b***r
发帖数: 149 | 1 问这个circuit的gain
我当时的回答：
if it is ideal op-amp
=〉equal potential at + and - terminals
=> + terminal virtual ground
=> (Vin-0)/R1 = (0-Vout)/R2
=> Vout = - (R2/R1) * Vin
then the interviewer asked me if i was sure that the gain is negative?
i said if the op-amp is ideal, then yes. however, this configuration forms a
positive feedback. if the output voltage increases a bit, the + positive
also increases a bit, then the output increases more. therefore, this
circuit can oscillate.
he put it aside an |
f*****0
发帖数: 489 | 2 there is no negative feedback. what you have is positive feedback, thus the
gain for an ideal opamp is infinite: the output is either the maximum
positive when Vin > 0 and maximum negative when Vin < 0.
you need to learn more common sense. |
f*****0
发帖数: 489 | 3 if there is negative feedback (as in a typical textbook), the input is on
the non-inverting end of the opamp so the output has to be in phase with the
input (thus gain > 0), not out-of-phase (gain < 0) as you suggested. |
f*****0
发帖数: 489 | 4 for this analysis (with negative feedback), an ideal opamp has two
properties you need to use:
a) inverting end and non-inverting end have the same potential;
b) the input impedance of inverting and non-inverting end is infinite (thus
no current going in / out of those two terminals).
once you remember them, your life will be much easier. |
b***r
发帖数: 149 | 5 thank you for the replies
i agree that the circuit is in positive feedback
my confusion is:
does this equation still hold for "ideal" op-amp? (Vin-0)/R1 = (0-Vout)/R2
clearly i can't derive Vout=infinity from this equaiton
the
【在 f*****0 的大作中提到】
: there is no negative feedback. what you have is positive feedback, thus the
: gain for an ideal opamp is infinite: the output is either the maximum
: positive when Vin > 0 and maximum negative when Vin < 0.
: you need to learn more common sense.
|
r*****e
发帖数: 620 | 6 definitely not, in positive feedback vin+ is not equal to vin-
your equation is based on the assumption that vin+ = vin-, so it is wrong
【在 b***r 的大作中提到】
: thank you for the replies
: i agree that the circuit is in positive feedback
: my confusion is:
: does this equation still hold for "ideal" op-amp? (Vin-0)/R1 = (0-Vout)/R2
: clearly i can't derive Vout=infinity from this equaiton
:
: the
|
r*****e
发帖数: 620 | 7 all your answers to this question are all wrong...
a
【在 b***r 的大作中提到】
: 问这个circuit的gain
: 我当时的回答：
: if it is ideal op-amp
: =〉equal potential at + and - terminals
: => + terminal virtual ground
: => (Vin-0)/R1 = (0-Vout)/R2
: => Vout = - (R2/R1) * Vin
: then the interviewer asked me if i was sure that the gain is negative?
: i said if the op-amp is ideal, then yes. however, this configuration forms a
: positive feedback. if the output voltage increases a bit, the + positive
|
le
发帖数: 190 | 8 cannot agree it more
【在 r*****e 的大作中提到】
: definitely not, in positive feedback vin+ is not equal to vin-
: your equation is based on the assumption that vin+ = vin-, so it is wrong
|
b***r
发帖数: 149 | 9 想明白了
many thanks
【在 r*****e 的大作中提到】
: definitely not, in positive feedback vin+ is not equal to vin-
: your equation is based on the assumption that vin+ = vin-, so it is wrong
|
m*********8
发帖数: 6 | 10 你倒是懂了，能不能说说到底是怎么回事。是不是一个wave generator啊 |
i****g
发帖数: 19 | 11 suggest you to read the book by Sedra/Smith "Microelectronic circuits"
bistable circuit part.
It is a schmitt trigger.
a
【在 b***r 的大作中提到】
: 问这个circuit的gain
: 我当时的回答：
: if it is ideal op-amp
: =〉equal potential at + and - terminals
: => + terminal virtual ground
: => (Vin-0)/R1 = (0-Vout)/R2
: => Vout = - (R2/R1) * Vin
: then the interviewer asked me if i was sure that the gain is negative?
: i said if the op-amp is ideal, then yes. however, this configuration forms a
: positive feedback. if the output voltage increases a bit, the + positive
|
h*******y
发帖数: 896 | 12 make sure R2>R1 for the sch trigger |
f*****0
发帖数: 489 | 13 "make sure R2>R1 for the sch trigger"
no need to do that. |
f*****0
发帖数: 489 | 14 the output of the opamp will be either +max if the voltage on the non-
inverting end is greater than that on the inverting end (in this case, it is
zero), or -max if otherwise.
the voltage on the non-inverting end is the weighted average of Vin and Vout
. so the opamp will flip to +max if Vin is such that it causes the weighted
average to be positive; or the opamp will flip to -max if the weighted
average goes negative.
this causes the system to have a small hysteria (thus noise resistance)
depe |
