CAV ammeter spec - please

[Tony Brooks]

This one is probably for Arnot or Sir Nibble.

 

On the Canal Boat Forum I have just been asked about the shunt is for a 100 amp, CAV, moving coil ammeter I expect its the one that was fitted to many trucks.

 

I have been unable to find the mV rating for full scale deflection so that info would be gratefully received.

 

Cheers

[WotEver]

Hi Tony, all I can find is that the CAV shunt for that ammeter was the type Z5913/30, but I can't find anyone offering either that shunt or a replacement for it.

 

Perhaps the questioner will have to do some experimenting?

 

Tony

[Tony Brooks]

Hi Tony, all I can find is that the CAV shunt for that ammeter was the type Z5913/30, but I can't find anyone offering either that shunt or a replacement for it.

 

Perhaps the questioner will have to do some experimenting?

 

Tony

 

 

Thanks for that. I have already suggested he gets it hooked up to a variable power supply and gradually turn the volts up until he gets full scale deflection and that most secondary schools and colleges probably have them in their science labs. More than that we can not do.

 

I fear it may be an Ebay purchase gone sour because there are other CAV meters on there without the shunt.

 

Anyway, thanks for your efforts.

[paulcatchpole]

Will the resistance of the meter coil not allow you to work it out?

 

(Thinking aloud, and after a pint... )

 

PC

[WotEver]

Will the resistance of the meter coil not allow you to work it out?

 

(Thinking aloud, and after a pint... )

 

PC

Ummm... no

[paulcatchpole]

Ummm... no

 

Now there's what a pint does for you...

 

To flog a dead horse though, why not?

 

All meter movements have resistance, because the coil uses many turns of fine wire. The resistance varies from perhaps 200 ohms or so (1mA movement) up to around 3.5k for a 50uA movement. These figures can vary quite widely though, depending on the exact technique used by the manufacturer.

 

The above from: http://sound.westhost.com/articles/meters.htm

 

And this for calculating it:

 

4.0 - Current Shunt

 

The situation is a little more complex when calculating a shunt for current measurement. Not so much because the calculations are difficult, but because you will be working with very low resistance values. It is also important to ensure that the meter is connected directly to the shunt - even a small length of wire in series may make readings uselessly inaccurate. The schematic diagram below shows not only the electrical connection, but also the physical connection to the shunt.

 

In most cases, it is easier to calculate (or measure) the voltage across the meter movement for FSD. If you don't know the resistance, it can be measured with a digital multimeter. The current from most digital multimeters is low enough not to cause damage to the meter, but the pointer may swing rather violently. Connect with reverse polarity to minimise the risk of bending the pointer.

 

Unless you are measuring low currents (less than 1A or so), the shunt resistance can be worked out using Ohm's law, and will be accurate enough for most purposes. This is covered below.

 

Figure 4

Figure 4 - Shunt Resistor for Current Measurement

 

Assuming a 1mA movement with an internal resistance of 200 ohms, as an example we wish to measure 5A. This means that 4.999A must pass through the shunt, with the remaining 1mA passed by the meter movement. The shunt resistance can be found with the following formula ...

 

Rs = Rm / ( Is / Im) where Rs is the shunt resistance, Rm is the meter resistance, Is = shunt current, Im = meter current

 

So for our example,

 

Rs = 200 / ( 5A / 1mA ) = 0.04Ω

 

If we use only Ohm's law (having determined that there will be 200mV across the movement - 1mA and 200 ohms), the shunt can be calculated as ...

