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IVC102 tests

K

Kingcosmos

Jan 1, 1970
0
I am looking at the IVC102 from Texas Instruments, specifically the
first page:

http://focus.ti.com/lit/ds/symlink/ivc102.pdf

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!"
 
C

Chris

Jan 1, 1970
0
Kingcosmos said:
I am looking at the IVC102 from Texas Instruments, specifically the
first page:

http://focus.ti.com/lit/ds/symlink/ivc102.pdf

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
 
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