ELL 100 - Introduction to Electrical Engineering LECTURE 23: BIPOLAR JUNCTION TRANSISTOR (BJT)
1 CONTENTS
• Introduction
• Construction & working principle of BJT
• Common-base and common-emitter circuits
2 BJT INTRODUCTION
• Bipolar: Both electrons and holes (positively charged quasi- particles, absence of electron) contribute to current flow
• Junction: Consists of an n- (or p-) type silicon sandwiched between two p- (or n-) type silicon regions
• Transistor: 3-terminal device (Base, Emitter, Collector) Two types: npn & pnp
3 Bipolar Junction Transistor (BJT)
4 Various Types of General-Purpose or Switching
Transistors:
(a) Low Power (b) Medium Power (c) Medium to high power
5 First Working Transistor, Invented in 1947.
Earliest Raytheon Blue transistors, Replica of 1st device made by first appearing in 1955 Shockley, Bardeen & Brattain at Bell Labs 6 Applications: Transistor As Switch
7 Applications: Transistor As Relay (Switch)
Single channel 5V relay breakout board
8 Applications: Transistor As Amplifier
TDA7297 Power Amplifier & Audio amplifier Module 9 Applications: Transistors in Radio
10 Applications: Transistor in Oscillator
Colpitts Oscillator
11 Applications: Transistor in Oscillator
Astable Multivibrator
12 Transistors in Timer Circuits
13 Transistor As NOT Gate
14 Transistors in NAND Gate
15 Bipolar Junction Transistor (BJT) The three terminals of the BJT are called the Base (B), the Collector (C) and the Emitter (E).
n p p n n p
npn transistor pnp transistor
16 Bipolar Junction Transistor (BJT)
npn pnp
17 p-n-p Transistor
IE, IB, IC – Emitter, Base and Collector currents
VEB, VCB, VCE – Emitter-Base, Collector-Base and Collector-Emitter voltages
18 n-p-n Transistor
IE, IB, IC – Emitter, Base and Collector currents
VEB, VCB, VCE – Emitter-Base, Collector-Base and Collector-Emitter voltages
19 BJT Construction
20 Bipolar Junction Transistor (BJT)
Basic Principle
The voltage between two terminals (B and E) controls the current through the third terminal (C).
21 Bipolar Junction Transistor (BJT) : Modes
Mode VBE VBC Forward active Forward Bias Reverse Bias Amplifier Reverse active Reverse Bias Forward Bias
Saturation Forward Bias Forward Bias Switch ON (VCE ~ 0) Cut off Reverse Bias Reverse Bias Switch OFF (IC,E ~ 0)
22 Biasing of transistors in “Forward Active” mode
B-E forward-biased and B-C reverse-biased
23 (a) B-E forward-bias => large current (b) B-C reverse-bias => small current carried by majority carriers (IE) carried by minority carriers (ICBO)
24 BJT carrier flow in Forward-Active mode
25 Mode V ~ 0.7 V V < 0 Basic Working BE BC Forward Active Forward Bias Reverse Bias
26 BJT carrier flow in Forward-Active mode
(IC = αIE, Diffusion Drift α ~ 0.95-0.99)
Thus IC ~ IE and IB << IE,C
Also, IC fairly independent
of VCB
27 BJT Currents in Forward-Active Mode I I IE = IB+IC E C C IC ~ αIE 0.95 < α < 0.995 npn α is the emitter-to-collector gain B In addition, due to the reverse IB biased C-B junction, a small reverse saturation current flows, IC = αIE+ICBO = α(IC+IB) + ICBO ICBO (collector-to-base leakage current with emitter open) => (1- α)IC= αIB+ICBO => I = (α/1- α)I + (1/1-α)I β = α/(1-α) is the base-to-collector gain C B CBO
20 < β < 200 => IC = βIB + (β+1)ICBO
28 BJT Amplifier/Circuit Configurations
Common-Base Configuration Common-Emitter Configuration Common-Collector Configuration
29 Common-Base Configuration
Vin R E C RLL
VVEB VCB BE B
The base terminal is common to the input and output ports
30 The arrows in the graphics define the direction of current through the device.
pnp transistor npn transistor
31 Input dc characteristics in Common-Base configuration
Vin R E C RLL
VVEB VCB BE B
Notation:
Input current (IE) Input voltage (VBE) Output voltage (VCB)
32 Output dc characteristics in Common-Base configuration Vin R E C RLL
VVEB VCB BE B
Notation:
Output current (IC) Output voltage (VCB) Input current (IE) 33 Modes of operation of npn transistor in Common-Base configuration
Cutoff region is defined as where the collector current IC = ICBO ~ 0 A The base–emitter and collector–base junctions are both reverse-biased.
