Chapter 1:
Bipolar Junction Transistor (BJT)
Dr. Pham Nguyen Thanh Loan
October, 2022
Contents
2
Structure and operation of BJT
Different configurations of BJT
Characteristic curves
DC biasing method and analysis
Base bias
Collector-feedback bias
Voltage divider bias
AC signal analysis
Impact of other parameters (temperature, leakage
currents)
The content of these slides are based on the book titled “Electronics Devices and Circuit theory of
Robert Boylestad”
Structure and operation of BJT
3
❖BJT structure
BJT :Bipolar Junction Transistor
2 kinds of BJT: NPN & PNP
3 terminals: E, B và C
E: Emitter; B: Base, C: Collector
Base located in the middle:
thinner than E & C; and lower
dope
Structure and operation of BJT
4
❖ Bias condition for 2 junctions: JBE & JBC
Junction BE in forward bias: electrons
(e) move from E region to B region to
create the current IE (diffusion current;
flow of majority carriers)
Junction BC in reverse bias: e that
moved from E to B then move from B to
C to create the current IC (drift current,
flow of minority carriers)
The combination of some electrons with
holes in B region creates the current IB
So: IE = IC + IB
Structure and operation of BJT
5
3 terminals: B, E và C
❖BJT symbol
Arrow instructs the current
IC
direction between B & E
Conventional current is the
IB flow of positive charges
IE (holes)
NPN: B → E
PNP: E → B
❖Explain the symbol of BJT?
Technical parameters
6
IC = αIE + ICBO
⚫ IE = IC + IB
⚫ IC = βIB IC ≈ αIE (neglect leakage ICBO)
⚫ β = 100 ÷ 200 (may be higher) α = 0.9 ÷0.998.
⚫ β is DC current gain α is DC current transfer coefficient
=
+1
BJT as an amplifier
Different amplifier configurations
7
3 configurations
Common emitter (CE)
Common base (CB)
Common collector (CC)
Look at the input and output to distinguish these
configurations
Configuration Input Output
BC E C
EC B C
CC B E
BJT as an amplifier
8
CE with voltage CC with voltage CB with fixed
divider biasing divider biasing biasing
C
C
B B
E E
CE configuration
9
E is used in common for
in and out
Input: re is considered as
AC resistor of diode BE
re=26mV/IE
Output: Ic= βIb
❖ Definition of internal emitter re:
9
CE configuration – small signal
10
Zi = Ube/Ib ≈ βIbre/Ib ≈ βre
(~ n100Ω – nKΩ)
Zo = r o → ∞
(ignore in re model)
Av = - RL/re (ro→ ∞)
Ai = Ic/Ib = β
Characteristics
+ Zi, Zo average
+ Av, Ai high
10
Characteristic curves: CE
11
Input and output characteristic curves of CE configuration
❖Explain why VCE increases, IB decreases for a constant VBE?
11
Characteristic curves: CE
12
0<VCE<0.7V: Junction BE
starts moving to forward
bias→ IC increases gradually
VCE >0.7V: Junction BE is in
FB and Junction BC in reverse
→ IC = β*IB
CB configuration
13
B is used in common for
in and out
Input: re is considered as
AC resistor of diode BE
re=26mV/IE
Isolation between in and
out
Output: Ic=αIe
13
CB configuration
14
1) Z i = re (nΩ-50 Ω)
2) Zo = ro ≈ ∞ (nMΩ)
3) Av = αRL/re ≈ RL/re quite big, Uo & Ui in phase
4) Ai = -α ≈ 1
14
Characteristic curves: CB
15
Input and output characteristic curves of CB configuration
CC configuration
16
Similar to CE configuration
Refer to Electronic Devices – Thomas Floyd
16
Limits of operation
17
Two limits:
cut-offregion
Saturation region
Cutoff and saturation
18
Cutoff state Saturation state
DC loadline & Q point
19
❖ Q_point deplacement when Rc, Vcc, IB vary respectively
Variation of RC Variation of VCC Variation of IB
19
DC power vs. AC signal
20
DC power AC signal
Example of CE configuration
21
❖ Output and input signal is out of phase
❖ Output signal is amplified
22
DC bias:
DC operating point & DC load line
DC bias
23
A transistor must be properly biased in order to operate as
an amplifier
DC bias can be considered as supply power to BJT so that
◼ NPN: VE < VB < VC (JE: in Forward; JC: in Reverse bias)
◼ PNP: VE > VB > VC
DC bias is characterized by Q-point (DC operating point)
and DC load line
DC bias
24
NOTES: REMEMBER some equations:
VBE ≈ 0,6 ÷ 0,7V (Si) ; 0,2 ÷ 0,3(Ge)
IE = IC + IB IC = βIB IC ≈ αIE
There 3 types of bias circuits
Base bias
Collector-feedback bias
Voltage divider bias
Question: How many amplifier circuits can be
designed?
