Paul wrote:
> Hi,
>
> As you know, the *input* offset voltage is the voltage required across
> the op-amp's input terminals to drive the output voltage to zero.
> Although it has been my experience that for most op-amps the input
> offset voltage is due to the "-" input pin for the *most* part. For
> example, according to Spice the input offset voltage on the "+" input
> pin on a LMC660A op-amp for a non-inverting amp circuit is a few
> nanovolts, disregarding thermoelectric effects mind you, but a few
> millivolts on the "-" input pin. Although as you know the input signal
> is not applied to the "-" input pin for a non-inverting amp circuit,
> which means there's just a few nanovolts on the input of such a
> circuit if we disregard thermoelectric effects.
>
> I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
> 25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to be
> 0.5mV on the input of this Instrumentation op-amp circuit with 3fA
> bias current that the DUT input impedance would have to be 0.50mV /
> 3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
> say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
> = 0.6nV, disregarding thermoelectric effects?
>
> INA116PA datasheet:
> http://focus.ti.com/lit/ds/symlink/ina116.pdf
>
> Regards,
> Paul
The last time I checked, the offset voltage would be the difference
between the (-) and (+) input with the Op-amp in (-) loop back mode.
So, if you were to put an op-amp in (-) loop back and lets say 5 volts
on the (+) input, the (-) input should be offset no more than what the
spec's state.
Or, I guess if you were using a +/- to common supply, you can simply
tie the (+) to common with op-amp in (-) loop back. You should be seeing
that offset factor at the (-)/output..
Maybe things have changed but that is what I go by..
http://webpages.charter.net/jamie_5"
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