A couple of things to suggest:

Use a smaller shunt resistor (.005 ohms - 40 mv at 8 amps) and set up the
opamp for a gain of 100 to give you 4 volts out. This will reduce your
heating - now  you are dissipating .04*8 = 0.32 watts.

Finding such a small value resistor may be a real pain, however. You can
wind your own using copper wire tables to determine what gauge and length
you need. Problem here is that copper has a poor temperature co-efficient.

Use a low-offset rail-to-rail opamp such as the LMC6462 - .25 mv offset, vs
2 mv offset of the 358. They are a couple of bucks each from Digikey, as i
recall.

If you want to proceed as per your original design, the 0.1 5 watt in
parallel should work OK. Looking at the Dale web site for power resistors,
it appears that the temperature coeffiecient is around 600 ppm per deg
C.  If the resistor heats up by 50 degrees, that will be 30,000 ppm or 3%.
I don't know if you will get that much of a temperature rise using a 10
watt resistor dissipating only 3 watts or so.

If you want to use the LM358, just use a simple charge pump such as an
ICL7662 or equivalent with 2 10uf tantalum capacitors and you have instant
-ve voltage for the 358.

You don't really need to worry about the 2 mv offset of the 358 - you can
compensate for that in your program. Same with the resistor tolerance - you
can calibrate the beast.

Have fun!

Larry


At 01:09 PM 2/17/2003 +1100, you wrote:
>I am building a 0 to 8A current sensor as part of a PIC16F877 project.  The
>voltage supply is 12V.
>
>I am wondering whether I should really be using proper shunt resistors to
>create the voltage drop, or is it OK to use normal, easily obtained 5W or
>10W ceramic wirewound resistors ?
>
>I was figuring on using 2 x 5W 0.1R wirewound resistors in parallel for
>0.05R at 10W, with the following calculations...
>
> >From my calculations, the power dissipation is I2R = 64*0.05 = 3.2W - so 2 x
>5W resistors better than one to avoid overheating.
>
>Now, the voltage developed across the resistors is = IR = 8 x 0.05 = 400mV.
>I am planning to op-amp this to get about 0 to 3V out for the PIC AD input.
>This is easily achieved using a cheap op-amp such as an LM358, esp. if a
>negative rail is supplied to get the output down to zero.  If I forego the
>negative rail, I cannot quite get down to zero (not that that would be any
>great hassle anyway) - I figured I would need a more expensive rail-to-rail
>op-amp for that.
>
>I have SPICE'd the differential amplifier and it seems to work ... but I am
>not sure if SPICE takes component heating into account in it's simulations
>(I am using Protel CircuitMaker).  I can supply a circuit if anyone feels
>the need for it to help me answer my question, which is ...
>
>My main question of concern before I go ahead and prototype this is, should
>I be too concerned about the voltage drop changing across the resistor as it
>heats up, or is the difference negligable ?  I am not too concerned with
>high resolution - even 1/8 or 1/4 amp resolution is fine for my application.
>
>Any help would be appreciated.
>
>Rgs
>Ian
>
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Larry Bradley
Orleans (Ottawa), Ontario, CANADA

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