Electronic Devices - Operational Amplifiers - Discussion
Discussion Forum : Operational Amplifiers - General Questions (Q.No. 5)
5.
For an op-amp having a slew rate SR = 5 V/ms, what is the maximum closed-loop voltage gain that can be used when the input signal varies by 0.2 V in 10 ms?
Discussion:
8 comments Page 1 of 1.
Seetha said:
6 years ago
SR = A(dvin/dt).
Chippuzi said:
7 years ago
Here; dvo = sr*dt /vid =Acl.
Samantha said:
8 years ago
Closed loop gain Acl = SR/(dvi/dt).
= 5/(0.2/10ms),
= 250.
= 5/(0.2/10ms),
= 250.
Meghana said:
1 decade ago
SLEW RATE = dV0/dt(MAX)= 5.
VOLTAGE GAIN = OUTPUT VOLTAGE / DIFFERENTIAL i/p VOLTAGE.
= 5/0.02.
= 250.
VOLTAGE GAIN = OUTPUT VOLTAGE / DIFFERENTIAL i/p VOLTAGE.
= 5/0.02.
= 250.
Suruchi said:
1 decade ago
Thanks ajin!
Ajin job said:
1 decade ago
Total effect = 0.2/10ms
=.02V/ms
So voltage gain= 5V/.02V/ms
=.02V/ms
So voltage gain= 5V/.02V/ms
(1)
Amit said:
1 decade ago
Slew rate = 2piefv
Pramod said:
1 decade ago
Use
1. SR=(V(out)-V(in))/t
and
2. Gain=V(out)/V(in)
1. SR=(V(out)-V(in))/t
and
2. Gain=V(out)/V(in)
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