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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:

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This Lecture Current and Voltage Analysis of BJT – A Review 2

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Types of Bipolar Junction Transistors npnpnp npn B C pnp E B Cross Section B C E Schematic Symbol B C E Collector doping is usually ~ 10 6Collector doping is usually ~ 10 6 Base doping is slightly higher ~ 10 7 – 10 8Base doping is slightly higher ~ 10 7 – 10 8 Emitter doping is much higher ~ 10 15Emitter doping is much higher ~

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BJT Equations B C E IEIEIEIE ICICICIC IBIBIBIB - + V BE V BC V CE B C E IEIEIEIE ICICICIC IBIBIBIB - + V EB V CB V EC npn I E = I B + I C V CE = -V BC + V BE pnp I E = I B + I C V EC = V EB - V CB 4

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DC Beta and DC Alpha DC Beta ( dc ) : The ratio of the dc collector current (I c ) to the dc base current (I B ) 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 (I c ) to the dc emitter current (I E ). – Typically values of dc range from 0.95 to 0.99, but it is always less than unity. 5

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Relationship between dc and dc For an NPN transistor Dividing each term by I C we get or Similarly, we can prove that 6

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Problems on dc and dc 1.Determine dc and I E for I B = 50 A and I C = 3.65 mA. Solution: 7

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Problems on dc and dc 2.What is the dc when I C = 8.23mA and I E = 8.69 mA. Solution: 3.A certain transistor exhibits an dc of Determine I C when I E = 9.35 mA. Solution: 8

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Current and Voltage Analysis I B : dc base current I E : dc emitter current I C : dc collector current V BE : dc voltage across base-emitter junction V CB : dc voltage across collector-base junction V CE : dc voltage from collector to emitter Transistor bias circuit. 9

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Current and Voltage Analysis 10 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 R B is By Ohm’s law; Solving for I B

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Current and Voltage Analysis 11 The voltage at the collector is; The voltage drop across R C is V CE can be rewritten as The voltage across the reverse-biased CB junction is

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Problems 12 Determine I B, I C, I E, V BE, V CB and V CE in the circuit. The transistor has a dc = 150. Solution:

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Problems A base current of 50 A is applied to the transistor in the adjacent Fig, and a voltage of 5V is dropped across R C. Determine the dc and dc of the transistor. Solution: 13

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Problems 14 Find V CE, V BE and V CB in the given circuit. Solution:

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Problems: Homework 1.Find I B, I E and I C in Fig.1. dc = Ans: I E = 1.3 mA, I B = 30 , I C = 1.27 mA. 2. Determine the terminal voltages of each transistor with respect to ground for circuit in Fig. 2. Also determine V CE, V BE and V BC. Ans. V B = 10 V, V C = 20 V, V E = 9.3 V, V CE = 10.7, V BE = 0.7 V, V BC = -10 V. 15 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 circuit 3.Cutoff - BJT acts like an open circuit 16 BJT is used as a switch By switching between these two regions.

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More about Transistor Regions Cutoff: In this region, I B = 0 and V CE = V CC. 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 I CE0 due mainly to thermally produced carriers. It is usually neglected in circuit analysis. 17

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More about Transistor Regions 18 Saturation: When the Base-emitter junction is forward biased and the base current is increased, The collector current also Increases (I C = dc I B ) and V CE Decreases (V CE = V CC – I C R C ). When V CE reaches its saturation, there is no further change in I C.

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DC Load Line The bottom of the load Line is at ideal cutoff where I C = 0 and V CE = V CC. The top of the load line is at saturation where I C = I C(sat) and V CE = V CE (sat). 19

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Quiescent-Point (Q-Point) Operating point of an amplifier to state the values of collector current (I CQ ) and collector-emitter voltage (V CEQ ). Determined by using transistor output characteristic and DC load line. Quiescent means quiet, still or inactive. 20

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Example The transistor shown in Figure (a) is biased with variable voltages V CC and V BB to obtain certain values of I B, I C, I E and V CE. The collector characteristic curves are shown in Figure (b). Find Q-point when: (a) I B = 200 A (b) 300 A (c) 400 A. 21

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Solution: (a)I C = dc I B = 100 200 = 20 mA V CE = V CC – I C R C = 10 – 20 220 = 5.6 V This Q-Point is shown as Q 1. (b) I C = dc I B = 100 300 = 30 mA V CE = V CC – I C R C = 10 – 30 220 = 3.4 V This Q-Point is shown as Q 2. (c) I C = dc I B = 100 400 = 20 mA V CE = V CC – I C R C = 10 – 40 220 = 1.2 V This Q-Point is shown as Q 3. 22

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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 I B = 20 A. To what voltage must you change the V BB supply. What are I C and V CE at the Q-point, given that dc = 50. V BE =0.7 23

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Problem Solution: (a)Horizontal intercept V CE = V CC = 20 V Vertical intercept (b) V BB = I B R B + V BE = 20 1 = 2.7 V I C = dc I B = 50 20 = 1 mA V CE = V CC – I C R C = 20 - (1 10 1000) = 10 V 24

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