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

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics