Ammeter Shunts

[maxi77]

I am considering upgrading my alternator from the present 35A to about 70A. The current ammeter only reads to 35A so one suspects its shunt will be upset with double the current and will thus need changing. The question is if I simply replace the shunt will the meter now at full scale deflection read the value of the new shunt or does the meter have to match the shunt.

I also note that affordable meteres seem to stop at 60A though 100A shunts seem readily available.

Thanks in advance

 

[wooslehunter]

Essentially, the meter has to match the shunt or the meter reading will be wrong.

It's all ohms law. V = I x R. Volts dropped (V) is current (I) x the shunt resistance (R).

Your ammeter isn't an ammeter, it's a voltmeter. It measures the voltage drop across the shunt which is a resistor. So, the meter & the shunt have to be matched. The resistance of the shunt has to be small for 2 reasons. First it can't drop too much voltage otherwise it would disrupt the circuit. Second, the higher the resistance, the more power it dissipates & the hotter it gets.

If you are lucky, you may get away with just changing the shunt. For this to work, you will have to work out what voltage the shunt drops at full load. Ideally if you can the same rated voltage drop at 70A, then you simply multiply the meter reading by 2 & you have the actual current. By convention shunts drop 50mV, 75mV or 100mV at rated current. You may have something different.

Run up the system & measure the voltage across it with a volt meter.

Let's say you're ammeter measures 20A & you measure a voltage drop of 57mV. 0.057 x 35 / 20 = 0.1. So you have a 100mV shunt. If you can get a 70A, 100mV shunt, you're in business. The restance of this will be half that of the 35A shunt & you can simply multiply the meter reading by 2. If you get a 75mV shunt, then replace it with a 75mV, 70A shunt. Other combinations are possible but it will require a different multipler. At least try to keep the voltage drop the same. i.e. if you replace with a 100A shunt of the same voltage drop, then the multiplier will be 100 / 35 = 2.86. Not sure I could do that in my head. If you can get to the scale card inside the meter, you could make a new one.... Now we're getting silly.

Don't what ever you do think of putting another similar shunt in parallel with the existing one. The resistance will be right & you can multiply the meter by 2 but the shunts will get too hot & burn out or worse.

 

[ianj99]

I am considering upgrading my alternator from the present 35A to about 70A. The current ammeter only reads to 35A so one suspects its shunt will be upset with double the current and will thus need changing. The question is if I simply replace the shunt will the meter now at full scale deflection read the value of the new shunt or does the meter have to match the shunt.

I also note that affordable meteres seem to stop at 60A though 100A shunts seem readily available.

Thanks in advance

Most mechanical meters use 75mv shunts. So just buy a higher rated 75mv shunt. Eg if you fit a 100amp 75mv shunt, the meter will read 100amps full scale so will need the scale replacing - many have removeable scales so it could be modified.

Alternatively, buy a 100mv/100amp shunt (i.e 1mv per amp) and a cheap digital meter off ebay for £10-15. Most of these read 1999 (ie 199.9mv) full scale so 1 amp will give a reading of 1.0 and 100.5amps a reading of 100.5

If you fit a 2 or 3 position switch and have 2 or 3 shunts, one meter can monitor several batteries an/or the alternator for example.

This is what I've done but using a NASA BM1 Compact battery monitor.

I

 

[Billjratt]

your 35 amp meter will show the same, 'cos that's the numbers on it's face - unless you change them. You could try creating a new face on your computer, printing it, and sticking it over the existing numbers...100-0-100 is what mine shows
As for the shunt - is it in the negative lead common to all batteries at one end and with all inputs and loads at the other end? If so, try removing the existing shunt and use the whole cable as a shunt. (you are effectively displaying the voltage drop across the cable).
You can then add a small preset potentiometer to the meter and calibrate it to read an accurate known load (car headlamp bulb for instance).
Removing the shunt cuts down the joints and a little of the resistance in the cable so is better for the circuit in general.

 

[Ubergeekian]

Essentially, the meter has to match the shunt or the meter reading will be wrong.

It's all ohms law. V = I x R. Volts dropped (V) is current (I) x the shunt resistance (R).

Your ammeter isn't an ammeter, it's a voltmeter. It measures the voltage drop across the shunt which is a resistor.

Er-hem.

All electromagnetic meters are current meters. The only voltmeters are electrostatic ones. I have one here: it's two feet high, weighs four stone and is calibrated up to 100kV.

The ammeter, in otherwords, is an ammeter. Since it's not up to measuring the full output of the alternator, the shunt is used to give most of the electricity an easier pass: the relative resistances of ammeter and shunt determine what proportion of the current goes through the ammeter and gets measured.

