4 Common Emitter Amplifiers CH5

8/11/2019 4 Common Emitter Amplifiers CH5

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Common-Emitter Amplifiers


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Objectives

DescribeAV,AI,APassociated with the threeamplifier configurations.

Describe the input/output voltage and currentphase relationships.

Calculate the ac emitter resistance of a transistor. Discuss two roles of the capacitors in the circuits.

Derive the ac equivalent circuit for a givenamplifier.

Explain how the voltage gain drift due totemperature occurs.

Discuss the relationship between the loadresistance and voltage gain of a CE amplifier.


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Objectives (Cont.)

Calculate Zin(base)and Zinfor a CE amplifier Discuss the effects of swamping on the ac

characteristics of a CE amplifier

List and describe the four ac h-parameters


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Common-emitter input-output phaserelationship.

RCR

B

Q1

Ai= 100

15 A10 A 5 A

VB

IB

1.5 mA1 mA

500 A

VC

IC

8 V6 V4 V

+VCC


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CE transistor characteristics, dc load line, andsinusoidal variation in base current, collector current,

and collectoremitter voltage

5


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Example

21 2

2.2k10V 1.8V

12.2kB CC

RV V

R R

0.7V1.1mA

1k

E B

EE

V VI

R

25mV 25mV22.7

1.1mAe

E

rI

+10 V

hFE= 300

RE

1k

RC4k

R1

10k

R2

2.2k

RL

15k


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Graphical determination of acemitter resistance.

BE

eE

V

r I

IE

VBE

VBE VBE

Q2

IE

IE

Q1


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The determination of ac beta.

IB

IC

Q

IB

IC

C cac

B b

I i

I i

hFE

= dc beta

hfe = ac beta


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Roles of Capacitors in Amplifiers

1. Acoupling capacitorpasses an acsignal from one amplifier to another,while providing dc isolation between thetwo.

2. A bypass capacitor

is used to shortcircuit an ac signal to ground while notaffecting the dc operation of the circuit.

1

2CX

fC The higher the freq.,

the lower the capacitor impedance.


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Coupling capacitors in amultistage amplifier.

Load

VCC

CC1

CC2

CC3


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AC coupling.

Load

GND

CC1

CC2

CC3


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DC isolation.

Load

VCC

CC1

CC2

CC3


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Capacitive vs. direct coupling.

VCC

CC

VCC


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Bypass capacitors.

For AC analysis For DC analysis

VCC

CB

VCC

CB

GND

CB


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Typical common-emitter amplifiersignals.

VCC

0V

1.8V

5.6V

1.1VDC

1.1VDC

1.8V

0V

5.6V


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Deriving the CE ac-equivalentcircuit.

RE

RCR

1

R2

RE

RCR

1

R2

RS= 0 (ideally)VCC

RE

RCR

1

R2

(a) (b) (c)


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Example

VCC

Q1

RE

RCR1

R2

RL

CC1

CC2

CB


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Example Cont.

GND

Q1

RE

RCR1

R2

RL


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Example Cont.

Q1

RC

R1

R2

RL

Q1

R1||R

2

RC||RL

(c)

(d)


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Voltage Gain of CE

out

in

c C Cv

e e e

v i r r A

v i r r

Q1

R1||R

2

RC||R

Lib

ic= i

b

r'e

vin

vout

Q1

R1||R

2

RC||R

L

in e ev i r

LCCCc

LCcou t

RRrwhereri

RRiv

||

||


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Example (1)

+20 V

hFE

= 200

RE

2.2k

RC

12k

R1

150k

R220k

RL

50k

Transform the base circuit to itsThevenin equivalent.

2.353V 0.7V

( 1) 17.65k 201 2.2k

3.595A

th BE B

th FE E

V VI

R h R


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Example (2)

+20 V

hFE

= 200

RE

2.2k

RC

12k

R1

150k

R220k

RL

50k

200 3.595A 718.9A

1 201 3.595A 722.5A

C FE B

E FE B

I h I

I h I

20V 718.9A 12k 722.5A 2.2k

9.784V active

CE CC C C E E V V I R I R

9.677k279.7

34.6

Cv

e

rA

r


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Example (1)

Transform the base circuit to itsThevenin equivalent.

