An AD620, I've got 1 k ohm in the input lead (v-). When the output voltage rails, I find that the input is drawing 'significant' current (a few mA.) I can change the amount of current by changing the 'gain' resistor. And also the voltage difference on the input... the larger the difference the larger the current. At highest gains (1k) (lowest R_g = 50 ohm)) the current is the highest. (~mA) The sign of the current depends on the sign of the output voltage. Is this behavior expected? I've read the spec sheet again, (quickly) and find no mention of this. I'll freely admit I don't really understand the input of this IC.
I replaced the IC thinking I might have fried the input somehow, but no change. And in 'normal' operation it works just fine.
V_supply is +/- 15V V_input ranges from 0 to 10V on each input.
I'm going to get my white proto board out and play on the bench with only the AD620.
Any thoughts?
George H.