EDIT: STUPID QUESTION ALERT! DON'T BOTHER READING UNLESS YOU WANT TO WASTE YOUR TIME.
...but do take the time to check out Warren's interesting article.
Good evening all,
I have read with interest Warren Young's article called "Working With Cranky Op-Amps". You can find it here. Having read the article I have just one question, and it's with regard to the highlighted part of the following quote:
The highlighted statement refers to the circuit shown here. Sorry about all the page links by the way; I did try to attach pre-prepared explanatory images into this post but when I try that I just get a server complaint - I think it might be because I'm a new user. So external links is the only way I can reference Warren's material.
My question is, can someone explain to me how the D.C. resistance at the inverting input becomes R3 & R4 in parallel? I've had a little think about it and the only way I can imagine that being the case is if the op-amp output end of R4 was tied to ground (which would be stupid, obviously). Otherwise I can't understand how the inverting input resistance becomes the parallel combination of R3 and R4.
To my mind, the resistance at the inverting input is 1K in parallel with (10K + output load). The load isn't shown, but if it's a set of headphones (<32R?) then we can just forget the load and call it 10K in parallel with 1K which is indeed the 909R stated in the article.
...I think I've just answered my own question! Doh!
Edited by BJH1 - 1/23/13 at 1:51pm