ECE 271
Electronic Circuits I
Topic 6
Analog Circuits
Ideal Operational Amplifiers
One Transistor Amplifier
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 1
Example of Analog Electronic System:
FM Stereo Receiver
•
•
•
•
•
Much information in the world (temperature, pressure, light intensity, sound,
etc.) is analog in nature.
In electric form those signals are transformed by different linear and nonlinear
functions.
Linear functions: Radio and audio frequency amplification, frequency selection
(tuning), impedance matching, local oscillator
Nonlinear functions: DC power supply(rectification), frequency conversion
(mixing), detection/demodulation
The characteristics of signals are most often manipulated with linear
amplifiers.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 2
Amplification: Introduction
Example: Audio amplifier. The input is a form of complicated periodic signal.
i
i
v
i
R
i
A complex periodic signal can be represented as the sum of many individual
sine waves, one component of which has amplitude Vi = 1 mV and
frequency ws with 0 phase (signal is used as reference):
vi  Vi sinwst
After amplification, linear amplifier output is sinusoidal with same
frequency but different amplitude VO and phase :
vo  Vo (sinwst  )
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 3
Amplification: Introduction (cont.)
Amplifier output power is:
PO 





Vo
2




1
2 RL
If we need to deliver PO = 100 W and have RL = 8 W, then

Vo  2PO RL  21008  40V
This power results in output current:
where Io 
io  Io (sinwst  )
Vo 40V

 5A
RL 8W
Input current is given by (Vi = 1 mV )

-3 V
Vi
10
Ii 

1.82108 A
Ri  Rin 5kW 50kW
and the phase is zero because circuit is purely resistive.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 4
Amplification: Gain
The main parameter of an amplifier is the gain. Using phasor
representation v V  , we can introduce three types of gain.
v V  Vo
 
• Voltage Gain (complex number): Av  vo  o
Vi
V

0
i
i
Magnitude and phase of voltage gain are given by
Vo
Av  
Av 
and
Vi
Vo 40V
Av   3  4104
For our example,
Vi 10 V
• Current Gain: A  io  Io   Io 
i
ii Ii 0 Ii
Magnitude of current gain is given by
Io
5A
8
Ai  

2.75

10
Ii 1.8210-8 A
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 5
Amplification: Gain (cont.)
Vo
PO
A

• Power Gain: P P  V2
i
i
2
For our example,
Io
2  Vo I o  A A
v
i
I i Vi I i
2
405
13
AP  3
1.1010
10 1.82108

The gain is often expressed in decibel scale:
AvdB  20logAv
AidB  20logAi
APdB 10logAP

NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 6
Two-port Model for Amplifier
Type of input-output model, black box that relates outputs to inputs.
input
x1
Amplifier
output
x2
• Simplifies amplifier-behavior modeling in complex systems.
• Two-port (four terminal) models are linear network models, valid only
under small-signal conditions.
• Represented by g-, h-, y- and z-parameters.
• (v1, i1) and (v2, i2) represent signal components of voltages and
currents at the network ports.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 7
Two-port Model Parameters
•
•
Theoretically, two-port model could be described fully by a 4x4 matrix or 16
parameters.
In practice, several sets of 4 parameters are used.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 8
Two-port Model Parameters
•
•
Theoretically, two-port model could be described fully by a 4x4 matrix or 16
parameters.
In practice, several sets of 4 parameters are used.
Impedance (z)-parameters
NJIT ECE 271 Dr. Serhiy Levkov
Admittance (y)-parameters
Topic 6 - 9
Two-port Model Parameters
•
•
Theoretically, two-port model could be described fully by a 4x4 matrix or 16
parameters.
In practice, several sets of 4 parameters are used.
Impedance (z)-parameters
Hybrid (h)-parameters
NJIT ECE 271 Dr. Serhiy Levkov
Admittance (y)-parameters
Inverse hybrid (g)-parameters
Topic 6 - 10
Two-port Model Parameters
•
•
Theoretically, two-port model could be described fully by a 4x4 matrix or 16
parameters.
In practice, several sets of 4 parameters are used.
Impedance (z)-parameters
Hybrid (h)-parameters
Admittance (y)-parameters
Inverse hybrid (g)-parameters
Transfer (A,B,C,D) -parameters
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 11
g-parameters
i1 g11v1  g12i 2
v 2  g21v1  g22i 2
i
g11  v1
1
g12 
i2 0
i1
i2 v 0
1
NJIT ECE 271 Dr. Serhiy Levkov
Use open-circuit (i = 0) and short-circuit (v = 0)
terminal conditions
Open-circuit input
conductance
v
g21  v2
Reverse short-circuit
current gain
g22 
1
i2 0
v2
i2 v 0
1
Forward open-circuit
voltage gain
Short-circuit output
resistance
Topic 6 - 12
g-parameters
i1 g11v1  g12i 2
v 2  g21v1  g22i 2
i
g11  v1
1
g12 
i2 0
i1
i2 v 0
1
If g12 = 0
NJIT ECE 271 Dr. Serhiy Levkov
Use open-circuit (i = 0) and short-circuit (v = 0)
terminal conditions
Open-circuit input
conductance
v
g21  v2
Reverse short-circuit
current gain
g22 
1
i2 0
v2
i2 v 0
1
Forward open-circuit
voltage gain
Short-circuit output
resistance
Norton
transformation
Topic 6 - 13
g-parameters: Example
Problem: Find g-parameters.
Approach: Apply specified boundary
conditions for each g-parameter, use
circuit analysis.
For g11 and g21: apply voltage v1 to
input port and open circuit output port.
For g12 and g22: apply current i2 to
output port and short circuit input port.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 14
g-parameters: Example
i
g11  v1

1i 0
2
1
210 4 W 51(200kW)
g11  9.79108S
v
g21  v2
 g11(51)(200kW) 0.998
1 i 0
2
Problem: Find g-parameters.

Approach: Apply specified boundary
conditions for each g-parameter, use
circuit analysis.
For g11 and g21: apply voltage v1 to
input port and open circuit output port.
For g12 and g22: apply current i2 to
output port and short circuit input port.
NJIT ECE 271 Dr. Serhiy Levkov
g22 
v2
i2 v
i1
i2 v
1
1
 391W
1
51

200kW 20kW
391W

 0.0196
20kW
0
1
g12 

0
i 1 9.79108 v1 1.96102 i2
v2  0.998v1  3.91102 i2
Topic 6 - 15
Ideal Voltage Amplifier
Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and
load resistances and calculate the voltage gain from the source voltage to load voltage.
RL
Rout  RL
Rin
v 1 vs
RS  Rin
V
Rin
RL
Av  o  A
Vs
RS  Rin Rout  RL
v o  Av1
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 16
Ideal Voltage Amplifier
Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and
load resistances and calculate the voltage gain from the source voltage to load voltage.
RL
Rout  RL
Rin
v 1 vs
RS  Rin
V
Rin
RL
Av  o  A
Vs
RS  Rin Rout  RL
v o  Av1
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 17
Ideal Voltage Amplifier
Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and
load resistances and calculate the voltage gain from the source voltage to load voltage.
RL
Rout  RL
Rin
v 1 vs
RS  Rin
v
Rin
RL
Av  o  A
vs
RS  Rin Rout  RL
v o  Av1
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 18
Ideal Voltage Amplifier
Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and
load resistances and calculate the voltage gain from the source voltage to load voltage.
RL
Rout  RL
Rin
v 1 vs
RS  Rin
v
Rin
RL
Av  o  A
vs
RS  Rin Rout  RL
v o  Av1
Conclusion: full gain depends on
source and load parameters.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 19
Ideal Voltage Amplifier
Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and
load resistances and calculate the voltage gain from the source voltage to load voltage.
How to make it non- dependent?
RL
Rout  RL
Rin
v 1 vs
RS  Rin
v
Rin
RL
Av  o  A
vs
RS  Rin Rout  RL
v o  Av1
Conclusion: full gain depends on
source and load parameters.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 20
Ideal Voltage Amplifier
Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and
load resistances and calculate the voltage gain from the source voltage to load voltage.
How to make it non- dependent?
Av  A
Rin >> Rs and Rout<< RL,

RL
Rout  RL
Rin
v 1 vs
RS  Rin
V
Rin
RL
Av  o  A
Vs
RS  Rin Rout  RL
v o  Av1
Ideal voltage amplifier: Rout = 0, Rin 

Conclusion: full gain depends on
source and load parameters.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 21
Ideal Voltage Amplifier
Consider amplifier with g12 = 0 (typical voltage amplifier) that includes source and
load resistances and calculate the voltage gain from the source voltage to load voltage.
How to make it non- dependent?
Av  A
Rin >> Rs and Rout<< RL,

RL
Rout  RL
Rin
v 1 vs
RS  Rin
V
Rin
RL
Av  o  A
Vs
RS  Rin Rout  RL
v o  Av1
Conclusion: full gain depends on
source and load parameters.
NJIT ECE 271 Dr. Serhiy Levkov
Ideal voltage amplifier: Rout = 0, Rin 
For current:

Vo
V R R
R R
I
RL
Ai  o 
 o S in  Av S in
Vs RL
RL
Vs
I1
RS  Rin
Topic 6 - 22
Other Amplifier Types
Voltage amplifier
Ri , Ro  0
v
Avo  vo
(V/V)
i i 0
o
NJIT ECE 271 Dr. Serhiy Levkov
Ri  0, Ro 
Ais 
io
ii v
(A/A)
o
0
Transresistance amplifier
Transconductance amplifier
Ri , Ro 
Current amplifier
i
Gm  vo
i v 0
o
(A/V)
Ri  0, Ro  0
Rm 
vo
ii i
o
(V/A)
0
Topic 6 - 23
Other Amplifier Types
Voltage amplifier
Ri , Ro  0
v
Avo  vo
(V/V)
i i 0
o
NJIT ECE 271 Dr. Serhiy Levkov
Ri  0, Ro 
Ais 
io
ii v
(A/A)
o
0
Transresistance amplifier
Transconductance amplifier
Ri , Ro 
Current amplifier
i
Gm  vo
i v 0
o
(A/V)
Ri  0, Ro  0
Rm 
vo
ii i
o
(V/A)
0
Topic 6 - 24
Other Amplifier Types
Voltage amplifier
Ri , Ro  0
v
Avo  vo
(V/V)
i i 0
o
NJIT ECE 271 Dr. Serhiy Levkov
Ri  0, Ro 
Ais 
io
ii v
(A/A)
o
0
Transresistance amplifier
Transconductance amplifier
Ri , Ro 
Current amplifier
i
Gm  vo
i v 0
o
(A/V)
Ri  0, Ro  0
Rm 
vo
ii i
o
(V/A)
0
Topic 6 - 25
Other Amplifier Types
Voltage amplifier
Ri , Ro  0
v
Avo  vo
(V/V)
i i 0
o
NJIT ECE 271 Dr. Serhiy Levkov
Ri  0, Ro 
Ais 
io
ii v
(A/A)
o
0
Transresistance amplifier
Transconductance amplifier
Ri , Ro 
Current amplifier
i
Gm  vo
i v 0
o
(A/V)
Ri  0, Ro  0
Rm 
vo
ii i
o
(V/A)
0
Topic 6 - 26
Other Amplifier Types
Voltage amplifier
Ri , Ro  0
v
Avo  vo
(V/V)
i i 0
o
NJIT ECE 271 Dr. Serhiy Levkov
Ri  0, Ro 
Ais 
io
ii v
(A/A)
o
0
Transresistance amplifier
Transconductance amplifier
Ri , Ro 
Current amplifier
i
Gm  vo
i v 0
o
(A/V)
Ri  0, Ro  0
Rm 
vo
ii i
o
(V/A)
0
Topic 6 - 27
Operational Amplifier (Op Amp)
• Op amp is a fundamental building block in electronic design. They are cheap mass
produced IC and can be used to build many different types of electronic devices like
instrumentation amplifiers, active filters, rectifiers, A/D and D/A converters, and
others.
• Typical integrated circuit amplifier: LM386N (~ $1.00 each)
• The practical op amp is a form of differential amplifier: responds to a difference of
two input signals.
Typically: VCC >0, VEE <0 – so the
voltages are symmetric 5V, 10V, 18V.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 28
Differential Amplifier – Signal Amplification
The typical voltage transfer characteristic for a differential amplifier biased by two
symmetric power supplies:
A - open-circuit voltage gain,
A =10, Adb =20log(10) = 20 db
vID = (v+-v--) - differential input signal
vID = VID + vid = 1 + 0.25sin t, where
VID – dc value, vid – signal component
vO = VO + vo = 5 + 2.5sin t
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 29
Distortion in Amplifiers
•
•
•
If the ac input signal exceeds 0.5V, the output signal will be clipped off (see vO2 ) and
get distorted.
Different gains for positive and negative values of input also cause distortion in output.
Total Harmonic Distortion (THD) is a measure of signal distortion that compares
undesired harmonic content of a signal to the desired component.

v(t)VO V1(sinwot 1)V2(sin2wot 2 )V3(sin3wot 3)...
dc
desired
output
NJIT ECE 271 Dr. Serhiy Levkov
2nd harmonic
distortion
THD100% 
3rd harmonic
distortion

2
Vi
i2
V1
Topic 6 - 30
Differential Amplifier Model: Basic
The basic model of differential amplifier is represented by:
A - open-circuit voltage gain
vid = (v+-v-) = differential input
signal voltage
Rid - amplifier input resistance
Ro - amplifier output resistance
Signal developed at amplifier output
is in phase with the voltage applied at
+ input (non-inverting) terminal and
180o out of phase with that applied at input (inverting) terminal.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 31
Differential Amplifier Model:
With Source and Load
RL = load resistance
RS = Thevenin equivalent resistance of
signal source
vs = Thevenin equivalent voltage of
signal source
RL
v  Av
o
id Ro  RL
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 32
Differential Amplifier Model:
With Source and Load
RL = load resistance
RS = Thevenin equivalent resistance of
signal source
vs = Thevenin equivalent voltage of
signal source
RL
v  Av
o
id Ro  RL
NJIT ECE 271 Dr. Serhiy Levkov
and
v
id
v
R
id
s R R
id
s
Topic 6 - 33
Differential Amplifier Model:
With Source and Load
RL = load resistance
RS = Thevenin equivalent resistance of
signal source
vs = Thevenin equivalent voltage of
signal source
RL
v  Av
o
id Ro  RL
and
v
id
v
R
id
s R R
id
s
v
R
R
o
id
L
A 
A
v v
R R R R
s
id
S o
L
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 34
Differential Amplifier Model:
With Source and Load (Example)
•
•
Problem: Calculate voltage gain
Given Data: A=100, Rid =100kW, Ro = 100W, RS =10kW, RL =1000W
v
R
R
o
id
L
A 
A
v v
R R R R
s
id
S o
L
100kW
1000W



 100 

  82.6  38.3dB
10k
W

100k
W
100
W

1000
W



NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 35
Differential Amplifier Model:
With Source and Load (Example)
•
•
Problem: Calculate voltage gain
Given Data: A=100, Rid =100kW, Ro = 100W, RS =10kW, RL =1000W
v
R
R
o
id
L
A 
A
v v
R R R R
s
id
S o
L
100kW
1000W



 100 

  82.6  38.3dB
10k
W

100k
W
100
W

1000
W



•
Ideal amplifier’s output depends only on input voltage difference and not on
source and load resistances. This can be achieved by using resistance
condition (Rid >> RS or infinite Rid and Ro << RL or zero Ro ):
v
A  o A
v v
id
A - open-loop gain (maximum voltage gain available from the device)
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 36
Ideal Operational Amplifier
•
Ideal op amp is a special case of ideal differential amplifier with:
– A   - infinite gain,
– Rid   - infinite Rid ,
– Ro  0 - zero Ro .
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 37
Ideal Operational Amplifier
•
•
Ideal op amp is a special case of ideal differential amplifier with:
– A   - infinite gain,
v
– Rid   - infinite Rid ,
v  o , lim v  0
id A
id
A

– Ro  0 - zero Ro .
Thus:
1. Input voltage is zero, vid =0 for any finite output voltage (since A is infinite).
2. Input currents are zero, i+ =0 and i-=0 (since Rid is infinite).
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 38
Ideal Operational Amplifier
•
•
•
Ideal op amp is a special case of ideal differential amplifier with:
– A   - infinite gain,
v
– Rid   - infinite Rid ,
v  o , lim v  0
id A
id
A

– Ro  0 - zero Ro .
Thus:
1. Input voltage is zero, vid =0 for any finite output voltage (since A is infinite).
2. Input currents are zero, i+ =0 and i-=0 (since Rid is infinite).
Additional properties:
–
–
–
–
–
–
•
infinite common-mode rejection (rejects input signals common to both inputs)
infinite open-loop bandwidth (frequency magnitude response is flat everywhere with zero phase shift)
infinite output voltage range ( in practice limited by VS+ and VS- )
infinite output current capability
infinite slew rate (rate of change of output voltage)
zero input-offset voltage (when the input terminals are shorted, vout = 0)
The major benefit of ideal op-amp: the same standard op-amp allows to build various
circuits with very specific properties by using external to op-amp circuit elements. The
classic op-amps circuits are considered further.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 39
Inverting Amplifier: Configuration
• Non-inverting input is grounded.
• Feedback network (resistor R2 ) is connected between
inverting input and output
• Input network (resistor R1 ) is connected between
inverting input and signal source.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 40
Inverting Amplifier:Voltage Gain
KVL: - vS  is R1  i2 R2  vO  0
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 41
Inverting Amplifier:Voltage Gain
KVL: - vS  is R1  i2 R2  vO  0
KCL:  is  i2  i  0  i2  is
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 42
Inverting Amplifier:Voltage Gain
KVL: - vS  is R1  i2 R2  vO  0
KCL:  is  i2  i  0  i2  is
NJIT ECE 271 Dr. Serhiy Levkov
since i  0 by 2
Topic 6 - 43
Inverting Amplifier:Voltage Gain
KVL: - vS  is R1  i2 R2  vO  0
KCL:  is  i2  i  0  i2  is
vs v vs
iS 

