Electronic Devices - Op-Amp Applications - Discussion

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

Thanks for explaining.

Why they took 2R1 in the formula?

Anyone explain me.

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.

Asking a normal q here. When amplifier at v1 and v2 have supply voltage +10 &-10v then how v1 and v2 amplified out can use on the third opamp?

v0 =A * (V2 - V1),

A = 1 + (2 * 5k)/500.

@Arnesh Sen answer is correct .

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

Is it instrumentation amplifier?

@Arnesh.

How is the three resistances are connected in series?