Circuit for converting ammeter to voltmeter.

[blkfrd]

I have a shunt ammeter that has about 0.3 ohms of resistance and is made up internally with an electromagnet (a metal post with wire wound around the post and both ends of the wire connected to external terminals). The needle sits in the middle and fully deflects in either direction for about 2A of current depending on the direction of current flow.

I want to convert it to a voltmeter with an 8 to 16V range. A 12V stable reference power supply came to mind using a SEPIC DC-DC converter powered by the same battery whose voltage I am measuring. The SEPIC supply is connected to one side of the ammeter and a current limiting resistor and the battery is connected to the other side.

When the battery voltage is +/-4V different than the 12V reference and with 2 ohms of circuit resistance, 2A would be flowing and the ammeter would be fully deflected in one direction or the other. Current and subsequent ammeter deflection would be linear for < +/-4V.

Problem here is that a SEPIC converter can't sink current. I could add a current sink transistor switch and resistor possibly that turns on only when 12V < battery voltage.

4 switch Synchronous DC-DC converters can sink current, but this solution is expensive and i'm trying to find the lowest cost solution.

What do you guys recommend? I thought of using a Power Op Amp, but I think it would require a V+ and V- power input. I'm a bit rusty on Op Amps so I can use some help though.

 

[CDRIVE]

Analog voltmeters usually employ a 50uA meter movement with series multipliers. You're 2A meter is not suitable for use with multipliers. A center 0 ammeter is nice to have. I wouldn't screw it up if I were you.

Chris

 

[blkfrd]

Analog voltmeters usually employ a 50uA meter movement with series multipliers. You're 2A meter is not suitable for use with multipliers. A center 0 ammeter is nice to have. I wouldn't screw it up if I were you.

Chris

These old ammeters relied on wire size differences to shunt a small amount of current through them while the main current path took the brunt of the current...not a very robust design and worse yet...too much current goes thru them due to one reason or another and poof.

Lots of guys complain about them either not registering or burning out and they switch to aftermarket voltmeters. With the right design, they will be more useful while also employing a PCB circuit breaker or fuse to protect it. It may not be easy, but there has to be a reasonable solution. If not, then so be it.

 

[CDRIVE]

These old ammeters relied on wire size differences to shunt a small amount of current through them while the main current path took the brunt of the current...not a very robust design and worse yet...too much current goes thru them due to one reason or another and poof.
.

That's my point. Multipliers can't be used with the movement shunted. Since the shunt is internal you'll probably damage the movement when trying to remove it.

Chris

 

[blkfrd]

That's my point. Multipliers can't be used with the movement shunted. Since the shunt is internal you'll probably damage the movement when trying to remove it.

Chris

Guess i'm not explaining this clearly. The ammeter is not going to be modified. It takes 2A to deflect it fully from center 0 in either direction depending on the direction of current flow. All 2 amps of current flows through the ammeter and there is only one path thru the ammeter. I'd say its 18 to 20 ga wire that goes from terminal A then many turns around a metal core and then to terminal B...thats it.

I've already been able to make this work in the lab with lab power supplies. It is possible to make it work with a 4 switch synchronous buck/boost converter, but the cost is an issue so i'm investigating simpler approaches.

 

[CDRIVE]

So you're saying the shunt is not internal, it's visible and across the meter terminals? Does the movement say what the movement uA, mA rating is? It's usually in small print on the lower part of the white face.

Chris

 

[(*steve*)]

Sure, it's easy to convert a 2A meter into a voltmeter. For this, I'm going to assume the resistance of the meter is zero (it isn't, but if you can measure it, simply subtract that value from the resistors I will specify.

Now let's assume you want a 20V FSD. That's better than 16V because the existing markings on the meter will be usable. If you want a different value then just do the calculations with different values.

So we need a resistor that will drop 20V at 2A. Ohms law tells us that the resistance is V/I = 20/2 = 10 ohms

So simply place a 10 ohm resistor in series with the meter and it will become a 0 to 20V voltmeter.

Easy!

BUT! It will draw a current of 2A at 20V. That's about 40,000 times more current than a decent "cheap" voltmeter, and about 1,000,000 times the current of a cheap digital voltmeter.

