Resistance of high power and small resistivity

You could alternatively use a clamp on magnetic sensor around the cable.

A shunt resistor 20A over 0.1 ohm then I^2R is 40W of waste heat. thermal effects will alter the load resistance unless you have a good way to keep it cool. A smaller value resistor would help.

0.1R 10W is a nominal stock item in hobbyist shops like Maplin Two of those in parallel on a good heatsink would just about handle your 20A (and cost less than £1).

Why make your own when you can buy them off the shelf?

Regards, Martin Brown

Yes, I would if I could - but I don't have one.

Of cours, this was just an example.

I know that already. I saw 0,01 Ohm 100W resistors somewhere.

Because the shelf is miles away.

Plus - I like the fun of doing things by myself.

I use stainless steel wire for such purposes. I recently needed a load to test a 70 Amp, 14 Volt power supply. That's 980 watts. I made a resistance element winding a few feet of stainless wire, in two parallel paths on a junk 50 watt resistor. The resistor was a few hundred ohms and was used only as a ceramic support for the stainless wire. It's resistance was insignificant. The wound wire totaled 0.2 Ohms and was carefully spaced to avoid shorts between turns of the bare stainless. The wire ends were screwed to the resistor terminals to provide a solid electrical connection where copper leads were attached. The wound resistor was placed in a 3 lb coffee can filled with water. The water provides heat sinking for the resistor and easily dissipated the 980 watts produced. The water got warm but did not boil during the few minutes of the test.

If you don't have stainless, steel bailing wire, nichrome, nickel or other junk wire can be used. The thing is to get rid of the excess heat if there is sufficient power dissipation.

to

e

nly

d to

wed

e

nd

ther

e

Hey, that is useful for my inspiration - thanks for your help.

Regular, commercial meter shunt resistors are available, of course.

Resistance of stainless steel is relatively well behaved, and shim stock (thin sheets) is a common mechanical-trade item.

A square of .002" ( 0.005 cm) thick in alloy #304 stainless steel has

Sheet resistance =3D 72 *10**-6 ohm-cm / (5*10-3 cm) =3D 0.360 ohms per square

So a square chunk at 20A will dissipate a little over 100W. Several squares in parallel, or thicker material, can do the job. I'd allow at least a square inch of surface per watt of energy dissipated.

It has to be soldered (at the edges) to a better conductor (sheet of copper?) to make a good electrical connection to the wires, and has to be supported for good airflow (and so it isn't a shock hazard).

Graphite you get from almost anywhere is very unreliable in resistivity and rather unstable to say the least. I do not know what could be worse. Copper wire is better because then it is very easy to solder on kelvin measuring taps; just be aware of the temperature coefficent..

^^^^^^^^^^ Right here is going to be your problem.

Since you have no way already to measure 20 amps, how do you plan to calibrate your homemade shunt? Resistors (shunts) for this current magnitude tend to get hot. If you make it out some ordinary conductor, copper or stainless steel, its resistance will change substantially as it gets hot. A good shunt needs to be made of some alloy with a very low temperature coefficient of resistance, such as manganin. Where will you get the thick sheet of manganin to make it? And, once it's made, how will you calibrate it? Shunts are one thing it's probably worth while to buy. I would get this one:

Item # 120342761683 on ebay.

r

Thanks.

Really ... can you tell me why ?

brate

d to get

teel, its

You're perfectly right. The measure will take less than thw seconds.

such

? =A0And,

Yes, accurate measure of a very low resistance is tricky. I think I will use an accurate AC/DC power supply in constant current mode, and measure the voltage across the resistance.

Are YOU selling this one ?

get

its

How will you know what current the power supply is delivering, since you have no way to measure currents of the order of 20 amps?

No, I'm going to keep all my shunts. I need them to measure high currents of my own.

Graphite from lead pencils have a lot of clay and are porous, from Leclanche (old fashioned "zinc") cells have less clay but are also porous, from welding supplies are better but still porous, from arc lamps ditto. Maybe if you ground up a supply and then compacted the pile one might have a less reliable and less stable resistivity... ...gotta work hard...

its

Actually, if you use a sheet, wire or rod stock the resistivity (at a given temp) would be known, and so the length for a given resistance can be calculated; placing kelvin contacts for a milliammeter would be fairly easy.

get

its

no

my

Well, clibration could be done at 200mA or 1000mA...

calibrate

get

its

such

And,

no

my

He said he wanted to "...measure accurately..."

calibrate

get

steel, its

such

And,

no

my

....and.....??? Calibration can be done at any level where the output signal is at least 100X that of noise, for 1% worst case accuracy..... ...With proper instrumentation and setup, 2mA could do the job.....

Sure thing. The point is, 2mA across 0,1 Ohm =3D 0,2 mV (or 0,01 Ohm -> 20 uV) and I guess my voltmeter is not accurate enough on this scale. That's why I thought 1A will be better : 100 mV (or 10 mV) is OK.

Yes, I think the same.

--- If 10mV is OK, then you're saying that +/- 1 digit plus the accuracy of the meter is OK.

If that's true, then with 20A through it and 10mV across it, the resistance of the shunt would have to be:

E 0.01V R = --- = ------- = 0.0005 ohms = 500 µohms I 20A

and the power the shunt would dissipate would be:

P - IE = 20A * 0.01V = 0.2 watt

#10AWG copper magnet wire has a resistance of 0.9989 ohm per 1000 feet, so a 6 inch length of it would get you damn close to 0.005 ohms and if your test was of short duration that 0.2 watt would certainly not heat the shunt enough to matter.

If you wanted to get better accuracy by using a 100mV shunt, then its resistance would have to be:

E 0.1V R = --- = ------- = 0.005 ohms = 5 milliohms I 20A

and the power it would dissipate would be:

P - IE = 20A * 0.1V = 2 watts

In this case, if you wanted to use #10AWG wire you'd need 60" of it, which could easily be wound around something like a 1-5/8" diameter paper towel core, which would yield a 5 turn coil.

For a dissipation of 2 watts I don't think you'd have a problem with it, but you could certainly tack the turns together with epoxy and dunk the thing in a bowl of water if it turned out to be problematical.

JF