Jaeger, Chapter 5 “BJT “
Jaeger Chapter 13 “Small Signal Analsys”
Jeff Davis
ECE3040
1
References
• Prof. Alan Doolittle’s Notes
– users.ece.gatech.edu/~alan/index_files/ECE3040index.htm
• Prof. Farrokh Ayazi’s Notes
– users.ece.gatech.edu/~ayazi/ece3040/
• Figures for Require Textbooks
– (Pierret and Jaeger)
2
Quick Circuit Review
3
Thevenin’s Equivalent Circuits
A
Configuration of resistors,
Voltage sources,
or current sources
Rth
A
B
+
-
Vth
B
ACCORDING TO
THEVENIN THIS
CAN BE REPLACED
WITH ….
4
Thevenin’s Theorem
R1 = 10k
R3 = 1k
Rth
A
+
-
R2 = 5k
+
-
Vdd = 12V
VOC
B
1) Thevenin voltage is the open circuit voltage seen at terminals AB
2)Thevenin resistance is equivalent resistance seen by AB with
all voltage sources shorted and all current sources replaced by
open circuits
5
Thevenin’s Theorem
1) Thevenin voltage is the open circuit voltage seen at terminals AB
R1 = 10k
R3 = 1k
A
+
-
R2 = 5k
Current is zero
here!
Vdd = 12V
B
R1 = 10k
R3 = 1k
A
+
-
Output is
voltage across
R2
R2 = 5k
Vdd = 12V
B
6
Thevenin’s Theorem
1) Thevenin voltage is the open circuit voltage seen at terminals AB
R1 = 10k
R3 = 1k
A
+
-
Output is
voltage across
R2
R2 = 5k
Vdd = 12V
B
VOC
R2
= Vdd
R1 + R2
VOC
5k
= 12
= 4V
15k
7
Thevenin’s Theorem
2)Thevenin resistance is equivalent resistance seen by AB with
all voltage sources shorted and all current sources replaced by
open circuits
R1 = 10k
R3 = 1k
Rth
A
+
-
R2 = 5k
+
-
Vdd = 12V
VOC
B
8
Thevenin’s Theorem
2)Thevenin resistance is equivalent resistance seen by AB with
all voltage sources shorted and all current sources replaced by
open circuits
R1 = 10k
R3 = 1k
A
R2 = 5k
B
Rth = R3 + R1 || R2
R1 R2
Rth = R3 +
R1 + R2
10k * 5k
50e6
Rth = 1k +
= 1k +
10k + 5k
15e3
= 1e3 + 3.33e3 = 4.33k
9
Thevenin’s Theorem
R1 = 10k
R3 = 1k
Rth =4.33k
A
A
+
-
+
-
R2 = 5k
Vdd = 12V
B
VOC=4V
B
10
Norton Equivalent Circuits
A
Configuration of resistors,
Voltage sources,
or current sources
A
B
Rth
Voc/Rth
B
ACCORDING TO
NORTON THIS CAN
BE REPLACED WITH
….
11
Norton Equivalent Circuit
R1 = 10k
R3 = 1k
Rth =4.33k
A
+
-
R2 = 5k
A
+
-
Vdd = 12V
VOC=4V
B
B
A
Rth
Rth =4.33k
Voc/Rth
B
I=0.923e-3 A
12
Back to Current Equations for BJT
13
Large Signal Model of a BJT (Ebers-Moll Model)
PNP
& VCB VT #
& VEB VT #
I E = I F 0 $$ e
' 1!! ' ( R I R 0 $$ e
' 1!!
%
"
%
"
& VCB VT #
& VEB VT #
I C = ( F I F 0 $$ e
' 1!! ' I R 0 $$ e
' 1!!
