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Recommended Books. Robert Boylestad and Louis Nashelsky , “ Electronic Devices and Circuit Theory ”, Prentice Hall, 7 th Edition or Latest. Thomas L. Floyd, “ Electronic Devices ”, Prentice Hall, 7 th Edition or Latest, ISBN: 0-13-127827-4. This Lecture. - PowerPoint PPT Presentation
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1
Recommended Books
• Robert Boylestad and Louis Nashelsky, “Electronic Devices and Circuit Theory”, Prentice Hall, 7th Edition or Latest.
• Thomas L. Floyd, “Electronic Devices”, Prentice Hall, 7th Edition or Latest, ISBN: 0-13-127827-4
2
This Lecture
Current and Voltage Analysis of BJT – A Review
3
Types of Bipolar Junction Transistors
npn pnp
n p n
B
C p n pE
BCross Section
B
C
E
Schematic Symbol
B
C
E
Schematic Symbol
• Collector doping is usually ~ 106
• Base doping is slightly higher ~ 107 – 108
• Emitter doping is much higher ~ 1015
4
BJT Equations
B
CE
IE IC
IB
-
+
VBE VBC
+
-
+- VCE
B
CE
IE IC
IB-
+
VEB VCB
+
-
+ -VEC
npnIE = IB + IC
VCE = -VBC + VBE
pnpIE = IB + IC
VEC = VEB - VCB
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DC Beta and DC Alpha• DC Beta (dc) : The ratio of the dc collector current (Ic) to the dc base
current (IB) is the dc beta. It is also called the dc current gain of a transistor.
– Typical values of dc range from less than 20 to 200 or higher.
– If temperature goes up, dc goes up and vice versa.
• DC Alpha (dc): It is the ratio of dc collector current (Ic) to the dc emitter current (IE).
– Typically values of dc range from 0.95 to 0.99, but it is always less than unity.
B
cdc I
I
E
cdc I
I
6
Relationship between dc and dc
For an NPN transistor
Dividing each term by IC we get
or
cBE III
C
C
C
B
C
E
I
I
I
I
I
I
1I
I
I
I
C
B
C
E
111
dcdc
dc
dc
dc
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dc
dcdc 1
Similarly, we can prove that
dc
dcdc 1
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Problems on dc and dc
1. Determine dc and IE for IB = 50A and IC = 3.65 mA.
Solution:
731050
1065.3
I
I6
3
B
Cdc
mA70.3A1070.3
1065.31050III3
36CBE
8
Problems on dc and dc
2. What is the dc when IC = 8.23mA and IE = 8.69 mA.
Solution:
3. A certain transistor exhibits an dc of 0.96. Determine IC when IE = 9.35 mA.
Solution:
947.01069.8
1023.8
I
I3
3
E
Cdc
E
Cdc I
I mA976.835.996.0II EdcC
9
Current and Voltage Analysis
IB: dc base current
IE: dc emitter current
IC: dc collector current
VBE: dc voltage across base-emitter junction
VCB: dc voltage across collector-base junction
VCE: dc voltage from collector to emitter
Transistor bias circuit.
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Current and Voltage Analysis
• When the BE junction is forward-biased, it is like a forward-biased diode. Thus; (Si = 0.7, Ge = 0.3)
• From KVL, the voltage across RB is
• By Ohm’s law;
• Solving for IB
V7.0VBE
BEBBR VVVB
BBR RIVB
B
BEBBB R
VVI
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Current and Voltage Analysis
• The voltage at the collector is;
• The voltage drop across RC is
• VCE can be rewritten as
• The voltage across the reverse-biased CB junction is
CCR RIVC
CRCCCE VVV
CCCCCE RIVV
BECECB VVV
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Problems
Determine IB, IC, IE, VBE, VCB and VCE in the circuit. The transistor has a dc = 150.
Solution:A430
000,10
7.05
R
VVI
B
BEBBB
mA5.64
10430150II 6BdcC
V55.3)100)(105.64(10RIVV 3CCCCCE
V85.27.055.3VVV BECECB
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Problems
A base current of 50A is applied to the transistor in the adjacent Fig, and a voltage of 5V is dropped across RC. Determine the dc and dc of the transistor.