2.070V 0.7V

( 1) 3.727k 31 1.2k

33.49A

th BE B

th FE E

V VI

R h R

+10 V

hFE

= 30

hfe= 200

RE

1.2k

RC

1.5kR118k

R24.7k

RL

5k


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Example (2)

30 33.49A 1.005mA

1 31 33.49A 1.038mAC FE B

E FE B

I h II h I

10V 1.005mA 1.5k 1.038mA 1.2k

7.247V active

CE CC C C E E V V I R I R

1.154k47.91

24.08

Cv

e

rA

r

+10 V

hFE

= 30

hfe= 200

RE

1.2k

RC

1.5kR118k

R24.7k

RL

5k


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CE Current Gain

Ai

is always less than hfe

due to two factors:

1.The input ac current is divided betweenthe transistor and the biasing network.

2.The output collector current is divided

between the collector resistor and theload.

out

in

i

iA

i


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Power Gain

p i vA A A

Example The amplifier shown in Fig. has values of

Av= 45.3 and Ai= 20. Determine the power gain(Ap) of the amplifier and the output power whenPin= 80 W.

20 45.3 906p i vA A A

out in 906 80W 72.48mWpP A P


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The Effects of Loading

The lower the load resistance is, the lower the voltage gain.

RC

3kR

L

6k

R1||R

2r'e=25

RC

3kR

L

12k

R1||R

2r'e=25


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Example

The load in previous Fig. is open. Calculatethe open-load voltage gain of the circuit.

3kC Cr R

3k120 (max. gain)

25

Cv

e

rA

r


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Calculating AmplifierInput Impedance

Zin(base)

Zin

R1

R2

Biasing circuit

VCC

IN(base)DC: 1FE E

FE E

R h R

h R

in(base)AC: 1fe efe e

Z h r

h r


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Example

in(base)

200 24 4.8kfe e

Z h r

+10 V

hFE= 30hfe

= 200

RE

1.2k

RC

1.5kR118k

R2

4.7k

RL5k

Zin(base)

Zin

Q1

R2

4.7k

R1

18k

RC

1.5k

RL

5k

r'e= 24

hfe= 200

Zin(base)

Zin


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Calculating the Value of Ai

out

in

i

iA

i cfe

b

ih

i

in in in in(base)

in(base)in

in in

1

b

b fe eb

v i Z i Z

i h ri ZiZ Z

in

in(base)

Ci fe

L

Z rA h

Z R

Q1

RC

R1

R2

RL

icib

Zin(base)

iout

vout

iin

vin

Zin


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Example

Calculate the value ofAifor the circuit shown below

in(base) 200 24 4.8kfe eZ h r inin(base)

2.1k 1.15k200

4.8k 5k

20.2

Ci fe

L

Z rA hZ R

Q1

R2

4.7kR

1

18k

RC

1.5kR

L

5k

r'e= 24

hfe

= 200

Zin(base)

Zin


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Multistage Amp.Gain Calculations

1 2 3vT v v vA A A A

1 2 3iT i i iA A A A

pT vT iTA A A

Procedure:

1. Do dc analysis

2. Find refor each stage

3. Find rCfor each stage

4. Using reand rCto findAv

for each stage

Input impedance of next stage is the load of current stage.

(Zinof next stage is RLof current stage)


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Example (1)

( ) 1 201 17.4 3.497kin base fe eZ h r DetermineAvof the 1

st

stage. Assume that refor the 1ststage is 19.8 and refor the 2

ndstage isfound to be 17.4 . Forthe 2ndstage, hfeis 200.

The input impedance of the 2

nd

stage:

+15V

CC1

CC2

CC3

R1

22k

R2

3.3kR

4

1k C

B1

R8

1k C

B2

R3

5k R515k

R6

2.5k

R7

5k

RL

10kQ

1 Q

2

hFE

= 150

hfe

= 200


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Example (2)

+15V

CC1

CC2

CC3

R1

22k

R2

3.3kR

4

1k C

B1

R8

1k C

B2

R3

5k R522k

R6

3.3k

R7

5k

RL

10kQ

1 Q

2

hFE

= 150

hfe

= 200

Finally,Avfor the 1ststage is found as

1.05k53.03

19.8

Cv

e

rA

r


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Example (1)

Determine the value ofAvTfor the amplifier inprevious Figure.

rCfor the 2ndstage can be found as

3.33k

191.3817.4

C

v

e

r

A r

Avfor the 2ndstage is found as

31 2 53.03 191.38 10.15 10vT v vA A A

Th d CE lifi d it


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The swamped CE amplifier and its acequivalent ckt.

