Transcript Chapter 13 Small-Signal Modeling and Linear Amplification

ELCN 201
Analog & Digital Electronics
Dr. Ahmed Nader
Dr. Ahmed Hussein
Fall 2013
4/12/2017
Faculty of Engineering Cairo
University
Grading
Final Exam
Term work
Quizzes
Project(s)
Mid-Term Exam
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60
20
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Text book:
Microelectronic Circuit Design by Richard C. Jaeger & Travis N. Blalock
Website: http://scholar.cu.edu.eg/anader/
Email: [email protected]
TA: Eng. Mazen Soliman
Office hours:
Sunday 4:00 – 5:00 pm
Monday 4:00 – 5:00 pm
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Introduction
Electronic Circuits can be divided into 2 main categories
1- Analog
 Operational Amplifiers
 Applications (Linear and Non-linear)
 Waveform generation (Oscillators), analog multiplier (Mixer),
phase detection (PLL)
2- Digital
 Logic gates (TTL, ECL, CMOS)
 Applications (Flip Flops, Counters, Memory, )
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Chap 13 - 3
Syllabus
Week
1
2
3
4
5
6
7
8
9
10
11
12
Lecture/Studio Topic
Small signal model (BJT + MOS)
Linear amplification
Single stage amplifiers (CE/CS,CB/CG,CC/CD)
Differential amplifiers
Multistage amplifiers
Frequency Response: transfer function
Short circuit time constant method
Miller effect and HF analysis
Current Sources
Advanced Current Sources
Digital logic
Digital logic
Digital logic
13
Digital logic
14
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Assignment
1
1
2
3
4
5
5
Lecture 1
Small-Signal Modeling and Linear
Amplification (Chapter 13)
Dr. Ahmed Nader
Adapted from presentation by Richard C. Jaeger
Travis N. Blalock
4/12/2017
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Chap 13 - 5
Chapter Goals
Understanding of concepts related to:
• Transistors as linear amplifiers
• dc and ac equivalent circuits
• Use of coupling and bypass capacitors and inductors to modify dc and
ac equivalent circuits
• Small-signal voltages and currents
• Small-signal models transistors (BJT and MOS)
• Amplifier characteristics such as voltage gain, input and output
resistances and linear signal range
• Identification of common-source and common-emitter amplifiers
• Rule-of-thumb estimates for voltage gain of common-emitter and
common-source amplifiers.
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Chap 13 - 6
Review: Operation Regions of Bipolar
Transistors
Base-Emitter
Junction
Base-Collector Junction
Reverse Bias
Forward Bias
Forward Bias
Forward-Active
Region
(Good Amplifier)
Saturation
Region
(Closed Switch)
Reverse Bias
Cutoff Region
(Open Switch)
Reverse-Active
Region
(Poor Amplifier)
Binary Logic
States
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Chap 13 - 7
i-v Characteristics of Bipolar Transistor:
Common-Emitter Output Characteristics
For iB = 0, transistor is cutoff. If iB > 0, iC also
increases.
For vCE > vBE, npn transistor is in forward-active
region, iC = bF iB is independent of vCE.
For vCE < vBE, transistor is in saturation.
For vCE < 0, roles of collector and emitter
reverse.
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Chap 13 - 8
i-v Characteristics of Bipolar Transistor:
Common-Emitter Transfer Characteristic
Defines relation between collector current
and base-emitter voltage of transistor.
Almost identical to transfer characteristic
of pn junction diode
Setting vBC = 0 in the collector-current
expression yields


S 




 BE 


 T 
v
iC  I exp
V





1
Collector
current expression has the same

form as that of the diode equation
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Chap 13 - 9
Common-Emitter Voltage
Transfer Characteristic
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Chap 13 - 10
NMOS Transistor: Saturation Region
•
•
•
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If vDS increases above triode region limit,
channel region disappears, also said to be
pinched-off.
Current saturates at constant value,
independent of vDS.
Saturation region operation mostly used for
analog amplification.
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Chap 13 - 11
NMOS Transistor: Saturation Region
(contd.)
K W 
2
n

i 
V 
v
D
TN 
 GS
2 L

v
v V
DSAT GS TN

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v v V
DS GS TN
for

is also called saturation or pinch-off
voltage
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Chap 13 - 12
Introduction to Amplifiers
• BJT is used as an amplifier when biased in the forward-active (active)
region
• FET can be used as amplifier if operated in the saturation (pinch-off)
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
• Q-point also determines
– Small-signal parameters of transistor
– Voltage gain, input resistance, output resistance
– Maximum input and output signal amplitudes
– Power consumption
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Chap 13 - 13
BJT Amplifier
BJT is biased in active region by dc voltage source VBE. Q-point is set at
(IC, VCE) = (1.5 mA, 5 V) with IB = 15 mA.
Total base-emitter voltage is:
vBE VBE vbe
Collector-emitter voltage is:
vCE 10iC RC