R1
R1
NJIT ECE 271 Dr. Serhiy Levkov
since i  0 by 2
Topic 6 - 44
Inverting Amplifier:Voltage Gain
KVL: - vS  is R1  i2 R2  vO  0
KCL:  is  i2  i  0  i2  is since i  0 by 2
vs v vs
since v  0 and vid  0 by 1
iS 

R1
R1
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 45
Inverting Amplifier:Voltage Gain
KVL: - vS  is R1  i2 R2  vO  0
KCL:  is  i2  i  0  i2  is since i  0 by 2
vs v vs
since v  0 and vid  0 by 1
iS 

R1
R1
Substituting iS and i2 into KVL:
vS  vS (R1  R2 )  vO  0   vS R1  vS (R1  R2 )  vO R1  0  vS
R1
NJIT ECE 271 Dr. Serhiy Levkov
vo
R2
R
 vO or A 
 2
v v
R1
R1
s
Topic 6 - 46
Inverting Amplifier:Voltage Gain
KVL: - vS  is R1  i2 R2  vO  0
KCL:  is  i2  i  0  i2  is since i  0 by 2
vs v vs
since v  0 and vid  0 by 1
iS 

R1
R1
Substituting iS and i2 into KVL:
vS  vS (R1  R2 )  vO  0   vS R1  vS (R1  R2 )  vO R1  0  vS
R1
vo
R2
R
 vO or A 
 2
v v
R1
R1
s
Conclusions:
• The gain of the inverting op-amp circuit is determined by R1 and R2 .
• The gain is a result of imposing the negative feedback.
• Gain is greater than 1 if R2 > R1. Gain is less than 1 if R1 > R2
• Negative voltage gain implies 1800 phase shift between input and output signals.
• Inverting input of op amp is at ground potential (not connected directly to ground) and is
said to be at virtual ground.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 47
Inverting Amplifier: Input and Output
Resistances
vs
Rin 
 Rin  R1
is
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 48
Inverting Amplifier: Input and Output
Resistances
vs
Rin 
 Rin  R1
is
Rout is found by applying a test current
(or voltage) source to amplifier output,
turning off all independent sources and
determining the voltage (or current):
KVL:  vx  i2 R2  i1R1  0
But i1=i2 (i- =0) 
vx  i1(R2  R1)
Since v- = 0, i1=0 and vx = 0
irrespective of the value of ix ,
Rout  vx  0
i
x
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 49
Inverting Amplifier: Example
• Problem: Design an inverting
amplifier with
Av=40 dB, Rin =20kW,
• Assumptions: Ideal op amp
• Convert the gain from db:
Av[db]  20log A  40  A 1040/ 20  100
v
v
• Input resistance is controlled by R1 and voltage gain is set by R2 / R1.
Thus
R1  Rin  20kW
and
R
A  2  R2  100R1  2M W
v R
1
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 50
Summing Amplifier
Since i-=0, i3= i1 + i2,
vid
vo  
Since negative amplifier input is
at virtual ground (vid = 0):
i1 
v1
R1
NJIT ECE 271 Dr. Serhiy Levkov
i2 
v2
R2
i3  
vO
R3
R3
R
v1  3 v2
R1
R2
• Scale factors for the 2 inputs
can be independently adjusted
by proper choice of R2 and R1.
• Any number of inputs can be
connected to summing
junction through extra
resistors.
• This is an example of a simple
digital-to-analog converter.
Topic 6 - 51
Non-inverting Amplifier: Configuration
• Input signal is applied to the non-inverting input terminal.
• Portion of the output signal is fed back to the negative input terminal.
• Analysis is done by relating voltage at v1 to input voltage vs and output
voltage vo .
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 52
Non-inverting Amplifier: Analysis
R
Since i-=0, v1  vo R 1R (voltage divider)
i-
+
1
2
vid
i+
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 53
Non-inverting Amplifier: Analysis
R
Since i-=0, v1  vo R 1R
i-
+
1
vid
2
From KVL vS  vid  v1  0
i+
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 54
Non-inverting Amplifier: Analysis
R
Since i-=0, v1  vo R 1R
i-
+
1
vid
i+
NJIT ECE 271 Dr. Serhiy Levkov
2
From KVL vS  vid  v1  0 , since vid =0 
vs  v1
and
vo  vs
R1  R2
R1

Topic 6 - 55
Non-inverting Amplifier: Analysis
R
Since i-=0, v1  vo R 1R
i-
+
1
vid
i+
2
From KVL vS  vid  v1  0 , since vid =0 
vs  v1
and
vo  vs
R1  R2
R1

vo R  R
R
A   1 2 1 2 .
v v
R1
R1
s
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 56
Non-inverting Amplifier: Analysis
R
Since i-=0, v1  vo R 1R
i-
+
1
vid
From KVL vS  vid  v1  0 , since vid =0 
i+
Since i+=0,
NJIT ECE 271 Dr. Serhiy Levkov
2
vs  v1
vs
Rin    .
i
and
vo  vs
R1  R2
R1

vo R  R
R
A   1 2 1 2 .
v v
R1
R1
s
Topic 6 - 57
Non-inverting Amplifier: Analysis
vo R  R
1
2  1 R2
A


R
R1
R1
Since i-=0, v1  vo R 1R v vs
i-
+
1
vid
From KVL vS  vid  v1  0 , since vid =0 
i+
Since i+=0,
2
vs  v1
vs
Rin    .
i
and
vo  vs
R1  R2
R1

vo R  R
R
A   1 2 1 2 .
v v
R1
R1
s
Rout is found by applying a test current source to amplifier output and
setting vs = 0 and is identical to the output resistance of inverting amplifier
i.e. Rout =0.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 58
Non-inverting Amplifier: Analysis
vo R  R
1
2  1 R2
A


R
R1
R1
Since i-=0, v1  vo R 1R v vs
i-
+
1
vid
From KVL vS  vid  v1  0 , since vid =0 
i+
Since i+=0,
2
vs  v1
vs
Rin    .
i
and
vo  vs
R1  R2
R1

vo R  R
R
A   1 2 1 2 .
v v
R1
R1
s
Rout is found by applying a test current source to amplifier output and
setting vs = 0 and is identical to the output resistance of inverting amplifier
i.e. Rout =0.
R
A  1 2 , Rin , Rout  0
v
R1
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 59
Non-inverting Amplifier: Example
•
i-
+
vid
io
•
•
•
Problem: Determine the characteristics of given
non-inverting amplifier.
Given Data: R1= 3kW, R2 =43kW, vs=+0.1 V
Assumptions: Ideal op amp.
Analysis:
i+
R
43kW
A  1  2  1
 15.3
v
R
3kW
1
vo  A vs  (15.3)(0.1V ) 1.53V
v
Since i-=0,
iO 
NJIT ECE 271 Dr. Serhiy Levkov
vO
1.53V

 33.3 A
R2  R1 43k W3k W
Topic 6 - 60
Voltage Follower (Unity-gain Buffer)
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 61
Voltage Follower (Unity-gain Buffer)
• A special case of non-inverting amplifier, also called voltage follower
with infinite R1 and zero R2. Hence Av 1 R2 / R1 1
• Provides excellent impedance-level transformation while maintaining
signal voltage level.
• Ideal voltage buffer does not require any input current and can drive
any desired load resistance without loss of signal voltage.
• Unity-gain buffer is used in many sensor and data acquisition systems.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 62
Difference Amplifier
A combination of inverting and non-inverting amplifier.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 63
Difference Amplifier
A combination of inverting and non-inverting amplifier.
vo  v-  i2 R2  v-  i1R2 (since i  0)
 R R 
(v1 v- )
R
 v- 
R2   1 2  v-  2 v1

R1 
R1
R1

NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 64
Difference Amplifier
A combination of inverting and non-inverting amplifier.
vo  v-  i2 R2  v-  i1R2 (since i  0)
 R R 
(v1 v- )
R
 v- 
R2   1 2  v-  2 v1

R1 
R1
R1

R2
Also, v 
v (voltage division)
R1  R2 2
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 65
Difference Amplifier
A combination of inverting and non-inverting amplifier.
Since v-= v+,
R
vo   2 (v  v )
R 1 2
1
For R2= R1,
vo  (v  v )
1 2
vo  v-  i2 R2  v-  i1R2 (since i  0)
 R R 
(v1 v- )
R
 v- 
R2   1 2  v-  2 v1

R1 
R1
R1

R2
Also, v 
v (voltage division)
R1  R2 2
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 66
Difference Amplifier
A combination of inverting and non-inverting amplifier.
•
Since v-= v+,
R
vo   2 (v  v )
R 1 2
1
For R2= R1,
vo  (v  v )
1 2
The circuit amplifies difference between input
signals (differential subtractor).
vo  v-  i2 R2  v-  i1R2 (since i  0)
 R R 
(v1 v- )
R
 v- 
R2   1 2  v-  2 v1