%
"
%
"
NPN
" VBE VT
%
" VBC VT
%
IE = IF0 $ e
! 1' ! ( R I R0 $ e
! 1'
#
&
#
&
" VBE VT
%
" VBC VT
%
IC = ( F I F 0 $ e
! 1' ! I R0 $ e
! 1'
#
&
#
&
14
Further Model Simplifications
(useful for circuit analysis)
# VVEB
&
# VVCB
&
I C = ! F I F 0 % e T " 1( " I R0 % e T " 1(
$
'
$
'
Assume in the forward active mode
# VVEB
&
I C = ! F I F 0 % e T " 1( " I R0
$
'
Neglecting small terms…
IC = ! F I F 0e
Change in notation…
IC = IS e
VEB
VT
VEB
VT
I S = ! F I F 0 = "transistor saturation current"
15
Simplified Forward-Active Region Model
(NPN)
!
#
#
#"
$
v
i = I exp BE &&
S
C
V &%
T
!v
$
IS
i =
exp ## BE &&
E '
#" V
&%
T
F
!v
$
IS
i =
exp ## BE &&
B (
#" V
&%
T
F
i =! i
C
FE
i =" i
C
FB
iB = iE # iC = iE # "FiB
i = (" +1)i
E
B
F
Fundamental forward active behavior suggests a voltage- controlled
current source.
16
Simplified Forward-Active Region Model
(PNP)
!
#
#
#"
$
v
i = I exp EB &&
S
C
V &%
T
!v
$
IS
i =
exp ## EB &&
E '
#" V
&%
T
F
!v
$
IS
i =
exp ## EB &&
B (
#" V
&%
T
F
i =! i
C
FE
i =" i
C
FB
iB = iE # iC = iE # "FiB
i = (" +1)i
E
B
F
Fundamental forward active behavior suggests a voltage- controlled
current source.
17
Modeling the “Early Effect” (non-zero
slopes in IV curves)
IC
iB1< iB2< iB3
Early voltage!
i B3
al)
u
t
c
(a
)
ctual
a
(
i B2
ual)
i (act
B1
VA
iB3 (theory)
iB2 (theory)
iB1 (theory)
VCE
•Base width changes due to changes in the basecollector depletion width with changes in VCB.
•This changes αT, which changes IC, αDC and BF
18
Early Effect and Early Voltage
•
•
•
•
As reverse-bias across collector-base junction increases, width of collector-base
depletion layer increases and width of base decreases (base-width modulation).
In practical BJT, output characteristics have a positive slope in forward-active region,
collector current in not independent of vCE.
Early effect: When output characteristics are extrapolated back to point of zero iC,
curves intersect at common point vCE = -VA (Early voltage) which lies between 15 V
and 150 V.
Simplified equations (including Early effect):
NPN
PNP
'
!v
$*'
v
BE
, )1 + CE
i = I ) exp #
&
C
S)
#" V &% , )
V
T +(
A
(
"
v
$1 + CE
! =!
F
FO $
V
#
A
i
I
C
S
i =
=
B !
!
F
FO
*
,
,+
%
'
'&
(
"v
%+
BE
* exp $
'$# V '& *
T ,
)
'
!v
$*'
*
v
EB
EC
,
i = I ) exp #
& , )1 +
C
S)
#" V &% , )
V ,
T +(
A +
(
"
%
v
EC
$1 +
'
! =!
F
FO $
V '
#
A &
i
I
C
S
i =
=
B !
!
F
FO
(
"v
%+
EB
* exp $
'$# V '& *
T ,
)
19
Biasing for BJT
•
•
•
•
•
Goal of biasing is to establish known Q-point which in turn
establishes initial operating region of transistor.
In BJT, Q-point is represented by (IC, VCE) for npn transistor or (IC,
VEC) for pnp transistor.
Q-point controls values of diffusion capacitance,
transconductance, input and output resistances.
In general, during circuit analysis, we use simplified mathematical
relationships derived for specified operation region and Early
voltage is assumed to be infinite.
The practical biasing circuits used for BJT are:
– Four-Resistor Bias network
– Two-Resistor Bias network
•
BEFORE WE LOOK AT PRACTICAL LET’S LOOK AT EASY!
