# Operational amplifier question

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```					Operational amplifier exam question

a) In the circuit of Fig. 1a, the differential amplifier has infinite gain, infinite input impedance
and negligible output impedance. Derive an expression for the circuit’s transfer function
V0 / V i .
b) i) Show that the input impedance of the circuit in Fig. 1b is given by:
V R1 R
Z IN          1  jCR2 
I 2R2
ii) The circuit is used to simulate a real inductor having inductance of 10 mH and effective
series resistance of 1 k. If R = 10 k suggest suitable values for R1, R2 and C.

Z2

-
Z1
Vi                     +            V0

Fig. 1a

R
R2
R
I
+                              -
C

-                      R1
R                 +
V

R

Fig. 1b
Solution

a)         In the circuit of Fig 1a, for the ideal differential amplifier the inverting input is a ‘virtual
earth’. So the current through Z1 is Vi/Z1 and the current through Z2 is Vo/Z2 and applying
Kirchhoff’s Current Law at the inverting input:
V i Vo                                     Vo   Z
   0                   so:                2
Z1 Z 2                                     Vi   Z1

b)
i)        Consider the circuit in two stages with input and output voltages as defined in the
diagram below:

I2

I1                          R
R2
R
I
+                                  C
-
-                  R1
R                      +
V
V1                        V0
R

For the first stage the potential of the non-inverting input is defined by the voltage V, while the
inverting input potential is equal to the stage output voltage V1 potential divided by the
resistances R:
V = V1/2                  so: V1 = 2V
The second stage is an inverting amplifier, to which the result from part a) can be applied:
1    1         1  jCR2                                     R2
    jC                                   Z2 
Z 2 R2             R2                                    1  jCR2

and      Z1  R1
V0   Z    R                   1                    V0   2R           1    
so:            2  2                                          2                
V1   Z1   R1             1  jCR2                 V     R1     1  jCR2 
Now
V0   Z    R       1                             V0   2R       1    
 2  2                                       2            
V1   Z1   R1 1  jCR2                          V     R1 1  jCR2 
and from the circuit above:
V  V1  V  V0  V 2V V 2VR2                                1    
I  I1  I2                                                             
 R   R  R R R R1 R                                    1  jCR2 
so:
V R1 R
     1  jCR2   R1 R  jC R1 R
I 2 R2                2 R2        2

R1 R      RR
ii)         The inductor has complex impedance: Z L  RL  jL               jC 1
2 R2       2
so with RL = 1 k, L = 10 mH, R = 10 k :

R1   1                                             R1 .10 4
           5R1  R2       and     10 2  C               CR1  2s
2 R2 10                                                 2

alternative values:
R1               R2                  C

100             500                20 nF

1 k              5 k               2 nF          *

2 k             10 k               1 nF          *

10 k            50 k              200 pF
* most suitable values for use with a general-purpose differential amplifier

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