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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.

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Recommended books
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


This lecture
This Lecture

Current and Voltage Analysis of BJT – A Review


Types of bipolar junction transistors
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
BJT Equations

IE

IC

IE

IC

-

VCE

+

+

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 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
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
Problems on dc and dc

  • Determine dc and IE for IB = 50A and IC = 3.65 mA.

    Solution:


Problems on dc and dc1
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 0.96. Determine IC when IE = 9.35 mA.

    Solution:


Current and voltage analysis
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 analysis1
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 analysis2
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
Problems

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

Solution:


Problems1
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:


Problems2
Problems

Find VCE, VBE and VCB in the given circuit.

Solution:


Problems homework
Problems: Homework

  • 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


Modes of operation
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.


More about transistor regions
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.


More about transistor regions1
More about Transistor Regions

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
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
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
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

    VCE = VCC – ICRC =

    10 – 3010-3220 = 3.4 V

    This Q-Point is shown as Q2.

    (c) IC = dcIB = 100400  10-6

    = 20 mA

    VCE = VCC – ICRC

    = 10 – 4010-3220 = 1.2 V

    This Q-Point is shown as Q3.


Problem
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


Problem1
Problem

Solution:

  • Horizontal intercept

    VCE = VCC = 20 V

    Vertical intercept

    (b) VBB = IBRB + VBE

    = 2010-6 1 106 + 0.7

    = 2.7 V

    IC = dcIB = 502010-6 = 1 mA

    VCE = VCC – ICRC = 20 - (110-3101000) = 10 V


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