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
This is the load line equation.
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Chap 13 - 14
BJT Amplifier (cont.)
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:
Vce 1.65180
Av 

 206180206
V
0.0080
8 mV peak change in vBE gives 5 mA
be
change in iB and 0.5 mA change in iC.
Minus sign indicates 1800 phase shift
 V
0.5 mA change in iC produces a 1.65
between input and output signals.
change in vCE .
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Chap 13 - 15
MOSFET Amplifier
MOSFET is biased in active region by dc voltage source VGS. Q-point is set
at (ID, VDS) = (1.56 mA, 4.8 V) with VGS = 3.5 V.
Total gate-source voltage is:
vGS VGS vgs
1 V p-p change in vGS gives 1.25 mA p-p change in iD and 4 V p-p change

in vDS.
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Chap 13 - 16
Coupling and Bypass Capacitors
C1 and C3 are large-valued coupling
capacitors or dc blocking capacitors
whose reactance at the signal frequency
is designed to be negligible.
• AC coupling through capacitors is
used to inject ac input signal and
extract output signal without
disturbing Q-point
• Capacitors provide negligible
impedance at frequencies of interest
and provide open circuits at dc.
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C2 is a bypass capacitor that provides a
low impedance path for ac current from
emitter to ground, thereby removing RE
(required for good Q-point stability)
from the circuit when ac signals are
considered.
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Chap 13 - 17
dc and ac Analysis
• DC analysis:
– Find 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:
– Find 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.
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Chap 13 - 18
dc Equivalent for BJT Amplifier
• All capacitors in original amplifier circuits are replaced by open
circuits, disconnecting vI , RI , and R3 from circuit.
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Chap 13 - 19
ac Equivalent for BJT Amplifier
RB  R1 R2 10k 30k
R RC R3  4.3k100k

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Chap 13 - 20
DC and AC Equivalents
for MOSFET Amplifier
dc equivalent
Full circuit
ac equivalent
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Simplified ac equivalent
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Chap 13 - 21
Small-Signal Operation of Diode
• The slope of the diode characteristic at the
Q-point is called the diode conductance
and is given by:
gd 
gd 
iD
v D
ID
VT
Q  point


IS
VD  ID  IS

 exp 
VT
VT
VT 
ID
0.025V
 40ID for ID  IS
• gd is small but non-zero for ID = 0 because

slope of diode equation is nonzero at the
origin.
1
r

• Diode resistance is given by:
d
gd
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
Chap 13 - 22
Small-Signal Operation of Diode (cont.)

v  
V  v  

iD  I exp D 1
ID id  IS exp D d 1

V  
 V  

 T  
 T  



V  



2

3

v
v
v
v


1
1
ID id  IS exp D 1 IS exp D  d   d    d  ...

V  
V 2 VT  6 VT 

VT 

 T

 T  




S 


Subtracting ID from both sides of the equation,



d
S 
 T

2
3

v 1 vd  1 vd 

id (ID  I )       ...
V 2 VT  6 VT 


For id to be a linear function of signal voltage vd , vd  2VT  0.05V or vd  5 mV
 requirement for small-signal operation of the diode.
This represents the


 d 
S 

 T 
id (ID  I )
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
v
= gdvd  iD  ID  gdvd
V
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Chap 13 - 23
Current-Controlled Attenuator
Magnitude of ac voltage vo developed
across diode can be controlled by value
of dc bias current applied to diode.
From ac equivalent circuit,
From dc equivalent circuit ID = I,
For RI = 1 k, IS = 10-15 A,
r
1
 vi
r R
R
1 I
rd
1
vo  v
i (I  I )R
S
I
1
VT




d
i 

 d
I 
vo  v

If I = 0, vo = vi, magnitude of vi is
limited to only 5 mV.
If I = 100 mA, input signal is
attenuated by a factor of 5, and vi
can have a magnitude of 25 mV.
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Chap 13 - 24
Small-Signal Model of BJT
y12 
y21 
Using 2-port y-parameter network,
ib  y11vbe  y12vce
ic  y vbe  y22vce
y22 
21
The port variables can represent either
time-varying part of total voltages and
currents or small changes in them away
from Q-point values.