20
Simplified Forward-Active Region Model
(Example 1)
• Problem: Estimate terminal currents and base-emitter voltage
• Given data: IS =10-16 A, αF = 0.95, VBC =VB - VC= -5 V, IE =100 µA
• Assumptions: Simplified transport model assumptions, room
temperature operation, VT = 25.0 mV
• Analysis: Current source forward-biases base-emitter diode, VBE > 0,
VBC < 0, we know that transistor is in forward-active operation region.
21
Simplified Forward-Active Region Model
(Example 1)
I = ! I = 0.95 "100µA = 95µA
C
F E
"F
! =
F 1#"
I
We know IE!!
%
v
'
BE
i =
exp
E !
V ''&
T
F
IS
"
$
$
$#
B
=
=
F
0.95
=19
1#0.95
100µA
= 5 µA
! F +1
20
IE
=
! I
V
=V ln F E = 0.69V
BE
T
I
S
22
Simplified Forward-Active Region Model
(Example 2)
• Problem: Estimate terminal currents, base-emitter and base-collector
voltages.
• Given data: IS =10-16 A, αF = 0.95, VC= +5 V, IB =100 µA
• Assumptions: Simplified transport model assumptions, room
temperature operation, VT = 25.0 mV
• Analysis: Current source causes base current to forward-bias baseemitter diode, VBE > 0, VBC <0, we know that transistor is in forward-active
operation region.
23
Simplified Forward-Active Region Model
(Example 2)
Common emitter gain known from
example 1!
I = ! I =19 "100 µA =1.90mA
C
F B
I
We know the
base current!
E
!
#
#
#"
$
v
&
BE
i = I exp
S
C
V &&%
T
= (! +1)I = 20 "100µA = 2.00mA
B
F
V
V
BC
I
=V ln C = 0.764V
BE
T I
S
=V !V =V
!V = !4.24V
B
BE
C
C
24
What about this type of Circuit?
vBE
VT
IS
iE =
e
!F
vBE + iE R + "VEE = 0
We could solve
numerically!!! OR…
25
Simplified Circuit Model for Forward-Active
Region
• Current in base-emitter diode is amplified by commonemitter current gain βF and appears at collector,base and
collector currents are exponentially related to baseemitter voltage.
26
Simplified Circuit Model for Forward-Active
Region
•
•
Base-emitter diode is replaced by constant voltage drop model(VBE =
0.7 V) since it is forward-biased in forward-active region.
Dc base and emitter voltages differ by 0.7 V diode voltage drop in
forward-active region.
27
Simplified Forward-Active Region Model
(Example 3)
• Problem: Find Q-point
• Given data: βF = 50, VBC =VB - VC= -9 V
• Assumptions: Forward-active region of operation, VBE = 0.7 V
28
Simplified Forward-Active Region Model
(Example 3)
29
Simplified Forward-Active Region Model
(Example 3)
V
+ 8200I ! V
=0
BE
E
EE
8.3V
I =
= 1.01mA
E
8200!
Now we know a current!
!VBE
I
E = 1.02mA = 19.8 µ A
B ! +1
51
F
I = ! I = 0.990mA
C
F B
V
=V
" ("V
) = 9 + 0.7 = 9.7V
CE
CC
BE
I
=
30
Four-Resistor Bias Network for BJT
What is the Q-point (IC, VCE)?
Chap 5 - 31
31
Four-Resistor Bias Network for BJT
Use Thevenin equivalent circuit to simplify!
12V
+
-
+
-
12V
Chap 5 - 32
32
Four-Resistor Bias Network for BJT
Use Thevenin equivalent circuit to simplify!
RR
36 *18 *10 6
1
2
R
= R1 || R2 =
=
= 12k
TH
R +R
54 *10 3
1
2
R
18
1
V
=V
= 12
= 4V
OC
CC R + R
54
1
2
12V
+
-
Rth =12k
A
A
+
-
B
VOC=4V
B
Chap 5 - 33
33
Four-Resistor Bias Network for BJT
If we solve for the base current, then the hard work is done!