R1 
R1
R1

R2
Also, v 
v (voltage division)
R1  R2 2
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 67
Difference Amplifier
A combination of inverting and non-inverting amplifier.
•
•
vo  v-  i2 R2  v-  i1R2 (since i  0)
 R R 
(v1 v- )
R
 v- 
R2   1 2  v-  2 v1

R1 
R1
R1

R2
Also, v 
v (voltage division)
R1  R2 2
NJIT ECE 271 Dr. Serhiy Levkov
Since v-= v+,
R
vo   2 (v  v )
R 1 2
1
For R2= R1,
vo  (v  v )
1 2
The circuit amplifies difference between input
signals (differential subtractor).
To understand better, use superposition:
For v2=0, Rin1= R1, the circuit reduces to an
inverting amplifier with the gain – R2 /R1 :
vO1= (– R2 /R1 ) v1
For v1=0, v2 is attenuated by the voltage
divider and then amplified by non-inverting
gain 1+R2 /R1:
vO2= (R2 /(R1 +R1 ))(1+ R2 /R1 ) v2
Then vO = vO1 + vO2, 
vo  
R2
(v  v )
R1 1 2
Topic 6 - 68
Difference Amplifier: Example
Do example on the board
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 69
Difference Amplifier: Example
•
•
•
•
Problem: Determine Vo, V+, V-, Io, I1, I2, I3 .
Given Data: R1= 10kW, R2 =100kW, V1= 5V,
V2=3V
Assumptions: Ideal op amp.
Hence, V-= V+ and I-= I+= 0.
Analysis:
R2
100k W
V  V 
V  3V
 2.73V
- R R 2
10k W100k W
1 2
Io  I  227 A
2
V V 5V -2.73V
I I  1 - 
 227 A
1 2
R1
10k W
V2 0
3V
I 

 27 A
3 R  R 110k W
1 2
Vo  V  I R  I R  5V  (227 A)(110k W)  20.0V
1 11 2 2
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 70
Integrator
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 71
Integrator
• Feedback resistor R2 in the inverting
amplifier is replaced by capacitor C.
• The circuit uses frequency-dependent
feedback.
dvo
vs
ic  C
is 
dt
R
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 72
Integrator
Since ic= is ,
dvo
1

vs
RC
dt
• Feedback resistor R2 in the inverting
amplifier is replaced by capacitor C.
• The circuit uses frequency-dependent
feedback.
dvo
vs
ic  C
is 
dt
R
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 73
Integrator
Since ic= is ,
dvo
1

vs
RC
dt
1
dv


vsd
 o 
RC
t
vo (t )   1  vs ( )d  vo (0)
RC 0
vo (0) V (0)
c
• Feedback resistor R2 in the inverting
amplifier is replaced by capacitor C.
• The circuit uses frequency-dependent
feedback.
dvo
vs
ic  C
is 
dt
R
NJIT ECE 271 Dr. Serhiy Levkov
Voltage at the circuit’s output at
time t is given by the initial
capacitor voltage plus integral of
the input signal from start of
integration interval, here, t=0.
Topic 6 - 74
Integrator
Since ic= is ,
dvo
1

vs
RC
dt
1
dv


vsd
 o 
RC
t
vo (t )   1  vs ( )d  vo (0)
RC 0
vo (0) V (0)
c
• Feedback resistor R2 in the inverting
amplifier is replaced by capacitor C.
• The circuit uses frequency-dependent
feedback.
dvo
vs
ic  C
is 
dt
R
NJIT ECE 271 Dr. Serhiy Levkov
Voltage at the circuit’s output at
time t is given by the initial
capacitor voltage plus integral of
the input signal from start of
integration interval, here, t=0.
Integration of an input step signal
results in a ramp at the output.
Topic 6 - 75
Differentiator
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 76
Differentiator
• Input resistor R1 in the inverting
amplifier is replaced by capacitor C.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 77
Differentiator
v
i  o
R
R
dvs
is  C
dt
Since iR= is
dvs
vo  RC
dt
• Input resistor R1 in the inverting
amplifier is replaced by capacitor C.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 78
Differentiator
v
i  o
R
R
dvs
is  C
dt
Since iR= is
dvs
vo  RC
dt
• Input resistor R1 in the inverting
amplifier is replaced by capacitor C.
NJIT ECE 271 Dr. Serhiy Levkov
Output is scaled version of
derivative of input voltage.
Topic 6 - 79
Differentiator
v
i  o
R
R
dvs
is  C
dt
Since iR= is
dvs
vo  RC
dt
• Input resistor R1 in the inverting
amplifier is replaced by capacitor C.
Output is scaled version of
derivative of input voltage.
Derivative operation emphasizes
high-frequency components of
input signal, hence is less often
used than the integrator.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 80
Op-Amps – What’s Next
•
Frequency dependent feedback (cont. ch 10):
– high, low and band pass amplifiers (filters)
– integrators and differentiators
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 81
Op-Amps – What’s Next
•
Frequency dependent feedback (cont. ch 10):
– high, low and band pass amplifiers (filters)
– integrators and differentiators
•
•
Analysis of non-ideal amplifiers (ch. 11)
Amplifier frequency response and stability (ch. 11)
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 82
Op-Amps – What’s Next
•
Frequency dependent feedback (cont. ch 10):
– high, low and band pass amplifiers (filters)
– integrators and differentiators
•
•
•
Analysis of non-ideal amplifiers (ch. 11)
Amplifier frequency response and stability (ch. 11)
Op-amps applications (ch. 12)
–
–
–
–
–
–
instrumentation amplifiers
active filters
D/A and A/D converters
Oscillators
Nonlinear circuits (precision rectifiers)
Circuits with positive feedback (comparators, triggers)
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 83
Op-Amps – What’s Next
•
Frequency dependent feedback (cont. ch 10):
– high, low and band pass amplifiers (filters)
– integrators and differentiators
•
•
•
Analysis of non-ideal amplifiers (ch. 11)
Amplifier frequency response and stability (ch. 11)
Op-amps applications (ch. 12)
–
–
–
–
–
–
•
instrumentation amplifiers
active filters
D/A and A/D converters
Oscillators
Nonlinear circuits (precision rectifiers)
Circuits with positive feedback (comparators, triggers)
Single transistor amplifiers – linear amplification and small signal model (ch. 13)
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 84
Op-Amps – What’s Next
•
Frequency dependent feedback (cont. ch 10):
– high, low and band pass amplifiers (filters)
– integrators and differentiators
•
•
•
Analysis of non-ideal amplifiers (ch. 11)
Amplifier frequency response and stability (ch. 11)
Op-amps applications (ch. 12)
–
–
–
–
–
–
•
•
instrumentation amplifiers
active filters
D/A and A/D converters
Oscillators
Nonlinear circuits (precision rectifiers)
Circuits with positive feedback (comparators, triggers)
Single transistor amplifiers – linear amplification and small signal model (ch. 13)
Single transistor amplifiers – detailed design (ch. 14)
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 85
Op-Amps – What’s Next
•
Frequency dependent feedback (cont. ch 10):
– high, low and band pass amplifiers (filters)
– integrators and differentiators
•
•
•
Analysis of non-ideal amplifiers (ch. 11)
Amplifier frequency response and stability (ch. 11)
Op-amps applications (ch. 12)
–
–
–
–
–
–
•
•
•
instrumentation amplifiers
active filters
D/A and A/D converters
Oscillators
Nonlinear circuits (precision rectifiers)
Circuits with positive feedback (comparators, triggers)
Single transistor amplifiers – linear amplification and small signal model (ch. 13)
Single transistor amplifiers – detailed design (ch. 14)
Op-amps design (multistage amplifiers) (ch. 15)
What do we study in this course?
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 86
Op-Amps – What’s Next
•
Frequency dependent feedback (cont. ch 10):
– high, low and band pass amplifiers (filters)
– integrators and differentiators
•
•
•
Analysis of non-ideal amplifiers (ch. 11)
Amplifier frequency response and stability (ch. 11)
Op-amps applications (ch. 12)
–
–
–
–
–
–
•
•
•
instrumentation amplifiers
active filters
D/A and A/D converters
Oscillators
Nonlinear circuits (precision rectifiers)
Circuits with positive feedback (comparators, triggers)
Single transistor amplifiers – linear amplification and small signal model (ch. 13)
Single transistor amplifiers – detailed design (ch. 14)
Op-amps design (multistage amplifiers) (ch. 15)
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 87
Transistors as Amplifiers
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 88
Transistors as Amplifiers
In this section we will study
• The general techniques for employing individual transistors as amplifiers
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 89
Transistors as Amplifiers
In this section we will study
• The general techniques for employing individual transistors as amplifiers
• Operation of common source MOSFET amplifier
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 90
Transistors as Amplifiers
In this section we will study
• The general techniques for employing individual transistors as amplifiers
• Operation of common source MOSFET amplifier
• Operation of common emitter BJT amplifier
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 91
Transistors as Amplifiers
In this section we will study
• The general techniques for employing individual transistors as amplifiers
• Operation of common source MOSFET amplifier
• Operation of common emitter BJT amplifier
What do we know:
• MOSFET (FET) can be used as amplifier if operated in saturation region (SR)
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 92
Transistors as Amplifiers
In this section we will study
• The general techniques for employing individual transistors as amplifiers
• Operation of common source MOSFET amplifier
• Operation of common emitter BJT amplifier
What do we know:
• MOSFET (FET) can be used as amplifier if operated in saturation region (SR)
• BJT can be used as an amplifier when biased in the forward-active region (FAR)
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 93
Transistors as Amplifiers
In this section we will study
• The general techniques for employing individual transistors as amplifiers
• Operation of common source MOSFET amplifier
• Operation of common emitter BJT amplifier
What do we know:
• MOSFET (FET) can be used as amplifier if operated in saturation region (SR)
• BJT can be used as an amplifier when biased in the forward-active region (FAR)
• We will refer to the FAR and SR as to the “active region”
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 94
Transistors as Amplifiers
In this section we will study
• The general techniques for employing individual transistors as amplifiers
• Operation of common source MOSFET amplifier
• Operation of common emitter BJT amplifier
What do we know:
• MOSFET (FET) can be used as amplifier if operated in saturation region (SR)
• BJT can be used as an amplifier when biased in the forward-active region (FAR)
• We will refer to the FAR and SR as to the “active region”
• In these regions, transistors can provide high voltage, current and power gains
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 95
Transistors as Amplifiers
In this section we will study
• The general techniques for employing individual transistors as amplifiers
• Operation of common source MOSFET amplifier
• Operation of common emitter BJT amplifier
What do we know:
• MOSFET (FET) can be used as amplifier if operated in saturation region (SR)
• BJT can be used as an amplifier when biased in the forward-active region (FAR)
• We will refer to the FAR and SR as to the “active region”
• In these regions, transistors can provide high voltage, current and power gains
• Bias is provided to stabilize the operating point in a desired operation region
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 96
Transistors as Amplifiers
In this section we will study
• The general techniques for employing individual transistors as amplifiers
• Operation of common source MOSFET amplifier
• Operation of common emitter BJT amplifier
What do we know:
• MOSFET (FET) can be used as amplifier if operated in saturation region (SR)
• BJT can be used as an amplifier when biased in the forward-active region (FAR)
• We will refer to the FAR and SR as to the “active region”
• In these regions, transistors can provide high voltage, current and power gains
• Bias is provided to stabilize the operating point in a desired operation region
• The Q-point also determines
–
–
–
–
Small-signal parameters of transistor
Voltage gain, input resistance, output resistance
Maximum input and output signal amplitudes
Power consumption
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 97
BJT Amplifier
Consider a BJT (Is 1015, F 100, VT  0.025) , biased in active region by dc
voltage source VBE = 0.7V.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 98
BJT Amplifier
Consider a BJT (Is 1015, F 100, VT  0.025) , biased in active region by dc
voltage source VBE = 0.7V.
From the simplified FAR model we get:
IB 
IS
F
VBE
VT
e
15μA  IC  F I 1.5mA  VCE 10  3.3K W IC 10  3.31.5  5V
NJIT ECE 271 Dr. Serhiy Levkov
B
Topic 6 - 99
BJT Amplifier
Consider a BJT (Is 1015, F 100, VT  0.025) , biased in active region by dc
voltage source VBE = 0.7V.
From the simplified FAR model we get:
IB 
IS
F
VBE
VT
e
15μA  IC  F I 1.5mA  VCE 10  3.3K W IC 10  3.31.5  5V
NJIT ECE 271 Dr. Serhiy Levkov
B
Topic 6 - 100
BJT Amplifier
Consider a BJT (Is 1015, F 100, VT  0.025) , biased in active region by dc
voltage source VBE = 0.7V.
From the simplified FAR model we get:
IB 
IS
F
VBE
VT
e
15μA  IC  F I 1.5mA  VCE 10  3.3K W IC 10  3.31.5  5V
NJIT ECE 271 Dr. Serhiy Levkov
B
Topic 6 - 101
BJT Amplifier
Consider a BJT (Is 1015, F 100, VT  0.025) , biased in active region by dc
voltage source VBE = 0.7V.
From the simplified FAR model we get:
IB 
IS
F
VBE
VT
e
15μA  IC  F I 1.5mA  VCE 10  3.3K W IC 10  3.31.5  5V
B
Thus the Q-point is set at (IC, VCE) = (1.5 mA, 5 V)
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 102
BJT Amplifier
Now let’s inject the sin signal with the amplitude 8 mV by adding the voltage source vbe
The total base-emitter voltage becomes:
vBE VBE vbe

NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 103
BJT Amplifier
Now let’s inject the sin signal with the amplitude 8 mV by adding the voltage source vbe
The total base-emitter voltage becomes:
vBE VBE vbe
From the FAR model for VBEh = 0.708V
and VBEl = 0.692V we will get I Bh  20μA and I Bl 10μA

NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 104
BJT Amplifier
Now let’s inject the sin signal with the amplitude 8 mV by adding the voltage source vbe
The total base-emitter voltage becomes:
vBE VBE vbe
From the FAR model for VBEh = 0.708V
and VBEl = 0.692V we will get I Bh  20μA and I Bl  10μA

Thus 8 mV peak change in vBE :  5 A change in iB  and 0.5 mA change in iC.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 105
BJT Amplifier
Now let’s inject the sin signal with the amplitude 8 mV by adding the voltage source vbe
The total base-emitter voltage becomes:
vBE VBE vbe
From the FAR model for VBEh = 0.708V
and VBEl = 0.692V we will get I Bh  20μA and I Bl  10μA

Thus 8 mV peak change in vBE :  5 A change in iB  and 0.5 mA change in iC.
From the load line equation vCE 10  iC 3300 this will produce the 1.65V change in vCE .
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 106
BJT Amplifier
If changes in operating currents and voltages are small enough, then iC and vCE
waveforms are undistorted replicas of input signal.
Small voltage change at base causes large voltage change at collector. Voltage gain is
given by:
V
1.65180
Av  ce 
 206180 206
V
0.0080
be
Minus sign indicates 1800 phase shift between input and output signals.

NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 107
MOSFET Amplifier
Consider MOSFET biased in active region by dc voltage source VGS = 3.5 V, which
Will set the Q-point at (ID, VDS) = (1.56 mA, 4.8 V).
Then we introduce the signal vGS and total gate-source voltage becomes vGS VGS vgs
Doing similar analisis as for BJT, we get :

1 V change in vGS  1.25 mA change in iD  4 V change in vDS.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 108
Coupling and Bypass Capacitors
•
The constant voltage biasing is not a good method, 4 resistor is better.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 109
Coupling and Bypass Capacitors
•
•
The constant voltage biasing is not a good method, 4 resistor is better.
We need to introduce the input signal without disturbing the bias point.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 110
Coupling and Bypass Capacitors

•
•
•
•
The constant voltage biasing is not a good method, 4 resistor is better.
We need to introduce the input signal without disturbing the bias point.
AC coupling through capacitors is used to inject ac input signal and extract output
signal
Capacitors provide negligible impedance at frequencies of interest and provide open
circuits at dc.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 111
Coupling and Bypass Capacitors

•
•
•
•
The constant voltage biasing is not a good method, 4 resistor is better.
We need to introduce the input signal without disturbing the bias point.
AC coupling through capacitors is used to inject ac input signal and extract output
signal
Capacitors provide negligible impedance at frequencies of interest and provide open
circuits at dc.
C1 and C3 are large-valued coupling capacitors or dc blocking capacitors whose
reactance at the signal frequency is designed to be negligible.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 112
Coupling and Bypass Capacitors

