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Re: [Amps] meter shunt

To: "amps@contesting.com" <amps@contesting.com>
Subject: Re: [Amps] meter shunt
From: "Dr. David Kirkby" <drkirkby@ntlworld.com>
Date: Sun, 21 Sep 2003 05:59:04 +0100
List-post: <mailto:amps@contesting.com>
carl seyersdahl wrote:
> 
>  I am repairing an FL2100B in which the powersupply went south, and I have
> found what appears to be the meter shunt burned out.  (IP)  It is a
> wirewound unit , and the manual doesn't tell me what it's value is.!!
>   Can anyone give me a clue ??? I'm guessing it is likely to be .1 ohm , as
> some others are.
>   Any help out there?? Thanks !!
>  carl / kz5ca

If the resistor has burt out, there is a good chance the mater has too,
so I would check that first. 


The value depends very much on the meter you have - its FSD and internal
resistance as well as the new FSD you want for the meter. However, if
you can find out:

a) The FSD of the meter in amps (say 0.001 if 1 mA) - call it Im. That
would be easy to measure.
b) The internal resistance of the meter in Ohms - call it Rm. Again it
is easy to measure. 

c) The current you want the meter to read in Amps - call it Iw. 

you can do it all with Ohm's Law. 

Your meter without it's shunt is a volt meter with a full scale voltage
of Vm = Im * Rm. 

Your shunt needs to take the difference (excess current) to between the
meter's current range and that of FSD you want, whilst at the same time
developing a voltage Vm across it. 

So your shunt resistance (call it Rs) needs to take the current (call it
Is) o

Is = Iw - Im

As you need to develop Vm across this shunt. So the resistance of the
shunt Rs is:

Rs = Vm/Is = (Im * Rm) / (Iw - Im)

The power dissipation of the shunt will need to be Is^2 * Rs. 

As an example. You want a current of 100 mA FSD, using a meter with an
FSD of 1 mA and a resistance of 2500 Ohms. 

Vm = Im * Rm = 0.001 * 2500 = 2.5 V
Is = Iw - Im = 0.1 - 0.001 = 0.099 A

Rs = Vm/Is = 2.5/0.099 = 25.3 Ohms. 

Power dissipation for shunt = Is^2 * Rs = 0.099*0.099*25.3 = 0.24 W

 
Hence unless you are measuring lots of current, the dissipation of the
shunt is not the high. However, the wire-wound design might be for arc
protection. Since it seems not to have worked, perhaps putting a higher
power one back might not be a bad idea. 

I expect you will find other ways of working out the shunt resistance on
the web. 
-- 
"The day Microsoft makes something that doesn't suck is probably 
the day they start making vacuum cleaners." -Ernst Jan Plugge.

Dr. David Kirkby,
Senior Research Fellow,
Department of Medical Physics,
University College London,
11-20 Capper St, London, WC1E 6JA.
Website: http://www.medphys.ucl.ac.uk/~davek
Author of 'atlc' http://atlc.sourceforge.net/
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