Hello,

When I connect V+ and V- both to ground to effectively turn off the op. amp,
the voltage between the resistors in the voltage divider is 1.3V (measured
with an X10 probe and paid more attention this time).  I am indeed using a
R-R op. amp that can handle inputs close to both the positive and negative
rails.  The reference voltage is connected to the non-inverting input on the
amplifier.

What I can't figure out is why the voltage between the resistors in the
voltage divider is 1.3V (I when the op. amp has V+ and V- are ground.

             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?

Even more interesting, is that when I have V+ at 5V and V- at ground.  If I
turn the +5V supply off and then on, the current being drawn by the op. amp
decreases from around 11uA when I first turn on the power to 0uA (less than
my DMM can measure) in about 30sec.  Does an op. amp. have a "warm up" time?
Is the results I'm seeing expected?  How is this "warm up" time connected
with my voltage divider.

Note: that the problem is easily solved by replacing the 200K resistors with
25K resistors, but I wish to understand why this is happening.

Regards,
Donovan Parks

> I'm not sure from your description where the 1.7 volt is I thought the
> ref was connected to V= or V- But if you expect to use an an op amp on a
> single supply and have inputs close to ground you must use an op amp
> who's input range common mode goes to ground most of them don't check
> the common mode input range in their spec's
>
> PC
>
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