john merryman wrote:
> I have a 50ua meter with a resistance 1.085 kilo ohm reading from a
> digital multimeter. I need to get this meter to read 300ma full
> scale, I am having trouble coming up with a correct shunt calculation
> using the formula in the ARRL hanbook, it is not explained in a great
> amount of detail. Can anybody help me with the exact formula to obtain
>
> the shunt value. I think i might be mixing or screwing up the ua to ma
>
> conversion.
>
> John Merryman
The meter you have has a full scale voltage of V=IR=50x10^-6*1085 =
0.05425 V (I would stick to Ohms, Volts and Amps, and forget about using
uA, mA or kOhms) Your current shunt needs to take 0.3 - 0.00005A =
0.29995 A. Hence the resistance needs to be R=V/I=0.05425 V / 0.29995 A
= 0.181 Ohms. The power handling of the shunt needs to be P=V*I=0.05425
V *0.29995 A = 16 mW.
There are other methods, that could be a little more convenient, that
allow the use of pot to trim the full scale, as getting a resistor of
exactly 181 mOhms wont be easy. You could try putting a variable
resistor of say 1kOhms in series with your 50uA meter, then working out
the shunt assuming a meter resistance of 1.5 K, instead of 1.085 k. That
way, you will have some adjustment available, and the shunt will not
have to be that accurate - about 40 % error will be fine. The downside
of this is that the voltage drop across the meter will be somewhat more
than the 54 mV calculated in the last paragraph, but that usually dont
matter too much.
Dave Kirby G8WRB.
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