Current shunt application, use with an ammeter or a voltmeter?

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I have a onemili ohm current shunt rated for 50A and I have two ways of measuring its output.

I could use a 100 ohm resistance =B5A meter to yield 10=B5A/A reading or a 10Mohm input voltmeter to yield 1mV/1A reading.

In theory, either works well, but which is the preferred method for a real life application if it matters at all?

ammeter specifications:

500=B5A F.S. 0.01=B5A resolution (1mA resolution) 0=2E25% + 20 dgt

Voltmeter specifications:

50mV F.S. 1=B5V (1mA resolution, but LSD is too jumpy to be of real use) 0=2E1% + 20 dgt
Reply to
itsme.ultimate
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On 22 Jul 2006 09:27:09 -0700, snipped-for-privacy@gmail.com Gave us:

Wrong. One measures voltage across a current shunt to get a table of reading that correspond to amperage in the resistor/circuit.

See above. BTW, your method changes the circuit.

One reads volts across a current shunt and extrapolates the corresponding current value for the voltage presented across the resistor. Basic Ohm's Law.

Reply to
Phat Bytestard

Practically, use the voltmeter method. The accuracy of the ammeter method depends directly on the specified 100 ohm resistance, but you have not provided an accuracy for that parameter. As a corollary, the voltmeter method will be acceptably accurate with any commercially available voltmeter, the ammeter method will work only with a 100 ohm ammeter, which may or may not always be available.

Reply to
BFoelsch

I have a onemili ohm current shunt rated for 50A and I have two ways of measuring its output.

I could use a 100 ohm resistance µA meter to yield 10µA/A reading or a 10Mohm input voltmeter to yield 1mV/1A reading.

In theory, either works well, but which is the preferred method for a real life application if it matters at all?

ammeter specifications:

500µA F.S. 0.01µA resolution (1mA resolution) 0.25% + 20 dgt

Voltmeter specifications:

50mV F.S. 1µV (1mA resolution, but LSD is too jumpy to be of real use) 0.1% + 20 dgt

---------- Either will work. B Foelsch suggests the voltmeter. Assuming resistances are exact (not true) Ammeter At 50A, current in shunt =49.9995 A current in meter =499.995 microamp Error is negligable Voltmeter At 50A current in shunt =50.0 as near as damn current in meter 0.005 microamps. near as damn B Foelsch is right but on the other hand, if you want a continuously available reading, you could use the ammeter and save the voltmeter for moving around taking other measurements. The error is negligable. Take your pick

Phats is wrong. Chances are that your ammeter is made as a 50mvFS voltmeter (say several megohms) with an internal shunt of about 100 ohms.

Reply to
Don Kelly

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One advantage I thought might be present with the current method is that 100 ohm shunt is inside the meter. I measured the resistance of the meter using a different meter and its within 1% of 100 ohm.

Since the voltage is in the order of 10=B5V, I thought maybe the exposed, unshielded high impedance loop to the meter might be suspecible to magnetic interference and things like thermoelectric voltage at connections.

Basically, the question is whether use the =B5A range or the mV range on the same multimeter. Using one of the other doesn't free the meter for other purpose.

Reply to
itsme.ultimate

mV is the preferred method. Using the ammeter assumes that you are able to reliably share the current between the shunt and the meter (forming a current divider). We use this sort of calibrated shunt in many applications where I work.

One advantage I thought might be present with the current method is that 100 ohm shunt is inside the meter. I measured the resistance of the meter using a different meter and its within 1% of 100 ohm.

Since the voltage is in the order of 10µV, I thought maybe the exposed, unshielded high impedance loop to the meter might be suspecible to magnetic interference and things like thermoelectric voltage at connections.

Basically, the question is whether use the µA range or the mV range on the same multimeter. Using one of the other doesn't free the meter for other purpose.

Reply to
no_one

Just one small point which doesn't seem to have been raised. The possible effects of component or circuit failure.

For example, if the shunt could become open circuit with the meter in place, then the outcome could be rather different if a low impedance ammeter is used instead of a high impedance voltmeter.

It is a rather essential part of any engineering project to look at abnormal, as well as normal, conditions.

Reply to
Palindr☻me

One advantage I thought might be present with the current method is that 100 ohm shunt is inside the meter. I measured the resistance of the meter using a different meter and its within 1% of 100 ohm.

Since the voltage is in the order of 10µV, I thought maybe the exposed, unshielded high impedance loop to the meter might be suspecible to magnetic interference and things like thermoelectric voltage at connections.

Basically, the question is whether use the µA range or the mV range on the same multimeter. Using one of the other doesn't free the meter for other purpose.

Since it is the same meter-used the voltage range. On the ammeter range, with the external shunt, a failure of the internal shunt would not be a problem except for wrong data being displayed, but as the meter* is basically a voltmeter with "add on" current measurements-why bother.

  • it is a digital meter which, as opposed to an analogue meter, is basically a voltmeter with high impedance bridged by an internal shunt of 100 ohms.

Apology to phat- he was not wrong-but half right.

Reply to
Don Kelly

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