4 = 12k * I
B
+V
+R I
BE
E E
Use constant voltage drop model!
4 = 12, 000I + 0.7 + 16, 000(! + 1)I
B
F
B
4V - 0.7V
"I =
= 2.68 µA
B 1.23 # 10 6 $
I
= ! I = 201µA
C
F B
I = (! +1)I = 204 µA
E
B
F
Chap 5 - 34
34
Four-Resistor Bias Network for BJT
Solve for the VCE!
I
C
= ! I = 201µA
F B
I = (! +1)I = 204 µA
E
B
F
V
=V
!R I !R I
CE
CC
C C
E E
#
R &
=V
! % R + F ( I = 4.32V
CC % C " ( C
$
F'
Q-point is (201 µA, 4.32 V)
35
Two-Resistor Bias Network for BJT: Example
• Problem: Find Q-point for pnp transistor in 2-resistor bias circuit with
given parameters.
• Given data: βF = 50, VCC = 9 V
• Assumptions: Forward-active operation region, VEB = 0.7 V
36
Two-Resistor Bias Network for BJT: Example
Find the base current!
9 =V
+18,000I +1000(I + I )
EB
B
C
B
IC = !F I B
9 =V
+18,000I +1000(51)I
EB
B
B
I
B
=
9V!0.7V
=120µA
69,000"
37
Two-Resistor Bias Network for BJT: Example
Find the Q-Point current!
I
I
B
C
=
9V!0.7V
=120µA
69,000"
= 50I
B
= 6.01mA
V
= 9 ! 1000(I + I ) = 2.88V
EC
C
B
Q-point is : (6.01 mA, 2.88 V)
38
BJT Amplifier
Amplification!
39
BJT Amplifier/small Signal
40
Small Signal Model of a BJT
•Just as we did with a p-n diode, we can break the BJT up into
a large signal analysis and a small signal analysis and
“linearize” the non-linear behavior of the Ebers-Moll model.
•Small signal Models are only useful for Forward active mode
and thus, are derived under this condition.
•Small signal models are used to determine amplifier
characteristics (Example: “Gain” = Increase in the magnitude
of a signal at the output of a circuit relative to it’s magnitude at
the input of the circuit).
•Warning: Just like when a diode voltage exceeds a certain
value, the non-linear behavior of the diode leads to distortion
of the current/voltage curves (see previous lecture), if the
inputs/outputs exceed certain limits, the full Ebers-Moll model
must be used.
41
Small Signal Model for BJT… What is a linear
Model for a 3-terminal device?
2-terminal Small Signal Model for Diode!
+
vd
-
1
rd =
gd
id = gd vd
But what about a BJT is 3-terminal device?
42
Linear relationship in common emitter
Configuration
ib=y11vbe + y12vce
ic=y21vbe + y22vce
43
Consider the BJT as a two-port Network
Two Port Network
General “y-parameter” Network
BJT “y-parameter” Network
i1=y11v1 + y12v2
ib=y11vbe + y12vce
i2=y21v1 + y22v2
ic=y21vbe + y22vce
44
Equivalent circuit for two port network admittance Matrix
Let’s just look at what the circuit might look like
for some particular values of v1 and v2!
i1=y11v1 + y12v2
i2=y21v1 + y22v2
If.. v2 = 0
i1 = y11v1
i2 = y21v1
45
Equivalent circuit for two port network admittance Matrix
Let’s just look at what the circuit might look like
for some particular values of v1 and v2!
i1=y11v1 + y12v2
i2=y21v1 + y22v2
If..
i1 = y12 v2
v1 = 0
i2 = y22 v2
46
Equivalent circuit for two port network admittance Matrix
Superimposing both conditions gives final equivalent circuit!
i1=y11v1 + y12v2
i2=y21v1 + y22v2
47
For the common-emitter Configuration of BJT
ib=y11vbe + y12vce
ic=y21vbe + y22vce
ic
ib
Vbe
y12Vce
y21Vbe
Vee
48
How to calculate the small signal
Parameters?