In this case you want to divert a higher proportion of the current through the shunt. There are two ways of doing this: lower the shunt resistance (by fitting a new one) or increase the meter resistance (by adding a suitable resistor in series). The former will be simpler, the latter very much cheaper.

To size the series resistor, let's assume that the shunt has a resistance Rs, the meter has a resistance Rm, the maximum current in the current setup is Imax and the full scale deflection current is If. Since the voltage across the shunt must be the same by either path,

If x Rm = (Imax - If) x Rs

In the new setup we increase the maximum current to 2 Imax and add a ballast resistor Ib to the meter. Now

If x (Rm + Rb) = (2 Imax - If) x Rs

Subtract the first from the second to get

If x Rb = Imax x Rs = If (Rm + Rs)

using the first equation again. Since the shunt will have very low resistance compared to the meter (obviously) Rm + Rs ~ Rm and, to cut a long story short, all you need to do is add a resistor with the same resistance as the meter in series with itand you will have near-as-dammit halved the sensitivity. Maplin will sell you the resistor for pence.

Caveat: Make sure the existing shunt can handle the greater current safely

 

[JohnTH]

This is just like being back at school boys. Will we have homework to do later? Hi.

73s de

Johnth

 

[halcyon]

Er-hem.
The ammeter, in otherwords, is an ammeter. Since it's not up to measuring the full output of the alternator, the shunt is used to give most of the electricity an easier pass: the relative resistances of ammeter and shunt determine what proportion of the current goes through the ammeter and gets measured.

Er-hem

Unless it is a solid state shunt, which measures the magnetic field and drives the ammeter via voltage output.

Brian

 

[Billjratt]

This place is full of teachers and theorists. We know moving coil meters measure current, but the current needs a PD to drive it and it is that PD across the cable or shunt we are interested in as it is proportional to the current flowing in the main conductor.
I KNOW my method works, you can come and watch it if you like. NO CALCULATIONS WERE USED in the conception and installation of the setup. I used a sensitive centre-off uammeter wired across the neg lead, and trimmed it to suit with a pot. simples.

 

[Ubergeekian]

Unless it is a solid state shunt, which measures the magnetic field and drives the ammeter via voltage output.

That's not a shunt. But yes, you are right, I have used Hall-effect current sensors and should perhaps have measured them. If they drive a mechanical meter, though, it's still measuring current!

 

[halcyon]

That's not a shunt. But yes, you are right, I have used Hall-effect current sensors and should perhaps have measured them. If they drive a mechanical meter, though, it's still measuring current!

If a solid state shunt drives a meter, and indicates current flowing in a electric cable, what else do I call it ?

So they measure only magnetic field, which can then be coverted to clone any resistive shunt to drive analogue, digital or bar-graph, and have the big advantage of infinite current over-load.

But the ammeter is still only driven by voltage, were as a resistive shunt uses part of the current to drive the meter. So not all ammeters are current driven.

Brian

 

[Ubergeekian]

If a solid state shunt drives a meter, and indicates current flowing in a electric cable, what else do I call it ?

This is all getting a bit abstruse, but might I suggest "Hall effect current sensor" or "Contactless current sensor" or "Inductive current sensor"

But the ammeter is still only driven by voltage, were as a resistive shunt uses part of the current to drive the meter. So not all ammeters are current driven.

You'll note that I only said that all electromagnetic meters are current driven.

Anyway, my point stands. Stick a resistor in series with the meter and leave the shunt alone.

 

[William_H]

Amp meter shunts

Essentially, the meter has to match the shunt or the meter reading will be wrong.

Don't what ever you do think of putting another similar shunt in parallel with the existing one. The resistance will be right & you can multiply the meter by 2 but the shunts will get too hot & burn out or worse.

I can't imagine why another shunt in parallel would get hot and burn. Total current will be halved into each shunt total power dissipated in each shunt half of total. Infact I would suggest this approach if another shunt was to hand.

Back to the op. You will need to relabel your amp meter. This is not as difficult as you might imagine. If you can get the cover off the meter. I use correcting fluid /white paint to cover existing numbers and a black pen to pen in new numbers. Not very professional but kind of rustic.
A very simple way to then set the amp meter up for twice as much current is to fit a potentiometer in series with the meter. About 2000 ohms max would be a good guess. Set it at zero ohms then get a decent current in the system. Adjust the potentiometer to half of the original reading. This should then be what you want. Potentiometers are not so stable over time so you can measure the resistance once set and replace with a fixed resistor of same resistance.
The concern here is that the shunt will be twice the resistance really needed so more volt drop /power dissipated though I don't think this will be a problem and then only at currents in excess of that it was made for. Check the shunt temperature when under big charge load.
Or spend the cash and try to find a shunt of half resistance.

i like the idea of using wiring itself as the shunt but you need another amp meter (clamp type) to calibrate the system to. good luck olewill