Download Introduction to Operational Amplifiers

E E 2415
Lecture 05 - Introduction to
Operational Amplifiers
Operational Amplifiers
• Ideal operational amplifiers are easily
analyzed
• Assumptions for ideal operation are realistic
• Inexpensive commercial Op-Amps available
as integrated circuits
• Practical applications for instrumentation,
signal processing and control circuits
mA741 Pinout
Offset Null
1
2
Non-inverting Input 3
Vcc- 4
Inverting Input
mA741
8
7
6
5
NC
Vcc+
Output
Offset Null
Op Amp Terminals
Inverting
Input
Noninverting
Input
Positive supply
voltage
+Vcc
Output
-Vcc
Negative supply
voltage
Op Amp Terminal Currents
ic+
in
+Vcc
ip
io
-Vcc
icVcc
Vcc
Common Node
Op Amp Terminal Voltages
+Vcc
+
vn
+
vp
-
-
-Vcc
Vcc
+
vo
-
Common Node
Vcc
Equivalent Circuit
ip
Ri
in
A(Vp-Vn)
io
Ro
+
vo
-
Assumptions for Ideal Op Amp
•
•
•
•
•
Open loop Gain, A    vp - vn  0
Then vp - vn  0  Ri  
Ri    ip  0 and in  0
Gain is linear up to the saturation voltage
Saturation voltage equals power supply
voltage
Open-Loop Characteristics
+Vcc
vo
+Sat
A(vp-vn)
Linear
vp-vn
-Sat
-Vcc
Inverting Amplifier
Rg
vg
+ in
vn
-
Rf
+Vcc
-Vcc
+
vo
-
0
Vn = 0 and in = 0. Then KCL at node n yields:
0  vg
Rg
0  vo

00
Rf
and
vo  
Rf
Rg
vg
1V
Saturated Mode
vg
5V
t
-1 V
vo
t
-5 V
• Inverting Amplifier with +Vcc = 5 V and Vcc
= -5V.
• Rf = 10 k and Rg = 1 k
Non-Inverting Amplifier
vn = vp = v g
Rf
+Vcc
0
n
+
vn
Rg
Rs
vg
-
0
p
+
vp
-
-Vcc
+
vo
-
KCL at node n:
R f  Rs
vg vg  vo

 0 vo  vg
Rs
Rs
Rf
Summing Amplifier (1/2)
Rf
Ra
Rb
Rc
+Vcc
-Vcc
va
vb
vc
+
vo
-
Summing Amplifier (2/2)
va vb vc

 
Ra Rb Rc
vo  
Rf
Ra
va 
Rf
Rb
vo

0
Rf
vb 
Rf
Rc
vc 
Difference Amplifier (1/3)
in  i p  0
Ra
n
Rc p
va
vb
KCL @ n:
Rb
+Vcc
Rd
+
vp
-
-Vcc +
vo
-
vn  va vn  vo

0
Ra
Rb
Difference Amplifier (2/3)
Ra
Rb
+Vcc
n
Rc p
va
vb
Rd
Voltage Divider @ p:
+
vp
-
-Vcc +
vo
-
Rd
vn  v p  vb
Rc  Rd
Difference Amplifier (3/3)
Substitute:
into:
and simplify:
If:
Ra Rc

Rb Rd
Rd
vn  v p  vb
Rc  Rd
vn  va vn  vo

0
Ra
Rb
Rd  Ra  Rb 
Rb
vo  vb
 va
Ra  Rc  Rd 
Ra
Then:
Rb
vo 
 vb  va 
Ra
Analyzing an OpAmp Circuit
Rg
+Vcc
-Vcc
vg
Rf
Rx
RL
Ry
Analyzing an OpAmp Circuit
Analyzing an OpAmp Circuit
Rx
vp 
vg
Rx  Rg
vo  1
1 
 
 vp

R f  Ry R f 
v p v p  vo
0 
Ry
Rf
 Rf

Rx
vo  
 1 
vg
R
 R R
g
 y
 x
 R f  Ry
vo  
 R
y


Rx
vg
 
 Rx  Rg