 

Rs = Vm / I where Rs is shunt resistance, Vm is meter voltage at FSD, and I is the current

Rs = 0.2 / 5 = 0.04Ω

 

This method will work to within 1% accuracy provided the measured maximum current is more than 100 times the meter current. One thing we have to be careful of with shunts is that the voltage 'lost' across them is not excessive. This will reduce the voltage supplied to the load, and can result in significant errors, especially at low currents. For example, if we only need to measure 1mA, we can use the meter directly, but we lose 200mV across the meter. In the case of the 0.04 ohm shunt calculated above, we lose

 

V = R * I = 0.04 * 5A = 200mV

 

... exactly the same voltage loss! It's not a great deal, but can be critical in some exacting tests or at very low voltages. 200mV is almost nothing with a 50V supply (0.4%), but is very significant if the applied voltage is only 1V (a full 20% loss). The voltage drop can be reduced slightly by using a more sensitive movement. For a 50uA movement with 3,500 ohms resistance, the loss is

 

V = R * I = 3500 * 50uA = 175mV

 

... not much of a gain, but not many alternatives. Current measurement will always lose some voltage, so it is important that the voltmeter is always connected after the ammeter, so that the 'lost' voltage is taken into consideration. Where extremely low voltage drop is important, one must resort to amplification. An opamp can be used to amplify the voltage across a much smaller value shunt, but at the expense of circuit complexity and temperature drift. Digital panel meters are better than analogue movements for current measurements.

 

The idea of a shunt is all well and good, but where does one obtain an 0.04 ohm resistor? It can be made up of a number of wirewound or metal film resistors in parallel, or a dedicated shunt may be available. Obtaining high accuracy at such low resistances is very difficult though, and shunts are generally cut, machined or filed to remove small amounts of metal until the exact value needed is achieved. The shunt must be made from metal having a low temperature coefficient of resistance to prevent the reading being affected by changes in temperature - either ambient, or caused by the load current heating the shunt..

 

There is an easier way, as shown in Figure 5. The voltage drop will be a bit higher than it should be, but you only need a few millivolts extra to be able to use the technique.

 

Figure 5

Figure 5 - Variable Shunt Resistor

 

Now it is possible to use 2 x 0.1 ohm resistors in parallel, giving 0.05 ohm. The voltage drop at 5A will be 250mV, but you have the advantage of being able to use standard tolerance resistors, which can represent a significant saving. The power is only 1.25W at full current, so a pair of 5W resistors will barely get warm. The trimpot can be adjusted to give an accurate reading, without having to resort to close tolerance resistors with impossible values. As an example for the above 5A meter, we could use a 100 ohm trimpot in series with the meter. The value is not particularly important, but needs to be within a sensible range.

 

What is 'sensible' in this context? Easy. We already know that the meter needs 200mV for full scale and that we will get 250mV across a 0.05Ω shunt, so we need a resistance that will drop 50mV at 1mA.

 

R = V / I = 0.05 / 0.001 = 50 ohms

 

Since we are using a pot, it is advisable to centre the wiper under ideal conditions to give maximum adjustment range (to allow for worst case tolerance), so a 100 ohm pot is ideal.

 

???

 

PC

[WotEver]

To flog a dead horse though, why not?

Because a moving coil meter will give FSD at a certain voltage.

 

Dependant upon the size of the wire used in the coil and the number of turns, the resistance of the coil could be almost anything (but in practice is commonly in the region of around 200R). The sensitivity of the meter and therefore the voltage at which it deflects fully is determined by the relationship between the coil and the permanent magnet.

 

The two bold bits are a clue

 

Tony

 

Quick edit to add that your big quoted thingy included the same clue...

We already know that the meter needs 200mV for full scale

 

And another edit to say that once we know the voltage at which the meter will give FSD (as suggested by Tony B above) it's a simple matter to use Ohms law to determine the value of the shunt to drop the required voltage at 100A, which is basically what your quote was all about.

Edited May 18, 2010 by WotEver

[paulcatchpole]

Because a moving coil meter will give FSD at a certain voltage.

 

Dependant upon the size of the wire used in the coil and the number of turns, the resistance of the coil could be almost anything (but in practice is commonly in the region of around 200R). The sensitivity of the meter and therefore the voltage at which it deflects fully is determined by the relationship between the coil and the permanent magnet.

 

The two bold bits are a clue

 

Tony

 

Quick edit to add that your big quoted thingy included the same clue...

 

 

And another edit to say that once we know the voltage at which the meter will give FSD (as suggested by Tony B above) it's a simple matter to use Ohms law to determine the value of the shunt to drop the required voltage at 100A, which is basically what your quote was all about.

