>
>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.
>
>---+--Meter---+--- (50uA, 1.085kohm)
> | |
> +--VVVVV---+ (299.95mA, R ohm)
>
> |<-- E -->|
>
>>From E = IR, you can see 0.00005 x 1085 = 0.29995 x R
>
> -> R = 0.18 [ohm].
>
>But, is the value '1.085 kilo ohm' correct? I can hardly believe it.
>
If the meter has 1095 ohms of R, yes. However, most 50uA meters have a
bit more R. . I would use a 1.0 ohm shunt and put a series resistor
between the shunt and the meter. At 300mA, there will be 300mV across
the shunt. The meter needs 50uA x 1085 ohms = 54mV to read fs. The
series resistor would be 300 -54 mV divided by 50uA - or 246mV/50uA =
4.9k ohms. However, it is better to calibrate with a DMM by adjusting
the series resistor to whatever is needed to produce a 300mA fs reading.
- later, Sugi
Rich...
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
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