Electronic Devices - Op-Amp Applications - Discussion

vo=v1+v2.

Vo=(1+(Rf/R1)(Rf2/R2)[V1-V2]

Vo= (1+(2Rfa/R1a))*(Rfb/R1b)(V2-V1)

= (1+(2*5k/500))*(5k/5K)*(2.5-2.25)

= (1+20)*1*0.25)

= 21*0.25

= 5.25

@ Bipin S.

You did a wonderful work but I would like to make slight modification to your formula used. Instead of being (V2-V1) this should have been (V1-V2). Though you have used it correctly in the following step by saying (2.5 - 2.25).

Nothing offensive. Just wanted to keep it clear.

Output Voltage = (Rf/R1)(1+(2R'/R))(v1-v2).

Vo = (v2-v1)(1+2R1/Rgain)(R3/R2), R1=R2=R3=5K, Rgain=500,

= (.25)(21)(1).

= 5.25V.

What is Rf and R1? I don't get it.

The formula for finding output of op-amp is = A(V2-V1).

Output is = v1+v2 = 4.75 then = 10-4.75 = 5.25.

v1 = 2.5v and v2 = 2.25v. Then for virtual short v1(-) = 2.5v and
v2(-) = 2.25.

Thus the voltage difference across 500 ohm is 2.5v-2.25v = 0.25v.

Then current through the 500 ohm resistor is (0.25/500) = (5x10^-4 Amp).

Now 500 ohm, and two 5k resistors are connected in series. So the current through this total branch is (5x10^-4) amp.

That's why the total voltage difference at this branch is (5k+5K+.5k)*(5x10^-04) = 5.25volt.

Now the 2nd part is simply a differential circuit whose output is the differences of the outputs of the two op amp which is 5.25volt.

If any one have any question please reply me.