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Re: [Amps] meter shunt

To: TexasRF@aol.com
Subject: Re: [Amps] meter shunt
From: Carey Lockhart <kc5gtt@gmail.com>
Date: Thu, 25 Aug 2011 11:02:52 -0500
List-post: <amps@contesting.com">mailto:amps@contesting.com>
thanks Gerald that makes much more sense than mos of what i  found on the
web.

Carey, kc5gtt

On Thu, Aug 25, 2011 at 6:25 AM, <TexasRF@aol.com> wrote:

> **
> Hi Carey, there are number of ways to design meter shunts. A method used by
> many is to install a low value resistor in series with the circuit and
> measuring the voltage drop across the resistor with the meter.
>
> Meter over current protection is easily added by shunting the low value
> resistor with opposite connected diodes. That clamps the voltage to +/- .7
> volts or multiples of that with diodes in series.
>
> Starting with a 1 ohm resistor and a target of 500 mA full scale for
> example, the voltage would be I X R. At 500 mA, there would be .5 volts.
> Your meter being 1 mA, needs a resistor in series to limit the current to 1
> mA when there is .5 volts present. Normally that would be .5 / .001 = 500
> ohms. But, the meter has resistance that has to be included in the 500 ohms.
> Personally, I determine the series resistor experimentally while measuring
> the voltage drop with a known good DVM.
>
> The 1 ohm resistor needs to handle full current reliably and for a long,
> stable life should be rated for several times the current expected. At 500
> mA that would be .5 X .5 X 1 or .25 watts. 2 to 5 watt resistors would give
> plenty of head room and are small enough to fit most anywhere.
>
> At a lower current, like 10 mA for example, the low value resistor is not
> exactly low value. If you want to use the diode protection scheme, then a
> value of about 50 ohms will have a voltage drop of .01 X 50 = .5 volts. The
> meter resistor then would be .5 / .001 = 500 ohms less the meter resistance.
>
> The meter and it's resistor are in parallel with the low value resistor.
> That means the 1 mA of current flowing through the meter is not flowing
> through the low value resistor. So, the 500 mA example would actually be
> 499 mA full scale and the 10 mA example would be 9 mA full scale. To
> compensate, the meter resistor would need to be adjusted to a slightly lower
> value for accurate measurements.
>
> All of this is just using Ohm's Law; no rocket science at all!
>
> 73,
> Gerald K5GW
>
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>  In a message dated 8/24/2011 10:47:28 P.M. Central Daylight Time,
> kc5gtt@gmail.com writes:
>
> i have a 1ma meter that i would like to rescale and use with a little gi7 2
> meter amplifier. what size shunt should i use or what is the calculation i
> should use?
>
> Carey, kc5gtt
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>
>


-- 
Carey Lockhart, KC5GTT
Boerne, Tx. 78006
www.kc5gtt.com
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