IVC102 tests

I am looking at the IVC102 from Texas Instruments, specifically the first page:

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A colleague of mine is testing it under two conditions with pin 2: floating and shorted to ground. Here is how I see it working:

With switch 1 open the summing junction of the integrator is not connected externally (assuming ideal switch) so it does not matter if pin 2 is floating or grounded in this case. If switch 2 is closed then that shorts the feedback capacitors and you basically have a voltage follower. I would expect to see the offset voltage on the output. If switch 2 is open then I say the output will saturate to one of the rails because of the input bias current (+/-) will charge the feedback capacitors. If switch 1 is closed and pin 2 floats then the above should happen as well. If switch 1 is closed and pin 2 is shorted to ground while switch 2 is closed then the output voltage should be close to 0V. I see the output and summing junction shorted to ground. In this case I can also see lots of current draw because of that fact: the output is shorted to the summing junction shich is shorted to ground by virtue of both switches being closed and pin 2 connected to ground. If switch 1 is closed, pin 2 is shorted to ground, but switch 2 is open I still see a ~ 0V on the ouput because the summing junction is shorted to ground. I don't see the input bias current charging the cap here.

My colleague does not see the last scenario (SW1 closed, pin 2 ground, SW2 open) and he has not confirmed the other scenarios yet. He says he sees a saturation voltage when switch 1 is closed, pin 2 grounded, and switch 2 is open.

Am I all wet or can I say "Check your connections and logic voltages!"

Hi, King. First off, a couple of observations....

The input pin of the IVC102 is *not* supposed to be left floating. I'd guess that it's the input to an analog switch, and both sides of the switch are supposed to be at a definite potential, whether the switch is on or off. You can't guarantee operation if the input side is floating in the breeze -- it's not the same as a mechanical switch.

The IC is made for a three-part cycle -- reset (SW1 open, SW2 closed -- voltage follower discharging both sides of the cap to GND), measure (SW1 closed, SW2 open -- integrating input current over time), and hold (SW1 open, SW2 open -- keeps the op amp output stable for A-to-D conversion). I don't think many people are really too concerned about SW1 closed, SW2 closed. And I'm not too sure the manufacturer is really that concerned, either.

Having said that, if you've got SW1 closed, SW2 closed, and the input pin 2 grounded, any offset on the op amp (and there is some -- ) will cause the output to go into current limit, held in check only by the resistance of the switches. Again, I don't know for sure what you can expect from this situation, but it probably isn't too good.

It's a nice little IC -- an elegant little solution to the challenge of getting a current/charge over time solution. You might want to develop another set of tests to see if it's still working.

Good luck Chris