Saturation region is defined as where VCB < 0 V The base–emitter and collector–base junctions are both forward-biased.
Active region is defined as where VCB > 0 V The base–emitter junction is forward-biased and the collector–base junction is reverse-biased.
34 Common-Emitter Configuration
RRLL
B C
VCE VBE E
The emitter terminal is common to the input and output ports
35 The arrows in the graphics define the direction of current through the device.
npn transistor pnp transistor 36 Input dc characteristics in Common-Emitter configuration
RLRL
B C
VCE VBE E
Notation:
Input current (IB) Input voltage (VBE) Output voltage (VCE)
37 Output dc characteristics in Common-Emitter configuration
RLRL
B C
VCE VBE E
Notation:
Output current (IC) Output voltage (VCE) Input current (IB) 38 Saturation region is defined as where VCE < VCE(sat) ~ 0.7-1 V (i.e. VCB < 0 V) The base–emitter and collector–base junctions are both forward-biased.
Active region is defined as where VCE > VCE(sat) ~ 0.7-1 V (i.e. VCB > 0 V) The base–emitter junction is forward-biased and the collector–base junction is reverse-biased.
Cutoff region is defined as where the collector current IC = ICEO = βICBO The base–emitter and collector–base junctions are both reverse-biased.
39 Common-Collector Configuration
RLRL
B CE - - + VVCEEC VCBBE + EC
The common-collector configuration is used primarily for impedance- matching purposes: High input impedance and low output impedance
40 The arrows in the graphics define the direction of current through the device.
pnp transistor npn transistor 41 • The collector is tied to ground even though the transistor is connected in a manner similar to the common-emitter configuration
• The output characteristics of the common-collector configuration are same as for the common-emitter configuration Common-collector configuration used for impedance-matching purposes.
42 BJT Configurations in pnp Transistor
43 BJT amplifier configurations : Summary
Characteristics Common Base Common Emitter Common Collector Input Impedance Low Medium High Output Impedance Very high High Low Phase shift 00 1800 00 Voltage gain High Medium Low Current Gain Low Medium High Power gain Low Very High Medium Application Preamplifier Most Common Buffer
44 Numerical Problems
45 Q1: In an npn transistor, 108 holes/µs move from the base to the emitter region while 1010 electrons/µs move from the emitter to the base region.
An ammeter reads the base current as IB = 16 µA. Determine the emitter current IE and the collector current IC
46 Solution:
8 IB = 16 μA (corresponding to 10 positive electronic charges/sec)
10 IE = (100 × 16) μA (corresponding to 10 negative electronic charges/sec) + 16 μA (corresponding to 108 positive electronic charges/sec)
= (101 x 16) μA = 1616 μA = 1.616 mA
IC = IE − IB = 1.6 mA
47 Q2: For the previous question (data given below), find the transistor α and β values, if the reverse collector-base leakage current is considered negligible
Transistor: npn transistor
Base current: IB = 16 µA Emitter current: IE = 1.616 mA Collector current: IC = 1.6 mA
48 Solution:
If we assume ICBO = ICEO = 0, then
Emitter-collector current gain α = IC / IE = 1.6/1.616 = 0.99
Base-collector current gain β = IC / IB = 1.6/0.016 = 100
49 Q3: A BJT has α = 0.99, IB = 25 µA, ICBO = 200 nA.
Find the following parameters: a) the dc collector current, b) the dc emitter current, and c) the percentage error in the emitter current when the leakage current is neglected.
50 Solution: 0.99 a) 99 11 0.99 69 IIImACBCBO(1)99(25 10 ) (99 1)(200 10 ) 2.495
b) IE = IC + IB = 2.495 + 0.025 = 2.52 mA c) Neglecting the leakage current, we have giving an emitter current error of 6 IImACB 99(25 10 ) 2.475 2.52 2.5 100% 0.793% I 2.475 ImAC 2.5 2.52 E 0.99
51 Q4: A Si transistor connected in the common-emitter configuration (see fig.) has base current of 40 µA and ICBO = 0. If VBB = 6 V, RE = 1 kΩ, and β = 80, find (a) IE, (b) RB , (c) If VCC = 15 V and RC = 3 kΩ, find VCE .