49
Calculating two-port Network Parameters
i1=y11v1 + y12v2
i2=y21v1 + y22v2
Two Port Network
50
Calculating two-port Network Parameters
ib=y11vbe + y12vce
ic=y21vbe + y22vce
Two-port network
For small signal
BJT
51
Calculating two-port Network Parameters
First let vce be zero! This gives the following relationship
for y11 and y21 in terms of ib and ic and vbe (for vce =0)
ib=y11vbe
ic=y21vbe
Two-port network
For small signal
BJT
52
Calculating two-port Network Parameters
First let vbe be zero! This gives the following relationship
for y22 and y12 in terms of ib and ic and vce (for vbe =0)
ib= y12vce
ic= y22vce
Two-port network
For small signal
BJT
53
Consider the BJT as a two-port Network
!v
$
BE
i = I exp #
&
C
S
#" V &%
T
vCE = 0
iC
diC
dvBE
VBE
vBE =VBE
vBE
54
Consider the BJT as a two-port Network
55
Y12 calculation
i
I
C
S
i =
=
B !
!
F
FO
(
"v
%+
BE
* exp $
'$# V '& *
T ,
)
In forward active region approximation, base current is not a
function of vCE!
56
Y21 calculation
Use the active mode current equations with Early Voltage Effect!
'
!v
$*'
v
BE
, )1 + CE
i = I ) exp #
&
C
S)
#" V &% , )
V
T +(
A
(
*
,
,+
57
Y22 calculation
Use the active mode current equations with Early Voltage Effect!
'
!v
$*'
v
BE
)
,
)
i = I exp #
1 + CE
&
C
S)
#" V &% , )
V
T +(
A
(
*
,
,+
58
Y22 calculation
Use the active mode current equations with Early Voltage Effect!
'
!v
$*'
v
BE
)
,
)
i = I exp #
1 + CE
&
C
S)
#" V &% , )
V
T +(
A
(
*
,
,+
59
Y11 calculation
(
"v
%+
i
I
C
S
BE
* exp $
i =
=
'B !
$# V '& !
*
F
FO )
T ,
!i
B = ! # iC &
!vBE
!vBE %$ " F ('
'
!v
$*'
v
i = I ) exp # BE & , )1 + CE
C
S)
#" V &% , )
V
T +(
A
(
*
,
,+
IF βF is approximately constant, then this expression becomes:
(BOOK SHOWS SOLUTIONS WHEN THIS IS NOT TRUE!)
!i
B = ic
!vBE
" FVT
Let β0 = βF , which is referred to as the “small signal gain”
60
Hybrid-pi Model
Remember DC bias value!
Transconductance g m = y 21 =
IC
! 40 I C
VT
" oVT " o
1
=
=
Input Resistance r# =
y11
IC
gm
V A + VCE
1
=
Output Resistance ro =
y 22
IC
Y-parameter Model
Hybrid-pi Model
v1
61
Alternative Representations
gm vbe
vbe = ib r!
gm vbe = gm r! ib = " oib
"o
r! =
gm
62
Alternative Representations
g m vbe = g m r" ib = ! o ib
Voltage Controlled Current
source version of Hybrid-pi
Model
Current Controlled Current
source version of Hybrid-pi
Model
63
Small-Signal Model for pnp BJT is the same
as an npn BJT
Chap13 - 64
64
Small-Signal Model for pnp BJT is the same
as an npn BJT
How can this BE! The DC currents in active mode are in opposite
directions!
65
Small-Signal Model for pnp BJT is the same
as an npn BJT
Consider the currents for a moment
npn
iB = I B + ib
iC = I C + ic = ! I B + ! ib
pnp
iB = I B ! ib
iC = I C ! ic = " F I B ! " F ib
Increase in ac
base current
increases
collector
current!
Increase in ac
base current
decreases
collector
current!
66
Small-Signal Model for pnp BJT is the same
as an npn BJT
A decrease in total collector current is
equivalent to an increase in the
ac collector current!!!
Constant
iC = I C ! ic
decreasing
Means ic
must
increase!
