Ammeter shunt - positive or negative side of load?

[PaulMcC]

Hi All,
I have an ammeter and shunt that I intend to wire in so that I can measure amps going in to my batteries from the alternator and mains charger. It is one of the cheap digital ones from DEOK bought on Ebay which are dual voltmeter/ammeters and has a common wire, a sensing wire to the other side of the shunt to allow an amps reading and a sensing wire to allow a voltage reading against common. It is designed so that the shunt should go in the negative return wiring with the common wire at the battery side of the shunt so most negative of the the three wires.
Is it simple to come up with a wiring solution that will allow me to put the shunt in the wiring on the positive side of the battery and still have both volts and amps working? (This would allow neat wiring, that's the only reason!)
I can't see how to make this work as I end up with the voltage sensing wire less positive than the common and the voltmeter won't read negative voltages.

I can imagine a polarity reversing switch in the meter wiring which would allow either the ammeter or voltmeter to work (no idea whether the meter would survive this!) but can't think of a solution that allows both to work simultaniously.

Cheers!

 

[ianj99]

Hi All,
I have an ammeter and shunt that I intend to wire in so that I can measure amps going in to my batteries from the alternator and mains charger. It is one of the cheap digital ones from DEOK bought on Ebay which are dual voltmeter/ammeters and has a common wire, a sensing wire to the other side of the shunt to allow an amps reading and a sensing wire to allow a voltage reading against common. It is designed so that the shunt should go in the negative return wiring with the common wire at the battery side of the shunt so most negative of the the three wires.
Is it simple to come up with a wiring solution that will allow me to put the shunt in the wiring on the positive side of the battery and still have both volts and amps working? (This would allow neat wiring, that's the only reason!)
I can't see how to make this work as I end up with the voltage sensing wire less positive than the common and the voltmeter won't read negative voltages.

I can imagine a polarity reversing switch in the meter wiring which would allow either the ammeter or voltmeter to work (no idea whether the meter would survive this!) but can't think of a solution that allows both to work simultaniously.

Cheers!

If its designed to work in the negative lead, it must be connected this way or you will probably damage it or at the very least, the amps reading won't work . This is because it measures the voltage drop across the shunt with respect to the battery negative - ie from 0v to 0.075v (for a 75mv shunt).

The other way to look at it is the voltages (both battery voltage and shunt voltage) use the battery -ve as a reference as it doesn't change, whereas the +ve can't be used as it varies with battery state and charging voltage.

 

[Billjratt]

The reason for using the common negative lead is to measure total current regardless of the position of the 1: both:2: off switch.

 

[ianj99]

The reason for using the common negative lead is to measure total current regardless of the position of the 1: both:2: off switch.

No it isn't, that's just a benefit of using the negative as a reference for the reasons I explained.

 

[VicMallows]

Agree with Ianj99. It is perfectly possible to design such a device which will work in the positive feed (or even either), but from your description your device is not one of them.

 

[ianj99]

It would be possible to use it in the positive but not to measure volts as well.

You will need to power the module from an isolated dc-dc converter such as http://uk.farnell.com/multicomp/mca12d12d/dc-dc-converter-1w-dual-o-p/dp/2079676

The two sensing wires are then connected to the shunt in the positive lead. The voltage display would be that of the dc-dc output supplying the module, not the battery voltage. (sensing wire to the load side of the shunt, common to the battery side so its is negative with respect to the load side)
By having an isolated power supply, there is no conflict between supply and sensing voltage.

You could rig up a 3pole switch to enable it to be switched from amps to battery volts, or just buy another module that just reads volts, (or ignore the sense wire)
Ian

 

[theoldsalt]

If its designed to work in the negative lead, it must be connected this way or you will probably damage it or at the very least, the amps reading won't work . This is because it measures the voltage drop across the shunt with respect to the battery negative - ie from 0v to 0.075v (for a 75mv shunt).

The other way to look at it is the voltages (both battery voltage and shunt voltage) use the battery -ve as a reference as it doesn't change, whereas the +ve can't be used as it varies with battery state and charging voltage.