3 types of baising
25
Base bias Voltage divider bias
Collector feedback bias
Example of DC bias
26
Q1. What are the amplifier configuration of these circuits?
Q2. What kind of DC bias? And then draw DC equivalent circuit.
(a) (b) (c)
Question 3: How many amplifier circuits can be designed?
Base bias
27
Consider the analysis for only EC configuration (similar
analysis can be obtained for BC and CC)
Base bias
28
BE loop:
Vcc – IBRB – UBE = 0
➔ IB= (Vcc - UBE)/RB
IC=β*IB
CE loop:
➔ UCE = Vcc - ICRC
Voltage divider bias
29
Method 1: Thevenin equivalent circuit:
* Group R1, R2 and Vcc can be considered as
follows:
RBB=R1//R2
VBB= Vcc * R2/(R1+R2)
➔ Now it is similar to base-bias
analysis
Current and voltage do not Method 2: Approximative analysis
depend on β
If β*R2 ≥ 10R2 -> I2 ≈ I1
VB=Vcc*R2/(R1+R2)
VE=VB-VBE ➔ Ic ≈ IE=VE/RE
VCE=Vcc- IC(RC+RE)
Collector-feedback bias
30
BE loop:
(1) Vcc- Ic‘ RC – IBRB – UBE – IERE =0
(2) IC= β *IB ; IE IC
(3) KCL at C: IC= IB + Ic‘ → Ic‘ = IC - IB
= (β-1)IB
(1)+(2)+(3)
→ IB= (Vcc - UBE)/[RB+ β(Rc+Re)]
CE loop:
UCE = Vcc – IC (RC+RE)
Quite stable
Example 1
31
❖Analyze the following circuit and then determine its Q-
point and DC loadline?
Example 1 (cont’d)
32
Example 2
33
❖Analyze the following circuit and then determine its Q-point
and DC loadline?
Example 2 (cont’d)
Analysis by method 1
34
Homework
35
❖Analyze the following circuit and then determine its Q-point
and DC loadline?
Homework
36
❖Analyze the following circuit and then determine its Q-point
and DC loadline?
Homework
37
❖Analyze the following circuit and then determine its Q-point
and DC loadline?
38
Working on your
1st in-class assignment
39
AC analysis
(Small signal analysis)
40
Small signal analysis
Small signal analysis:
Small signal refers to AC signal with small amplitude that take up
a relatively small percentage of an amplifier’s operation range
(compared to DC power supply)
The operation region on amplifier should be in linear
BJT model for small signal analysis
Represent the BJT by an equivalent circuit that allows to visualize
and analyze the operation of BJT as an amplifier
Example of CE configuration
41
❖ Output and input signal is out of phase
❖ Output signal is amplified
Gain and impedances
42
AC equivalent circuit
43
1. Setting all DC sources to zero
2. Replacing all capacitors by a
short-circuit equivalent (wire)
3. Regrouping all elements
(resistors) in parallel (introduced
by step 1 and 2)
4. Redrawing the network in a
more convenient and logical form
AC analysis
44
BJT amplifier is considered linear → be able to be
analyzed DC and AC separately (using
superposition theorem)
Different approaches
Using graphical determination method
Using equivalent circuits
◼T model
◼ rE model
◼ Hybrid equivalent model (quite popular in the past)
AC analysis methods
Graphical Analysis
45
❖ Q-point and DC load-line
Quiescent point (Q-point) is fixed on the output characteristic curve
and corresponding to a fixed collector-to-emitter voltage (VCE)
DC load-line is used to describe the DC operation of BJT, a straight
line from saturation point (IC=ICmax, y-axe) to cutoff point
(VCE=VCEmax, x-axe)
➔ Q-point : intersect between DC load line and
characteristic curve
❖ DC load line vs. AC load line
DC load line: VCE = VCC – ICRC
AC load line: VCE = VCC - Ic(RC//RL)
AC analysis methods
Graphical determination
46
❖ Input and output characteristic curves of EC config.