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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
b oVT
bo is the small-signal commonemitter current gain of the BJT.
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Chap 13 - 25
Hybrid-Pi Model of BJT
Transconductance:
IC
gm  y21   40IC
VT
Input resistance:

r 
• The hybrid-pi small-signal
model is the intrinsic
representation of the BJT.
• Small-signal parameters are
controlled by the Q-point and
are independent of geometry of
the BJT
1 b oVT b o


y11 I C
gm
Output resistance:
1 VA VCE VA
ro 


y22
IC
IC

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Chap 13 - 26
Small-Signal Current Gain and
Amplification Factor of BJT
Amplification factor is given by:
bo  gmr  





1 I

bF



C 

 F


F 


C Q  point
1 b
b i
bo > bF for iC < IM , and bo < bF
for iC > IM , however, bF and bo







I VA VCE  VA VCE
mF  g ro 


V
I
VT


C
For VCE << VA,
VA
mF   40VA
VT
m



C 


T 
mF represents
maximum voltage gain

individual BJT can provide and doesn’t
change with operating point.
are assumed to be equal.
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Chap 13 - 27
BJT Small Signal Parameters
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Chap 13 - 28
Equivalent Forms of Small-Signal Model
for BJT
• Voltage -controlled current source gmvbe can be transformed into
current-controlled current source,
vbe ibr
gmvbe  gmibr  boib
v
ic  boib  ce  boib
ro
• Basic relationship ic = bib is useful in both dc and ac analysis when
BJT is in forward-active region.

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Chap 13 - 29
Small-Signal Operation of BJT
v 
iC  I exp BE 
V 
 T 



S 




C 




 BE 


 T 
V
iC  IC ic  IS exp
V
2


 be 


 T 
exp
v
V

3
vbe 1 vbe  1 vbe 

IC ic  I 1       ...
VT 2 VT  6 VT 




be
C 
 T

2
3


v
vbe  1 vbe 

1

ic  iC  IC  I
      ...
V 2 VT  6 VT 


For linearity, ic should be proportional to vbe with vbe 2VT or vbe 0.005V


vbe 
I

ic  IC 1  IC  C vbe  IC  gmvbe
VT 
VT


Change in ic that corresponds to small-signal operation is:
ic gm
vbe 0.005

 vbe  
 0.200
IC IC
VT 0.025
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
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Chap 13 - 30
Small-Signal Model for pnp BJT
• For pnp transistor
iB  IB -ib
iC  IC -ic  bF IB  bF ib
• Signal
current injected into base

causes decrease in total collector
current which is equivalent to
increase in signal current entering
collector.
• Identical to that of npn transistor
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Chap 13 - 31
Small-Signal Analysis of Complete C-E
Amplifier: ac Equivalent
• Ac equivalent circuit is
constructed by assuming that all
capacitances have zero
impedance at signal frequency
and dc voltage sources are ac
ground.
• Assume that Q-point is already
known.
RB  R1 R2

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Chap13 - 32
Small-Signal Analysis of Complete C-E
Amplifier: Small-Signal Equivalent
Input applied to Base
Output appears at Collector
Emitter is common (through
RE) to both input and output
signal - Common-Emitter
(CE) Amplifier.
RL  RC R3
AvtCE is the terminal voltage
gain of the CE amplifier.
v v v
v
AvtCE  vo  vo vb  AvtCE vb
i
b
i
i


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Chap 13 - 33
Common-Emitter (CE): Terminal
Voltage Gain
Solving for ib and substituting,
bo R
o
L
Avt   
v
b o /gm (b o 1)R
E
b
v
For bo 1 and bo  gmr
Using alternate small-signal model form
and test source vb to drive the base
terminal of the transistor, neglecting ro,
g R
m L


ACE
vt
1 g R
m E
v o  b oib RL
v b  ib r  (ib  b oib )RE
 ib [r  (bo  1)RE ]
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
What is the current gain?
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Chap 14 - 34
Common-Emitter (CE): Input Resistance
and Signal Source Voltage Gain
(bo 1)RE is the impedance in
the emitter side of the transistor
reflected to the base side.