•
•
•
•
The constant voltage biasing is not a good method, 4 resistor is better.
We need to introduce the input signal without disturbing the bias point.
AC coupling through capacitors is used to inject ac input signal and extract output
signal
Capacitors provide negligible impedance at frequencies of interest and provide open
circuits at dc.
C1 and C3 are large-valued coupling capacitors or dc blocking capacitors whose
reactance at the signal frequency is designed to be negligible.
C2 is a bypass capacitor that provides a low impedance path for ac current from emitter
to ground thereby removing RE from the circuit when ac signals are considered.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 113
DC and AC Analysis
• DC analysis:
– Make dc equivalent circuit by replacing all capacitors by open circuits
and inductors by short circuits.
– Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 114
DC and AC Analysis
• DC analysis:
– Make dc equivalent circuit by replacing all capacitors by open circuits
and inductors by short circuits.
– Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model.
• AC analysis:
– Make ac equivalent circuit by replacing all capacitors by short circuits,
inductors by open circuits, dc voltage sources by ground connections
and dc current sources by open circuits.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 115
DC and AC Analysis
• DC analysis:
– Make dc equivalent circuit by replacing all capacitors by open circuits
and inductors by short circuits.
– Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model.
• AC analysis:
– Make ac equivalent circuit by replacing all capacitors by short circuits,
inductors by open circuits, dc voltage sources by ground connections
and dc current sources by open circuits.
– Replace transistor by small-signal model
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 116
DC and AC Analysis
• DC analysis:
– Make dc equivalent circuit by replacing all capacitors by open circuits
and inductors by short circuits.
– Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model.
• AC analysis:
– Make ac equivalent circuit by replacing all capacitors by short circuits,
inductors by open circuits, dc voltage sources by ground connections
and dc current sources by open circuits.
– Replace transistor by small-signal model
– Use small-signal ac equivalent to analyze ac characteristics of amplifier.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 117
DC and AC Analysis
• DC analysis:
– Make dc equivalent circuit by replacing all capacitors by open circuits
and inductors by short circuits.
– Find Q-point from dc equivalent circuit by using appropriate largesignal transistor model.
• AC analysis:
– Make ac equivalent circuit by replacing all capacitors by short circuits,
inductors by open circuits, dc voltage sources by ground connections
and dc current sources by open circuits.
– Replace transistor by small-signal model
– Use small-signal ac equivalent to analyze ac characteristics of amplifier.
– Combine end results of dc and ac analysis to yield total voltages and
currents in the network.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 118
DC Equivalent for MOSFET Amplifier

Full circuit
dc equivalent
Make dc equivalent circuit by replacing all
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 119
DC Equivalent for MOSFET Amplifier

Full circuit
dc equivalent
Make dc equivalent circuit by replacing all
- capacitors by open circuits
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 120
DC Equivalent for MOSFET Amplifier

Full circuit
dc equivalent
Make dc equivalent circuit by replacing all
- capacitors by open circuits
- inductors by short circuits
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 121
DC Equivalent for MOSFET Amplifier
Full circuit
dc equivalent
Make dc equivalent circuit by replacing all
- capacitors by open circuits
- inductors by short circuits
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 122
AC Equivalent for MOSFET Amplifier

Full circuit
dc equivalent
Make ac equivalent circuit by replacing all
- capacitors by short circuits
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 123
AC Equivalent for MOSFET Amplifier

Full circuit
dc equivalent
Make ac equivalent circuit by replacing all
- capacitors by short circuits
- inductors by open circuits
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 124
AC Equivalent for MOSFET Amplifier

Full circuit
dc equivalent
Make ac equivalent circuit by replacing all
- capacitors by short circuits
- inductors by open circuits
-dc voltage sources by ground connections
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 125
AC Equivalent for MOSFET Amplifier

Full circuit
dc equivalent
Make ac equivalent circuit by replacing all
- capacitors by short circuits
- inductors by open circuits
-dc voltage sources by ground connections
-dc current sources by open circuits
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 126
AC Equivalent for MOSFET Amplifier

Full circuit
dc equivalent
Make ac equivalent circuit by replacing all
- capacitors by short circuits
- inductors by open circuits
-dc voltage sources by ground connections
-dc current sources by open circuits
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 127
AC Equivalent for MOSFET Amplifier

Full circuit
dc equivalent
Simplify the ac equivalent circuit
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 128
DC and AC Equivalents
for MOSFET Amplifier

Full circuit
dc equivalent
ac equivalent
NJIT ECE 271 Dr. Serhiy Levkov
Simplified ac equivalent
Topic 6 - 129
DC Equivalent for BJT Amplifier
All capacitors in original amplifier circuits need to be replaced by
open circuits, disconnecting vI , RI , and R3 from circuit.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 130
DC Equivalent for BJT Amplifier
All capacitors in original amplifier circuits need to be replaced by
open circuits, disconnecting vI , RI , and R3 from circuit.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 131
DC Equivalent for BJT Amplifier
All capacitors in original amplifier circuits are replaced by open
circuits, disconnecting vI , RI , and R3 from circuit.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 132
AC Equivalent for BJT Amplifier
Make ac equivalent circuit by replacing all
- capacitors by short circuits
- inductors by open circuits
- dc voltage sources by ground connections
- dc current sources by open circuits
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 133
AC Equivalent for BJT Amplifier
Simplify the ac equivalent circuit
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 134
AC Equivalent for BJT Amplifier
Simplify the ac equivalent circuit
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 135
AC Equivalent for BJT Amplifier
RB  R1 R2 160kW 300kW
RL  RC R3  22kW 100kW
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 136
Small-Signal Modeling
• For AC analysis (or general time varying analysis) we would need to use
phasor , time domain, Laplace Transform (or similar) methods.
• Those are working better with linear systems.
• Need to make the linear model – small signal model.
• Assume that the time varying components are small signals and
construct the two port model, which is linear.
• The concept of small signal is device dependent.
• We start from developing small signal model for a diode and then switch
to transistor
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 137
Small-Signal Operation of Diode
•
The slope of the diode characteristic at the Qpoint is called the diode conductance and is
given by:
ID and VD represent the DC bias point
Q-point, vD and iD are small changes
around Q-point
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 138
Small-Signal Operation of Diode
•
The slope of the diode characteristic at the Qpoint is called the diode conductance and is
given by:
gd 
iD
vD
v 
   v   I
   I S exp D 1   S exp  D 
V 
V
vD    VT   


 i I
I  v 
 S exp D 11  D S
VT   VT 
VT


gd 
•
Qp
T

T 
iD
iD

 40I D for iD  I S
VT 0.025V
gd is small but non-zero for ID = 0 because slope
of diode equation is nonzero at the origin.
ID and VD represent the DC bias point
Q-point, vD and iD are small changes
around Q-point
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 139
Small-Signal Operation of Diode
•
The slope of the diode characteristic at the Qpoint is called the diode conductance and is
given by:
gd 
iD
vD
v 
   v   I
   I S exp D 1   S exp  D 
V 
V
vD    VT   


 i I
I  v 
 S exp D 11  D S
VT   VT 
VT


gd 
•
•
Qp

T
T 
iD
iD

 40I D for iD  I S
VT 0.025V
gd is small but non-zero for ID = 0 because slope
of diode equation is nonzero at the origin.
Diode resistance is given by: rd  1
gd

ID and VD represent the DC bias point
Q-point, vD and iD are small changes
around Q-point
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 140
Small-Signal Operation of Diode
•
The slope of the diode characteristic at the Qpoint is called the diode conductance and is
given by:
gd 
iD
vD
v 
   v   I
   I S exp D 1   S exp  D 
V 
V
vD    VT   


 i I
I  v 
 S exp D 11  D S
VT   VT 
VT


gd 
•
•
•
ID and VD represent the DC bias point
Q-point, vD and iD are small changes
around Q-point
NJIT ECE 271 Dr. Serhiy Levkov
Qp

T
T 
iD
iD

 40I D for iD  I S
VT 0.025V
gd is small but non-zero for ID = 0 because slope
of diode equation is nonzero at the origin.
1
Diode resistance is given by: rd 
gd
Thus, linear model in the vicinity
of ID :

iD  I D  id , id  gd vd  small signal model
when vd 2VT 0.05V or vd 5 mV

Topic 6 - 141
Small-Signal Model of BJT
•
BJT is a three terminal device and to build
similar model to a diode, we would need to
use 2-port y-parameter network.
Using 2-port y-parameter network,
y21 
ib  y11vbe  y12vce
ic  y vbe  y22vce
21
y22 
The port variables can represent either time
varying
part of total voltages and currents or
small changes in them away from Q-point
values.
NJIT ECE 271 Dr. Serhiy Levkov
y12 

y11 
ib
v ce

v  0
be
ic
v be

v  0
ce
ic
v ce

v  0
be
ib
v be

v  0
ce
i B
vCE
0
Q  point
iC
v BE

IC
VA VCE
Q  point
i B
v BE
IC
VT
Q  point
iC
vCE


Q  point
IC
 oVT
Topic 6 - 142
Small-Signal Model of BJT
•
BJT is a three terminal device and to build
similar model to a diode, we would need to
use 2-port y-parameter network.
Using 2-port y-parameter network,
y21 
ib  y11vbe  y12vce
ic  y vbe  y22vce
21
y22 
The port variables can represent either time
varying
part of total voltages and currents or
small changes in them away from Q-point
values.
NJIT ECE 271 Dr. Serhiy Levkov
y12 

y11 
ib
v ce

v  0
be
ic
v be

v  0
ce
ic
v ce

v  0
be
ib
v be

v  0
ce
i B
vCE
0
Q  point
iC
v BE

IC
VA VCE
Q  point
i B
v BE
IC
VT
Q  point
iC
vCE


Q  point
IC
 oVT
Topic 6 - 143
Small Signal Hybrid-Pi Model of BJT
•
•
The hybrid-pi small-signal model is most
widely accepted model for BJT amplifier.
Small-signal parameters are controlled by the
Q-point
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 144
Small Signal Hybrid-Pi Model of BJT
•
•
•
The hybrid-pi small-signal model is most
widely accepted model for BJT amplifier.
Small-signal parameters are controlled by the
Q-point
VA - Early Voltage
Trans-conductance:
I
gm  1  C  40IC
y21 VT
Input resistance:
 V
r  1  o  o T
y11 g m
IC
Output resistance:
ro 
NJIT ECE 271 Dr. Serhiy Levkov
1 VA VCE VA


