Amp Meter

most likely the meter is already designed to be used inline directly. other wise you would need to know the R of the meter coil and voltage of the meter to calculate for a shunt resistor.

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A more detailed review of ammeters:

--
John Popelish

you will need to determine the coil R (resistance), 120 maybe it. lets assume for the moment that it is. your scale is 0..60, we will use that for amp scale.

Imeter = 0.050/120 = 416 uA ( does not seem likely etc).

Rshunt = 0.050/(60.0 - 0.000416) = 0.00083 ohms

so this would mean you need a shunt resistor on the meter terminals of 0.00083 which is more like a bar of copper across the terminals of the meter. if the 120 number you are reading is actually indicating I of the meter then you use 120 uA or 120 mA instead of the 416 uA in the first calculation.

in either case, you will need to refer to a wire&meteral chart to get teh ohms per 1K' (per 1k feet) to determine what you need to create a shunt .

these are only my thoughts off the top of my head going by the info your passing.

It sounds like the meter mechanism inside is a 50mV full scale instrument, but it is very likely that it has a series or parallell resistor inside the cover to make it suitable for 48 Volt or a few amperes.

Open the cover and try to find a series or parallell resistor. Or simply test the meter very carefully, using a battery and a high value resistor in series, to limit the current to less than 10mA to begin with. If there is no reaction change to a ten times lower resistor until it works.

--
Roger J.

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Let's imagine a circuit in our minds. The instrument is in parallell with a big shunt, we are seeing it deflect to the full 60 Amp, as you wanted it.

What do we know about this circuit?

The instrument is a 65mV 148uA full scale instrument. (1.5V/10400 Ohm = 148uA, and wasn't it a 65mV instrument?)

The voltage over the shunt must be 65mV too.

This shunt must have a resistance of 65mV divided by 20 Amps, which is

0.00325 Ohm.

So you need to find a piece of metal with this resistance from end to end, then connect your instrument in parallell to it, and you have a 60 Amp current meter.

Make it a little longer than needed, and adjust the connection points of the meter to the shunt to adjust it. Connect the outer current at the real ends of the shunt.

That would create an instrument which does not need a power supply, no battery.

If we add a battery we can choose a shunt much more freely, as we can use an opamp to scale the result to fit to the instrument. It will give adjustment possibilities and it is a lot easier to find a shunt.

--
Roger J.

I see now that you wanted a full scale of 6 Amp. Well, adjust the calculations accordingly.

--
Roger J.

There is also a practical way to solve this problem without calculations.

Try different pieces of metal, pieces of rebar, metal bars of all kinds.

Send a known current through it. Put the probes from the instrument together at the middle of the metal bar. Pull the probe tips apart slowly and watch how the instrument shows higher indications. When at the full scale, or the number you want, mark the two points.

Cut off the metal bar right outside the marks. Connect the probes at their marked points, and the outer current from the outer ends of the metal bar.

--
Roger J.

increment.

meter

--------------- You will have to do a bit better. If you have a voltmeter, measure the voltage at the battery- it is unlikely to be 1.5V. You could also measure the voltage across the potentiometer and determine the actual current. It appears to be a 50mv meter which seems to be reading full scale at

120ma.giving a resistance of 0.417 ohms. If this is so -then for 6A full scale, the meter will need an external shunt of 0.0085 ohms. You will need a wire large enough to carry 6A and long enough to get 0.0085 ohms (4 inches of #24 wire may do). However- check the actual current and voltage as above as the FS reading may not be 120ma. I assume you are not looking for great accuracy. See calculations below V=meter FS voltage. Im=meter FS current. I=desired full scale current Rshunt =V/(I-Im) In your case V=50mv, I=6A and, if the assumption above is correct, Im =0.120A then Rshunt =0.05/(6-0.12) =0.0085 ohms #24 wire should carry 6A and has a resistance at 20degrees C of 25.7 ohms/1000ft so 4 inches should do.

Someone please check the above.

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

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Thanks guys If I have a circuit lets say 1.5 volt bat and a 1.5 volt lamp and I want to measure the current I would put the meter parallel to a shunt resistor in the circuit right. But does the shunt resistor not cut down on the voltage to the lamp

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that looks to be a 150 uA meter at full scale. if memory serves,.i think you said the meter was a 0.050 mv?

to get a 6 amp scale. 6.00 - .000150 = 5.999850

Shunt = 0.050 / 5.999850 = 0.0083 ohms

i think that is correct,i normally measure the coil R and use the I (current) to do all of this how ever, this should still work. this is just off the top of my head, i am sure some one will correct me.

I think you are right.

In practical terms: Take a piece of copper wire, diameter 1-2mm. Send 1 Amp through it. Put the wires from the meter together on that wire. Pull them slowly apart until the meter reads 1 Amp. Mark these measuring points and solder the instrument wires to them. Cut off the thick wire outside the measuring marks and attach the current connections to the ends of the wire.

--
Roger J.