Oh, and that series resistor will be dissipating 40W at full scale deflection.

So yeah, easy. But also impractical.

 

[CDRIVE]

Steve is just having fun with you. That's not saying it isn't true but only if you can't remove the shunt.

Chris

Edit: I just re-read this quote and Steve is NOT having fun with you! What you're wanting to do is totally impractical unless this meter is going to be used to check your cordless drill batteries!

Guess i'm not explaining this clearly. The ammeter is not going to be modified. It takes 2A to deflect it fully from center 0 in either direction depending on the direction of current flow. All 2 amps of current flows through the ammeter and there is only one path thru the ammeter. I'd say its 18 to 20 ga wire that goes from terminal A then many turns around a metal core and then to terminal B...thats it.

I've already been able to make this work in the lab with lab power supplies. It is possible to make it work with a 4 switch synchronous buck/boost converter, but the cost is an issue so i'm investigating simpler approaches.

 

[Miguel Lopez]

It seems as it is not a d'Arsonval movement. For the description it seems to be a moving iron movement. The only way to convert it to a voltmeter it's to re-wind the coil with many turns of thin wire, which in practice, should not be "practical".

 

[CDRIVE]

The most common use for moving vane meters is for AC measurement. Besides that I see no purpose for a center zero AC meter. Since the vane moves in the same direction regardless of polarity how would a center zero moving vane meter deflect to the left? On a side note, a moving vane meter would not have a linear scale

Chris

 

[blkfrd]

Sure, it's easy to convert a 2A meter into a voltmeter. For this, I'm going to assume the resistance of the meter is zero (it isn't, but if you can measure it, simply subtract that value from the resistors I will specify.

Now let's assume you want a 20V FSD. That's better than 16V because the existing markings on the meter will be usable. If you want a different value then just do the calculations with different values.

So we need a resistor that will drop 20V at 2A. Ohms law tells us that the resistance is V/I = 20/2 = 10 ohms

So simply place a 10 ohm resistor in series with the meter and it will become a 0 to 20V voltmeter.

Easy!

BUT! It will draw a current of 2A at 20V. That's about 40,000 times more current than a decent "cheap" voltmeter, and about 1,000,000 times the current of a cheap digital voltmeter.

Oh, and that series resistor will be dissipating 40W at full scale deflection.

So yeah, easy. But also impractical.

Voltage differential does not need to be that high. delta V of 3V or 4V depending on the working range of the meter (9 to 15 or 8 to 16) results in the R power being either 6W or 8W...no problem. This works in the lab just fine with lab supplies. Case mount wire wound 10W power resistors are small.

Only rarely will the meter be anywhere near full deflection.

If I had a power supply that can read the delta V between the 12V ref and the battery and produce a variable voltage between 0 and 4V and source up to 2A and I can switch the supply between terminals of the ammeter depending on whether 12V > battery voltage or 12V < battery voltage (with hysteresis so it doesn't flip flop at delta 0), then i've got it.

Won't a power Op amp do that or something else?

 

[CDRIVE]

This just basically repeats Steve's comments...>>

Your meter scale is -2A - 0 - +2A. If full scale = 2V then R would be 1 Ohm and Pd @ full scale would be 4W. If full scale = 20V then R would be 10 Ohms and Pd @ full scale would be 40W!.

What you want to do makes no sense at all. I can find no use for something like this short of being used as a battery load tester and they are meant for intermittent duty.

Chris

 

[Miguel Lopez]

Yep, you're right (I missed that detail ), but still I think it is not a d'Arsonval movement. Maybe of the Moving coil-Fixed coil type, the one used in wattmeters.

 

[CDRIVE]

That would be an Electrodynamometer. One coil measures voltage while the other measures current. They're expensive and I doubt that's what he has. Regardless of what type of meter he has he should use it for what it was intended. What he wants to do is exceedingly impractical

Chris

 

[blkfrd]

This just basically repeats Steve's comments...>>

Your meter scale is -2A - 0 - +2A. If full scale = 2V then R would be 1 Ohm and Pd @ full scale would be 4W. If full scale = 20V then R would be 10 Ohms and Pd @ full scale would be 40W!.

