Electronic Devices - Basic Op-Amp Circuits - Discussion
Discussion Forum : Basic Op-Amp Circuits - General Questions (Q.No. 11)
11.
Refer to the given figure. If Vin = 5 V, the rate of change of the output voltage in response to a single pulse input is:
15.2 mV/
s
s1.52 V/s
s1.52 mV/s
s15.2 V/s
sDiscussion:
9 comments Page 1 of 1.
Lamchek said:
6 years ago
Why is there 1E(-6)?
The correct answer is 1515. The unit is mV/microS.
Which is why it should be 1.515
The correct answer is 1515. The unit is mV/microS.
Which is why it should be 1.515
Totoy said:
7 years ago
(1/(15E3 * 0.22E(-6)) * 5 * 1E(-6) = 1.52E(-3).
Abasiakan said:
9 years ago
15E3 * 0.22E(-6) = 3.3E(-03).
1/3.3E(-03) * 5 = 1515.
1/3.3E(-03) * 5 = 1515.
(1)
Prami said:
9 years ago
I need correct answer for this question. Someone explain it.
Das said:
1 decade ago
If we solved we get Vo = 1515 v. But he asked in volts per microseconds, hence option in milli volts.
Dandelion said:
1 decade ago
Integration = Rate of change. This one is an integrator op amp.
(1)
Sachin said:
1 decade ago
Answer asked for rate of change of output voltage.
i.e dvo/dt = (1/RC)*Vin since Iin = If.
i.e dvo/dt = (1/RC)*Vin since Iin = If.
Dhanashri said:
1 decade ago
For integrator:
Vo = Vin/ (RC).
Vo = Vin/ (RC).
Ovii said:
1 decade ago
Expalanation please.
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