L
Lostgallifreyan
- Jan 1, 1970
- 0
http://www.allaboutcircuits.com/vol_3/chpt_8/9.html
Error in article, surely? Their analysis of the differential amp says that
the output is V2-V1, where V2 goes to the noninverting input. So far, so
good.
Then they say that "the analysis is essentially the same as that of an
inverting amplifier, except that the noninverting input (+) of the op-amp
is at a voltage equal to a fraction of V2, rather than being connected
directly to ground".
Isn't this wrong? Assuming all resistors to be equal, that's unity gain, so
1V on the inverting side gets -1V out, but the 1V into the divider on the
non-inverting side gets 0.5V on the output, so a total of -0.5V out.
This looks like an analysis failure like that old game about three man and
a hotal porter and a missing dollar, but I can't see where it goes wrong.
Their analysis can't account for the missing half volt in the non-
inverting input, yet we know there is really no half volt missing.
Error in article, surely? Their analysis of the differential amp says that
the output is V2-V1, where V2 goes to the noninverting input. So far, so
good.
Then they say that "the analysis is essentially the same as that of an
inverting amplifier, except that the noninverting input (+) of the op-amp
is at a voltage equal to a fraction of V2, rather than being connected
directly to ground".
Isn't this wrong? Assuming all resistors to be equal, that's unity gain, so
1V on the inverting side gets -1V out, but the 1V into the divider on the
non-inverting side gets 0.5V on the output, so a total of -0.5V out.
This looks like an analysis failure like that old game about three man and
a hotal porter and a missing dollar, but I can't see where it goes wrong.
Their analysis can't account for the missing half volt in the non-
inverting input, yet we know there is really no half volt missing.