Electronic Devices - Basic Op-Amp Circuits - Discussion
Discussion Forum : Basic Op-Amp Circuits - General Questions (Q.No. 14)
14.
An op-amp has an open-loop gain of 90,000. Vsat = ±13 V. A differential voltage of 0.1 V p-p is applied between the inputs. What is the output voltage?
Discussion:
5 comments Page 1 of 1.
Junayd said:
1 decade ago
The user expects the answer in same unit of which it has given input Vp-p as 13V is peak value. So peak to peak will be 26V.
Nitu kumari said:
1 decade ago
Answer should be 13 V. Why it is 26V p-p?
(1)
Dhanashri said:
1 decade ago
Output = Gain* Input.
Vo= 90,000*0.1 = 9,000V.
But output is exceeding the Vsat= 13V (Applied Voltage).
Output cannot exceed Vsat.
So Ans is 13 V i.e. 26 V p-p.
Vo= 90,000*0.1 = 9,000V.
But output is exceeding the Vsat= 13V (Applied Voltage).
Output cannot exceed Vsat.
So Ans is 13 V i.e. 26 V p-p.
Ovii said:
1 decade ago
Need proper expalantion of theis with reasons.
Gethanjali said:
1 decade ago
Can anyone explain this?
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