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Distortion
+V =+5V
cc
+
V
V
+5V
o
d

0
5V
V =5V
cc
The output voltage never excess the DC voltage
supply of the Op-Amp
Ref:080114HKN
Operational Amplifier
1
Common-Mode Operation
+
• Same voltage source is applied
at both terminals
o
• Ideally, two input are equally
amplified
• Output voltage is ideally zero
due to differential voltage is
zero
• Practically, a small output
signal can still be measured
V

V
i
~
Note for differential circuits:
Opposite inputs : highly amplified
Common inputs : slightly amplified
 Common-Mode Rejection
Ref:080114HKN
Operational Amplifier
2
Common-Mode Rejection Ratio (CMRR)
Differential voltage input :
Vd  V  V
Noninverting
+
Input
Output
Common voltage input :
1
Vc  (V  V )
2
Output voltage :
Vo  Gd Vd  GcVc
Gd : Differential gain
Gc : Common mode gain
Ref:080114HKN
Inverting
Input

Common-mode rejection ratio:
CMRR 
Gd
G
 20 log 10 d (dB)
Gc
Gc
Note:
When Gd >> Gc or CMRR 
Vo = GdVd
Operational Amplifier
3
CMRR Example
What is the CMRR?
100V
100V
+
60700V
80600V
20V

+
40V

Solution :
Vd 1  100  20  80V
(1)
Vd 2  100  40  60V
100  20
100  40
 60V
Vc 2 
 70V
2
2
From (1)
Vo  80Gd  60Gc  80600V
Vc1 
From (2)
Gd  1000
(2)
Vo  60Gd  70Gc  60700V
and
Gc  10
 CMRR  20 log( 1000 / 10)  40dB
NB: This method is Not work! Why?
Ref:080114HKN
Operational Amplifier
4