resistor?

#### ymg200

Joined Oct 2, 2015
36
I see a lot of instructions how to use a potentiometer as a variable resistor, but I can't find how to do the opposite.
I have a system that is controlled by the potentiometer. The potentiometer has three terminals: Low, High, Out (wiper). Lets say the resistance between low and high of the potentiometer is Rpot (which is constant no matter where the wiper is).
I need to control the output by using a variable resistor (Rvar), while keeping the resistance between low and high at a constant value - just like the potentiometer does (Rpot).
If I introduce the Rvar between the Low and Out, or between High and Out of the potentiometer, that would control the Output voltage, but at the same time it would keep changing Rpot, which I want to avoid.
Ideally, I would like to replace the original potentioimeter with a voltage divider in combination with a variable resistor, but I can't find where to place Rvar so that it controls the Out, while keeping Rpot constant.
Is there a way to do this?

#### sailorjoe

Joined Jun 4, 2013
363
If you connect Out to either Low or High, you get a variable resistor. Does that do what you want?

#### ymg200

Joined Oct 2, 2015
36
No, that would be using a pot as a var resistor. I need to use a var resistor as a pot.

I got a matrix keyboard 4x3 with 7 leads (4 rows, 3 columns)
I want to use this keyboard to connect different resistors (think of it as a variable resistor), thus mimicking the work of the pot.

#### SLK001

Joined Nov 29, 2011
1,548
If you put a resistor (say 500 ohm) on each of the ends of the columns and rows (total of 7 resistors) and tied the other end of the row resistors together (the INPUT leg of the pot) and tied the other end of the column resistors together (the OUTPUT leg of the pot), each single push of a key would give you 1000 ohms. Pushing two keys on the same row would give you 750 ohms. Pushing two keys NOT on the same row or column would give you 500 ohms. Pressing multiple keys would give you other resistance values. You would have to map the keyboard (or simply calculate) the resistances for your results.

The results would be very coarse using a single value for all resistors. Perhaps if you would tell us what you want to do, it would help refine solutions.

#### sailorjoe

Joined Jun 4, 2013
363
Have you ever seen a variable resistor? I ask because I only want to be sure we're talking about the same thing. When I google variable resistor, I get potentiometer.
Can you describe more about how you want to use the seven wires on your keypad to control the variable resistor/pot?
Check out this example. Figure 26 shows a voltage controlled resistor application.
Here is perhaps another option.
http://cdn.sparkfun.com/datasheets/Components/General IC/22060b.pdf
http://www.nteinc.com/specs/7000to7099/pdf/nte7072.pdf

#### ymg200

Joined Oct 2, 2015
36

Ok, the picture is worth a thousand words.

I have a "black box" circuit that is controlled by a pot. I need to replace the pot with a set of fixed resistors. Each resistor I need to pick by using a keyboard (one key press at a time).
Here are images of the keyboard and the circuit that I came up so far.
The matrix keyboard had 4 rows (I marked them as 1-2-3-4) and 3 columns (A-B-C).
The attached circuit works just like the pot would by using terminal A as an output.
Keys S1, S2, S3, S4 on the keyboard would pick the "pot" value. These keys are actually "1", "4", "7" and "*" on the keyboard.

This design works, but it only uses column A. Terminals B and C would connect to same rows 1-2-3-4, and give identical output values as A.

Is there a better way to do this so that I can use all 12 keys of the keypad, not just 4 like in the circuit above?

#### ebeowulf17

Joined Aug 12, 2014
3,282
I need to control the output by using a variable resistor (Rvar), while keeping the resistance between low and high at a constant value - just like the potentiometer does (Rpot).
Why are you concerned with keeping Rpot the same? Does your black box circuit actually require that? More often than not, I see pots setup with high and low connected to power supply voltage and ground, and only the voltage out on the wiper ends up being relevant. If you just need your keypad to deliver a choice of voltages to the former wiper location based on which key you press, that's easy - you're on the right track, and there's room for refinement and enhanced functionality if you're interested. If you actually need Rpot to stay constant for your black box circuit, and need the wiper output connection to have the corresponding resistances to high and low terminals, that will be a much trickier problem.

So, do you know if you actually need that, or is it possible you just need an appropriate voltage to be supplied where the wiper would be?

#### ymg200

Joined Oct 2, 2015
36
I don't have the circuit for the "black box" and I'm afraid to damage it. Therefore I'm looking to match my custom component to properties of the original pot, namely keeping the same Rpot.

#### ebeowulf17

Joined Aug 12, 2014
3,282
How many distinct voltages to you need?

I think we can get three distinct voltage levels from that keypad with 3 pairs of resistors mimicking the original pot almost perfectly, but if you pressed any of the other buttons, you'd get unmatched Rpot values.

#### ErnieM

Joined Apr 24, 2011
8,177
I have not worked out the details but my intuition tells me getting all 16 voltages out may be possible, though you may have to renumber the buttons due to the constraints of where your switches are. This is due to the common connections of the switch matrix.

