I have a drawer full of standalone voltmeter displays which I would like to repurpose into simple ammeters. The circuit I came up with to sense the current is basically a differential amp with a 1-ohm resistor between the inputs, all powered by a 9V battery. (The resistor is of the chunky 100W variety which is typically used in doorbells.) The general idea is by using that particular resistor value, I can take advantage of Ohm's law in such a way that the voltage output will be more or less precisely the number of amps being pulled by the external circuit.



Looks good on paper. But I do wonder if even a single ohm might disrupt the operation of certain circuits? Also, I am planning to use the rugged LM324 (which I have plenty of, and works well in single-supply configuration) although I did notice during simulation that it bottoms out as a current-sensor below 34mA or so (ie. 34mV at the output). That shouldn't be too much of an issue because I can't imagine I'd need to measure such a small current (not to mention that the last digit of the display corresponds to tens of millivolts) but it does make me wonder if I should be using a better quality opamp (and if so, which might be best)?

Finally, are there any other issues with the design that stand out?

 

You don’t need an opamp, unless the shunt resistor produces a voltage drop too low for the voltmeter’s full scale range.
But most of the low cost 7106 based digital voltmeter have a full scale sensitivity of +/-200 mV, which is sensitive enough to read directly from the shunt resistor in most instances.

 

You don’t need an opamp, unless the shunt resistor produces a voltage drop too low for the voltmeter’s full scale range.
But most of the low cost 7106 based digital voltmeter have a full scale sensitivity of +/-200 mV, which is sensitive enough to read directly from the shunt resistor in most instances.

Of course, I didn't even think of that! Well that simplifies things quite a bit (to say the least). I'll try it out.

 

Is VCC & VSS not connected to the 24V supply in any way?
The 1M resistors could be 100k or something. I worry about input current and offset current on the amp. 1M does not give you much current.

 

Is VCC & VSS not connected to the 24V supply in any way?

No. I was planning on running the current-sensing bit (and hence VCC & VSS) off of a battery. The 24V supply was just an example to show an external circuit being measured.

The 1M resistors could be 100k or something. I worry about input current and offset current on the amp. 1M does not give you much current.

Ah okay. I was thinking the higher the impedance, the better. But yeah that sounds pretty reasonable. Thanks for the tip!

 

With a bias current of 250nA you would get an error of 250nA * 1MΩ = 250mV which could be a problem.
Provided the input resistance of the amplifier is > 100 times the source impedance (which is 1Ω) you will get an error of <1%.
With a floating supply common-mode rejection is likely to be a problem, so you would need 0.1% tolerance resistors.
If you added some gain, you would not need to dissipate such a large amount of power in the shunt resistor, but then the large offset voltage of the LM324 would become a problem.
Soon you will have built a INA181 but without the precision matched resistors and low bias current.

 

Just as an aside, the LM324 has difficulty getting its output to ground. I found that a 1K resistor from the output to ground makes that all better. For high-accuracy, low-drift applications I don't bother with the LM324/LM358 but go for the modern bipolar or bifet op-amps.

 

With a bias current of 250nA you would get an error of 250nA * 1MΩ = 250mV which could be a problem.

Provided the input resistance of the amplifier is > 100 times the source impedance (which is 1Ω) you will get an error of <1%.

With a floating supply common-mode rejection is likely to be a problem, so you would need 0.1% tolerance resistors.

Got it, I hadn't even considered the fact that the error rate was dependant upon the bias current. I will remember to keep that in mind.

If you added some gain, you would not need to dissipate such a large amount of power in the shunt resistor, but then the large offset voltage of the LM324 would become a problem.

Soon you will have built a INA181 but without the precision matched resistors and low bias current.

The main concern I had going with a larger valued shunt is that it might interfere with the operation of the circuit being measured. The resistor I will be using is rated at a 100W so I think it should be okay. Also the short duration of measurement means I probably won't need to use a heatsink either.

Just as an aside, the LM324 has difficulty getting its output to ground. I found that a 1K resistor from the output to ground makes that all better. For high-accuracy, low-drift applications I don't bother with the LM324/LM358 but go for the modern bipolar or bifet op-amps.

Wouldn't that mess with the reading though? The simulator seems to confirm.

 

You need a totaly isolated supply for the meter, and keep the inputs separate from the power connections. AND one ohm is a quite high resistance ammeter shunt.

 

Wouldn't that mess with the reading though?

Shouldn't.

The simulator seems to confirm.

How?

I would go with no higher than 10kΩ resistors for the op amp resistors as they are still 10k orders of magnitude higher than the sense resistor and will given low offset from the bias current.

But you can simplify the circuit by just using the op amp in the common inverting or non-inverting configuration since, with a separate, floating op amp supply, you don't need a differential input.

 

ALWAYS you need to use a differential input when reading a voltage across a shunt resistance! AND, for any sort of lower voltage higher current application, one ohm for a shunt resistance is LARGE.
Really, without more information about the application, it is difficult to provide any applicable comments.

 

ALWAYS you need to use a differential input when reading a voltage across a shunt resistance!

NOT ALWAYS!
Why do you need a differential input if the sensing circuit has an isolated, floating supply?

 

if those are panel meters, they should have selectable range with lowest range being +/-200mV. you load current is under an Amp so suitable range would be 0-2A (max value to show 1.99 Amp). then direct shunt would be 200mV/2000mA=0.1 Ohm.
max power dissipation for shunt is simply P=I^2 * R = 2A * 2A * 0.1 Ohm = 0.4W but i would pick 1W

 

NOT ALWAYS!
Why do you need a differential input if the sensing circuit has an isolated, floating supply?

A differential input is aleays needed because to be accurate the current shunt voltage must be the only voltage presented to the measuring input. Any other connection scheme has the probability of adding some other voltage to the input, because of additional currents flowing in some sared part of the path. between the shunt and the input.

 

Any other connection scheme has the probability of adding some other voltage to the input

Probable yes, but that can be made negligible if the impedance is make large enough.
For example, a 100kΩ input impedance of the measuring circuit will have no practical effect on the voltage measured across a 1Ω shunt.

 

Certainly NONE of my clients over the years would have approved an installation that did not use a differential pair for the designed connection of a current shunt to the current measurement instrumentation input. Of course, I would never have considered any other scheme, either. Any other scheme, except for wireless or some fiber-optic package is probably going to need a lot of very good luck!!
In some istances yoou can get away without a differential pair. The bigbattery chargers that roll on two wheels often do not use them, but also, the ammeter is only a few inches away, and the accuracy specifications are often less demanding.

 

Certainly NONE of my clients over the years would have approved an installation that did not use a differential pair for the designed connection of a current shunt to the current measurement instrumentation input.

So apparently NONE of your clients are very technically savvy.

 

Adifferential input connection does not need to be zero resistance, but it does need to exclude other voltages that would affect the measurement voltage, which is exclusively the voltage across the meter terminals of the shuntresistance. How else could you produce an accurate current reading using a series shunt resistor??

 

How else could you produce an accurate current reading using a series shunt resistor??

What is not clear about a current sensor with an isolated power source not needing a differential input to get an accurate voltage reading across the shunt?

 

OK, somebody else neds to explain!!
A SERIES-SHUNT RESISTOR is part of an unknown and unedefined external circuit, which may have connections to "other things" . To measure the voltage across the meter terminals of the shunt resistor requires a voltage measurement connection that does not include connections to other circuitry that may include any voltage other than the voltage developed across the sunt resistor. THAT SORT OF CONNECTION IS DEFINED AS A DIFFERENTIAL CONNECTION!!!