Download Almost Ideal Op Amps

Objective of Lecture
 Apply the ‘almost ideal’ op amp model in the following
circuits:
 Inverting Amplifier
 Noninverting Amplifier
 Voltage Follower
 Summing Amplifier
 Difference Amplifier
 Cascaded Amplifiers

Chapter 5.4-5.8 Fundamentals of Electric Circuits
Almost Ideal Op Amp Model
Ri = ∞ W and Ro = 0 W
i2 = 0
i1 = 0
Linear Region:
When V+< vo< V- , vo is
determined from the closed
loop gain Av times v2 as
v1 = v2 (vd = 0 V).
Saturation:
When Av v2 ≥ V+, vo = V+.
When Av v2 ≤ V-, vo = V-.
Example #1: Inverting Amplifier
if
is
i1 = 0
i
i2 = 0
V+ = 15V
V- = -10V
Example #1 (con’t)
if
is
i1 = 0
i
i2 = 0
V+ = 15V
V- = -10V
Example #1
 Closed loop gains are dependent on the values of R1
and Rf.
 Therefore, you have to calculate the closed loop gain for
each new problem.
Example #1 (con’t)
is  i f  i1  i f
is  VS / R1
if
i f  Vo / R f
Av  Vo Vs   R f R1
is
i1 = 0
R f  10kW
i
R1  1kW
i2 = 0
vo
Av  10
Example #1 (con’t)
 Since AV = -10
 If Vs = 0V, V0 = -10(0V) = 0V
 If Vs = 0.5V, Vo = -10(0.5V) = -5V
 If Vs = 1V, Vo = -10(1V) = -10V
 If Vs = 1.1V, Vo = -10(1.1V) < V-, Vo = -10V
 If Vs = -1.2V, V0 = -10(-1.2V) = +12V
 If Vs = -1.51V, Vo = -10(-1.51V) > V+, Vo = +15V
Example #1 (con’t)
 Voltage transfer characteristic
Slope of the voltage transfer
characteristic in the linear
region is equal to AV.
Example #2: Noninverting Amplifier
Example #2 (con’t)
Example #2 (con’t)
V2  V1  VS
Example #2 (con’t)
is   VS R1
i f  VS  Vo  R f
Vo  VS  R f i f
Example #2: Noninverting Amplifier
i f  is
Vo  VS  R f is
VS
Vo  R f
 VS
R1
Rf
Vo
Av 
 1
VS
R1
Example #2 (con’t)
 AV = +11
 If Vs = 0V, V0 = 11(0V) = 0V
 If Vs = 0.5V, Vo = 11(0.5V) = +5.5V
 If Vs = 1.6V, Vo = 11(1.6V) > V+,
Vo = +15V
 If Vs = -0.9V, V0 = 11(-0.9V) = -9.9V
 If Vs = -1.01V, Vo = 11(-1.01V) < V-
Vo = -10V
Example #2 (con’t)
 Voltage transfer characteristic
Slope of the voltage transfer
characteristic in the linear
region is equal to AV.
Example #3: Voltage Follower
A voltage follower is a
noninverting amplifier
where Rf = 0W and R1 = ∞W.
Vo /Vs = 1 +Rf/R1 = 1 + 0 = 1
Example #4: Summing Amplifier
V+ = 30V
V-=-30V
A summing amplifier is an inverting amplifier with multiple inputs.
Example #4 (con’t)
if
i1 = 0
iA
v1
i2 = 0
iB
v2
iC
We apply superposition to obtain a relationship
between Vo and the input voltages.
Example #4 (con’t)
A virtual ground
Example #4 (con’t)
KCL :
i A  iB  iC  i f
Example #4 (con’t)
iC  0V  0V  / RC
Note that the voltages at
both nodes of RC are 0V.
iC  0 A
Example #4 (con’t)
i A  iB  iC  i f
iA  i f
i A  VA  0V  RA
i f  0V  Vo  R f
Vo  
Rf
RA
VA
Example #4 (con’t)
i A  iC  0
iB  i f
iB  VB  0V  RB
i f  0V  Vo  R f
Vo  
Rf
RB
VB
Example #4 (con’t)
i A  iB  0
iC  i f
iC  VC  0V  RC
i f  0V  Vo  R f
Vo  
Rf
RC
VC
Example #4 (con’t)
Rf
Rf 
 Rf
Vo    VA 
VB 
VC 
RB
RC 
 RA
10kW
10kW 
10kW
Vo   
VA 
VB 
VC 
2kW
4kW 
 5kW
Vo  2VA  5VB  2.5VC 
Vo  2(1V )  5(3V )  2.5(2V )
Vo  12V
Example #4 (con’t)
 Do not apply the limits on vo (V+≥ vo ≥ V-) until after
adding the results from the three circuits together.
Example #5: Difference Amplifier
Example #5 (con’t)
if
iA
i1 = 0
v1
i2 = 0
iB
iC
v2
Example #5 (con’t)
if
iA
i1 = 0
v1
i2 = 0
iB
iC
v2
iB  i2  iC
i2  0 A
iB  iC
Example #5 (con’t)
if
iA
i1 = 0
v1
i2 = 0
iB
iC
v2
RC
v2 
VB
RB  RC
v2  v1
RC
v1 
VB
RB  RC
Example #5 (con’t)
if
iA
RC
v1 
VB
RB  RC
i A  VA  v1  RA
iA  i f
i f  v1  Vo  R f
Example #5 (con’t)
Vo 
R f 1  RA R f 
RA 1  RB RC 
VB 
Rf
RA
VA
30kW 1  6kW 30kW 
30kW
Vo 
(3V ) 
(2V )
6kW 1  8kW 4kW 
6kW
Vo  4V
Example #5 (con’t)
 If RA/Rf = RB/RC
Vo 
Rf
RA
VB  VA 
 And if RA = Rf
Vo  VB  VA
Example #6: Cascading Op Amps
Example #6 (con’t)
 Treat as two separate amplifier circuits
Example #6 (con’t)
First Circuit
Second Circuit
Example #6 (con’t)
 It is a noninverting
amplifier.
 Rf 1 
VS
Vo1  1 
R1 

Vo1  R f 1 

AV 1 
 1 
VS 
R1 
Example #6 (con’t)
 It is a inverting amplifier.
Vo  
Rf 2
R2
V01
Rf 2
Vo
AV 2 

V01
R2
Example #6 (con’t)
 The gain of the cascaded amplifiers is the
multiplication of the two individial amplifiers
Rf 2  Rf 1 
1 
VS
Vo  
V01  
R2
R2 
R1 
Vo  AV 1 AV 2VS  AV VS
Rf 2
Summary
 The ‘almost ideal’ op amp model:
 Ri = ∞W.

i1 = i2 = 0A; v1 = v2
 Ro = 0W.

No power/voltage loss between the dependent voltage source and vo.
 The output voltage is limited by the voltages applied to the positive
and negative rails.

V+ ≥ vo ≥ V-
 This model can be used to determine the closed loop voltage
gain for any op amp circuit.
 Superposition can be used to solve for the output of a summing
amplifier.
 Cascaded op amp circuits can be separated into individual
amplifiers and the overall gain is the multiplication of the gain of
each amplifier.