Okay, we seem to be have gotten off topic into the general subject of current measurement.
The TS stated:

I have a drawer full of standalone voltmeter displays which I would like to repurpose into simple ammeters.

So, if those displays are powered by their own isolated supply (e.g. a battery) then you don't need a differential circuit to sense the voltage across the shunt resistor.
Make sense?

 

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 needs to be differential connection when it is not isolated (as long as common mode limits are not exceeded).
but once it is isolated, common mode is no longer an issue so anything will do... even single-ended measurement os free game because now measurement circuit is free to float.

 

Most GOOD digital panel meters provide connections for both sides of the measurement input voltage that is separate from other connections. This does not include those DPMs that use one terminal for common of both power and signal and whatever else, and are priced as hobby devices. I am referring to the products with separate power and signal inputs.
Given the low resistance of most ammeter shunts, any current passing thru a shared conductor would contribute an additionalvoltage drop and create some error to the reading.

 

Given the low resistance of most ammeter shunts, any current passing thru a shared conductor would contribute an additionalvoltage drop and create some error to the reading.

I'm not sure why I need to say this again, but with an isolated power source for the meter, no current will flow from the meter to the shunt.

 

I'm not sure why I need to say this again, but with an isolated power source for the meter, no current will flow from the meter to the shunt.

That would depend on the specific conection scheme. An isolated power source is only part of an answer. IT assumes that there is no other circuitry involved.

 

That would depend on the specific conection scheme. An isolated power source is only part of an answer. IT assumes that there is no other circuitry involved.

I'm flatly and unequivocally stating there's no other circuitry involved in the TS's battery operated voltmeter turned ammeter circuit.
That's not an assumption.
Clear enough?

By the way, were you a politician at some point in your career?

 

There is a great deal of unknown circuit between the various symbols that denote a connection to "ground"! Just because we do not see it does not mean it is not there!! Without the opamp there still needs to be adequately defined connections for both the positive and negative connection to the actual meter positive and negative sense input, which constitute a differential input arrangement.
Whatever portionof the connection is not completely defined is undefined, subject to random signals.
MOST digital panel meters include quite a few connections, they are not like those wonderful megnetic movement meters that we love and use in equipment, nor are they like our wondefu friends, the battery powered digital multimeters that have only a positive and negative signal connection.

 

It is exactly like a battery powered multimeter.

But you win.
I realize I'm beating a dead horse.

 

Most GOOD digital panel meters provide connections for both sides of the measurement input voltage that is separate from other connections. This does not include those DPMs that use one terminal for common of both power and signal and whatever else, and are priced as hobby devices. I am referring to the products with separate power and signal inputs.
Given the low resistance of most ammeter shunts, any current passing thru a shared conductor would contribute an additionalvoltage drop and create some error to the reading.

Hi,

With a differential input I would think we get better CMRR. That could be important depending on the environment.

 

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.

View attachment 360980

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?

Hi,

As others have pointed out, the 1M resistors are too high. Go with 100k max, but the lower the better.
Also the 1 Ohm shunt is rather high, much higher than usual. It will definitely interfere with some circuits.
The max is usually 0.2 Ohms but 0.1 is my standard. With that you don't steal as much power from the 24v test source.

For the op amp, the LM324 is only good if you don't mind some errors due to input offset and stuff like that.
I always recommend a chopper stabilized op amp they are very good for this kind of thing. They have very very little input offset voltage.

There is also the "zero Ohm" current shunt, which is an electronic circuit that make it look as though the sense resistor is actually very close to zero resistance. It's a lot more complicated though requiring a good power supply of its own. Really just mentioning it in passing.

In the real world, we often use a four terminal resistor for the sense (shunt) resistor.

Remember there are the Hall effect current sensors also. They make it even easier just with a little less accuracy unless you can go expensive.

 

It is exactly like a battery powered multimeter.

But you win.
I realize I'm beating a dead horse.

Hi,

I think you both made some good points.

What do you think about CMRR for each of the two solutions (single ended vs differential) ?

 

What do you think about CMRR for each of the two solutions (single ended vs differential) ?

Not a factor if the sense circuit is isolated on battery power.

Do you think a battery powered multimeter has a differential input circuit for CMR?

 

Radiated noise from nearby circuits can be a problem even for battery powered isolated instruments. But those conditions usually cause humans to flee.

 

Not a factor if the sense circuit is isolated on battery power.

Do you think a battery powered multimeter has a differential input circuit for CMR?

Hi,

Well the short answer is, a multimeter is designed for that purpose. That's really the key point, except for noise. I don't think multimeters can handle noise very well unless they have a true symmetrical differential input. But the design is critical for a single ended input not so much for a differential input. We should look briefly at the situation. We can call the device to be tested the DUT with output Vout and GDUT its ground, and the measurement circuit MC with input Vin and ground GMC.

What else we have to consider is the interactions with free space, and the initial direct connection from the DUT ground to the input ground. But first we have to consider that the MC is far from the DUT, and then we gradually bring them closer and closer together, and the effect free space has on their reactions before (when they get close) and after (when the direction grounds connection is made).
What we want to avoid is current from GMC through the input to Vin, and then from Vin through free space back to GDUT. This is if we consider the DUT to have some offset, like say 200v. This could mean current flow from the GMC through the input circuit to Vin, which can cause device failure.
A multimeter will have a voltage divider on the front end which will help with this, and that could be what makes a single ended input work. A differential input has resistors on both leads (input and ground) which limits surge current so that should take care of that, so maybe all that is needed is a resistor from GMC to a new MC ground that serves as the external MC ground.

I'd like to prove this and show what is required for each but it's going to take a while I think. There are various capacitances to consider with how free space reacts to the measurement circuit coming closer to the DUT. The exponential part of the response is what we would be dealing with, but I think there could be a simpler approximation to what happens and how to fix it properly like a multimeter would have.

We have to take into account the fact that the GMC has to be charged if there is a potential difference, and that would partly depend on the charge history of the MC ground. It seems like there is no path for current, but the properties of free space makes that path. The question then becomes how much current flows through the input circuitry of the measuring circuit.