67
Small Signal Operation of BJT
&
$
$
$
$
%$
,v
i = I exp* BE
C S
* V
+ T
)#!
'!
'!!
("!
Decompose into DC and AC components!
)
,v
)
V
'
*
'
-i = I + ic = I exp BE ' exp* be '
C C
S
V '
*V
'
T (
+ T (
&
#
2
3
,
)
,
)
$
!
v
*v
'
*v
'
1
1
$
= I $1+ be + * be ' + * be ' + ...!!
C$ V
2*V ' 6*V '
!
T
+ T (
+ T (
$
!
,
*
*
*
+
%
"
#
2
3
,v
)
,v
)
!
v
*
'
*
'
1
1
!
be
be
be
.ic = i - I = I
+ *
' + *
' + ...!
C C C V
2*V ' 6*V '
!
T
+ T (
+ T (
!
&
$
$
$
$
$
%
"
68
Small Signal Operation of BJT
2
3
(
+
"
%
"
%
v
v
v
1
1
*
i = i ! I = I * be + $ be ' + $ be ' + ...c C
C
C V
2 $# V '&
6 $# V '&
T
T
T
*)
-,
For linearity, ic should be proportional to vbe
2
1 ! vbe $
vbe
<<
#
&
2 " VT %
VT
v << 2V
be
T
Assuming that v
is a factor of 10 smaller then
be
v = 0.005V
be
69
Single Transistor Amplifier Analysis Example!
STEP 1: Determine DC operating point and calculate small signal parameters
In DC all capacitors are open circuit!
70
Single Transistor Amplifier Analysis Example!
STEP 1: Determine DC operating point and calculate small signal parameters
In DC all capacitors are open circuit!
71
Single Transistor Amplifier Analysis Example!
STEP 1: Determine DC operating point and calculate small signal parameters
10k
= 3V
30k + 10k
10k * 30k
Rth = 30k || 10k =
= 7.50k!
40k
VOC = 12
Thevenin equivalent circuit
Rth
+
-
VOC
72
Single Transistor Amplifier Analysis Example!
STEP 1: Determine DC operating point!
(Used later to calculate small signal parameters)
KVL
3V=IERE+Vbe+IBRTH
IC = ! I B
I E = ( ! + 1) I B
3V=IB (βo+1) Re+0.7V+IBRTH
KVL
3V= IB(100+1)1300+0.7+ IB7500
IB=16.6 uA
IC= IB βo=1.66 mA
IE=(βo+1) ΙΒ=1.67 mA
73
Coupling and Bypass Capacitors
Used to inject
AC input signal
•
•
Used to extract
AC output signal
AC coupling through capacitors control AC signal without
disturbing Q-point
Capacitors provide negligible impedance at frequencies of
interest and provide open circuits at dc.
74
Coupling and Bypass Capacitors
C1 and C3 are called coupling capacitors OR DC blocking capacitors
C2 is referred to as a bypass capacitor providing AC
short to ground removing RE for ac signal analysis
75
Single Transistor Amplifier Analysis Example!
STEP 2: Convert to the AC model only
•DC Voltage sources are shorts to ground
•DC Current sources are open circuits
•Large capacitors are short circuits
•Large inductors are open circuits
Short
capacitor!
Short to
ground!
Short
capacitor!
Short
capacitor!
76
Single Transistor Amplifier Analysis Example!
STEP 2: Convert to the AC model only
77
Single Transistor Amplifier Analysis Example!
STEP 3: Simplify AC circuits (Use Thevenin or Norton where
necessary)
78
Single Transistor Amplifier Analysis Example!
STEP 4: Replace transistor with small signal model
79
Single Transistor Amplifier Analysis Example!
STEP 5: Simplify circuit with hybrid-pi model as much as possible
RL = ro || Rc || R3
80
Single Transistor Amplifier Analysis Example!