46
AC load line determination
47
AC load line
(Slope_AC: 1/(Rc //Rtai)
DC load line
(slope= 1/Rc)
Q N
❖ AC loadline is steeper than DC loadline
❖Graphically: ON = OQ + QN where QN = IC-Q/Slope_AC =
IQ*(Rc//Rtai)
❖ A straight line through Q_point and N : AC load line
AC analysis methods
Graphical determination
48
❖ Q_point deplacement when Rc, Vcc, IB vary respectively
Variation of RC Variation of VCC Variation of IB
48
AC analysis methods
Graphical determination
49
Basing on input and output characteristic curves→ determine
small signal input and output waveform
49
AC analysis methods
Graphical determination
50
Δvbe → Δib
Δvce → Δic
Ai = io/ii = Δic/Δib
AV = vo/vi = Δvce/Δvbe
Zin = vi/ii = Δvbe/Δib
Zout = vo/io = Δvce/Δic
50
AC analysis methods
Graphical determination
51
❖ Impact of Q point on AC output
signal
Q closed to cutoff → BJT is closed to
OFF operation, with a very small AC
input amplitude → output voltage is
distorsed (is cut) at upper-part
Q closed to saturation → BJT is
closed to saturation operation, with a
very small AC input amplitude →
output volage is distorsed (is cut) at
lower-part
Large-signal may be cut at upper and
lower part
AC analysis
52
BJT amplifier is considered linear → be able to be
analyzed DC and AC separately (using
superposition theorem)
Different approaches
Using graphical determination method
Using equivalent circuits
◼T model
◼ rE model
◼ Hybrid equivalent model (quite popular in the past)
Two-port model
53
❖ Most used for small signal analysis
❖ Characterized by 2 input terminals and 2 output terminals (4
-terminals model)
❖ The common terminal is used for input and output
Remind: AC equivalent circuit
54
1. Setting all DC sources to zero
2. Replacing all capacitors by a
short-circuit equivalent (wire) `
`
3. Regrouping all elements
(resistors) in parallel (introduced `
by step 1 and 2)
4. Redrawing the network in a
more convenient and logical
form
Remind: AC equivalent circuit
55
❖ Equivalent circuit ❖ Equivalent circuit
after step 1 and 2 after step 3 and 4
?????????
?????????
AC analysis
56
BJT amplifier is considered linear → be able to be
analyzed DC and AC separately (using
superposition theorem)
Different approaches
Using graphical determination method
Using equivalent circuits
◼ rEmodel
◼ Hybrid equivalent model (quite popular in the past)
AC analysis methods
rE model
57
BJT is modeled by a diode and current source
Input : BE junction is characterized by a diode in Forward bias
Output: dependent current source where controlled current is input current
that is expressed by Ic = βIb or Ic=αIe.
3 configurations: EC; BC và CC
Common Base
(CE)
AC analysis methods
rE model
58
BJT is modeled by a diode and current source
Input : BE junction is characterized by a diode in Forward bias
Output: dependent current source where controlled current is input current
that is expressed by Ic = βIb or Ic=αIe.
3 configurations: EC; BC và CC
Common Emitter
(CE)
AC analysis methods
rE model
59
BJT is modeled by a diode and current source
Input : BE junction is characterized by a diode in Forward bias
Output: dependent current source where controlled current is input current
that is expressed by Ic = βIb or Ic=αIe.
