# Using a digital panel voltmeter to read current

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#### auriuman78

##### New Member
My project is to turn a digital panel voltmeter (0-200mVDC) into an ammeter capable of reading current to 100A DC max. Application is to measure total current draw between my vehicle's 12VDC battery and my DC-AC inverter. I know i need to use a shunt resistor and measure voltage drop across it. Differential op-amp? From there i'm lost as to what amp to use or how to scale my voltage drop to read an equivalent value in the range i'm working with. Any help?

#### ericgibbs

##### Well-Known Member
My project is to turn a digital panel voltmeter (0-200mVDC) into an ammeter capable of reading current to 100A DC max. Application is to measure total current draw between my vehicle's 12VDC battery and my DC-AC inverter. I know i need to use a shunt resistor and measure voltage drop across it. Differential op-amp? From there i'm lost as to what amp to use or how to scale my voltage drop to read an equivalent value in the range i'm working with. Any help?
hi,
Use the meter on the 200mV range.

Measure the voltage drop across a 100mV/100A = 0.001Ω resistor.

The meter display in mV's will represent 1amps ie: 1mV/A.
Maximum current = 200A.

Resistor Watts = 0.2V * 200 = 40Watts, I would use a 100W resistor.
If you limit the current to 100A I would use a 40 to 50W,,, gets HOT.!

#### on1aag

##### New Member
Hi Auriuman78,

Pure technically there's nothing wrong with Eric's explanation
but a 1mΩ 50W resistor will be impossible to find, you'll have
to look for a shunt. You'll find these in surplus stores or
at radio rallies. They will look like this 30A 30mV shunt.

on1aag.

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#### ericgibbs

##### Well-Known Member
Hi Auriuman78,

Pure technically there's nothing wrong with Eric's explanation
but a 1mΩ 50W resistor will be impossible to find, you'll have
to look for a shunt. You'll find these in surplus stores or
at radio rallies. They will look like this 30A 30mV shunt.

on1aag.
hi on1aag,

As you say, buying one could be difficult.

The way we made very low shunts was using aluminium block or sheet.

Knowing the CSA and the coeff of resistance of the aluminium its possible to get close to the required resistance.
Drilled holes in the block for attaching the external leads, which have to be VERY heavy duty cables for 100/200A.

##### Banned
www.mpja.com
They have 10 through 100 amp shunts.for between 5-10 dollars or so. Shipping is reasonable and they have enough other good stuff it's a great site to buy from.

#### lomash

##### New Member
hello how can take output in reading current & voltage

#### MrAl

##### Well-Known Member
Hi there,

Another idea is to simply measure the voltage drop across
one of the feed wires itself. A setup like this probably uses
#6 gauge wire already, and that has close to 0.001 ohms
for a 3 foot length (copper).
Since copper resistances rises with temperature, you may
wish to measure the temperature of the wire and compensate
with a little circuit that will subtract a small voltage from the

If instead it uses aluminum wire for the feed you'll have to
look up the resistance and do the math.

Alternately, use a good DC shunt in series with the feed line
and while measuring the actual DC current adjust your
meter circuit to read the same as the actual DC current.

The meter leads would be connected to the #6 wire as
follows:
One lead to the battery terminal where the wire connects
to the battery, the other lead down the wire about 3 feet.
You can go to 6 feet if you have that much wire lead,
and halve the reading to get the right current measurement.

At 100 amps the meter will read roughly 100mv before
calibration. After calibration it should read exactly 100mv
for 100 amps.

One way to measure the temperature of the wire is to use
a much smaller gauge wire wrapped around the #6 wire
for the entire 3 feet. The length and gauge of the wire
is calculated to drop the right amount of voltage from
constant over temperature. The meter is also
'loaded' so that the thinner wire actually drops the
voltage getting to the meter.

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#### Hero999

##### Banned
What's powering the panel meter?

I think a hall effect sensor might be a more sensible option because its output is isolated from the circuit being measured it also has the advantage of not taking any power from the circuit being measured.

#### wingerr

##### New Member
It's interesting how they make the current shunts for the cheap DMMs, just a thick copper wire that they appear to "calibrate" by pinching it with some wire cutters until the right voltage drop occurs with the current value.
A real current shunt would be best, but a few wires paralleled together probably would work on the cheap.

If the shunt was set up to give full scale of 200mV at 100A, it would draw 2W from the signal at full load; if that's a problem then maybe just amplify the signal to be able to use a smaller resistance for the shunt, so there's less insertion loss.

#### Roff

##### Well-Known Member
It's interesting how they make the current shunts for the cheap DMMs, just a thick copper wire that they appear to "calibrate" by pinching it with some wire cutters until the right voltage drop occurs with the current value.
A real current shunt would be best, but a few wires paralleled together probably would work on the cheap.

If the shunt was set up to give full scale of 200mV at 100A, it would draw 2W from the signal at full load; if that's a problem then maybe just amplify the signal to be able to use a smaller resistance for the shunt, so there's less insertion loss.
200mV*100A=20W, not 2W.

#### wingerr

##### New Member
Oops lost a decimal place there-

#### Roff

##### Well-Known Member
Oops lost a decimal place there-
Just didn't want it to turn into an LER (light emitting resistor). Or would that be an SER?

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