Measure 100 amps with 10 amp meter ?

[moto-chrome]

I have built a 12 volt lamp controller capable of 100 Amps. The problem is that the meters on the market can only measure 10 amps. I have heard it is possible to do using resistors. Does anyone know how to do this ?

[eprigge]

Your meter has an internal shunt (resistor) which the meter uses to measure the current. What you need to do is attach a second shunt resistor in parallel with the meter. If the second shunt resistor has 1/9th the resistance of the meters shunt, the meter will read 1/10th the actual current flowing. And this second shunt would have to be rated for 90A or more.

[moto-chrome]

Thanks a million, that's exactly the answer I needed.

[seanc]

eprigge:

If the second shunt resistor has 1/9th the resistance of the meters shunt, the meter will read 1/10th the actual current flowing.

Your values are off.

The formula for parallel resitance is: (R1 X R2)/(R1 + R2)

In order to read 10x the original amperage, you need 1/10 the original shunt resistance. Your numbers produce just under 1/5 the resistance, and the meter would be way off.

The correct value for the second resistor would be @ 0.11 x the first value

MotoChrome:

The problem is that the meters on the market can only measure 10 amps. I have heard it is possible to do using resistors. Does anyone know how to do this ?

You first have to determine the internal resistance of the meter, which may or may not be easy.

Is it a digital or analog meter?

All digital meters are shunt type, but some are sealed so that you can't take them apart to find the resistance.

Analog meters are either shunt or series.

A series analog meter does not use a shunt resistor, you have to determine the resistance of the internal coils. Not easy. But the resistance MIGHT be marked on the face plate.

A shunt analog meter may have an external or internal shunt. If it's external, disconnect it and measure its resistance. If it's internal, you will have to disassemble the meter to get at it.

Once you've determined the original resistance, find a resistor of 0.11 x that value, and use it in parallel with the meter. You want a precision resistor, of 1% tolerance or better, which is temperature stable.

The wattage will depend on your maximum PS voltage. As eprigge said, the new resistor will carry 90 amps. Multiply that by your max voltage, and that will give you the MINIMUM wattage for the resistor.

eg. if you'll be plating @ 12v, then 12x90= 1080 watts. That would be a HUGE and EXPENSIVE resistor.

Sean

[sdold]

There's an easier way to do it, just find a shunt resistor with a known (small) value of resistance, say 0.1 ohms. Then when you have current flowing, you measure the voltage across the shunt, with the meter in the volts position (not in the amps position).

E = Voltage
I = current
R = resistance

E=IR
R=E/I
I=E/R

So, to find current, if you read 10V on the voltmeter (example), the current is 100A (if the shunt is 0.1 ohms).

Steve

[sdold]

I forgot:

power = E x I

So the 0.1 ohm resistor would have to dissipate 100amps x 10V, = 1000W.

It would be an impressive resistor!

[eprigge]

Originally Posted by seanc

eprigge:

Your values are off.

The formula for parallel resitance is: (R1 X R2)/(R1 + R2)

In order to read 10x the original amperage, you need 1/10 the original shunt resistance. Your numbers produce just under 1/5 the resistance, and the meter would be way off.

The correct value for the second resistor would be @ 0.11 x the first value

No, my numbers are right, 0.11 is (close to) the 1/9th as I said... the formula for the second shunt resistor is Rs = Rm/(n-1) where n is the factor of division you want to get on your meter readings and Rm is the meter shunt resistance.

I would hazard a guess that a DMMs 10A range shunt resistance is around 0.01 ohms. Any more than that and you'd have too much power dissipating inside the meter. You might be able to find out from the meter docs or the manufacturer instead of measuring it yourself. If it was 0.01 ohms then the second shunt would need to be 0.00111 ohms and would need to handle 9 watts to carry 90A.

I would use a constant, known current source like a current controlled power supply to make the new shunt. It's such a low resistance you will probably just make it out of a length of copper wire. If you know you've got 10A flowing through the load, diddle with the shunt resistance until your meter reads 1.00. Voila. In the example above, about a foot of 10-12 gauge wire would be in the 0.001 ohm ballpark.

[seanc]

Originally Posted by eprigge

No, my numbers are right

My apologies, you were correct. My dsylexic brain was mis-reading your value as 9/10. Too early in the morning and not enough coffee yet. Sorry.

Sean

[seanc]

Steve:

Originally Posted by sdold

There's an easier way to do it, just find a shunt resistor with a known (small) value of resistance, say 0.1 ohms.

That's all a digital ammeter is: a voltmeter w/a built-in shunt.

I got a surplus 200mA digital panel meter, which used a 1-ohm shunt (the actual meter was a 200mV full scale "movement"). Wired in additional 0.1 & 0.01 ohm resistors through a switch, and now have a switchable 20A, 2A, 200mA dedicated ammeter.

Sean

[sdold]

Originally Posted by seanc

Steve:

That's all a digital ammeter is: a voltmeter w/a built-in shunt.

Sean

Right, so if moto-chrome is going to use a multimeter, he can build his ammeter with the multimeter set as a voltmeter, across an external shunt, and not have to do any of the parallel resistance calculations.

Steve