STEP 6: Calculate small signal parameter from DC bias points
From DC analysis we know that : IC= 1.66 mA
Transconductance
g m = y 21 =
Input Resistance r$ =
Output Resistance ro =
IC
" 40 I C = 0.0664 S
VT
! = 100
# V
#
1
= o T = o = 1506 !
y11
IC
gm
V + VCE
V
1
= A
" A = 45.2 K !
y 22
IC
IC
VA = 75V
Rth = 880 Ω
Vth = 0.88vs
RL = ro || Rc || R3
81
Single Transistor Amplifier Analysis Example!
STEP 6: Calculate small signal parameter from DC bias points
Output Resistance
ro =
1
V + VCE VA
= A
!
= 45.2K "
y22
IC
IC
RL = ro || Rc || R3
1
1
1
1
1
1
1
= +
+
=
+
+
= 2.65e " 4
RL ro RC R3 45.2k! 4.3k! 100k!
RL = 3778.12
82
Single Transistor Amplifier Analysis Example!
What is the terminal voltage gain?
“Terminal voltage gain” is the gain between the base and collector
terminals.
Rth = 880 Ω
Vth = 0.88vs
RL = 3.778k!
vc vo
Avt = "Terminal voltage gain"= =
vb vbe
83
Single Transistor Amplifier Analysis Example!
What is the terminal voltage gain?
“Terminal voltage gain” is the gain between the base and collector
terminals.
Rth = 880 Ω
Vth = 0.88vs
RL = 3.778k!
vo !gm vbe RL
Avt =
=
= !gm RL = !(0.0664 * 3778) = !250.8
vbe
vbe
84
Single Transistor Amplifier Analysis Example!
What is the overall voltage gain?
“Overall voltage gain” is the gain between the input signal and the
output voltage.
Rth = 880 Ω
Vth = 0.88vs
RL = 3.778k!
vo vo vbe
vbe
Av = "overall voltage gain"= =
= Avt
vs vbe vs
vs
85
Single Transistor Amplifier Analysis Example!
What is the overall voltage gain?
Rth = 880 Ω
Vth = 0.88vs
RL = 3.778k!
vo vo vbe
r!
1
Av =
=
= Avt (
)vth
vs vbe vs
r! + Rth
vs
r!
1
Av = Avt (
)(0.88vs )
r! + Rth
vs
r!
Av = "gm RL (
)(0.88)
r! + Rth
1506
Av = "(250.8)
0.88 = "139.3
880 + 1506
86
Single Transistor Amplifier Discussion!
RL = ro || Rc || R3
Limits to overall gain is terminal gain voltage..
ASSUMING input resistance is much greater than source
resistance.
r!
vbe =
vth
r! + Rth
Av ! Avt = "gm RL = "gm (ro || RC || R3 )
87
Single Transistor Amplifier Discussion!
RL = ro || Rc || R3
Terminal voltage gain is effected by RL!
I C RC
Av ! Avt = "gm (ro || RC || R3 ) ~ "gm RC = "
VT
I C RC < VCC
Av ! 10VCC
Very rough rule of thumb…
88
Upper Limit on C-E Amplifier Gain
Av ! Avt = "gm RL = "gm (ro || RC || R3 )
Any load in parallel will decrease gain… therefore max gain
Av,max = !gm ro = µ F
Amplification factor
Typically on the order of 1000s!
89
Amplifier Signal Range
What is the limit of the vo so that we will NOT have signal distortion?
Vc voltage can never
exceed VCC… this
limits amplitude of vo!
(IF VC =10V then
amplitude of vo < 2V)
VCB must be greater ZERO
to be in active mode…
this limits the amplitude
of vo!
90
Amplifier Signal Range
LIMIT #1: BJT must stay in active mode!
v ! 0V
CB
v
!v
" 0V
CE
BE
v
>v
CE
BE
v
>V
CE
BE
91
Amplifier Signal Range
LIMIT #1: BJT must stay in active mode!
v
>V
CE
BE
v
=V
+v
CE
CE
ce
v
=V
! V sin " t
CE
CE
M
Because negative gain!
