Electronic Devices - Op-Amp Applications - Discussion
Discussion Forum : Op-Amp Applications - General Questions (Q.No. 8)
8.
Determine the output voltage when V1 = V2 = 1 V.
Discussion:
7 comments Page 1 of 1.
Abhi k said:
3 years ago
I think the answer should be,
If we apply both inverting and non-inverting formulas.
Then multiply by (V1-V2).
(-Rf/R1) + 1 + (Rf/R2) * (V1-V2).
(-100/100)+ 1 + (100/20) * (1-1).
= 0.
If we apply both inverting and non-inverting formulas.
Then multiply by (V1-V2).
(-Rf/R1) + 1 + (Rf/R2) * (V1-V2).
(-100/100)+ 1 + (100/20) * (1-1).
= 0.
Rajan said:
5 years ago
I think the answer is 1v (by virtual short method).
Because both the input resistor and feedback resistor are unequal hence we cant apply standard differential op-amp formula Vo= Rf/Rin(V2-V1).
Because both the input resistor and feedback resistor are unequal hence we cant apply standard differential op-amp formula Vo= Rf/Rin(V2-V1).
Ezhil said:
9 years ago
Consider the nodes after two input resistance as v3.
Bby using node analysis method.
(v3-v1)/100k + (v3-vo)/100k =0-----(1).
(v3-v2)/20k + (v3/)20k =0--------(2).
Sub v1=v2=1 in above equatios.
We get v3= 1/2 sub this in equ 1.
ve get vo =0.
Bby using node analysis method.
(v3-v1)/100k + (v3-vo)/100k =0-----(1).
(v3-v2)/20k + (v3/)20k =0--------(2).
Sub v1=v2=1 in above equatios.
We get v3= 1/2 sub this in equ 1.
ve get vo =0.
Gurpreet said:
1 decade ago
@Sunil brother in actual practice v1 and v2 are not equal but nearly equal.
Sunil said:
1 decade ago
What we are talking about is op-amp?
Which amplifies the difference between two signals?
Here, difference between two signal is 0(v1-v2). And also, the CMRR Common Mode Rejection Ratio.
Which amplifies the difference between two signals?
Here, difference between two signal is 0(v1-v2). And also, the CMRR Common Mode Rejection Ratio.
(1)
Mallik said:
1 decade ago
Assume common voltage at two terminals to be vb; lets write KCL for the ckt
[(v1-vb)/100k]+[(V0-Vb)/100k]= 0;
[(V2-Vb)/20k] = Vb/20k;
=> V0 = V2-V1;
=0
[(v1-vb)/100k]+[(V0-Vb)/100k]= 0;
[(V2-Vb)/20k] = Vb/20k;
=> V0 = V2-V1;
=0
Divya said:
1 decade ago
The given ckt is a differential amplifier ckt.
Vo=(Rf/R1)*(V2-V1)
Vo=(100/100)*(1-1)=0
Vo=(Rf/R1)*(V2-V1)
Vo=(100/100)*(1-1)=0
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