Electronic Devices - Op-Amp Applications - Discussion
Discussion Forum : Op-Amp Applications - General Questions (Q.No. 2)
2.
Calculate the output voltage if V1 = –0.2 V and V2 = 0 V.
Discussion:
13 comments Page 1 of 2.
Mallik said:
1 decade ago
As the figure shows.. op-amp used here is inverting amplifier. From the circuit configuration we can conclude that it is an adder , where inputs are given to inverting port. the formula goes in the following way..
Vout = -[{(V1/R1)+(V2/R2)}/{(1/Rf)}]
Here V1= -0.2v; V2= 0v R1= 33k; R2=10k ; Rf=330k
=> Vout = -[(V1/R1)*Rf]
Vout= -[V1*(330k/33k)]
Vout = -[-0.2*10]
Vout = 2v
Vout = -[{(V1/R1)+(V2/R2)}/{(1/Rf)}]
Here V1= -0.2v; V2= 0v R1= 33k; R2=10k ; Rf=330k
=> Vout = -[(V1/R1)*Rf]
Vout= -[V1*(330k/33k)]
Vout = -[-0.2*10]
Vout = 2v
Amit said:
8 years ago
As we see in the circuit, -o.2 is applied at the negative terminal so the op-amp is inverting op-amp. By general description, we calculate drop across 33k by ohms law. We get 1/165 amp and in other branches we applied same ohms law and we get 0 amp coz v=0v. Now add both the currents and applied again ohms law at feedback branch 330k we get vo= 2volt.
(1)
Dipanjan said:
1 decade ago
Ii=(V1-0)/33 [Ii=input current through 33k)
If=(V0-0)/330 [If=feed back current through 330k)
I-=(Ii)+(If) [Kirchhoff law at opamp input node]
As input impedance of opamp is Infinite So Input current(I-) is zero.
(Ii)+(If)=0
Ii=-If
V1/33=-V0/330
V0=-10*V1=-10*(-0.2)=2V
If=(V0-0)/330 [If=feed back current through 330k)
I-=(Ii)+(If) [Kirchhoff law at opamp input node]
As input impedance of opamp is Infinite So Input current(I-) is zero.
(Ii)+(If)=0
Ii=-If
V1/33=-V0/330
V0=-10*V1=-10*(-0.2)=2V
(1)
Surya said:
9 years ago
If the voltage source is connected to the positive terminal of an op amp and ground is connected to the negative terminal of the op amp then it's called as "non-inverting amplifier". If the terminals are reversed then acts as inverting amplifier.
(2)
Vandana said:
1 decade ago
It is not a non inverting amplifier, it is an simple inverting amplifier. If it is a non inverting amplifier then formula for voltage output should be diffrent.
Amit bhattacharya said:
1 decade ago
As v2 is zero so that branch can be neglected
i.e. 10k can be neglected
Now the circuit is simple non inverting amp
and v0=(-Rf/R1)*v1
=-(330k/33k)*(-0.2)
=2v
i.e. 10k can be neglected
Now the circuit is simple non inverting amp
and v0=(-Rf/R1)*v1
=-(330k/33k)*(-0.2)
=2v
Indu said:
1 decade ago
Here given the amplifier is summing opamp and the solution of the given problem.
vo = -[330/33 (-0.2)] = 2 v.
vo = -[330/33 (-0.2)] = 2 v.
(1)
Seema Rajput said:
9 years ago
Vo = (-Rf/R1)V1 + (-Rf/R2)V2.
= (-330k/33k) * (- 0.2v) + (- 330/10k) * 0.
= (- 10) * (-0.2) + 0.
Vo = 2V.
= (-330k/33k) * (- 0.2v) + (- 330/10k) * 0.
= (- 10) * (-0.2) + 0.
Vo = 2V.
(9)
Neeharika said:
8 years ago
By nodal analysis (0 -V1)/33k+(0-V0)/330k = 0.
0.2/33k-V0/330k = 0.
V0 = 2v.
0.2/33k-V0/330k = 0.
V0 = 2v.
(4)
MMA said:
1 decade ago
Vout = -330k{(-0.2/33k)+(0/10k)}.
= -330k(-0.2/33k).
= 2 volt.
= -330k(-0.2/33k).
= 2 volt.
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