Electronics - Basic Op-Amp Circuits - Discussion
Discussion Forum : Basic Op-Amp Circuits - Filling the Blanks (Q.No. 2)
2.
The rate of change of the output voltage in the given circuit equals ______.
40 mV/
s
s5 V/s
s0 V/s
s40 V/s
sDiscussion:
5 comments Page 1 of 1.
Garima said:
1 decade ago
-1/CR(intrigrat of 0 to 1 of 5dt)=40v/microsecond
Devi said:
1 decade ago
How to solve? please explain I can't find this answer.
Shailesh said:
1 decade ago
Here is d answer->
slew rate=dv0/dt
but ,
v0=-1/RC*!{0 to 5)v1 dt + c(const)
taking derivative
dvo= v1/RC
=40V/us.
slew rate=dv0/dt
but ,
v0=-1/RC*!{0 to 5)v1 dt + c(const)
taking derivative
dvo= v1/RC
=40V/us.
Raj said:
1 decade ago
Here V0= 1/RC*Vin.
Hence,
V0 = 5*1/(25pf*5kohm).
V0 = 5*1000/(25*5) microsecond.
V0 = 40.
Hence,
V0 = 5*1/(25pf*5kohm).
V0 = 5*1000/(25*5) microsecond.
V0 = 40.
Modi said:
1 decade ago
It is the integrator:
Formula for integrator is:
dVout/dt = -Vin/RC, So put the values in it.
We get the answer i.e. 40V/microsec.
Formula for integrator is:
dVout/dt = -Vin/RC, So put the values in it.
We get the answer i.e. 40V/microsec.
(1)
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