>              200k    200k
>  +5V --/\/\/\----/\/\/\---ground
>                      |
>                      |         100k
>                      ------/\/\/\------ non-inverting input of op. amp.
>
> I would expect the voltage between the two 200k resistors to be 2.5V as
the
> op. amp should draw almost no current, but instead it is at 1.3V.  Setting
> V+ and V- to ground has somehow caused the op. amp to draw more current (I
> have measured this the current into the op. amp to be 11uA which accounts
> for the lost voltage).  Why?

I am still confused by your drawing and description.  Why is the lower 200K
resistor tapped part way?  Is this really a pot?  If so, the voltage should
be less than 2.5V.

You should expect the opamp to draw current on its inputs when they exceed
the supply range by more than one diode drop.  Turning on power during such
a condition could even result in strange behaviour or damage to the part.

By the way, I've noticed that the 6132 can draw significant input current
when the differential input exceeds a diode drop.


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