🚀 go-pugleaf

Hi...  I'm currently working on a project where I need an A/D converter to
read the (average) current going through a load.  The power supply for 
this project uses phase-angle control with a TRIAC to vary how much power
gets to the load (a filiament, in this case).  There's also a choke to
reduce the RF harmonics produced when the TRIAC switches on in mid-cycle.

As I said, what I need to do is measure the current going through the
filiament.  There is a shunt (if you can call it that) resistor that goes
after the load and to ground.  It is approximaterly .025 ohms.  By looking
at this on a scope, I can watch the power going through the filiament and
the load resistor.  Since V=IR, I can measure the voltage, I know the
resistance, and therefore calculate I at any point.  If I look at the
angle that the TRIAC fires at, I can integrate all this (v*t, in effect) and
get the average current going through the filiament.

HOWEVER... I need a computer to be able to read this.  So, I need some sort
of integrator circuit that's LINEAR with respect to current -- I need to 
able to feed the output of some circuit through an A/D converter.  My question,
then, is: How do I build such a circuit!?  Put a digital meter across the
shunt and ground works fine, so I know that you can build such a creature.

Ideally, I want this circuit to be able to use a single sided power supply...
I have some LN324's on the board with some of their op-amps free, so it'd
be ideal if I had a circuit that could use LM324's.  

Any ideas?  As I said, I know it can be done, but I don't know how -- so I'm
asking you!!!  It doesn't have to be perfect -- if it's 10% accurate, I'd be
happy!  

I should also mention that I tried just feeding the shunt resistor's output
through an op-amp to up the voltage a little, and then chopped off the
negative part of the waveform, and finally ran it into a capcitor.  This
gave me a reading, but it's FAR from linear!  I assume because the voltage
on a capcitor vs. time for an applied voltage is logrithmic, right?

Thanks to anyone who can help.  I really appreciate it!

				---Joel Kolstad
				kolstad@cae.wisc.edu