BJT Biasing (Complete)

7/28/2019 BJT Biasing (Complete)

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1. Structure of the Lesson

Intro

Class

Class end

Study

Assessment

Review

1. intro Overview of the lesson2. Learning objective present learning objective of the lesson3. Table of Content structure of the topics and subtopics in thelesson

4. Lecture (75-90 minutes) present the lecture in detailed topics that covers all the

learning objectives of the lesson.- each topics should be divided into subtopics

(5-15 min in length is recommended)- if a subtopic goes over 15 minutes divide the subtopic into

series of subtopics.


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Course Circuit theory and Laboratory

Lesson # Lesson 2

Title Bipolar Junction Transistor biasing

SME Dr. Nguyen Vu Thang


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Intro

The basic of electronic system is semiconductor device. The most famous and commonly used

active devices are BJTs (Bipolar Junction Transistors) and FET (Field Effect Transistor).

BJT can be use as amplifier and logic switches.

Pre-required knowledge: structure and operation of diode


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Learning Objectives Table of Content

At the end of this chapter, the students shoul

be able to:

Understand the basic transistor operation

& configuration

Discuss transistor parameters and

characteristics

Operating point

Bias circuit

Design Operations

Transistor switching network

PNP Transistors

Bias Stabilization


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Operating Point

Operating point

within the limits of

operation of

transistor

IC (MA)

ICMAX

Saturation

5

10

15

20

25

IB = 0 A

10 A

20 A

30 A

40 A

50 A

60 A

70 A

80 A

A

B

C

D

VCE (V)VCESAT Cutoff

VCEMAX

5 10 15 20

PCmax


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Bias Circuit

Fixed bias Circuit

Emitter stabilized bias Circuit

Voltage - divider bias Circuit DC Bias with Voltage Feedback

Miscellaneous bias configuration


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Fixed-bias Circuit

Vcc

RB RC IC

IB

B

C

E

+

__+

VBE

C1

C2

RB

RC

IC

IB

B

C

E

+

+

VBE

__

Vcc Vcc


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Fixed-bias Circuit

RB

Vcc IB

+

+

_

_VBE

Forward Bias of Base-Emitter

Base-Emitter loop

+ VccIBRB VBE= 0

=

Collector-Emitter loop

Collector-Emitter loop

+

IC

+

_

RC

Vcc

VCE

_

=

+ = 0

=


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Fixed-bias Circuit

=

Load Line Analysis

Vcc

RC

RB

IC

IB

VCE

_

+

+

_

(V)

0

IC (MA)

IB = 0 A

10 A

20 A

30 A

40 A

50 A

VCE (V)5 10 15


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Fixed-bias Circuit**


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Fixed-bias Circuit


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Fixed-bias Circuit


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Emitter stabilized bias circuit


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(V)

0


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For the emitter bias network, determine:


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Voltage -divider bias circuit


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Voltage -divider bias circuitDetermine the dc bias voltage and the current IC for the voltagedivider

configuration


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Solution

RTH = R1 R2 = 3.55k

ETH =2V

IB = 6.05 A

IC = IB= (140)(6.05 A)

=0.88mA

VCE = VCC IC ( RC + RE )

=22V( 0.85 mA )( 10 + 1.5)= 22v 9.8V

=12.22 V


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DC Bias with Voltage Feedback


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Determine the quiesent levels of ICQ and VCEQ

EXAMPLE


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repeat above example using a beta of 135 ( 50 % more than above example )


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Determine the dc level of IB and Vc


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ExampleFor the network of fig

a)Determine ICQ and VCEQ

b)Find VB ,VC , VE and VBC


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Miscellaneous bias configuration Solution

a)The absence ofREreduces the reflection of resistive levels to

simply that ofRCand the equation for IB reduces to

IB=;

= ;.

()(.)=

9.

.

=15.51 A

ICQ=IB =(120)(15.51 A)

= 1.86 mA

VCEQ =VCC- ICRC

= 20V(1.86Ma)(4.7k)

= 11.26VVB =VBE=0.7 V

VC =VCE=11.26 V

VE =0 V

VBC=VB VC=0.7 V - 11.26 V

=-10.56 V


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EXAMPLEDetermine VC and VB for the network of fig


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Miscellaneous bias configurationEXAMPLE

Determine VCEQand IE for the network of fig


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Miscellaneous bias configurationSolution:

Applying Kirchhoffs voltage law to

the input circuit will result in

-IBRBVBEIERE+VEE= 0

But IE= (+1)IB

And VEE-VBE-(+1)IBRB-IBRB = 0

With IB=;

()

Substituting values yields

IB=;.

(9)()

=

9.

=

9.

= 45.73 A

IC= IB

= (90)(45.73A)

= 4.12 mA

Applying Kirchhoffs voltage law to the output

circuit, we have

-VEE + IERE+ VCE=0

But IE =(+1)IBAnd VCEQ = VEE(+1)IBRE

=20 V(91)(45.73 A)(2k)

=11.68 V

IE =4.16 mA


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EXAMPLE

Determine the voltage VCB and the current

IB for the commom-base configuration


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Miscellaneous bias configurationSolution

Applying Kirchhorrs voltage law to the input circuit yields

-VEE+ IERE+ VBE= 0

And IE=;

Substituting values, we obtain

IE= ;.

. = 2.75 mA

Applying Kirchhoffs voltage law to the output circuit gives

-VCB + ICRCVCC= 0

And VCB = VCCICRCwith IC IE

=10 V(2.75 mA)(2.4k)

=3.4 V

IB =

=.

