This thread got me thinking about something:

https://forum.allaboutcircuits.com/...er-a-relay-when-voltage-drops-to-zero.138295/

Within that thread, there was some question as to the impedance of the 3.6V found on a doorbell which the OP wanted to get a signal from. Normal operation is that one leg of the doorbell was at 3.6V normally, and it dropped to 0V when the doorbell was pressed. The first attempt to tap into the signal used an opto-isolator, but that pulled the line low. So, it was determined that there must be a high source impedance. Based on that, a successful circuit was devised with a high impedance input that wouldn't cause false triggering of the doorbell.

Enough background. My question is, if you wanted to determine the source impedance, how would you go about it? I think I have a simple idea, but I wanted to get other opinions.

My thought was that you could take a high value resistor, maybe 10k, and connect it from the 3.6V line to ground, then measure the voltage on the high side of the resistor. With a low enough source impedance you'd still see 3.6V there, but if the source impedance was fairly high, you could calculate the impedance by using the voltage divider formula in reverse, plugging in the two voltages and one known resistance and calculating the other resistance. Does that make sense? Am I on the right track?

Or maybe even better is connect resistor to 3.6V source, and then measure current through DMM from resistor to ground and calculate based on current:
3.6V = Current * (source impedance + known resistance)

So:
Source impedance = 3.6V/Current - known resistance

 

I'd go with the first option, given a reasonably tight resistor tolerance. Or even a potentiometer, which you can turn until you get a 50% voltage drop, then measure the pot resistance, and that's the same as the output impedance.

 

I'd go with the first option, given a reasonably tight resistor tolerance. Or even a potentiometer, which you can turn until you get a 50% voltage drop, then measure the pot resistance, and that's the same as the output impedance.

Oh, nice. I like the 50% trim pot idea. It eliminates all that pesky math!