Solution:
mA51051000
5
R
VI 3
C
RC
C
1001050
105
I
I6
3
B
Cdc
99.01100
100
1dc
dcdc
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Problems
V7.0VBE
mA1.1109.3
7.05
R
VVI 3
B
BEBBB
mA55101.150II 3BdcC
V10.5180105515
RIVV3
CCCCCE
V40.410.57.0VVV CEBEBC
Find VCE, VBE and VCB in the given circuit.Solution:
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Problems: Homework1. Find IB, IE and IC in Fig.1.
dc = 0.98.Ans: IE = 1.3 mA, IB = 30,
IC = 1.27 mA.
2. Determine the terminal voltages of each transistor with respect to ground for circuit in Fig. 2. Also determine VCE, VBE and VBC.
Ans. VB = 10 V, VC = 20 V, VE = 9.3 V, VCE = 10.7, VBE = 0.7 V, VBC = -10 V.
Fig. 1
Fig. 2
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Modes of Operation
BJTs have three regions of operation:1. Active: BJT acts like an amplifier (most common use)2. Saturation - BJT acts like a short circuit3. Cutoff - BJT acts like an open circuit
BJT is used as a switchBy switchingbetween these two regions.
VCE (V)
IC(mA)
IB = 50 A
IB = 0
30
5 10 15 20 0
0
IB = 100 A
IB = 150 A
IB = 200 A
22.5
15
7.5
Saturation Region
Active Region
Cutoff Region
17
More about Transistor Regions
Cutoff: In this region, IB = 0 and VCE = VCC.
That is, both the base-Emitter and the base- collector junctions are reversed biased.
Under this condition, there is a very small amount of collector leakage current ICE0 due mainly to thermally produced carriers. It is usually neglected in circuit analysis.
18
More about Transistor Regions
Saturation: When theBase-emitter junction isforward biased and thebase current is increased,The collector current alsoIncreases (IC = dcIB) and VCE
Decreases (VCE = VCC – ICRC). When VCE reaches its saturation, there is no further change in IC.
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DC Load LineThe bottom of the loadLine is at ideal cutoffwhere IC = 0 and VCE = VCC.
The top of the load lineis at saturation whereIC = IC(sat) and VCE = VCE (sat).
20
Quiescent-Point (Q-Point)Operating point of an amplifier to state the values of
collector current (ICQ) and collector-emitter voltage (VCEQ).Determined by using transistor output characteristic and
DC load line.Quiescent means quiet, still or inactive.
21
ExampleThe transistor shown in Figure (a) is biased with
variable voltages VCC and VBB to obtain certain values of IB, IC, IE and VCE. The collector characteristic curves are shown in Figure (b). Find Q-point when:
(a) IB = 200A (b) 300A (c) 400A.
22
Solution: (a) IC = dcIB = 100200 10-6
= 20 mAVCE = VCC – ICRC =
10 – 2010-3220 = 5.6 VThis Q-Point is shown as Q1.
(b) IC = dcIB = 100300 10-3
= 30 mAVCE = VCC – ICRC =
10 – 3010-3220 = 3.4 VThis Q-Point is shown as Q2.
(c) IC = dcIB = 100400 10-6
= 20 mAVCE = VCC – ICRC
= 10 – 4010-3220 = 1.2 VThis Q-Point is shown as Q3.
23
Problem(a) Determine the intercept points of the dc load line on The vertical and horizontal Axes of the collector characteristic curves in the Fig. (b) Assume that you wish to bias the transistor with
IB = 20A. To what voltage must you change the VBB supply. What are IC and VCE at the Q-point , given that dc = 50. VBE =0.7
24
ProblemSolution:
(a) Horizontal interceptVCE = VCC = 20 V
Vertical intercept
(b) VBB = IBRB + VBE
= 2010-6 1 106 + 0.7 = 2.7 V
IC = dcIB = 502010-6 = 1 mA
VCE = VCC – ICRC = 20 - (110-3101000) = 10 V
mA210000
20
R
VI
C
CC)sat(C
