Electronic – Best practice when choosing a trim pot value

best practicecomparatoroperational-amplifierpotentiometer

I have a basic comparator circuit I'm working on designing. I want one of the reference voltages to be tunable, so using a trim pot. My Vcc is 12V, so assuming a 1/4 watt pot, 1k seems like the lowest I should go. (12 * 12 / 1000 = 0.144 watts dissipated) I chose 10k for now, but I don't know why I chose that value. It's a common value, and seems fine.

What is the best practice when choosing a value like this and why? Lowest possible value for lowest noise? Something else?

Best Answer

Normally you want the adjustment to accommodate the worst-case variability and not too much more. In other words if you add up all the worst-case tolerances of the reference setting and it might be +/-100mV off from your nominal 3V setting, you might allow adjustment from 2.85 to 3.15V. Since potentiometer element resistance generally has a lousy tolerance you would prefer to either connect it across a "stiff" voltage or parallel the element with a precision resistor.

In my hypothetical case, suppose I have a 4.096V precision reference, I might use a 10K pot and parallel it with 301 ohms so the divider current is 1mA nominal. Then the divider would look like this:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The source resistance looking into R1 wiper varies by +/-1250 ohms as the wiper is moved. That's not a problem in many situations but you should be aware of it.

General rules-

Do you really need a trimpot or can you do it digitally? It's 2020, dude, what are you thinking?
Avoid using potentiometers as rheostats wherever possible, minimize current through the wiper. If you must use a pot as a variable resistor, tie the unused end of the element to the wiper. And avoid use of < 100 ohm cermet pots as rheostats because of contact resistance variation.
Use medium resistance pots were possible, especially if you need a wirewound
pot. Usually 2K to 50K is optimal.
If your adjustments are interactive, design your circuit so that at least one of the pots is nominally independent. It's worth extra care and even a few extra parts to minimize iterations.
Make the pot rotations "make sense", generally clockwise should increase what the technician is observing.

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It sounds like you don't have an external reset signal to respond to, you just want to count to some number then go to zero as the next step. You could consider this a mod n counter, where n is one more than the maximum count in your counter.

So I'm not sure what you mean about "losing a signal" while doing the reset. If you had a 3-digit decimal counter and it rolled over from 999 to 000, you wouldn't consider it as "losing a signal", it would just be counting to the next value in mod-1000 arithmetic.

So if you do

always @(posedge clock) begin
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         counter <= 0;
    end
    else begin
        counter <= counter + 1;
    end
end

You'll have a counter that counts continuously in mod-some_number+1 arithmetic. Alternately you could say it counts to some_number, then resets to zero without ever "losing" a clock pulse.

If some_number+1 happens to be a power of 2, you don't even need the reset condition in your code. For example, for a mod-16 counter, you can just use a 4-bit counter, and the synthesized logic will count continuously and repeatedly from 0 to 15.

Electronic – How to pull an op-amp output low without transients

The maximum voltage at the non-inverting input of U1 is about 0.15Vcc, so you need only bring the inverting input higher than that to ensure the output is low.

Assuming a minimum Vcc of 5V, Vcc-4V is enough. So, you could add a series resistor as in your solution 3, and a 1N4148 from the comparator op-amp output to the inverting input and it should shut down nicely.

Best Answer

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