y22
IC
IC
Topic 6 - 145
Small Signal Hybrid-Pi Model of BJT
•
•
•
The hybrid-pi small-signal model is most
widely accepted model for BJT amplifier.
Small-signal parameters are controlled by the
Q-point
VA - Early Voltage
o is the small-signal commonemitter current gain of the BJT.
Trans-conductance:
I
gm  iC |Q  C  40IC
vBE
VT
Input resistance:
r 
o
gm

 oVT
IC
Output resistance:
1 VA VCE VA
ro 


y22
IC
IC
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 146
Small Signal Hybrid-Pi Model of BJT
•
•
•
The hybrid-pi small-signal model is most
widely accepted model for BJT amplifier.
Small-signal parameters are controlled by the
Q-point
VA - Early Voltage
o is the small-signal commonemitter current gain of the BJT.
Definition of small signal for BJT:
vbe  2VT or vbe  0.005V
NJIT ECE 271 Dr. Serhiy Levkov
Trans-conductance:
I
gm  iC |Q  C  40IC
vBE
VT
Input resistance:
r 
o
gm

 oVT
IC
Output resistance:
1 VA VCE VA
ro 


y22
IC
IC
Topic 6 - 147
Small-Signal Current Gain and
Voltage Gain of BJT
The important small signal parameter is
trans-conductance, or current-voltage
(iC/vBE ) gain. It shows how the collector
current changes in response to baseemitter voltage.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 148
Small-Signal Current Gain and
Voltage Gain of BJT
The important small signal parameter is
trans-conductance, or current-voltage
(iC/vBE ) gain. It shows how the collector
current changes in response to baseemitter voltage.
Other two important parameters are:
- small signal current gain
Small signal current gain
(iC/iB = iC/(vBE /r) - gain) :
o  gmr  
F



C



1  F 
1 I
 F iC 










Q  point 
In practice, however, F and o are often

assumed to be equal.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 149
Small-Signal Current Gain and
Voltage Gain of BJT
Intrinsic voltage gain (vCE/vBE – gain) :
The important small signal parameter is
trans-conductance, or current-voltage
(iC/vBE ) gain. It shows how the collector
current changes in response to baseemitter voltage.
m
F



C



1  F 
1 I
 F iC 










Q  point 
In practice, however, F and o are often

assumed to be equal.
NJIT ECE 271 Dr. Serhiy Levkov

F 
VA
VT
 40VA
F represents maximum voltage gain
Small signal current gain
(iC/iB = iC/(vBE /r) - gain) :
o  gmr  

I VA VCE  VA VCE
F  g ro 


V
I
VT


C
For V <<
CE
Other two important parameters are:
- small signal current gain
- intrinsic voltage gain


C 


T
VA,

individual
BJT can provide, doesn’t change
with operating point, and ranges from 1000
to 4000.
Small signal model for pnp-BJT is
similar
Topic 6 - 150
Small-Signal Model of MOSFET (nmos)
Using 2-port y-parameter network,
ig  y11vgs  y12vds
id  y21vgs  y22vds
The port variables can represent either

time-varying part of total voltages and
currents or small changes in them away
from Q-point values.
NJIT ECE 271 Dr. Serhiy Levkov
y11 
y12 
y21 
y22 
ig
v gs

v
ds
0
ig
v ds
gs
0
id
v gs

v
ds
0
id
v ds
vGS

v

v
gs
0
iG
0
Q  point
iG
v DS
0
Q  point
iD
vGS
2ID
VGS VTN

ID
Q  point
iD
v DS

Q  point
1

VDS
Topic 6 - 151
Small-Signal Model of MOSFET (nmos)
Using 2-port y-parameter network,
ig  y11vgs  y12vds
id  y21vgs  y22vds
The port variables can represent either

time-varying part of total voltages and
currents or small changes in them away
from Q-point values.
NJIT ECE 271 Dr. Serhiy Levkov
y11 
y12 
y21 
y22 
ig
v gs

v
ds
0
ig
v ds
gs
0
id
v gs

v
ds
0
id
v ds
vGS

v

v
gs
0
iG
0
Q  point
iG
v DS
0
Q  point
iD
vGS
2ID
VGS VTN

ID
Q  point
iD
v DS

Q  point
1

VDS
Topic 6 - 152
Small-Signal Model of MOSFET (nmos)
Trans-conductance:
•
•
•
Since gate is insulated from channel by
gate-oxide input resistance of transistor
is infinite.
Small-signal parameters are controlled
by the Q-point.
For same operating point, MOSFET
has higher transconductance and lower
output resistance that BJT.
Definition of small signal for MOSFET:
vgs  0.2VGS VTN 
NJIT ECE 271 Dr. Serhiy Levkov
gm  y21 
2I D
 2K n I D
VGS VTN
Output resistance:
ro 
1 1/  VDS
1


y22
ID
ID
Amplification factor (intrinsic voltage
gain) for VDS<<1:
2Kn
 f  gmro  1/  VDS  1
(VGS VTN )/2  I D
Where  – channel-length modulation
parameter.
Topic 6 - 153
Small-Signal Model of JFET
iG
1
 y11 
rg
vGS
gm  y21 
iD  I



DSS 


1
gm  2
2
vGS 
V
1 v DS 

 
 
P 
vDS  vGS VP

 v  

iG  ISG exp GS  1

V  

 T  
vGS
VT
Q  point
iD
ID
VGS VP
2

ID
Q  point
Q  point
 0 for IG  0

IDSS
(V V )
VP2 GS P
iD
1
 y22 
ro
v DS
for

I G  I SG
1

VDS


NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 154
Small-Signal Model of JFET
For small-signal operation, the input
signal limit is:
vgs 0.2VGS VP 

The amplification factor is given by:
Since JFET is normally operated
with gate junction reversebiased,
1
 f  gmro  2 
VDS
 2
VGS VP  VP
I ISG
rg 
G
IDSS
ID


NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 155
Summary of FET and BJT Small-Signal Models
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 156
Common-Emitter (C-E) Amplifier
•
Construct AC equivalent circuit assuming
Q-point is known.
Input is applied to Base
Output appears at Collector
Emitter is common (through RE) to
both input and output signal Common-Emitter (CE) Amplifier.
Developing a small signal model:
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 157
Common-Emitter (C-E) Amplifier
•
Construct AC equivalent circuit assuming
Q-point is known.
•
NJIT ECE 271 Dr. Serhiy Levkov
Use the small signal hybrid model for BJT
Topic 6 - 158
Common-Emitter (C-E) Amplifier
•
•
Simplify: RL  ro || RC || R3
NJIT ECE 271 Dr. Serhiy Levkov
Construct AC equivalent circuit assuming
Q-point is known.
•
Use the small signal hybrid model for BJT
Topic 6 - 159
Common-Emitter (C-E) Amplifier
•
•
Simplify: RL  ro || RC || R3
NJIT ECE 271 Dr. Serhiy Levkov
Construct AC equivalent circuit assuming
Q-point is known.
•
Use the small signal hybrid model for BJT
Topic 6 - 160
C-E Amplifier – Voltage Gain
•
Goal – develop the overall voltage gain
from vi to vo - AvCE .
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 161
C-E Amplifier – Voltage Gain
•
•
Goal – develop the overall voltage gain
from vi to vo - AvCE .
v
v
CE
o
A

 o
v
v
v
i  b
 v
 b
 v
 i
v
CE
where A
 o
vt
v
 b





v 
CE
 A
 b
vt

v 

 i 
- terminal gain -
gain between transistor terminals.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 162
C-E Amplifier – Voltage Gain
•
•
Goal – develop the overall voltage gain
from vi to vo - AvCE .
v
v
CE
o
A

 o
v
v
v
i  b
 v
 b
 v
 i
v
CE
where A
 o
vt
v
 b





v 
CE
 A
 b
vt

v 

 i 
- terminal gain -
gain between transistor terminals.
•
To find AvtCE , we connect a test source vb
to the base terminal (right circuit):
v   g R v or ACE   g R
o
m L b
vt
m L
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 163
C-E Amplifier – Voltage Gain
•
•
Goal – develop the overall voltage gain
from vi to vo - AvCE .
v
v
CE
o
A

 o
v
v
v
i  b
 v
 b
 v
 i
v
CE
where A
 o
vt
v
 b




•
For the input resistance RiB : RiB 
vb
r

ib

v 
CE
 A
 b
vt

v 

 i 
- terminal gain -
gain between transistor terminals.
•
To find AvtCE , we connect a test source vb
to the base terminal (right circuit):
v   g R v or ACE   g R
o
m L b
vt
m L
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 164
C-E Amplifier – Voltage Gain
•
•
Goal – develop the overall voltage gain
from vi to vo - AvCE .
v
v
CE
o
A