RE
52 Solution: a) IE = (β+1)IB = 81 × 0.04 = 3.24 mA b) KVL in input (B-E) loop:
VBB = IBRB + VBE + IERE RB = (VBB – VBE – IERE)/IB
= (6 – 0.7 – 3.24×1)/(0.04) kΩ RE = 51.85 kΩ b) KVL in output (C-E) loop:
VCC = ICRC + VCE + IERE (IC = βIB = 80 × 0.04 = 3.2 mA) => VCE = VCC – IERE – ICRC = 15 – (3.24×1) – (3.2×3) = 2.16 V
53 Q5: Using the characteristics shown below, determine a. the resulting collector current if IE = 3 mA and VCB =10 V. b. the resulting collector current if IE remains at 3 mA but VCB is reduced to 2 V. c. VBE if IC = 4 mA and VCB = 20 V.
54 Solution:
a. The output characteristics clearly indicate that IC ~ IE = 3 mA. b. The effect of changing VCB is negligible and IC continues to be 3 mA. c. From output characteristics, IE = IC = 4 mA. From input characteristics, for IE = 4 mA and VCB = 20 V, the resulting value of VBE is ~0.75 V.
55 Q6: Using the characteristics shown below, determine a. IC at IB = 30 μA and VCE = 10 V. b. IC at VBE = 0.7 V and VCE = 15 V.
56 Solution:
a. From output characteristics, for IB = 30 μA and VCE = 10 V, IC ~ 3.4 mA. b. From input characteristics, IB ~ 20 μA for VBE = 0.7 V and VCE = 15V From output characteristics, IC ~ 2.5 mA for IB = 20 μA and VCE = 15 V.
57 Q7: A high-power BJT controls power from a dc source of VS = 200 V to a resistive load of R = 4 Ω. For this condition, the transistor is in saturation with VCE(SAT) = 1.1 V. The base resistance RB = 0.5 Ω, the base source voltage VB = 10 V and VBE(SAT) = 1.8 V. For this on-state condition, determine (a) the forced current gain βF and (b) the power loss in the BJT.
58 Solution:
()VVs CE() sat (a) KVL in outer loop: IAc R (200 1.1) / 4 49.7
()VV KVL in inner loop: IABBE sat () (101.8) / 0.516.4 B RB
The forced current gain is IC 49.7 /16.43 F IB
(b) The BJT power dissipation is
PVIVICE()() sat C BE sat B 1.1*49.7 1.8*16.4 54.7 29.5 84.2 W
The power transfer efficiency η of the BJT is η = output/(output + losses) = (200-1.1)49.7/(198.9*49.7+84.2) = 0.99 = 99%
59 Unsolved Problems
60 Question 1: For a BJT, the common – base current gain 훼 = 0.98 and the collector base junction reverse bias saturation current 퐼퐶0 = 0.6휇퐴 . This BJT is connected in the common emitter mode and operated in the active region with a base drive current 퐼퐵 = 20휇퐴. The collector current 퐼퐶 for this mode of operation is?
Answer:1.01mA
61 Question 2: In the circuit shown in the figure, the BJT has a current gain (β) of 50. For an emitter – base voltage VEB = 600 mV, the emitter – collector voltage VEC (in Volts) is
Answer:2V
62 −15 Question 3: An npn BJT having reverse saturation current 퐼푆 = 10 A is biased in the forward active region with VBE = 700 mV and the current gain (β) may vary from 50 to 150 due to manufacturing variations. Find the maximum emitter current (in µA)?
Solution: 1475 µA
63 Question 4: For a BJT circuit shown, assume that the ‘β’ of the transistor is very large and VBE = 0.7 V. The mode of operation of the BJT is ?
Answer: Saturation R1 10KΩ
+ + 10V 2V - -
R2 1KΩ
64 References
1. R. S. Ramshaw,“ Power Electronics Semiconductor Switches”, Champman & Hall, 1993, ISBN0-412-28870-2. 2. N. Mohan, T. M. Undeland and W. P. Robbins, “Power Electronics, Converter, Application and Design”, Second Edition, John Willey & Sons, 1995, New York, ISBN 9971-51-177-0. 3. M. H. Rashid, “Power Electronics, circuits, Devices and Applications”, Second Edition, Prentice-Hall, 1995, India, ISBN 81-203-0869-7. 4. B. W. Williams, “Power Electronics: Devices, Drivers and Applications”,Wiley,1987.NewYork,ISBN: 0470206969. 5. W. C. Lander, "Power Electronics", 3rd Edition, McGraw-Hill, 1993, New York, ISBN: 0077077148S.
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