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Problems on Omp amp Worksheet

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User Generated
Subject
Engineering
Type
Worksheet
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Some solved problems on Op-amp
Problem 1.
Realize the output voltage V
o
= -V
1
+ 0.5V
2
-2V
3
using one op-amp and passive
components:
i.e., V
o
= 0.5V
2
– (V
1
+
2 V
3
)
Solution :
Assuming non-inverting input at ground , we have
V
o
=– (V
1
+
2 V
3
)
If V
1
= V
3
= 0 the following circuit is obtained
V
o
R = 10 Kohm
F
V
x
-
R = 5 Kohm
3
R =10 Kohm
1
+

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The final differential amplifier circuit will be as shown
*************************************************************************************
Problem 2 Realize the output voltage V
o
= -2V
1
+ V
2
+2V
3
using one op-amp and passive components
Solution :
i.e., V
o
= (V
2
+2 V
3
)-2V
1
Consider the difference amplifier circuit
V
3
7 Kohm
1 Kohm
V
o
V
1
+
R = 5 Kohm
3
R =10 Kohm
1
V
2
R = 10 Kohm
F
-
V
1
R = 10 Kohm
F
R = 5 Kohm
1
V
x
V
3
-
R
3
V
2
+
R
2
V
o

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Some solved problems on Op-amp Problem 1. Realize the output voltage Vo= -V1 + 0.5V2 -2V3 using one op-amp and passive components: i.e., Vo= 0.5V2 – (V1+ 2 V3 ) Solution : Assuming non-inverting input at ground , we have Vo=– (V1+ 2 V3 ) If V1 = V3 = 0 the following circuit is obtained R = 10 Kohm F R =10 Kohm 1 R = 5 Kohm 3 - + V x V o The final differential amplifier circuit will be as shown R = 10 Kohm F R =10 Kohm 1 V 1 R = 5 Kohm 3 V - 3 V V o 7 Kohm + 2 1 Kohm ************************************************************************************* Problem 2 Realize the output voltage Vo= -2V1 + V2 +2V3 using one op-amp and passive components Solution : i.e., Vo= (V2 +2 V3 )-2V1 Consider the difference amplifier circuit R = 10 Kohm F V R = 5 Kohm 1 R V 2 2 R V - 1 3 3 V x + V o and The complete circuit will be shown below: R = 10 Kohm F V V V R = 5 Kohm 1 - 1 R =2 Kohm 2 2 V x V o + R =1 Kohm 3 3 ****************************************************************************** Problem 3. In an Op-amp Integrator circuit, given that R=10 Kohm and C=1 uF .Find the value of output voltage attained after t=1 millisecond, when the circuit is applied with Positive DC input of +5V applied at t=0 sec Solution : Consider the op-amp based integrator circuit Consider the expression for Vo t −1 vo = vin dt RC 0 If 𝑉𝑖𝑛 = 5 𝑉 , then v = − 5t o RC Given that R=10 Kohm and C=1 uF ,then RC=0.01 sec − 5t = −500t 0. ...
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