I'm struggling a bit with op amps again! I'm trying to learn how to use an inverting precision amplifier, and I'm getting very confused about the operation of the op amp in a active rectifier configuration. I'm sure it's a simple answer and I'm overcomplicating it, so here goes.
With a unity gain inverting op amp as shown here, when Vin
goes negative, Vout
goes positive. Opamps will do whatever is necessary to keep the two inputs the same. Since our non-inverting terminal is connected to GND, the opamp will do whatever it can to keep the inverting terminal GND, giving us our virtual ground. So it raises Vout
to match the inverse of Vin
to keep our inverting terminal close to GND, yes? -1V(Vin
) + 1V(Vout
) = 0V. Okay sick.
Now here we have added two diodes to create an active half-wave inverting rectifier. D1 is the top diode.
When Vin
is positive, we get the expected inverted output at Vout
. However, when Vin
is negative, the voltage gets locked at one forward diode drop. So I understand that Vout
raises to + 0.7V (the diodes Vf
), D1 is now forward biased and our inverting terminal is at GND. Now what happens when we lower Vin
below -0.7V to say -5V? Why does the Vout
not equally & inversely raise to keep the inverting terminal at ground? Vout
right now is at +0.7V (or D1's Vf), but Vin
is at -5V. Wouldn't the voltage at our inverting terminal be -5V? What am I missing?
Thanks!