# Why can't a feedback network use a capacitor? [duplicate]

Possible Duplicate:
Basic Frequency Control Circuit

In the following schematic, why can't R4 be taken away and the feedback network pass output feedback through C2 and R2?

-
from what i learned by reading the answer to your other question, that would lead to infinite amplification for the DC part, thus overriding any AC signals. –  noah1989 May 15 '12 at 13:32
Have you read the answers? It's pretty well explained –  clabacchio May 15 '12 at 13:32
I have read the answers and am extremely grateful to those that provided such detailed explanations. I now know why DC feedback is necessary and that R4 provides the network, however, nobody explained why feedback cannot travel from the output to the input via the capacitor route. I thought that the question was sufficiently different to that asked previously and so I created another post. –  user1083734 May 15 '12 at 13:37
That's because you don't know what's the effect of the capacitor on DC current: you should look at it first: electronics.stackexchange.com/questions/18301/… –  clabacchio May 15 '12 at 14:13
Related question: electronics.stackexchange.com/questions/31888/… –  stevenvh May 15 '12 at 14:13

## marked as duplicate by stevenvh, clabacchio♦, markrages♦May 15 '12 at 19:35

You said it yourself "DC feedback is necessary". Capacitors block DC, so a capacitor in series with the feedback path eliminates DC feedback. For the purpose of DC analysis, think of a capacitor as a open circuit.

-
Can't the opamp's V- be DC biased by placing R4 between both inputs? –  Federico Russo May 15 '12 at 14:04
@federico: For a perfect opamp yes, but for a real opamp no. That is because real opamps have both some unpredictable offset voltage and high gain. For most opamps, shorting the two inputs and holding them near the middle will still cause them to slam to one supply rail or the other. For example, 1 mV offset error times 100k gain is 100 V, which ordinary opamps can't do. 1 mV offset and 100k gain are quite reasonable numbers. –  Olin Lathrop May 15 '12 at 14:28
• Edit - strictly speaking, the gain at dc is undefined ($\frac{\infty}{\infty}$)