Chapter 4 - Ele232 [Compatibility Mode]

8/13/2019 Chapter 4 - Ele232 [Compatibility Mode]

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Chapter : DC Analysis of BJTNorsabrina Sihab

Faculty of Electrical Engineering,

Universiti Teknologi MARA

Pulau Pinang

Tel : 04-3823355

. . .

1

Chapter 4 DC Analysis of BJT2

At the end of this chapter, students able to: Analyze the dc biasing of single stage BJT amplifier circuit.

Calculates DC volta es and DC currents.

Calculates maximum output voltage, VO(max).

Electronics 1Updated May 2011NorsabrinaSihab


BiasingBiasing refers to the DC voltages applied to a transistor in order toturn it on so t at it can amp i y t e AC signa .

The DC input establishes an operating or quiescent point called the

QQ--pointpoint.In DC analysis all capacitor act as open circuit.

DC Biasing circuits

1. Fixed-bias circuit

2. Emitter-stabilized bias circuit

3. Voltage divider bias circuit

Electronics 1Updated May 2011Norsabrina Sihab

.

5. Voltage feedback


Proper biasing circuit which it operate in linear region and circuithave centered Q-point or midpoint biased

mproper as ng cause

Distortion in the output signal

Produce limited or clipped at output signali o

A

Linear operation for an inverting amplifier


Non-linear operation for an inverting amplifier


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1) Fixed-bias circuit - DC analysis

Input LoopInput Loop Output LoopOutput Loop

12

KVL at loop 1: The collector current is given by:

Solving for the base current:

CC B B BE =

KVL at loop 2 :

)2(= BIIC


)1(R

VVI

B

BECCB

=

)3(=

=

CCCCCE

CECCCC

RIVV


Load Line for FixedLoad Line for Fixed--bias circuitbias circuit

)3(= RIVV

From equation (3) :

The end points of the load line

are:

VVC

CC

satC

CE

R

VI

0=

=

AICCsatCE CVV

0==

where the value of RBB sets the value of IBB

he Q-point is the particular operating point:


that sets the values of VCECE and ICC


Circuit Values Affect the QCircuit Values Affect the Q--PointPoint

Increasing level of IB

Increasing level of RC

Decreasingvalue of V



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-

Example : Q point at midway between cut-off and saturation

Csat

ICQQ-point


CE(Cut-off)CEQ


Determine a) IBQ and ICQ, b) VCEQ, c) VB and VC, d) VBC



2) Emitter2) Emitter--Stabilized Bias CircuitStabilized Bias Circuit

Adding a resistor (RE) to improve the stability that is DC biascurrent and voltage remain closer to where they set eventemperature c ange.



2) Emitter2) Emitter--Stabilized Bias CircuitStabilized Bias Circuit DC AnalysisDC Analysis

npu oopnpu oop u pu oopu pu oop

KVL at loop 1 :

1 2

0RI-V-RI- EEBEBBCC =+V

Since I = (+1)I :

KVL at loop 2 :

0RIVRIV EECECCCC =+

0R1)I(-RI-V EBBBCC =+ BEV

Solving for IB:

E C

-(2)-)R(RIVV ECCCCCE +=

Also: EEE RIV =


)1(1)R(R

V-VI

EB

BECCB

++=

EBEBBCCB

CCCCECEC

VVRIVV

RI-VVVV

+==

=+=


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Load Line for EmitterLoad Line for Emitter--bias circuitbias circuit

From equation (2) :

-(2)-)R(RIVV ECCCCCE +=

The end points of the load lineare:

VVEC

CCC

CE

sat RR

VI

0=+

=

AICCCE CVV

0==



Determine a) IB, b) IC, c) VCE, d) VC, e) VE, f) VB, g) VBC



3) Voltage Divider Bias3) Voltage Divider Bias

This is a very stable.

The currents and

independent ofvariations in .

There are two waysof analyzing thevoltage divider biascircuit :-1. Exact analysis2. Approximate


ana ys s


3a) Exact Analysis

Step 1 : Redraw circuit Step 3 : Replace thevenin equiv.circuit

BEThB

VVI

=

. ep : pp y o e erm ne Band VCE

ETh ++

( )ECCCCCE RRIVV +=


21RRRTh=21

2

2 RRVV CCRTh

+==


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Determine VCE and IC



Used for circuit that have a very smallIB due to large resistance between

.

If Ri R2, IB < I2. So approx. I1 I2

E 2.

21

CC2B

RR

VRV

=

V

BEBE VVV

EE

RI =

EECCCCCE RI-RI-VV =

II CE


)R(R-IVV ECCCCCE


Determine VCE and IC



4) DC Bias with Voltage Feedback

Another way toimprove the stabilityof a bias circuit is toadd a feedback pathfrom collector tobase.

In this bias circuitthe Q-point is only

on the transistor

beta, .



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KVL at input loop:

0= EEBEBBCCCC RIVRIRIV

B C, .

CBCC IIII += EC II

Knowing IC = IB and IE IC, theloop equation becomes:

= 1

VV

EBBEBBCBCC

Solving for IB:


)R(RRI

ECB

B++

=


KVL at output loop:IERE + VCE + ICRC VCC = 0

Since IC IC as IB=0 and IC = IB:I (R + R ) + V V =0

Solving for VCE:

VCE = VCC IC(RC + RE)



Determine IB and VC



1) COLLECTOR FEEDBACK

Determine ICQ, VCEQ, VB, VC, VE and VBC

Vcc = +20V

CB

B2

Vo



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2) Common Collector

Determine VCEQ and IE



3) Common Base

Determine VCB, IB and VCE

CB Co

VoVi

E C

EE CC


Chapter 4 DC Analysis of BJT

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Chapter 4 - Ele232 [Compatibility Mode]

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