Electronic Devices - Op-Amp Applications - Discussion

Let take.

R1=250k.
R2=100k.
R3=50k.
And Rf1=500k;Rf2=500k.
Then,
v0 = -Rf2[(-Rf1/R2R1)v1+v2/R3].

V0=(-Rf1/R1)V1 = (-500/250)*700=-1400mV = -1.4V; .
The O/P of first Amp is the I/P of a second.

Therefore. (-Rf2/R2)V1 =(-500/199)(-1.4)=7V.
When V2=700mV (-Rf2/R3)V2=(-500/50)(700)=-7000mV=-7V.
Therefore 7-7=0V.

Hi, how to solve this? Please explain me.

Hi friends there is another way to solve this circuit.

Consider the first stage it is an inverting op amp so the output is.
V01=- (rf/r1) v1=- (500/250) *0. 7=-1. 4v.
Consider the second stage it is another inverting op amp with two inputs i.e. v01 and v2.
So calculate the first output.
V0= (-rf/r1) v01= (-5) * (-1. 4) =7v.
Then calculate the second output.
Vo= (-rf/r1) v2= (-10) * (0. 7) =-7.
Then final output is.
V0=7-7=0v.

Consider the first stage op-amp.

G= -(500k/250k);
Vo1 (output of first stage op-amp)= -2 V1:

second stage op-amp has inputs as V2 and -(2*V1)

so Vo = - {[(-2*V1)*(500k/100k)]+(V2*(500k/50k))} (Summer)
Vo = - {-10V1+10V1}
Vo = 0V;