[sdiy] Buffering 0V when using a pot as a voltage divider?

Dave Manley dlmanley at sonic.net
Thu Jun 30 17:10:07 CEST 2011

On 6/30/2011 7:33 AM, Harry Bissell wrote:
> Truly, sir... ground does not exist !
> I make the analogy of the North Atlantic Ocean (a virtual water ground)
> If I drop a shot-glass of water into it, does the water level rise ?
> of course it does, the trick is that it rises an infinitesimal amount and you
> really don't care, its 'good enough'.
> You might also note that shot-glass of water raises the the South Atlantic Ocean as well,
> but somewhat later than the North Atlantic.
> Perhaps a "black hole" could be considered a virtual ground, but I suspect that for a big enough
> input (a galaxy perhaps ?)even that might move....
> Ohms law rules !
> <bfg>
> H^) harry

LOL.  What's the equivalent of the Moon causing the tides in a circuit board?

Here's a calculator:


Example: 10 mil trace, 1/2 oz copper, 10 inches long, what's the resistance?

The calculator gives 0.971 Ohms.

How big is your pot?  10K Ohms?  How much voltage do you have across it? 10 Volts?

10V/10K Ohms = 1 mA

Considering just the ground trace, how much is your ground going to be different than true ground?

1mA * 0.971 Ohms = .971 mV.

Is that good enough?

If you start hanging other circuitry off that 10 mil trace that draws current, then
the voltage drop across the trace will increase due to the increased current and
disturb your 'ground'.  As Harry mentioned, that's Ohm's law for you.


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