3 configurations: EC; BC và CC
Common Collector
(CC)
AC analysis methods
rE model
60
❖ EC ❖ BC ❖ CC
e c
c c
b e
e 60
AC analysis methods
rE model
61
❖Refer to T model as learnt in Electronics Devices Course➔
Determine Rin & Iout =f(Iin) to obtain re model
❖ EC ❖ BC ❖ CC
e c
c c
b e
e
Input: ib, vb Input: ie, ve Input: ib, vb
Output: ic, vc Output: ic, vc Output: ie, ve
Rin = vb/ib = βre Rin = ve/ie = re Rin = vb/ib = βre
iout = ic = βi = βib iout = ic = αiin = αie
AC analysis methods
Hybrid equivalent model
62
U & I relation: Iv Ir
Ui=h11Ii+h12Uo Uv 2 ports Ur
Io=h21Ii+h22Uo
hij is determined at a given operating point (can be
different from Q_point)
Index e (or b, c) illustrated for CE topology (or CB, CC)
Hybrid parameters for 2N4400
AC analysis methods
Hybrid equivalent model
63
Parameters EC BC CC
h11 (hi) 1kΩ 20Ω 1kΩ
h12 (hr) 2,5x10-4 3x10-4 ≈1
h21 (hf) 50 -0,98 -50
h22 (ho) 25μA/V 0,5μA/V 25μA/V
1/h22 40kΩ 2MΩ 40kΩ
AC analysis methods
Hybrid equivalent model
64
Other names of hij
Read part 7.6, chapter 7 for further understanding
hi
hfIin h0
hr Vi
AC analysis methods
rE model
66
Analyze EC
66
EC configuration with fixed biasing
67
❖ EC
67
EC configuration with fixed biasing
68
68
EC configuration with fixed biasing
69
69
EC configuration with fixed biasing
70
1) Zi = Rb||βre if Rb ≥ 10βre, Zi ≈ βre
2) Zo = Rc||ro if ro ≥ 10Rc, Z o ≈ Rc
3) Av = - (Rc||ro)/re ≈ - Rc/re
(β appered in re)
Ui & Uo out of phase180o
4) Ai = βRbro / [(ro+Rc)(Rb+βre)] ≈ β
(Ii current source. Io collector current)
EC configuration with different biasing
71
EC configuration with voltage divider
72
EC configuration with voltage divider
73
EC configuration with voltage divider
74
1) Zi = R1||R2||βre = R’|| βre
2) Zo = Rc||ro (If ro ≥ 10Rc, Zo ≈ Rc)
3) Av = - (Rc||ro)/re ≈ - Rc/re
Similar to EC with fixed biasing
≈ βR’/(R’+ βre) if ro ≥ 10Rc
≈β if R’ ≥ 10 βre
EC configuration with voltage divider
75
EC configuration with feedback biasing
76
1) Zi = re/(1/β+Rc/Rf) 4) Ai = βRf/(Rf+ βRc)
2) Zo = Rc//Rf ≈ Rf/Rc
3) Av = -Rc/re if βRc >> Rf
When ro≠∞ → ro in equation
EC configuration with feedback biasing
77
EC configuration with feedback biasing
78
EC configuration with feedback biasing
79
AC analysis methods
rE model
80
Analyze BC
80
BC configuration
81
T model (learnt in Electronics Devices Courses)
BC: small signal model
82
82
BC analysis
83
83
BC analysis
84
84
Analyze BC configuration
85
1) Zi = Re||re Trở kháng vào tương đối nhỏ
2) Zo = Rc Trở kháng ra lớn
3) Av = αRc/re ≈ Rc/re Tương đối lớn
Ui & Uo cùng pha
4) Ai = - α ≈ -1 Không khuếch đại dòng
85
86
Analyze CC
86
CC configuration with fixed biasing
87
87
87
CC configuration with fixed biasing
88
88
CC configuration with fixed biasing
89
Analyze output impedance
CC configuration with fixed biasing
90
Vo
CC configuration with fixed biasing
91
1) Zi = Rb || [βre+(β+1)Re] ≈ Rb || β(re+Re)
High input impedance
2) Zo = Re||re ≈ re where Re >> re
Low output impedance
3) Av = Re/(Re+re) ≈ 1
Inphase with input and smaller amplitude
=> “emitter connection”
4) Ai = - βRb/[Rb+ β(re+Re)]
Application: Buffer
Quizz
92
1. eq/. Circuit
2. Av; Ai
Homework 1: Determine Ai, Av, Zi, Zo?
93
Homework 1:
94
HW1: T-model
95
HW1 : 2-port model
96
Homework 2: : Determine Ai, Av, Zi, Zo?
97
98
Homework 3: : Determine Ai, Av, Zi, Zo?
99
Homework 3
100
Homework 4: Determine Ai, Av, Zi, Zo?
101
Example: Determine Ai, Av, Zi, Zo?
102
SUMMARY
103
SUMMARY
104
105
Kiểm tra nhanh (lần 1)
106
Đề 1 Đề 2
Vẽ mạch BC sử dung Vẽ mạch CC sử dung
bộ chia áp phân cực bang hồi tiếp
Vẽ dạng song của tín Vẽ dạng song của tín
hiệu tại B, C khi biết VB hiệu tại B, C khi biết VB
= 3.7V, vB có biên độ = 4.5 V, vB có biên độ
10mV, VC =9.3V với 5mV, VC =8.7V với
biên độ vC = 65mV. biên độ vC = 45mV.
Xác định Av của mạch Xác định Av của mạch
Xác định điểm hoạt Xác định điểm hoạt
động tĩnh Q biết IB = động tĩnh Q biết IB =
40uA, beta =95 20uA, beta =90 106