V
! V sin " t > V
CE
M
BE
V
!V >V
CE
M
BE
V <V
!V
=V
M
CE
BE
CB
92
Amplifier Signal Range
LIMIT #1: Power supply limit!
Vc voltage can
never exceed VC.
vc < Vcc
0 < Vcc ! vc = vRc
vRc > 0
93
Amplifier Signal Range
LIMIT #1: Power supply limit!
vRC = VRC + vrc > 0
v
R
C
(t) = I R ! !V sin " t # 0
C C
M
I R + V sin ! t " 0
C C
M
I R #V " 0
C C
M
V
!I R
M C C
94
Amplifier Signal Range
Must consider both conditions:
Power supply
Saturation limit
V ! min #%$ IC RC ,VCE "VBE &('
M
“The more drop then room for Vo”
“Vo can drop vC to drop out of
active mode”
95
Amplifier Signal Range
Must consider both conditions:
V ! min #%$ IC RC ,VCE "VBE &('
M
VCC
RC
What is amplitude of the
largest sine wave
assuming VCE = VCC/2 ?
VBE=0.7
V ! min [ 6, 6 " 0.7 ]
M
V ! 5.3
M
96
EXAMPLE 2: Calculate the
signal range
IB=16.6 uA
IC= IB βo=1.66 mA
IE=(βo+1) ΙΒ=1.67 mA
12V = 4.3k(1.66mA) + VCE + 1.3k(1.67mA)
VCE = 2.69
V ! min #%$ IC RC ,VCE "VBE &('
M
V ! min #%$1.66mA*4.3k,2.69"0.7 &('
M
V ! min #%$ 7.138,1.99 &('
M
V !1.99
M
Why do we care? Will we ever be limited by this voltage?
97
EXAMPLE 2: Calculate the
signal range
Why do we care? Will we ever be limited by this voltage?
V !1.99
M
Av = !139
The largest input signal that we can have is approximately 5mV
vo
max
= 5mV *139 = 695mV
Therefore our amplifier design is a good one!!!
98
EXAMPLE 3: Calculate the
signal range and Gain
What if I
change this
to 0.43k?
IB=16.6 uA
IC= IB βo=1.66 mA
IE=(βo+1) ΙΒ=1.67 mA
V ! min #%$ IC RC ,VCE "VBE &('
M
12V = 0.43k(1.66mA) + VCE + 1.3k(1.67mA)
VCE = 9.11
V ! min #$%1.66mA*0.43k,9.11"0.7 &'(
M
V ! min #%$ 0.714,8.41&('
M
Limited by power supply!
What about the gain?
99
EXAMPLE 3: Calculate the
signal range and Gain
What if I
change this
to 0.43k?
Rth =
880 Ω
Vth =
0.88vs
RL = ro || Rc || R3
1
1
1
1
1
1
1
= +
+
=
+
+
= 2.35e " 3
RL ro RC R3 45.2k! .43k! 100k!
RL = 424.1
This drop in RL decreases the gain!
Avt =
vo !gm vbe RL
=
= !gm RL = !(0.0664 * 424) = !27.3
vbe
vbe
RL = ro || Rc || R3
Drops from
250!
100
EXAMPLE 3: Calculate the
signal range and Gain
What if I
change this
to 0.43k?
Rth =
880 Ω
Vth =
0.88vs
RL = ro || Rc || R3
This drop in RL decreases the gain!
vo vo vbe
r!
1
Av =
=
= Avt (
)vth
vs vbe vs
r! + Rth
vs
1506
Av = "(27.3)
0.88 = "15.16
880 + 1506
Drops from
140!
101
EXAMPLE 4: Calculate the
signal range and Gain
What if I
change this
to 8.6k?
IB=16.6 uA
IC= IB βo=1.66 mA
IE=(βo+1) ΙΒ=1.67 mA
12V = 8.3k(1.66mA) + VCE + 1.3k(1.67mA)
VCE = !2.27
VCB = VCE ! VBE = !2.97
No longer in active mode in DC!
ALL MODELS ARE INVALID!
102