Rewriting the previous equation, we can
find the impedance looking into the base
terminal:
vb
Rib   r  (b o  1)RE
ib
Combining equations, the overall
voltage gain can be written as:
v b 
A  A  
ib 
 gm RL  RB || Rib 
 


1 gm RE RI  RB || Rib 
CE
v
 r (1 gm RE )
assuming b o  1 and b o  gm r
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vb
RB || Rib

ib RI  RB || Rib
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CE
vt
Chap 14 - 35
C-E Amplifier Voltage Gain Example with RE = 0
• Problem: Calculate voltage gain
• Given data: bF = 100, VA = 75 V, Q-point is (1.45 mA, 3.41 V),
R1 = 10 k, R2 = 30 k, R3 = 100 k, RC = 4.3 k,RI = 1k.
• Assumptions: Transistor is in active region, bO = bF. Signals are low
enough to be considered small signals.
• Analysis: gm  40IC  40(1.45mA) 58.0 mS RB  R1 R2  7.5k
RL  ro RC R3  3.83 k
ro 

VA VCE
IC
b V 100(0.025V )
r  o T 
1.72 k
IC
1.45mA







75V  3.14V
 54.1 k
1.45mA



L 


Av  gm R
RB r





 

RI  RB r

 130 or 42.3 dB

R (RB r )
 8.57 mV
vi  (0.005V ) I

RB r


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


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Chap 13 - 36
Small-Signal Model Simplification
•
For max gain RI  RB r and RE  0,
AvCEgm RL gm ro RC R3 
•  For maximum gain we set R3 >> RC and load resistor RC << ro. 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:

•
•
Av 10VCC
Also, if the load resistor approaches ro (RC and R3 infinite), voltage gain is
limited by amplification factor, mf of BJT itself.
gm R
R
CE
L
For large RE, voltage gain can be aproximated as:
Avt 
 L
1 gm R
R
E
E
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Chap 13 - 37
38
General Concept
Amplifier
Amplifier
4/12/2017
Amplifier
© Ahmed Nader, 2013
C-E Amplifier Output Resistance
•
Output resistance is the total equivalent resistance
looking into the output of the amplifier at coupling
capacitor C3. Input source is set to 0 and a test
source vx is applied at output.
vx  vr  ve  (ix  b i)ro  ve
0
ve  ix [(R  r )|| R ]
E
th
R
E
i  ix
(current division)
R  r  R
E
th


b R


vx
(R  r )|| R
0 E
R =
 r 1
ic i
0 R  r  R 
E
th 
x



E
th


b R



 r [1 g (R ||r )]
0
E
 r 1
m E 
0 R  r  R  0

E 
th 

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R  m r  b r
ic
00
f
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(if R >> r )
E
Chap 13 - 39
Sample Analysis of C-E Amplifier
Analysis: To find the Q-point, dc
equivalent circuit is constructed.
105 IB VBE (bF 1)IB(1.6104 ) 5

IB  3.71 mA
IC  65IB  241 mA
IE  66IB  245 mA
• Problem: Find voltage gain, input
and output resistances.
• Given data: bF = 65, VA = 50 V

• Assumptions: Active-region
5104 IC VCE (1.6104 )IE (5) 0
operation, VBE = 0.7 V, small
signal operating conditions.
VCE  3.67 V
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Chap 13 - 40
Sample Analysis of C-E Amplifier
(cont.)
Next we construct the ac
equivalent and simplify it.
Rin  RB r  6.23 k

gm  40IC  9.64103 S
boVT
r 
 6.64 k
IC
VA VCE
ro 
 223 k
IC

4/12/2017
Rout  RC ro  9.57 k
Av 

vo
vi



3 


 gm (Rout R )
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Rin





in 
RI  R
 84.0
Chap 13 - 41
Small-Signal Model for the MOSFET
y11 
y12 
Using 2-port y-parameter network,
ig  y11vgs  y12vds
id  y21vgs  y22vds
y21 
The port variables can represent either

time-varying
part of total voltages and
currents or small changes in them away
from Q-point values.
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

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Chap 13 - 42
Small-Signal Parameters of MOSFET
Transconductance:
gm  y21 
• 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 lower
transconductance and lower output
resistance that BJT.

2I D
VGS VTN
 2K n I D
Output resistance:
ro 
1 1 VDS 1


y22
I D
I D
Amplification factor for VDS<<1:

m f  gmro 
1 VDS
I D
2K n
 ID
1

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Chap 13 - 43
Small-Signal Operation of MOSFET
K n 
2


v
V
for vDS  vGS VTN | Letting vGS VGS  vgs :
2  GS TN 
K 
2
2 

iD  ID id  n VGS VTN   2v gs VGS VTN  v gs

2 

K 
2 
id  n 2v gs VGS VTN  v gs


2 
iD 
For linearity, id should be proportional to vgs:

vgs  0.2VGS VTN 
Since the MOSFET can be biased with (VGS - VTN) equal to several volts, it can

handle much larger values of vgs than corresponding
the values of vbe for the BJT.
Change in drain current that corresponds to small-signal operation is:
id
ID
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

gm
ID
vgs 
0.2(VGS VTN )
 0.4
VGS VTN
2
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Chap 13 - 44
Body Effect in Four-terminal MOSFET
Drain current depends on threshold voltage which in
turn depends on vSB. Back-gate transconductance
is:
i
i
gmb  D
 D
v BS
v SB
Q  point
Q  point
 i V 
gmb  D  TN 
(gm) gm
V v 
 TN  SB Q  point

