Electronic Devices - Op-Amp Applications - Discussion

For those, who don't want to remember the formula.

Consider the above op-amp to be 2 having o/p Vo2 and below op-amp to be 1 having o/p Vo1.

Since the op-amps are ideal we assume no currents flow into the input terminals, so drop across the 500 Ohm resistor is (V1-V2).

Current flowing through the 500 Ohm resistor is 0.5 mA.

Vo1 = V1 + drop across 5k Ohm = 5V (simple KVL path).
Vo2 = - (drop across 5k Ohm - V2) = -0.25 V (simple KVL path).

These o/ps Vo1 and Vo2 act as i/p to the third op-amp.

Apply superposition,

Vo = Vo1' + Vo2'

Consider Vo2 to be GND.

Vo1' = (1 + Rf/R1) * Vo1 * (5/10) = 5V
Vo2' = -Rf/R1*Vo2 = 0.25V

Vo = 5.25V

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.

Vo = (1+2R/Rp)v1-v2.

Thanks for explaining.

Why they took 2R1 in the formula?

Anyone explain me.

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

Is it instrumentation amplifier?

@Arnesh.

How is the three resistances are connected in series?

We can calculate the current between two voltage and the find drop across the resistor?

First apply the nodal equation and find the output of first two op-amp. The upper one has output -.25 volt. and the lower one has 5 volt.