On 9/21/2017 5:41 PM, T i m wrote:
On Thu, 21 Sep 2017 16:54:07 -0500, Foxs Mercantile
wrote:

Sigh....

The simple answer is put a 62K 5% resistor in series with the meter.
It's read about 4% low (240 uA instead of 250 uA.)

Ok, so with our source at 15V we get 240 uA and ~FSD on our meter.
That's 14.88V across the resistor and 0.12V across the meter,
therefore the meter is ~500 ohms?

No that's 15 volt across the resistor only so it's ~240 uA.
With the internal meter resistance it would be slightly less current.


At 10V we get 2/3 of 240 uA (160uA) and so our voltage across the
meter is now 0.0256 (~1/5th of FSD) and so the meter will read 1 volt
and not 0?

No, it's still assuming the meter resistance is close to zero.
at 160 uA, it will read 2/3 scale at 10 volts.

You wanna get a little closer? put a 100K and a 150K in parallel.
Wanna make it adjustable? Put a 100K trimpot in parallel with a 180K
resistor.

The problem is that the voltage across the meter will fall in
proportion to the voltage across the entire network, not just across
the meter component (which is why I ruled it out in my second post).

No, you're NOT measuring voltage across the meter. You're measuring
current through the meter.

It's a series circuit. (The meter and the series resistor.) As such
in this case, the series resistance is much higher than the internal
resistance of the meter, and that defines what the current will be for
a given voltage.

I.e. 60K ohm, 15 volts, 250 micro-amps.



--
Jeff-1.0
wa6fwi
http://www.foxsmercantile.com