0 <  < 1 is called back-gate
tranconductance parameter.
Bulk terminal is a reverse-biased diode. Hence, no
conductance from bulk terminal to other terminals.
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Chap 13 - 45
Small-Signal Model for PMOS
Transistor
• For PMOS transistor
vSG VGG - vgg
iD  ID -id
• Positive signal voltage vgg reduces sourcegate voltage of the PMOS transistor
causing decrease in total current exiting
drain, equivalent to an increase in the
signal current entering the drain.
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Chap 13 - 46
Summary of FET and BJT Small-Signal
Models
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Chap 13 - 47
Small-Signal Analysis of Complete C-S
Amplifier: ac Equivalent
• ac equivalent circuit is
constructed by assuming that all
capacitances have zero
impedance at signal frequency
and dc voltage sources represent
ac grounds.
• Assume that Q-point is already
known.
RG  R1 R2

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Chap 13 - 48
Small-Signal Analysis of Complete CS
Amplifier: Small-Signal Equivalent
Noting similarity to CE case,
Terminal voltage gain between
gate and drain is found as:
vd vo gm RL
A v v 
g
g 1 g R
m E
CS
vt

With r infinite, RiG is also infinite,
therefore overall voltage gain from
source vi to output voltage across
RL is:
 v 
v 
v
v
Av  vo  vo vg  AvtCS vg 
i
 g  i 


G

m L 

 I
G
Av  g R
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R
R R
 i 






Chap 13 - 49
C-S Amplifier Voltage Gain: Example
• Problem: Calculate voltage gain
• Given data: Kn = 0.5 mA/V2, VTN = 1V, = 0.0133 V-1, Q-point is
(1.45 mA, 3.86 V), R1 = 430 k, R2 = 560 k, R3 = 100 k, RD = 4.3
k,RI = 1 k.
• Assumptions: Transistor is in active region. Signals are low enough to
be considered small signals.
R  R R  243kΩ
• Analysis: gm  2K I (1 V ) 1.23mS
G
1 2
n DS
DS
1
V
ro   DS  54.5kΩ RL  ro RD R3  3.83kΩ
I
D


2I
R




D  0.48V
G


  4.69  13.4dB
v  0.2V V   0.2
Av   gm R 
i
TN 
L  R  R 
 GS
Kn
 I
G 
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Chap 13 - 50
Small-Signal Model Simplification
•
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 load resistor << 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 infinite), voltage gain is limited
by amplification
factor, mf of the MOSFET itself.

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Chap13 - 51
C-S Amplifier Input Resistance
• Input resistance of C-S amplifier is
much larger than that of
corresponding C-E amplifier.
vx  ix RG
Rin  RG

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Chap 13 - 52
C-S Amplifier Output Resistance
• Output resistance is calculated in a
manner similar to that of CE amplifier
with r infinite.
RiD  ro (1 gm RE )
Rout  RD || RiD
 RD || ro (1 gm RE )

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Chap 13 - 53
Sample Analysis of C-S Amplifier
Analysis: dc equivalent circuit is
constructed and analyzed
Since IG  0,
V
I1  DS 6
510
V 10 2104 (ID  I1)
K
ID  n (0.4VDS VTN )2
2
VDS  5 V
VGS  2 V ID  250 mA
 DS
•
•
Problem: Find voltage gain, input
and output resistances.
Given data: Kn = 500 mA/V2, VTN =
1V, = 0.0167 V-1
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Chap 13 - 54
Sample Analysis of C-S Amplifier
(cont.)
Next we construct the ac
equivalent and simplify it.
Rin  RG1 RG2 1 M

gm  2K n I DS (1 VDS )  5.20104 S
ro 

1 VDS
I D
Rout  ro RD RG3 18.2 k
 260 k
Av 
vo
vi



3 

 I
gm (Rout R )
Rin





in 
R R
 7.93

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Chap 13 - 55
Summary CE and CS Characteristics
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Chap 13 - 56
Signal Range Constraints
Minimum output voltage set by active
region constraints of Q, maximum set by
drop across RC. For a sine wave with peak
VM, we can express the limits as:
4/12/2017
VM  min[ IC RC ,(VCE VBE )]
BJT
VM  min[ ID RD ,(VDS  (VGS VTN ))]
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FET
Chap 13 - 57