As far as maintaining the same-same resistance of the original pot, I question if you need do that. The equivalent resistance seen by your receiver piece is the same as the pot shorted out. That means at 50% the resistance is half the pot end to end value, and when the pot is off center the eq resistance gets even owner, down to zero at either end. So if your pot goes to either zero or max you need not worry about matching.

#### ymg200

Joined Oct 2, 2015
36
I have not worked out the details but my intuition tells me getting all 16 voltages out may be possible, though you may have to renumber the buttons due to the constraints of where your switches are. This is due to the common connections of the switch matrix.

As far as maintaining the same-same resistance of the original pot, I question if you need do that. The equivalent resistance seen by your receiver piece is the same as the pot shorted out. That means at 50% the resistance is half the pot end to end value, and when the pot is off center the eq resistance gets even owner, down to zero at either end. So if your pot goes to either zero or max you need not worry about matching.
The resistance between the Low and the High of the pot is the same no matter where the wiper is. That's I have trouble matching if I want to have more than 4 voltages out from my circuit....

#### ymg200

Joined Oct 2, 2015
36
How many distinct voltages to you need?

I think we can get three distinct voltage levels from that keypad with 3 pairs of resistors mimicking the original pot almost perfectly, but if you pressed any of the other buttons, you'd get unmatched Rpot values.
The circuit in my pic gives out 4 voltages exactly like the pot would. I'm looking if I can get more than 4...

#### ErnieM

Joined Apr 24, 2011
8,177
The resistance between the Low and the High of the pot is the same no matter where the wiper is. That's I have trouble matching if I want to have more than 4 voltages out from my circuit....
Yes it is, but what matters is what source resistance is seen by what is using that voltage. That can be calculated by finding the equivalent thevenin resistance. You find that by shorting any fixed voltage source, which means the source resistance seen by the load will vary from zero, to half the end to end resistance, and back to zero as you rotate the pot from end to end.

#### ebeowulf17

Joined Aug 12, 2014
3,282
Not sure how fool proof this needs to be, but if you press the first and last buttons in your circuit simultaneously, you've created a dead short from high to low. If high and low depend on there being some resistance between them, this could be bad.

#### ebeowulf17

Joined Aug 12, 2014
3,282
The circuit in my pic gives out 4 voltages exactly like the pot would. I'm looking if I can get more than 4...
And yes, I looked at your drawing again and realized I had misread it the first time through. So you've already got 4, which is better than the 3 I imagined getting with a different method, but I'd still be wary of the short-circuit potential in your proposed setup.

#### ymg200

Joined Oct 2, 2015
36
This is the only option I can see that might work. The problem is that this is the wrong type of keyboard to use for this situation. Take a look, and see if this meets your needs.

View attachment 99142
I'm not getting this solution. Where do leads 1-2-3-4 go ? What do two diodes in the series do? I don't see how to tell columns A-B-C apart.

#### sailorjoe

Joined Jun 4, 2013
363
Let me look at it again and get back to you.

#### sailorjoe

Joined Jun 4, 2013
363
Yeah, that last effort wasn't right, so I deleted it. Try this.
Questions?

#### ymg200

Joined Oct 2, 2015
36
Yeah, that last effort wasn't right, so I deleted it. Try this.
Questions?

View attachment 99149
The solution above doesn't keep Rpot constant. Pressing key #3 leaves only one last R between Low and High.
Keeping Rpot constant is really tricky.

The attached picture is my first design. Two terminals were going to be a one leg of the voltage divider replacing a pot (between High and Out/wiper).
C (column) resistors are 0, 5K, 10K. R resistors are 500, 1K, 2K, 3K. This way any combination of R+C gives a unique value. I also need a "default" resistance (when no key is pressed), which is served by R0 in this circuit.
But again, the Rpot is not constant, which has disqualified this solution.

#### sailorjoe

Joined Jun 4, 2013
363
On your circuit I only see two connections to the black box, which I assume are the high and low sides of Rpot. I thought you wanted a pot with three connections, High, Low, and Wiper. So that's what I tried to make. That's also what you showed in your first 1 column circuit. I note that in your solution the resistance between the two output pins is not constant depending on which buttons you press. Compare pressing button 1 with pressing button 2, for example. And all of the resistances end up in parallel with R0, but perhaps that's OK.
I also sent you links to Integrated Circuits that provide programmable or controlled resistors, but the control side is more complicated than just a keypad. Maybe that's too complex for your needs. In the end, I convinced myself that there is no simple solution to your requirements given the type of keyboard you have.
The only possibility is to use active circuits to generate a voltage to the Wiper pin that is consistent with the keypad button presses. For example, an R-2R ladder network feeding an op amp buffer that then controls the voltage to the gates of a P-type FET stacked on top of an N-type FET, set to run so that they look like the amount of resistance in Rpot. Then when a button is pressed the FETs change so that one's resistance goes up while the other's goes down, but the total remains the same. Take the Wiper signal from between the FETs, and you have an adjustable Pot. That's more work than I want to do, so I wish you luck and success, but I can't help further.