SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II DMT 231 / 3 ELECTRONICS II Lecture IV AC Analysis II [BJT Common-Emitter Amplifier]

SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II

DMT 231 / 3ELECTRONICS II

Lecture IV AC Analysis II

[BJT Common-Emitter Amplifier]


RC

RB

vs

vO

VBB

VC

C

Example I

Given : = 100, VCC = 12V

VBE = 0.7V, RC = 6k,

RB = 50k, and VBB = 1.2V

Calculate the small-signal voltage gain.

Small-signal hybrid- equivalent circuit


1.

2.

3.

4.

5.

6.

AR

VVI

B

onBEBBBQ 10

50

7.02.1)(

mAAII BQCQ 1)10(100

VRIVV CCQCCCEQ 6)6)(1(12

kI

Vr

CQ

T 6.21

)026.0)(100(

VmAV

Ig

T

CQm /5.38

026.0

1

4.11

B

Cms

ov Rr

rRg

V

VA

Example I: Solutions


Basic Common-Emitter Amplifier Circuit

vs

RS

R1

R2

RC

CCvO

VCCExample II

Given : = 100, VCC = 12V

VBE(on) = 0.7V, RS = 0.5k,

RC = 6k, R1 = 93.7k, R2 = 6.3k

and VA = 100V.

Calculate the small-signal voltage gain.


Como RrVgV

sS

VRrRR

rRRV

21

21

Co

S

ms

ov Rr

RrRR

rRRg

V

VA

21

21

Coo RrR

rRRRi 21

B C

E

R1 \\ R2Vs

RS

RCrOr gmV

Vo

Ri Ro

Example II: SolutionSmall-signal equivalent circuit


• The basic common-emitter circuit used in previous analysis causes a serious defect :– If BJT with VBE(on) = 0.7 V is used, IB = 9.5 μA & IC = 0.95 mA– But, if new BJT with VBE(on) = 0.6 V is used, IB = 26 μA & BJT goes

into saturation; which is not acceptable Previous circuit is not practical

– So, the emitter resistor is included: Q-point is stabilized against variations in β, as will the voltage gain, AV

• Assumptions– CC acts as a short circuit– Early voltage = ∞ ==> ro neglected due to open circuit

Basic Common-Emitter Amplifier


Common-Emitter Amplifier with Emitter Resistor

CE amplifier with emitter resistor Small-signal equivalent circuit (with current gain parameter, β)

inside transistor


Common-Emitter Amplifier

with Emitter Resistor• ac output voltage

• Input voltage loop

• Input resistance, Rib

• Input resistance to amplifier, Ri

• Voltage divider equation of Vin to Vs

Remember: Assume VA is infinite, ro is neglected

Cbo RIV

Ebbbin RIIrIV

Eb

inib Rr

I

VR 1

ibi RRRR 21

sSi

iin V

RR

RV



with Emitter Resistor

Cont..

• So, small-signal voltage gain, AV

• If Ri >> Rs and (1 + β)RE >> rπ

Remember: Assume VA is infinite, ro is neglected

Si

i

E

Cv

sib

inC

s

Cb

s

ov

RR

R

Rr

RA

VR

VR

V

RI

V

VA

1

1

E

C

E

Cv R

R

R

RA

1


Example IIIGiven : = 100, VBE(on) = 0.7V,

VT = 26 mV and VA = ∞.Determine:

(i) Base-emitter input resistance, r

(ii) transconductance, gm

(iii) small-signal transistor output resistance, ro

(iv) Input resistance to the base, Rib

(v) Input resistance to the amplifier, Ri

(vi) Small-signal voltage gain, Av.

Common-Emitter Amplifier with Emitter Resistor


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 short circuit

to ground for the ac signals


with Emitter Bypass Capacitor

Small-signal hybrid-π equivalent circuit


Example IVGiven : = 100, VBE(on) = 0.7V,

VT = 26 mV and VA = 100.Determine:

(i) Quiescent value of base current, IBQ

(ii) Quiescent value of collector current, ICQ

(iii) Quiescent value of collector-emitter voltage, VCEQ

(iv) Base-emitter input resistance, r

(v) transconductance, gm

(vi) small-signal transistor output resistance, ro

(vii) Input resistance seen by the signal source, Rin

(viii) Output resistance looking back into the output terminal, Ro

(ix) Small-signal voltage gain, Av.

Common-Emitter Amplifier with Emitter Bypass

Capacitor


AC LOAD LINE ANALYSIS

• DC load line – Visualized the relationship between Q-point & transistor

characteristics

• AC load line – Visualized the relationship between small-signal response

& transistor characteristics

– Occurs when capacitors added in transistor circuit



Common-emitter amplifier with emitter bypass capacitor

Note: The next DC & AC load line analysis will be based on this circuit


• KVL on C-E loop

AC LOAD LINE ANALYSIS - DC Load Line

21

21

21

21

21

1 Slope

)(So,

11

1, when point,-QFor

)(1

1 when ,)(

1

)(

EEC

EECCQCEQ

EECCCCE

CEEECCECC

EEECECC

RRR

-

RRRIVVV

RRIRIVVV

IIVRRIVRI

VRRIVRIV


• KVL on C-E loop

AC LOAD LINE ANALYSIS - AC Load Line

1

11

1

1- Slope

)(

Assuming

0

EC

ECcEcCcce

ec

EeceCc

RR

RRiRiRiv

ii

RivRi

AC equivalent circuit


AC and DC Load Lines


RS

R1

R2 RE

RC

RL

vs

vO

CC1

CC2

VCC

vO

vs R1 R2

RS

RE

RC RL


Common-emitter amplifier with emitter resistor

AC equivalent circuit

Note: The DC & AC load line analysis will be based on these circuits


AC LOAD LINE ANALYSIS - DC Load Line

+ VCE

0

+ IC

ICQ

VCEQ

Q

CCV

EC

CC

RR

V

EC RR

1

Slope

• KVL at C-E loop

EC

ECCQCEQCC

CEQECCQCC

ECCECC

EECECCCC

RR

RRIVV

VRRIV

RIVRI

RIVRIV

1- Slope

assume point, -Q For

)(

)(

11

1


+ VCE0

+ IC

ICQ

VCEQ

Q

CCV

EC

CC

RR

V

i

v

LC

CEQ

RR

Vi

)( LCCQ RRIv )( LCCQCEQoffcut RRIVv

LC

CEQCQcsat RR

VIi

AC LOAD LINE ANALYSIS - AC Load Line


Self-ReadingChapter 6, page 399 - 402: Basic

Common-Emitter Amplifier Circuit

Chapter 6, page 402 - 409: Circuit with Emitter Resistor

Chapter 6, page 409 - 413: Circuit with Emitter Bypass Capacitor

Chapter 6, page 415 - 418: AC Load Line Analysis

Donald A. Neamen, MICROELECTRONICS Circuit Analysis and Design, Third Edition

SEM I 2008/09 LECTURE IV: C-E AC ANALYSIS II DMT 231 / 3 ELECTRONICS II Lecture IV AC Analysis II [BJT Common-Emitter Amplifier]

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