Chptr 1 - Basic BJT Amplifier(II)

8/10/2019 Chptr 1 - Basic BJT Amplifier(II)

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ANALOG ELECTRONICCIRCUITS 1

EKT 204

Basic BJT Amplifiers (Part 2)

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The basic common-emitter circuit used in

previous analysis causes a serious defect:

If BJT with VBE=0.7 V is used, IB=9.5 A & IC=0.95 mA

But, if new BJT with VBE=0.6 V is used, IB=26 A & BJTgoes into saturation; which is not acceptablePrevious

circuit is not practical

So, the emitter resistoris included: Q-point is stabilized

against variations in , as will the voltage gain, AV

Assumptions CCacts as a short circuit

Early voltage = ==> roneglected due to open circuit

Basic Common-Emitter

Amplifier

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Common-Emitter Amplifierwith Emitter Resistor

CE amplifier with emitter resistor Small-signal equivalent circuit

(with current gain parameter, )

inside

transistor

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ac output voltage

Input voltage loop

Input resistance, Rib

Input resistance to amplifier, Ri

Voltage divider equation of Vin to Vs

Remember: Assume VAis infinite,

rois neglected

Cbo RIV

Ebbbin RIIrIV

Eb

inib Rr

I

VR 1

ibi RRRR 21

s

Si

iin V

RR

RV

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Cont..

So, small-signal voltage gain, AV

If Ri>> Rsand (1 + )RE>> r

Remember: Assume VAis infinite,

rois neglected

Si

i

E

C

v

sib

inC

s

Cb

s

ov

RR

R

Rr

RA

VR

VR

V

RI

V

VA

1

1

EC

E

Cv

R

R

R

RA

1

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RS

R1

R2 RE

RC

vs

vO

CC

VCC

CE

B C

E

Vo

Vs RC

RS

r roR1|| R2 gmV

Emitter bypass capacitor, CE

provides a shor t c i rcu i tto

ground for the ac signals

Common-Emitter Amplifierwith Emitter Bypass Capacitor

Small-signal hybrid-

equivalent circuit

Emitter bypass capacitor is used to

short out a portion or all of emitter

resistance by the ac signal. Hence

no RE appear in the hybrid-

equivalent circuit

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DC & AC LOAD LINE

ANALYSIS

DC load line

Visualized the relationship between Q-point & transistorcharacteristics

AC load line

Visualized the relationship between small-signal response &

transistor characteristics

Occurs when capacitors added in transistor circuit

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Common Emitter Amplifier with

emitter bypass capacitor

Common-emitter

amplifier with

emitter bypass

capacitor

Example 1

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DC Load Line

KVL on C-E loop

21

21

21

21

21

1Slope

)(So,

1

1

1,whenpoint,-QFor

)(1

1

when,)(

1

)(

EEC

EECCQCEQ

EECCCCE

CEEECCECC

EEECECC

RRR

-

RRRIVVV

RRIRIVVV

IIVRRIVRI

VRRIVRIV

Solution...

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KVL on C-E loop

AC Load Line

1

11

1 0

EC

ECcEcCcce

ec

EeceCc

RR

RRiRiRiv

ii

RivRi

1-Slope

Assuming

)(

AC equivalent circuit

Solution...

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DC & AC Load Lines

Full

solution

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AC LOAD LINE ANALYSISDetermine the dc and ac load line. VBE=0.7V, =150,

VA=

Example 2

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DC Load Line

To determine dc Q-point, KVL around B-E loop

kRR

-

RIRIVVV

mAIImAII

ARR

VVI

RIVRIRIVRIV

EC

EEQCCQCEQ

BQEQBQCQ

EB

EBBQ

EBQEBBBQEEEBBBQ

15

11Slope

53.6)(point,-QFor

9.0)1(&894.0Then

96.5)1(

)1(

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AC Load Line

)//()//)((

/4.34

36.4

53.6;894.0

LCcLCmeo

CQ

Ao

T

CQm

CQ

T

ECQCQ

RRiRRvgvv

I

Vr

VmAVIg

kI

Vr

VVmAI

Small signal

hybrid-

equivalent

circuit

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DC & AC Load lines

Full

solution

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Maximum Symmetrical Swing

When symmetrical sinusoidal signal applied to the

input of an amplifier, the output generated is also a

symmetrical sinusoidal signal

AC load line is used to determine maximum output

symmetrical swing

If output is out of limit, portion of the output signal will beclipped & signal distortion will occur

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Maximum Symmetrical Swing

Steps to design a BJT amplifier formaximum symmetrical swing:

Write DC load line equation (relates ofICQ& VCEQ)

Write AC load line equation (relates ic,vce ;vce= - icReq,Req= effective ac resistance in C-E circuit)

Generally, ic= ICQIC(min), whereIC(min)= 0 or someother specified min collector current

Generally, vce

= VCEQ

VCE

(min), where VCE

(min) issome specified min C-E voltage

Combination of the above equations produce optimumICQ& VCEQvalues to obtain maximum symmetricalswing in output signal

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Maximum Symmetrical SwingExample 3

Determine the maximum symmetrical swing in the output voltage ofthe circuit given in Example 2.

Solution: From the dc & ac load line, the maximum negative swing in the Ic

is from 0.894 mA to zero (ICQ). So, the maximum possible peak-to-peak ac collector current:

The max. symmetrical peak-to-peak output voltage:

Maximum instantaneous collector current:

mA79.1)894.0(2(min))(2 CCQc IIi

V56.2)2||5)(79.1()||(|||||| LCceqcce RRiRiv

mA79.1894.0894.0||2

1

cCQC iIi

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Self-Reading

Textbook:Donald A. Neamen,

MICROELECTRONICS Circuit Analysis &

Design,3rd Edition, McGraw Hill

International Edition, 2007

Chapter 6:Basic BJT Amplifiers

Page:397-413, 415-424.

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Exercise

Textbook:Donald A. Neamen,

MICROELECTRONICS Circuit Analysis &

Design,3rd Edition, McGraw Hill

International Edition, 2007

Exercise 6.5, 6.6, 6.7,6.9

Exercise 6.10 , 6.11

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Chptr 1 - Basic BJT Amplifier(II)

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