Discussion :: Op-Amp Applications - General Questions (Q.No.1)
Calculate the output voltage for this circuit when V1 = 2.5 V and V2 = 2.25 V.
Answer: Option D
No answer description available for this question.
|Ram said: (May 31, 2011)|
|Chandan Natkar said: (Jul 30, 2011)|
|Bipin S said: (Sep 14, 2011)|
|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,
|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)|
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.
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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