From: John Popelish
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Subject: Re: Simple, general OPAMP question...
Date: Thu, 12 Sep 2002 01:38:27 GMT
NNTP-Posting-Date: Wed, 11 Sep 2002 21:38:27 EDT
> If a battery is applied to the +, - inputs of an OPAMP, where does the
> output go to?
> Suppose the battery is equal to the input offset voltage so that the output
> goes to "zero". Is this zero the mid-rail of the OPAMP?
> Thanks, S.
The two inputs of the opamp have what is called a common mode voltage
range. This is the range of voltage (between the two supply rails or
more rarely even a little outside the supply rail range) that allow
the two inputs to amplify the difference between the two input
voltages, while ignoring the average voltage of the two inputs.
So, if the opamp is powered between, say, 0 and +12 volts, the common
mode range may be from zero to +10.5 (for the LM324 quad or LM358
dual, for instance). That means that as long as both inputs are
within this voltage range, the output will be the amplified difference
of the two inputs. So if the + input is at +3 volts, the only voltage
on the - input that will not saturate the output is +3 volts, also.
If the - input is +3.01 volts (+ input .01 volt more negative than the
- input), the output will saturate to 0 volts, and if the - input is
at +2.99 volts (+ input .01 volt more positive than the - input), the
output will saturate to about +10.5 volts. (these saturation voltages
will also vary with the type of opamp).
So, for stable negative feedback designs, the feedback will force the
output to be whatever voltage makes the two inputs match, exactly
(with the accuracy of "exactly" dependent on the quality of the
opamp). If the design forces the inputs to have different voltages,
the output will either be saturated positive or negative, depending on
the sign of the difference.