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Hi
Please have a look on the attachment. Please help me with it. Thanks a lot.
Regards
PG
However, the output will follow the input up to the point where the zeners conduct,
according to the gain set by R1 and R2. Since the ratio is about 3:1, the output
will be 3 times the input until th zeners conduct.
The voltage between the non inverting and inverting terminals is considered to be 0v for these kinds of problems. This happens only when the op amp is operated in the linear mode, ....In the linear mode, the circuit is always arranged so that the output can drive the input difference to zero volts, or nearly so.
The strange thing about this circuit is that it is not operated as a linear circuit after all, so why they would say the differential voltage is negligible i cant say. This circuit is non linear and thus does not have a near zero differential input.
The differential voltage generally becomes 0v when there is negative feedback so that the op amp can operate as a linear device. The output tries to reduce the differential input to near zero and that's what makes it work. In the non linear mode anything is possible.
The differential voltage WILL be negligible in this circuit. Vin will largely be dropped across Ri.
The output will be adjusted so there is no appreciable difference between the inverting and non-inverting input. The circuit has both positive and negative feedback.
The gain is infinite before the diodes conduct, and unity afterwards.
So, even when both types of feedbacks present, the output would still adjust itself to make the differential voltage 0v (I was under the impression that it only happens when there is only negative feedback present).
Why is so? Why would Vin largely be dropped across Ri? Could you please tell me?
I think the diodes won't take long to conduct but why is the gain unity?
Hi again
Please help me with these queries, **broken link removed** and **broken link removed**, enclosed in the attachments. Thank you.
Regards
PG