How would I add min & max limits to an analog voltage controlled by a potentiometer?

Thread Starter

dembkomj

Joined Feb 26, 2019
4
Hello. I'm a mechanical engineer with a grasp on basic electrical concepts but need some guidance in what I feel is a simple circuit problem.

I have a project in which I have a 5Vdc signal wired into a 5 kohm potentiometer such that I can vary its output from 0 to 5Vdc. What is the simplest way to add basic passive components or IC's (op-amp) into the circuit such that i can add a minimum and maximum limit to my controlled output voltage. In other words, I want the minimum of the output of the pot's signal to be no less than 1.75V and no more than 2.10V (can be other ranges as well), such that you couldn't go outside those boundaries by over-turning the multi-turn pot. But you could achieve every discreet analog voltage in between 1.75 and 2.10 with the turning of the potentiometer.

The circuit ultimately feeds into an automotive style ECU, so there is very little current on it.

I really do not want to use a micro-controller for this design, but am thinking this can be achieved by op-amps, zener diodes, resistors, or something of the like.

Any guidance is appreciated, or if there are similar existing threads in this forum that would be helpful.

Thanks,
Mike
 

wayneh

Joined Sep 9, 2010
17,189
I'd probably just add a resistor in series with the pot to establish the minimum resistance setting. Adding a resistor in parallel with the pot can change (lower) the maximum resistance. I may not understand exactly what you're doing, but it would be far easier if you can accomplish it with passive components (resistors) and not involve active circuitry such as an opamp in an automotive environment.

You might find these links useful.

tailoring potentiometers.pdf
This and [update] seems to have moved to here
this, "The Secret Life of Pots", too.​
 
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Thread Starter

dembkomj

Joined Feb 26, 2019
4
Assuming the 5V is constant, this should do what you want:

View attachment 171089
Crutschow, thank you very much this is exactly what I'm looking for. Yes, it is a regulated 5Vdc supply voltage.

I understand the circuit you drew, however what would be the mathematical formula for deriving the values of R1 and R2? For example, if I have different limits for my min and max voltage values when I sweep the pot, the values of R1 and R2 will change. Basically I'm trying to calculate these two resistor values without solving via a computer model.

I realize somehow I have to consider this as a voltage divider circuit with so much desired voltage drop across the potentiometer, I'm just getting a little lost in the mathematics.
 

wayneh

Joined Sep 9, 2010
17,189
Max voltage requirement: (5-Vmax)/5 = R1/∑R where ∑R = R1 + R2 + RU1
Min voltage requirement: Vmin/5 = R2/∑R
Three equations, two unknowns. The rest is just algebra.
 

Thread Starter

dembkomj

Joined Feb 26, 2019
4
Max voltage requirement: (5-Vmax)/5 = R1/∑R where ∑R = R1 + R2 + RU1
Min voltage requirement: Vmin/5 = R2/∑R
Three equations, two unknowns. The rest is just algebra.
Sorry, am I missing what the 3rd equation is?

And for my knowledge, are these equations above basically those for a voltage divider?
 

sghioto

Joined Dec 31, 2017
3,652
Another way is to divide the maximum range you need by the value of the pot, 5K. This will get you the current required through the divider. It's all Ohms law.
Example: you want .5 volt range from 2 volts to 2.5 volts.
.5v / 5K = .0001a
2v / .0001a = 20K = R2
2.5v / .0001a = 25K = R1
SG
 
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crutschow

Joined Mar 14, 2008
31,097
what would be the mathematical formula for deriving the values of R1 and R2?
No formula, just Ohm's law.
I use the technique that sghioto showed in post #7, which calculates the current first, since it doesn't involve the solving of any simultaneous equations, only the calculation of a few simple Ohm's law equations from inspection of the circuit.

To calculate the current, we look at the one resistor whose value is known, the potentiometer.
So the voltage swing you want is 1.75V to 2.1V, which means the voltage difference between the top and bottom of the pot is 2.1V-1.75V = 0.35V.
Thus we know that the current through the pot (and the other two series resistors) must be 0.35V / 5k = 70μA.
Now we can easily calculate the two other resistor values.
Since the pot bottom voltage is 1.75V, the bottom resistor value to drop that voltage is 1.75V/70μA = 25kΩ.
The pot top voltage is 2.1V so the top resistor must drop (5V-2.1V) = 2.9V. Its value must then be 2.9V/70μA = 41.43kΩ.
 

Thread Starter

dembkomj

Joined Feb 26, 2019
4
No formula, just Ohm's law.
I use the technique that sghioto showed in post #7, which calculates the current first, since it doesn't involve the solving of any simultaneous equations, only the calculation of a few simple Ohm's law equations from inspection of the circuit.

To calculate the current, we look at the one resistor whose value is known, the potentiometer.
So the voltage swing you want is 1.75V to 2.1V, which means the voltage difference between the top and bottom of the pot is 2.1V-1.75V = 0.35V.
Thus we know that the current through the pot (and the other two series resistors) must be 0.35V / 5k = 70μA.
Now we can easily calculate the two other resistor values.
Since the pot bottom voltage is 1.75V, the bottom resistor value to drop that voltage is 1.75V/70μA = 25kΩ.
The pot top voltage is 2.1V so the top resistor must drop (5V-2.1V) = 2.9V. Its value must then be 2.9V/70μA = 41.43kΩ.

Yes, thank you. Ended up crunching the numbers myself as well and came to that conclusion. Appreciate the help to this novice!
 
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