What you want to do makes no sense at all. I can find no use for something like this short of being used as a battery load tester and they are meant for intermittent duty.

Chris

The ammeter is a current device. It is not a voltage device. Why is a 20V example with a 40W resistor being used when the voltage used can be much less. This is an unreasonable example...I would not build this if it took a 40W resistor.

Why is it impractical? a 6W or 8W resistor and a 2A power supply can make this work...these are plentiful. I really didn't come here to ask if it was practical or not...rather to ask if anyone had any ideas on how to make it work besides using a 4 switch sync buck/boost converter.

Millions of ammeters were used in this way in automotive apps years ago and they had limited usefullness and were prone to getting fried if the main current line between the alternator and battery became degraded or if jumper cables were attached backwards.

Someone will get the last word, but I have to respectfully disagree.

 

[CDRIVE]

Someone will get the last word, but I have to respectfully disagree.

I deleted the lengthy explanation that I was going to post. Instead I chose to agree with you. You're absolutely correct.... Really! Steve and I are know nothing morons. We're here because we like giving bad advice to people that obviously have far more experience than we do.

Chris

 

[(*steve*)]

OK, I thought I explained it slowly the first time. I'll try again...

(1) You have a meter that requires 2A to achieve full scale deflection (FSD).

(2) I am presuming you can't modify the meter to change this.

(3) So it will always require 2A to get the needle right the way across to the other side. (See [1])

(4) Not less than 2A (see [3])

(5) 2A (See [4])

OK?

(6) You want to measure voltage with this meter.

(7) So, you want the deflection of the meter's needle to be in proportion to the voltage across 2 points.

(8) Let's say you want FSD to be 20V.

(9) That means the meter will have it's needle at the 2A reading with 20V across it. (See [8] and [5])

(10) It means the meter will read 1A with 10V across it. (See [7])

(11) It still means the meter will have 2A flowing through it at FSD (see [5]) -- we can't change that (See [2]).

(12) So we need something to cause the voltage across the meter to be 20V (See [8]) when 2A is flowing through it (See[3]).

(13) That something is called a resistor (See wikipedia)

(14) The value of the resistor is given by Ohms Law (See wikipedia)

(15) the value of the resistor is V/I (See [14])

(16) V is 20V (See [8]).

(17) I is 2A (See [5]).

(18) R = 20/2 ohms (See [15])

(19) R = 10 ohms (See you grade 2 teacher (10 times table))

OK?

(20) Current flowing through a resistor dissipates heat (See fingers if you hold it for a while)

(21) The rate at which heat is dissipated is given as I^2 * R (See [14])

(22) I is 2A (See [17]).

(23) R = 10 ohms (See [19])

(24) P = (2 x 2) x 10 Watts (See [21])

(25) P = 40 Watts (Still on the phone to your grade 2 teacher?)

OK?

If you disagree, please refer to the numbered line.

 

[blkfrd]

Guys,

Your taking this the wrong way. We just have a disagreement.

I don't think anyone is a moron here. How can I. This is the first post on your forum. I don't know your background and you do not know mine. We could probably learn a lot from each other.

I should not have used the words shunt ammeter to describe the ammeter in my original post. It was an ammeter that was part of a shunt circuit.

I've already done proof of concept with power supplies and a 10W resistor. If I come up with a circuit that i'm satisfied with, i'll post my results. If I can't, i'll still post my results. Maybe i'll surprise you and maybe not.

 

[CDRIVE]

I should not have used the words shunt ammeter to describe the ammeter in my original post. It was an ammeter that was part of a shunt circuit.

Well we don't know what that means. Does it mean that this meter was used with an external shunt but is no longer being used with the shunt? If so it's a totally different ball game.

I asked you earlier if you see any small print on the meter face. It usually specs the full scale meter current. You didn't answer the question. I won't ask it again.

Chris

 

[(*steve*)]

Your taking this the wrong way. We just have a disagreement.

So which point do you disagree with?

If you've mislead us about point 1 or 2 then everything changes.

We're just starting from what you've told us and then continue following the laws of nature.