 

 

Aye, sorry Tony, in flogging the dead horse, I meant really that "Isn't it still possible to work it out"...

 

So by connecting a 1.5v cell, and a pot, to it, you could work out the FSD voltage?

 

My fault for not reading the posts properly, and you're still right that my resistance idea was a red herring... Sorry!

 

PC

[WotEver]

So by connecting a 1.5v cell, and a pot, to it, you could work out the FSD voltage?

Yup. Just hang a high impedance multimeter across the terminals, raise the voltage until it reads 100A and note that voltage.

 

If you're using a 1.5V cell and the meter wants say 100mV for FSD and if the meter has a 200R coil (lots of assumptions there) then you'd probably want something like a 3k3 or 4k7 linear pot. That'd be a good starting point, anyway.

 

Tony

 

Edit to say that Tony B's idea of borrowing a school's variable PSU for 5 minutes is probably easier and quicker.

Edited May 18, 2010 by WotEver

[Tony Brooks]

Thanks all.

 

Paul had me going for moment in case it was possible to calculate from the resistance though.

 

I am sure the only way someone would know the answer is from a long memory and a lot of experience. I always advise buying the meter and shunt as a pair.

 

I strongly suspect, from the questions I get, that the magazine forum readers are, in the main, not well equipped to use pots and batteries, simple though it is. The most likely outcome of advising that is that I would Eb asked where to buy such kit and what does linear mean

[WotEver]

The most likely outcome of advising that is that I would Eb asked where to buy such kit and what does linear mean

[Tony Brooks]

 

 

That Bl*** spell chucker missed one

[Timleech]

I'm reviving this thread because it's relevant to what I've just been doing.

 

I wanted to use a CAV 60-0-60 ammeter which I had kicking around in conjunction with the CAV AC7 alternator that I've just fitted, as described in the Gardner sub-forum. The ammeter was one with the shunt between the studs on the outside of the case. My first thought was to remove the shunt and fit it into the alternator wiring, but that would have been a bit untidy - need to make a case etc etc.

Next thought was to find what sort of shunt was needed, so set about finding the meter charachteristics with the shunt removed. It turned out to be a 5mA movement, and something like 14 ohms between the terminals. I did some back of the envelope sums and worked out that the cable between alternator and starter motor, about 1m of 25 sqmm, would give a voltage drop of the right order but slightly more than needed, so I got it set up and working, giving a pretty close indication compared with a clamp meter, with about 10 ohms of resistance in series with the meter, connected across the negative charging lead. So far so good, pleased with progress.

Next thing was that I wanted a matching meter for the planned second alternator. This, if it happens, will be a 70A unit so slightly bigger full-scale meter needed so I started looking at ebay. There's a steady trickle of CAV ammeters coming up on ebay, but what caught my eye was some New old stock 0-80A units. The way I want the meters connected they should never need to show a discharge, so I got a pair of these, replacing the rather tatty old unit I'd already fitted with one of these.

They came without shunts. I did my measurements again, and was a bit surprised to find that although they were still 5mA full scale deflection the resistancce was roughly double. This meant that there wasn't going to be enough volts drop across my charging lead to get one to read correctly. The second alternator will have a longer cable so probably will be OK.

Time to look for a proper shunt?

A bit of googling for 'CAV ammeter shunts' brought up this thread, which I don't remember reading when it was current (!)

Reading that, combined with the odd measurements, set me wondering what's really inside the ammeter case, and my curiosity revealed this:-

 

 

There's a small resistance either side of the meter coil, which adds up to somewhere around 14 ohm total, so the meter movement is pretty much the same as my original centre-zero unit. A few minutes with a soldering iron bypassed one of the two resistances. Fitting the meter into my charging circuit, it now reads a few percent low but plenty close enough for my purpose which is just to give a general indication of what is going on.

 

The resistance figures I've given are from memory, may not be spot on.

 

Tim

Edited November 24, 2012 by Timleech