Measuring DC current >10a?

[loply]

Hi folks,

I need to measure a DC current which I know is more than 10a, I'm thinking somewhere between 25 and 30a.

My multimeter only has a 10a range and I presume it will either break it, or just not work, if I try.

What's normal here? Do I just need a better multimeter? Or use one of those 'clamp' type tools I see sometimes?

Thanks for any tips,

Rich

[Bluechip]

IF the circuit allows, put a 0.1 ohm alloy clad 10W or so  resistor in series with the load, and measure the voltage across said resistor.

EG:  http://cpc.farnell.com/welwyn/wh5-0r1-ji/resistor-10w-5-0r1/dp/RE05393

So: 22.5A =  2.25V

Yes, get a better meter ... 

Donkeys years ago I had a 50A shunt for my AVO8, lent it to some miserable snivelling scrote and never got it back.




Dave

EDIT BTW a lot of those clamp meters ONLY do AC  look at the spec first 

[awemawson]

You use a 'shunt', which is a low value resistance in parallel with your meter.

Find something that draws pretty much your 10 amp FSD (full scale deflection). Parallel up low value resistances with your meter until it displays exactly half what it did before,  and you now have a 20 amp fsd meter. Make it read 1/3rd and you have a 30 amp fsd meter.

As your meter will (or should) have a very low resistance on the 10 amp range, probably a length of iron wire a few inches long will be of the right sort of resistance and you can adjust     its value by trimming it's length.

[philf]

IF the circuit allows, put a 0.1 ohm alloy clad 10W or so  resistor in series with the load, and measure the voltage across said resistor.

EG:  http://cpc.farnell.com/welwyn/wh5-0r1-ji/resistor-10w-5-0r1/dp/RE05393

So: 22.5A =  2.25V

A 10W resistor would very soon die. If I remember correctly W=IČR so 22.5 * 22.5 * 0.1 = 57.375 Watts!

Edit - Dave's right - the power should be 50.625 Watts! I multiplied 22.5 by 25.5  Phil

Phil.

[Bluechip]

Whoops .... 

You're absolutely right Phil.

Probably explains why I occasionally re-invent the Direct Arc Furnace ....

What I thought of was 0R01, then a delivery arrived and I lost track a bit , so for some reason I decided it was 0r1 and hence screwed it up ... 

It should have been one of these:

http://uk.farnell.com/welwyn/wh25-0r01ji/resistor-25w-5-0r01/dp/9507159

in which case 22.5A would read as .225V . and the dissipation is 5.06W

Sorreeeeeee .... Brain Fart ....

BTW I think your 57 .375 watts  is adrift  I get 50.625W.

Dave



Dave

[vtsteam]

Here's lots more info on making shunts:

http://www.sentex.net/~mec1995/gadgets/shunts/shunts.html

And finally, last but not least -- the easiest way:

http://www.rcgroups.com/forums/showthread.php?t=393591

[loply]

Hi folks,

Thanks for all the help! Really useful.

So basically if I split the current up so only a portion of it goes through my meter, but I know what percentage, I can measure it.

Is there any way I can measure the resistance of the multimeter's ammeter so that I can just buy/find an appropriate resistor, rather than working it out by finding something which draws exactly 10a?

Cheers,
Rich

[loply]

Actually having read those links properly I think it would be easier to measure the voltage drop over a shunt instead.

[BaronJ]

I've seen those cheap multimeters smoke with way less than ten amps going through the internal shunt.

But I agree with the posters suggesting that you measure the voltage across a suitable resistance, plus that is only what the meter is doing anyway.

[John Hill]

You can approach the principle of the shunt from another angle..

Stretch a length of heavy gauge bare copper wire between two supports, maybe two nails driven into a plank a foot or two apart.

Put two alligator clips on your meter leads and clip these close together on the wire with the meter set on voltage,  pass the current through the wire and move the clip leads apart until you get a convenient voltage reading.  Switch off and measure the distance between the clips and calculate the current using the information on the wire gauge chart referenced by vsteam above.

[Bluechip]

Or ....

http://cpc.farnell.com/multicomp/sd-60-0-30a/meter-moving-coil-30a/dp/PM11439

I have bought meters from here and they have the appropriate shunt fitted. Well, the 15A one does anyway. See no reason why this should be different.

This is the one I have:

http://cpc.farnell.com/multicomp/sd-60-0-15a/meter-moving-coil-15a/dp/PM11438

Dave

[PekkaNF]

I like clamp meters when I'm happy with approximate value and when I don't want to remove the wire (Cars, mains). But using them takes a little practice.

Shunts are simple in theory, but need careful consideration as well. Shunts are easier to use if you can use relatively high ohmic value. If you use smaller ohmic value, where and how you measure voltage comes into consideration. There are losses on connections/wires etc.

Kelvin-type resistors are easier to use if any accuracy is needed:
http://www.farnell.com/datasheets/575641.pdf

Most of the time we don't need to go this far, but it helps to know where inaccuracy comes in.

Pekka

[lordedmond]

Pekka
At the iron works we used to measure DC 220 at 2000  amps plus on many machines and rotary converters we always used a very low resistance shunt these were calibrated with a hacksaw by cutting into the side to reduce the CSA

No problem at all

As to the OP

You need to do as described but you would need to calibrate the shunt with a known load could be a couple of car headlamp lamps and a 12v battery

[DavidA]

A while back I had the same problem and got around it by using a length of 1 mm copper wire (wound into a coil) and then an op-amp driving a 50 Micro Amp panel meter.
The very high gain of the op amp used the small voltage drop on the coil to indicate  20 Amp on the meter.
The wire ,  being copper,  generated very little heat.
I calibrated it by placing a DVM (on 10 AMP range) in series and turning up the current until I had 5 Amp on the DVM.  Then moved the input across the coil until I got 25% of fsd on the 50 Micro Amp meter.

That equates to 20 Amp fsd . I then removed the DVM.

This was a rather bulky affair,  but it worked until I could afford to buy a 50 Amp meter.

Dave.

[drmico60]

The problem with pure metals, such as copper, is that they have a high temperature coefficient which can effect accuracy. It is better to use a resistance wire such as constantan (so called because it has only a small temperature coefficient). For a given AMP range the shunt resistance is the same regardless of the shunt material so the heat generated is the same.
Mike

[lordedmond]

Rich

Can I ask you a question how accurate do you want the reading to be

If it's approx then what's been said will work but if it's +- say 0.01amp then it's a different question .
For that you will need to get spendy

Stuart