Download Answer Part A - UniMAP Portal

Answer Part A
Solution Q1
a)
20K
i
10K
1
50
_
Vs
+
_
2M
+
_
Vd
O
AVd
+
Figure 1a
b) We solve the circuit in Figure 1a above by using nodal analysis. At node 1, KCL
gives:
vs  v1
v1
v v

 1 o3
3
3
10 x10
2000 x10
20 x10
Multiplying through by 2000x10 3 , we obtain
200vs  301v1  100vo
or
2v  vo
v1  s
(1)
3
At node 0,
v1  vo vo  Avd

20 x103
50
But vd  v1 and A  200,000v1 . Then
v1  vo  400(vo  200,000v1 )
(2)
Substituting v1 from Eq (1) into Eq (2) gives,
0  26,667,067vo  53,333,333vs
vo
 20 log( 1.9999699) = 6.02 dB
vs
c)
vo  2 x(1.9999699)  3.9999398V
from Eq (1) we obtain v1  20.066667V
Thus i 
v1  vo
 0.1999mA
20 x103
Q2.
Ideal Operational amplifier shown in Figure 1 below with input voltage is vin= 5 sin 3t
mV.
Rf
i
R1
_
va
i
vb
+
vout
vin
Figure 2
a. Name the type of Op Amp configuration, shown in Figure 2.
b. Derive the formula of close-loop gain, ACL.
c. Given vin= 5 sin 3t mV, R1 = 4.7 k  and Rf = 47 k  . Determine and draw the
output voltage waveform.
Solution Q2
a) Non Inverting Op-Amp
b) Used nodal analysis at node va :
v
v v
0  a  a out
R1
Rf
at node vb : vb  vin
at node a va  vin
v
v v
0  in  in out
R1
Rf
R
vout
 (1  f )
vin
R1
c) Output voltage vout  (1 
47
)vin = 55 sin 3t mV
4 .7
vout
55mV
5mV
vin
wt
Answer Part B
Q1
i)
Derive the formula to calculate Vout.
Applying KCL at inverting terminal:
I1 + I2 + I3 = If
V1  0 V2  0 V3  0 0  Vout



R1
R2
R3
Rf
V1 V2 V3 Vout



R1 R2 R3
Rf
Vout  V1 V2 V3 
 
 
Rf
 R1 R2 R3 
V V V 
Vout   R f  1  2  3 
 R1 R2 R3 
ii) Repeat part (i) using R1 = R2 = R3 = R.
V V V 
Vout   R f  1  2  3 
 R1 R2 R3 
R1  R2  R3  R
iii)
V V V 
Vout   R f  1  2  3 
R R R 
Rf
Vout  
V1  V2  V3 
R
Calculate the output voltage, Vout if V1= 6V, V2= 5V, V3= 8V and Rf = 50kΩ,
R1= 70kΩ, R2= 80kΩ, R3= 100kΩ.
V V V 
Vout   R f  1  2  3 
 R1 R2 R3 
5V
8V 
 6V
Vout  50k  



 70k  80k  100k  
Vout  11.4107V
iv)
Calculate the output voltage, Vout if R1 = R2 = R3 = 50kΩ.
V V V 
Vout   R f  1  2  3 
R R R 
Rf
Vout  
V1  V2  V3 
R
50k 
 6V  5V  8V
50k 
 19V
Vout  
Vout
v)

Find the value of R1, R2, R3 if the output voltage, Vout is equal to the average
value of the inputs. Assume Rf = 35kΩ.
Averaging Amplifier
Rf
R

1
; n = no of inputs
n
R  n  Rf
R  n   35k  
R  3   35k  
R  105k 
Q2
Q3.
Figure 3
Answer the questions below according to Figure 3.
i)
Derive the formula to calculate the rate of change of the output,
Q  Ic  t
Vout
.
t
Q  C  Vc
C  Vc  I c  t
I 
Vc   c  t
C
Vout  Vc
I 
Vout    c  t
C
I c  I in
Vin  0
R
V
I in  in  I c
R
I in 
ii)
Vout
Vout
 Vin 


  R t
 C 


V 
   in  t
 RC 
Vout
V 
   in 
t
 RC 
Vout
during the time that
t
capacitor charging and discharging. Assume an input to the circuit is a square
wave with maximum value of ±5V. The value of capacitor, C and resistor, R
are 15nF and 20kΩ.
Determine the output voltage’s rate of change,
Vout
V 
   in 
t
 RC 
 5V  (5V ) 
Vout
10V
 
 33.33mV /  S

t
300 S
 20k 15 F 
Vout
V 
   in 
t
 RC 
 5V  (5V ) 
Vout
10V
 
 33.33mV /  S

t
 20k 15 F  300 S
iii)
Calculate input and output frequencies when the output, Vout value is 10 Vp-p.
Draw the input and output waveforms.
Vout
 33.33mV /  S
t
Vout
t 
33.33mV /  S
10V
thalfcycle 
 300.03 S
33.33mV /  S
1
1
1
freq 


 1.6665kHz
t fullcycle thalfcycle  2 300.03 S  2
Q4.
Figure 4
Answer the questions below based on Figure 4.
i)
Derive the formula to calculate Vout.
I 
Vc   c  t
C
V 
Ic   c  C
 t 
Vc  Vin
I c  I in
V 
I in   in  C
 t 
V 
Vout   RI c   RI in   R  in  C
 t 
V 
 V 
Vout   RC  in    RC  in 
 t 
 t 
ii)
Determine the output voltage, Vout and the feedback resistance, R. Assume an
input to the circuit is a triangular wave with maximum value of ±4V and
frequency of 100 kHz. The value of capacitor, C is 0.0011 µF and time
constant, τ is 3.3 µS.
freq 
1
2  thalfcycle
thalfcycle 
1
1

 5 S
2  freq 2 100kHz
V 
 V 
Vout   RC  in    RC  in 
 t 
 t 
 V   4V  (4V ) 
slope _ when _ ch arg ing   in   
  1.6V /  S
5 S
 t  

 V
Vout   RC  in
 t

  3.3 S 1.6V /  S   5.28V

 V   4V  (4V ) 
slope _ when _ disch arg ing   in   
  1.6V /  S
5 S
 t  

 V
Vout   RC  in
 t

  3.3 S  1.6V /  S   5.28V

time _ cons tan t ,  RC

3.3 S
R 
 3k 
C 0.0011 F
iii)
Draw the input and output waveforms.