Certainly it is usual to put the shunt on the battery negative but only necessary if the measuring instrument uses the same battery for it's power. If a totally independantly powered instrument is used then the shunt could be on the positive terminal as the voltage drop across the shunt is all that is required to measure the current flow. The type I have do not use the battery negative as a reference. It is also recommended that a battery isolator unit is included if the same battery is used to power the instrument as is being measured. This also separates the instrument from the measured battery.

So I tend to agree with Billjratt.

View attachment 49992

 

[ianj99]

View attachment 49993

Certainly it is usual to put the shunt on the battery negative but only necessary if the measuring instrument uses the same battery for it's power. If a totally independantly powered instrument is used then the shunt could be on the positive terminal as the voltage drop across the shunt is all that is required to measure the current flow. The type I have do not use the battery negative as a reference. It is also recommended that a battery isolator unit is included if the same battery is used to power the instrument as is being measured. This also separates the instrument from the measured battery.

So I tend to agree with Billjratt.

If we stick to what the OP has bought, if he wants to use it in the positive lead the circuit attached will work as it does include the isolated dc-dc converter needed.

Otherwise I agree it is normal to have the shunt in the negative lead, but for the reasons I gave in post #2.

 

[pvb]

View attachment 49993

If we stick to what the OP has bought, if he wants to use it in the positive lead the circuit attached will work as it does include the isolated dc-dc converter needed.

Otherwise I agree it is normal to have the shunt in the negative lead, but for the reasons I gave in post #2.

So, in your suggested diagram, is the 12v battery supplying the isolated converter the same as the battery whose current and voltage are being measured? Or is this another battery which happens to be lying around?

 

[ianj99]

So, in your suggested diagram, is the 12v battery supplying the isolated converter the same as the battery whose current and voltage are being measured? Or is this another battery which happens to be lying around?

It can only measure the current, the voltage the meter is measuring is usually its own supply voltage with these cheap meters, so it will display the dc-dc convertor output. Since the convertor/s output (both + and -) is isolated from its input, there is no need for a separate battery - that's the point of using an isolated convertor.

If you were to measure the meters +ve terminal with respect to battery -ve, it will be 12v higher than battery voltage (12v across the meter + the battery volts because the meter -ve is connected to the battery +v)

But across the meter it is 12v, and across the shunt, 0-75mv (for a 75mv shunt) with respect to the meter's -ve wire. ergo it will work.

 

[PaulMcC]

Thanks all, particularly Ianj, the circuit you describe makes sense to me. I take it that it is not common to have an isolated dc-dc converter whether the output voltage mirrors the input voltage (so having the benefit of the output voltage floating but without it being regulated to a pre-fixed 12v)?

 

[Lon nan Gruagach]

Thanks all, particularly Ianj, the circuit you describe makes sense to me. I take it that it is not common to have an isolated dc-dc converter whether the output voltage mirrors the input voltage (so having the benefit of the output voltage floating but without it being regulated to a pre-fixed 12v)?

Indeed it is very uncommon, so much so that they arent sold. However a cirduit to do the job is not too difficule to build.
What I would do (shy of sticking to separate or analog meters) is get a DC-DC converter to produce 30V from the battery (being measured). Use that and batt 0V to power an op amp. The batt +12V goes to the op amp + input. The op amp out goes to base of a npn transistor (collector at +30V), emitter goes to power/sense of meter and a R-R voltage divider (to halve the V), that half V goes to the - input of the op amp.
Net result: the meter in the +ve lead has its ammeter shunt at battery V and its power/sense at 2* battery V.

Sorry if thats all a bit too tech, but I'm new here.

Hi btw

 

[ianj99]

Thanks all, particularly Ianj, the circuit you describe makes sense to me. I take it that it is not common to have an isolated dc-dc converter whether the output voltage mirrors the input voltage (so having the benefit of the output voltage floating but without it being regulated to a pre-fixed 12v)?

I doubt there is a voltage mirroring type manufactured - have a look on Farnell or RS' websites.

Also check the current needed by your meter the one I linked to was just an example. A single output type with appropriate current capability is needed.

I've used this daisy chain technique a couple of time to get a low current ~24v supply from 12v and only having a 12:12 isolated convertor to hand.
Ian