= 45.8 A


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Design Operations

Design of a Bias Circuit with an Emitte Feedback Resistor


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Design Operations

Determine the resistor values for the network for the indicated operating point and supplyvoltage


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Design Operations

VE =

VCC =

= 2V

RE =

=

=

mA= 1k

RC =

=;CE;

=;;

mA = 4k

IB =

=

mA

= 13.33A

RB =

=

;BE;

=

;.;

.= 1.3M


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Design OperationsDesign of Current-Gain-Stabilized (Beta-Independent) Circuit


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Design Operations

RE

_

R1

R2

RC C2

CE

C1

VCC = 20V

100F

10F

10F = 8

= 10

AC input

AC output

+

Determine the levels of RC, RE, R1, and R2, for the network of this Figure for

the operating point indicated


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Design Operations

RE

_

R1

R2

RC C2

CE

C1

VCC = 20V

100F

10F

10F

= 8

= 10AC

input

AC

output+

VE =

VCC =

= 2V

RE =

=

=

mA= 200

RC =

=

;CE;

=

;;

mA= 1k

VB = VBE + VE = 0.7 + 2 = 2.7V


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RB

RC

VCC = 5V

68

0.82

= 125

Vi

5V

0V

VC

t

5V

0V

0 0

6.1

= 5


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RB

RC

VCC = 5V

= 250

t

VC

10V

0V

10V

t

VC

10V

0V 0V

Example:Determine RB and RC for the transistor inverter of this Figure if ICsax = 10mA


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Transistor switching networkAt saturation:

ICsat=

R

and 10mA =

R

so that RC=

mA= 1k

At saturation:

IB=

=

mA

= 40AChoosing IB = 60A to ensure saturation and using:

IB =i;.

R

We obtain:

RB=i;.

I=

;.

A= 155k

Choose RB= 150k, which is a standard value. Then

IB =i;.

R=

;.

k= 62A

And IB = 62A

= 40A

Therefore, use

RB= 150k, R

C= 1k


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PNP Transistor

IC+

RC

_

+

_

+_

VCE

+

_

RE

_

IE

IB

-IERE +VBE - IBRB +VCC = 0

IB =CCBE

R()R

-IERE +VCE ICRC +VCC = 0

VCE = -VCC + IC(RE+RC)


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PNP Transistor

CB

E

+VBE_

10F

10F47

2.4

1.1

10

Vi

18

Example: Determine VCE for the voltage-divider bias configuration of thisFigure


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PNP TransistorTesting the condition: RE 10R2

Results in: (120)(1.1k) 10(10 k)

132 k 100 k (sastified)

Solving for VB, we have:

VB =RCC

R

R

= k (; )

k k= 3.16V

Note the similarity in format of the equation with the resulting negative voltage for

VB.

Applying Kirchhoffs voltage law around the base-emitter loop yields: VB - VBE VE = 0

and VE = VB VBE

Substituting values, we obtain

VE= -3.16 V (-0.7 V)

= -3.16 V + 0.7 V

= -2.46 V


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PNP Transistor


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Bias stabilization

: : 7.5 (

)

: 10


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Bias stabilization

Shift in dc bias point due to change in temperature: 25 C and 100 C


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Bias stabilization

=

=

=

=

+ ( + 1)

=

=( 1 + /)

( 1 + + /)

:

= ( + 1)1 + /

+ 1 + /

/ ( + 1): = + 1

= ( + 1) 1 + /

+ 1 + /

= ( + 1)1 + /

+ 1 + /


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Bias stabilization

ExampleCalculate the stability factor and the change in IC from 25

oC to 100oC for the transistor

defined by the Table for the following emitter-bias arrangements.

(a) RB/RE = 250 ( RB = 250 RE)

(b) RB/RE = 10 ( RB = 10 RE)

(c) RB/RE = 0.01 ( RE = 100 RB)


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Bias stabilization


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Bias stabilizationDetermine the stability factor S(VBE) and the change in IC from 25

oC to 100oC for the transistor defined by the

Table for the following bias arrangements.

(a) Fixed-bias with RB= 240k and = 100.

(b) Emitter-bias with RB=240k, RE=1k, and =100.

(c) Emitter-bias with RB=47k, RE=4.7k, and =100.


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Bias stabilization


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Bias stabilization

Determine the stability factor S(VBE) and the change in IC from 25

o

C to 100

o

C for the transistor defined bythe Table for the following bias arrangements.(a) Fixed-bias with RB = 240k and = 100.(b) Emitter-bias with RB=240k, RE=1k, and =100.(c) Emitter-bias with RB=47k, RE=4.7k, and =100.

Assessment Quiz 1


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Assessment Quiz 2


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Assessment Quiz 3


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Assessment Quiz 4


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IB, IC, VC?

Assessment Quiz 5


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Determine the following for the network of the above figure

(a) S(ICO)

(b) S(VBE

)

(c) S() using T1as the temperature at which the parameter values are specified at (T2) as 25% more than (T1).

(d) Determine the net change in IC if a change in operating conditions results in ICO increasing from 0.2 to 10A,

VBE drops from 0.7 to 0.5V, and increases 25%.

Review


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Summary: this chapter provide Operating point, Bias circuits, Transistor switching network,

Bias Stabilization.

Main Bias circuits are: Fixed bias Circuit , Emitter stabilized bias Circuit,

Voltage - divider bias Circuit

BJT Biasing (Complete)

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