Electronic Devices - Op-Amp Applications - Discussion

Discussion :: Op-Amp Applications - General Questions (Q.No.1)

1. 

Calculate the output voltage for this circuit when V1 = 2.5 V and V2 = 2.25 V.

[A]. –5.25 V
[B]. 2.5 V
[C]. 2.25 V
[D]. 5.25 V

Answer: Option D

Explanation:

No answer description available for this question.

Ram said: (May 31, 2011)  
vo=v1+v2.

Chandan Natkar said: (Jul 30, 2011)  
Vo=(1+(Rf/R1)(Rf2/R2)[V1-V2]

Bipin S said: (Sep 14, 2011)  
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

Mallik said: (Mar 27, 2012)  
@ 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.

Mathi said: (Jul 21, 2013)  
Output Voltage = (Rf/R1)(1+(2R'/R))(v1-v2).

Jinju said: (Aug 17, 2013)  
Vo = (v2-v1)(1+2R1/Rgain)(R3/R2), R1=R2=R3=5K, Rgain=500,

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

= 5.25V.

Masta said: (Jan 10, 2014)  
What is Rf and R1? I don't get it.

Suji said: (Feb 25, 2014)  
The formula for finding output of op-amp is = A(V2-V1).

Raji said: (Apr 20, 2014)  
Output is = v1+v2 = 4.75 then = 10-4.75 = 5.25.

Arnesh Sen said: (Jun 30, 2014)  
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.

Nisha said: (Sep 26, 2014)  
Why you have considered 2R1 in the formula?

Javed said: (Jan 10, 2015)  
We can calculate the current between two voltage and the find drop across the resistor?

Akshay said: (Aug 22, 2015)  
@Arnesh.

How is the three resistances are connected in series?

Wasim said: (Mar 18, 2016)  
Is it instrumentation amplifier?

Husain Kapadia said: (Mar 30, 2016)  
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

Prachi Tewari said: (Sep 10, 2016)  
@Arnesh Sen answer is correct .

Sachin said: (Nov 26, 2016)  
v0 =A * (V2 - V1),

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

Sanjay said: (Dec 26, 2016)  
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?

Harsh Tyagi said: (Nov 28, 2017)  
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.

Charan said: (Sep 9, 2018)  
Why they took 2R1 in the formula?

Anyone explain me.

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