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[AMPS] Meter Shunt Calculation

To: <amps@contesting.com>
Subject: [AMPS] Meter Shunt Calculation
From: measures@vc.net (Rich Measures)
Date: Tue, 6 Oct 98 04:02:01 -0800
>
>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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