# Recommended Books Robert Boylestad and Louis Nashelsky, “Electronic Devices and Circuit Theory”, Prentice Hall, 7th Edition or Latest. Thomas L. Floyd,

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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:

This Lecture Current and Voltage Analysis of BJT – A Review

Types of Bipolar Junction Transistors
npn pnp n p n p n p C E C C Cross Section B B B B Schematic Symbol Schematic Symbol E E Collector doping is usually ~ 106 Base doping is slightly higher ~ 107 – 108 Emitter doping is much higher ~ 1015

BJT Equations npn IE = IB + IC VCE = -VBC + VBE pnp IE = IB + IC
VEC - E C E C - - + + VBE VBC IB VEB VCB IB + + - - B B npn IE = IB + IC VCE = -VBC + VBE pnp IE = IB + IC VEC = VEB - VCB

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.

Relationship between dc and dc
For an NPN transistor Dividing each term by IC we get or Similarly, we can prove that

Problems on dc and dc Determine dc and IE for IB = 50A and IC = 3.65 mA. Solution:

Problems on dc and dc 2. What is the dc when IC = 8.23mA and IE = 8.69 mA. Solution: A certain transistor exhibits an dc of Determine IC when IE = 9.35 mA.

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.

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

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

Problems Determine IB, IC, IE, VBE, VCB and VCE in the circuit. The transistor has a dc = 150. Solution:

Problems A base current of 50A 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:

Problems Find VCE, VBE and VCB in the given circuit. Solution:

Problems: Homework Find IB, IE and IC in Fig.1. dc = 0.98. Fig. 1
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

Modes of Operation BJTs have three regions of operation:
Active: BJT acts like an amplifier (most common use) Saturation - BJT acts like a short circuit Cutoff - BJT acts like an open circuit BJT is used as a switch By switching between these two 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.

Saturation: When the Base-emitter junction is forward biased and the base current is increased, The collector current also Increases (IC = dcIB) and VCE Decreases (VCE = VCC – ICRC). When VCE reaches its saturation, there is no further change in IC.

DC Load Line The bottom of the load Line is at ideal cutoff where IC = 0 and VCE = VCC. The top of the load line is at saturation where IC = IC(sat) and VCE = VCE (sat).

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.

Example The 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 = 200A (b) 300A (c) 400A.

Solution: IC = dcIB = 100200  10-6 = 20 mA VCE = VCC – ICRC = 10 – 2010-3220 = 5.6 V This Q-Point is shown as Q1. (b) IC = dcIB = 100300  10-3 = 30 mA 10 – 3010-3220 = 3.4 V This Q-Point is shown as Q2. (c) IC = dcIB = 100400  10-6 VCE = VCC – ICRC = 10 – 4010-3220 = 1.2 V This Q-Point is shown as Q3.

Problem 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 = 20A. 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

Problem (b) VBB = IBRB + VBE = 2010-6 1 106 + 0.7 = 2.7 V
Solution: Horizontal intercept VCE = VCC = 20 V Vertical intercept (b) VBB = IBRB + VBE = 2010-6 1 = 2.7 V IC = dcIB = 502010-6 = 1 mA VCE = VCC – ICRC = (110-3101000) = 10 V

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