I am Grad student. I was reading basics of opamps & read that diff voltage is ideally zero. So that voltage at inverting terminal(v1) is equal to non-inverting terminal(v2) ideally. (mostly in all derivation this has been done). i.e v1=v2. But if I apply volatge at V1 only & ground v2 to use as comparator. Then how does it works if volatge is equal. I found this confusing.
->. In case of opamp as comparator,(besides it drawbacks), wll give high/low if v1 not equal to v2. So what about vid now.
As I have read vid = Vout/A. A being very large, vid ~=0. Isn't it contradiction b/w two points.
->. Even in differential opamp with feedback,
Vo= (-Rf/R1)(v1-v2) //terms carry their usual meaning
if v1=v2, then above vout gets zero.(Obviously this don't happen, just some poor assumptions of mine, as I didn't get that concept till now).
Kindly help me to understand this.
->. In case of opamp as comparator,(besides it drawbacks), wll give high/low if v1 not equal to v2. So what about vid now.
As I have read vid = Vout/A. A being very large, vid ~=0. Isn't it contradiction b/w two points.
->. Even in differential opamp with feedback,
Vo= (-Rf/R1)(v1-v2) //terms carry their usual meaning
if v1=v2, then above vout gets zero.(Obviously this don't happen, just some poor assumptions of mine, as I didn't get that concept till now).
Kindly help me to understand this.