 o
v
v
v
i  b
 v
 b
 v
 i
v
CE
where A
 o
vt
v
 b




•
For the input resistance RiB : RiB 
Now, since vb  vi
RB || Rib
,
RI  RB || Rib
vb
r

ib

v 
CE
 A
 b
vt

v 

 i 
we have
- terminal gain -
v 
 RB || r 
AvCE  AvtCE  b    g m RL 

 vi 
 RI  RB || r 
gain between transistor terminals.
•
To find AvtCE , we connect a test source vb
to the base terminal (right circuit):
v   g R v or ACE   g R
o
m L b
vt
m L
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 165
C-E Amplifier – Voltage Gain
•
•
Goal – develop the overall voltage gain
from vi to vo - AvCE .
v
v
CE
o
A

 o
v
v
v
i  b
 v
 b
 v
 i
v
CE
where A
 o
vt
v
 b




Now, since vb  vi
RB || Rib
,
RI  RB || Rib
vb
r

ib
we have
- terminal gain -
v 
 RB || r 
AvCE  AvtCE  b    g m RL 

 vi 
 RI  RB || r 
•
To find AvtCE , we connect a test source vb
to the base terminal (right circuit):
v   g R v or ACE   g R
o
m L b
vt
m L
NJIT ECE 271 Dr. Serhiy Levkov
For the input resistance RiB : RiB 

v 
CE
 A
 b
vt

v 

 i 
gain between transistor terminals.
•
•
•
We see that the overall voltage gain is the
terminal gain reduced by the voltage
division between R1 and equivalent
resistance at the base.
Thus AvtCE is the upper limit of voltage
gain.
Topic 6 - 166
C-E Amplifier - Simplifications and Limits
•
If RI  RB r
we obtain the max gain:
AvCE  gm RL gm ro RC R3 

v
AvCE  AvtCE  b
 vi
NJIT ECE 271 Dr. Serhiy Levkov

 RB || r 


g
R


m L 

 RI  RB || r 
Topic 6 - 167
C-E Amplifier - Simplifications and Limits
AvCE  gm RL gm ro RC R3 
•
If RI  RB r
•
For maximum gain we set R3 >> RC and RC << rO   ro RC .R3   Rc . If we


assume IC RC = VCC with 0 <  < 1: 
I R
Av  Avt   gm RC   C C  40 VCC
VT
we obtain the max gain:
v
AvCE  AvtCE  b
 vi
NJIT ECE 271 Dr. Serhiy Levkov

 RB || r 
   g m RL 


 RI  RB || r 
Topic 6 - 168
C-E Amplifier - Simplifications and Limits
AvCE  gm RL gm ro RC R3 
•
If RI  RB r
•
For maximum gain we set R3 >> RC and RC << ro   ro RC .R3   Rc . If we


assume IC RC = VCC with 0 <  < 1: 
I R
Av  Avt   gm RC   C C  40 VCC
VT
Typically,  = 1/3, since common design allocates one-third power supply across
RC. To further account for other approximations leading to this result, we use:
we obtain the max gain:
Av  10VCC
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 169
C-E Amplifier - Simplifications and Limits
AvCE  gm RL gm ro RC R3 
•
If RI  RB r
•
For maximum gain we set R3 >> RC and RC << ro   ro RC .R3   Rc . If we


assume IC RC = VCC with 0 <  < 1: 
I R
Av  Avt   gm RC   C C  40 VCC
VT
Typically,  = 1/3, since common design allocates one-third power supply across
RC. To further account for other approximations leading to this result, we use:
we obtain the max gain:
Av  10VCC
•
•
Also, if the load resistor approaches ro (RC and R3 are very large), voltage gain is
limited by amplification factor, f , of BJT itself.
For large RE, voltage gain can be approximated as:
A
CE 
vt
gm R
R
L  L
1 gm R
R
E
E

NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 170
C-E Amplifier – Example
•
•
•
Problem: Find voltage gain, input
and output resistances.
Given data: F = 65, VA = 50 V
Assumptions: Active-region
operation, VBE = 0.7 V, small signal
operating conditions.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 171
C-E Amplifier – Example
We start from finding the Q-point.
DC equivalent circuit:
The KVL for the input loop :
105 IB VBE  (F 1)IB (1.6104 )  5  0

I B  3.71  A
IC  65I B  241  A
•
•
•
Problem: Find voltage gain, input
and output resistances.
Given data: F = 65, VA = 50 V
Assumptions: Active-region
operation, VBE = 0.7 V, small signal
operating conditions.
I E  66I B  245  A
The KVL for the output loop (blue):
5 104 IC VCE  (1.6104 ) I E  5  0

VCE  3.67 V
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 172
C-E Amplifier – Example
Next we construct the ac
equivalent and simplify it.
Rin  RB r  6.23 kW

gm  40IC  9.64103 S
oVT
r 
 6.64 kW
IC
VA VCE
ro 
 223 kW
IC

NJIT ECE 271 Dr. Serhiy Levkov
Rout  RC ro  9.57 kW
Av 
vo
vi



3 

 gm (Rout R )
Rin




in 
RI  R
 84.0

Topic 6 - 173
Common-Source Amplifier
•
•
NJIT ECE 271 Dr. Serhiy Levkov
Construct AC equivalent circuit.
Assume that Q-point is known.
Topic 6 - 174
Common-Source Amplifier
•
•
•
NJIT ECE 271 Dr. Serhiy Levkov
Construct AC equivalent circuit.
Assume that Q-point is known.
Use the small signal hybrid model for BJT
Topic 6 - 175
Common-Source Amplifier
•
•
•
Simplify: RL  ro || RC || R3
NJIT ECE 271 Dr. Serhiy Levkov
•
Construct AC equivalent circuit.
Assume that Q-point is known.
Use the small signal hybrid model for MOSFET
Topic 6 - 176
Common-Source Amplifier
•
•
•
Simplify: RL  ro || RC || R3
NJIT ECE 271 Dr. Serhiy Levkov
•
Construct AC equivalent circuit.
Assume that Q-point is known.
Use the small signal model
Topic 6 - 177
C-S Amplifier - Terminal Voltage Gain
Terminal voltage gain between gate
and drain:
v v
Avt  vd  v o  gm RL
g
gs
Overall voltage gain from source vi to
output voltage across R3

v 
vo  vo   vgs 
Av  v   v   v   Avt  vgs 
 gs 
i
 i 
 i 

Voltage division
NJIT ECE 271 Dr. Serhiy Levkov



L

RG
Av   gm R
RI  RG




Topic 6 - 178
C-S Amplifier - Simplifications
•
If we assume RI << RG
Av  Avt gm RL gm ro RD R3 
This implies that total signal voltage at input
appears across gate-source terminals.

•
Generally R3 >> RD and RD << ro. Hence, total load resistance on drain is RD.
For this case, common design allocates half the power supply for voltage drop
across RD and (VGS - VTN ) = 1V
Av  gm RD  
I D RD
 VDD
VGS VTN
2
•
Also, if load resistor approaches ro, (RD and R3 are very large), voltage gain is
limited by amplification factor, f of the MOSFET itself.
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 179
C-S Amplifier - Input and Output Resistance
•
Input resistance of C-S amplifier is
much larger than that of corresponding
C-E amplifier.
vx  ix RG
Rin  RG
•
For comparable
bias points, output

resistances of C-S and C-E amplifiers
are similar.
In this case, vgs= 0.
Rout 
NJIT ECE 271 Dr. Serhiy Levkov
vx
 RD ro  RD
ix
for ro  RD
Topic 6 - 180
C-S Amplifier - Example
Do example on the board
•
•
Problem: Find voltage gain, input
and output resistances.
Given data: Kn = 500 A/V2, VTN =
1V,  = 0.0167 V-1
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 181
C-S Amplifier - Example
Construct dc equivalent circuit
Since IG  0,
V
I1  DS 6 ,
510
KVL:
•
•
Problem: Find voltage gain, input
and output resistances.
Given data: Kn = 500 A/V2, VTN =
1V,  = 0.0167 V-1
NJIT ECE 271 Dr. Serhiy Levkov
10 VDS  2104 (I D  I1)  0
K
I D  n (0.4VDS VTN )2
2

VDS  5 V , VGS  2 V ,
I D  250  A
Topic 6 - 182
C-S Amplifier - Example
Next we construct the ac equivalent and simplify it.
gm  2K n I DS (1 VDS )  5.20104 S
Rin  RG1 RG2 1 MW


RI RG

ro 
1 VDS
I D
 260 kW
Rout  ro RD RG3 18.2 k W

Rin 
vo

Av    gm (Rout R3 ) 
  7.93
 RI  Rin 
vi
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 183
Summary of CE / CS Amplifiers
NJIT ECE 271 Dr. Serhiy Levkov
Topic 6 - 184