Resistance of high power and small resistivity

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How to build by myself a resistance of high power and small
resistivity ?

I need to measure accurately a continuous current
( roughly over 20 Amps and probably more later )
but my amperemeter is limited to 2A.
So I will pass this current thru a very small
resistance, and measure the votlage across.

Using the graphite of a lead pencil ?
I actually measured 1 ohm on a distance of several millimeters :
using some of them in parallel, one can get as low as 0,1 Ohm.

Using a thin aluminium sheet ?
Since: R = rho * L / S
I can cut it in order to change L and then get R as small as I want.

Any suggestion that may help ?

 

[Martin Brown]

How to build by myself a resistance of high power and small
resistivity ?

I need to measure accurately a continuous current
( roughly over 20 Amps and probably more later )
but my amperemeter is limited to 2A.
So I will pass this current thru a very small
resistance, and measure the votlage across.

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
http://www.maplin.co.uk/Module.aspx?moduleno=2181
Two of those in parallel on a good heatsink would just about handle your
20A (and cost less than £1).

Using the graphite of a lead pencil ?
I actually measured 1 ohm on a distance of several millimeters :
using some of them in parallel, one can get as low as 0,1 Ohm.

Using a thin aluminium sheet ?
Since: R = rho * L / S
I can cut it in order to change L and then get R as small as I want.

Any suggestion that may help ?

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

Regards,
Martin Brown

 

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

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

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.

Of cours, this was just an example.

0.1R 10W is a nominal stock item in hobbyist shops like Maplinhttp://www.maplin.co.uk/Module.aspx?moduleno=2181
Two of those in parallel on a good heatsink would just about handle your
20A (and cost less than 1).

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

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

Because the shelf is miles away.

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

 

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I use stainless steel wire for such purposes. I recently needed a  loadto
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

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

 

[whit3rd]

I need to measure accurately a continuous current
( roughly over 20 Amps and probably more later )
but my amperemeter is limited to 2A.

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 = 72 *10**-6 ohm-cm / (5*10-3 cm) = 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).

 

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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 = 72 *10**-6 ohm-cm / (5*10-3 cm) =  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).

Thanks.

 

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   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.

Really ... can you tell me why ?

 

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                 ^^^^^^^^^^
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.  

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

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?  

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.

Shunts are one thing it's probably
worth while to buy.  I would get this one:
Item # 120342761683 on ebay.

Are YOU selling this one ?

 

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....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 = 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.

 

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 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.

Yes, I think the same.

 

[Archimedes' Lever]

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.

A 60" length seems to me that it would traverse more than five turns on
a 1.625" tube.

I got 11 and 3/4 turns, and I didn't account for the center of the wire
being greater than 1.625, so it would be maybe a tenth or two of a turn
less.

 

[Archimedes' Lever]

For a dissipation of 2 watts I don't think you'd have a problem with it,

#10? No way 2 watts through that will exhibit much heat... at all.