Swamped amplifier is anamplifier that uses a partiallybypassed emitter resistance toincrease ac emitter resistance.

Also referred to as a gain-stabilized amplifier.

RC

rE

RE

CB

R1

R2

C1

C2 R

L

+VCC

Q1

rE

Q1

rC

R1//R

2


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Avof Swamped Amp.

Cv

e E

rA

r r

rE

Q1

rC

R1//R

2

rC

R1//R

2

rE

r'e

ib

ic=

acib

Q1

vin

vout

out c C

v

in e e E

v i rA

v i r r


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Example (2)

RC

1.5k

rE

300

RE

910 C

B

R1

18k

R2

4.7k

RL

10k

+10V

hFE

= 200

hfe= 150

10V 1.110mA 1.5k 1.116mA 1210

6.985V (active)

CE CC C C E E E V V I R I r R

25mV 25mV 22.411.116mA

e

E

rI

1.304k4.046

22.41 300

Cv

e E

rA

r r


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Example

Determine the change in gain for the amplifier when redoubles in value.

1.304k3.782

44.82 300

Cv

e E

rA

r r

4.046 3.782 0.2639vA

Swamping improves the gain stability of a CEamplifier when rE>> re.


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The Effect of Swamping on Zin

in(base) 1fe efe e

Z h r

h r

in(base) 1fe e E

fe e E

Z h r r

h r r

b

c

eZin(base)

r'e

b

c

e

Zin(base)

r'e

rE


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Gain stabilization.

Av

-rC/ re -rC/ (re+rE)

Zin(base) hfere hfe(re+rE)

Advantage Higher values ofAvthanthe swamped amplifier.

Relatively stableAv.Much smaller distortion.

Disadvantage Relatively unstable valuesofAv.

LowerAvthan thestandard amplifier.

RE

RE

rE


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The Hybrid Equivalent Model

Hybrid model is derived from two-port system.

V1

I1 2

2'

I2

V2

1

1'

Two-PortSystem


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Six Circuit-Parameter Modelsfor Two-Port Systems

IndependentVariables

DependentVariables

Circuit Parameters

I1,I2 V1,V2 Impedance Z

V1,V2 I1,I2 Admittance Y

V1,I2 I1,V2 Inverse Hybrid g

I1,V2 V1,I2 Hybrid h

V2,I2

V1,I1

Transmission T

V1,I1

V2,I2

Inverse TransmissionT


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Equations for Hybrid Model

1 11 1 12 2

2 21 1 22 2

V h I h V

I h I h V

LetV1= Vi,I1= Ii,V2= Vo,andI2= Io.Then

11 12

21 22

i i o

o i o

V h I h V I h I h V

E i l t Ci it f


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Equivalent Circuit forHybrid Model

hi

hrV

oh

oh

fI

iV

i

Ii 2

2'

Io

Vo

1

1'

11 12

21 22

i i o i i r o

o i o f i o o

V h I h V h I h V

I h I h V h I h V


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h-Parameters

h11= hi= Input Resistanceh12= hr= Reverse Transfer Voltage Ratioh21= hf= Forward Transfer Current Ratioh22= ho= Output Admittance

11 12

21 22

0 0

0 0

i i

o ii o

o o

o ii o

V Vh hV II V

I Ih hV II V


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h-Parameters for CE Amp.

hie= the base input impedance

hfe= the base-to-collector current gain

hoe= the output admittance

hre= the reverse voltage feedback ratio

be ie b re ce

c fe b oe ce

v h i h vi h i h v

H b id M d l f


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Hybrid Model forCE Configuration

hie

hre

vce

hoe

hfe

ib

vbe

ib c

ic

vce

b

e

ie

e

cb

ie

ici

b

in (output shorted)

(output shorted)

ie

b

cfe

b

vhi

ih

i

(input open)

(input open)

coe

ce

bere

ce

ihv

vh

v

May beneglected.


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h-parameters of 2N3904


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Hybrid Model without hreand hoe

fe ach

in(base)1ie fe e fe eh h r h r Z

in Ci fe

ie L

Z rA h

h R

fe C

v

ie

h rA

h

e

ie

r'e

acib

ib

ic

b

c

hie

hfeib

b c

ib ic

ie

e

ie

ac+1)r'

e

acib

ib

ic

b

c


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Determining h-Parameter Values

Use geometric means if given max. and min.values.

(min) (max)ie ie ieh h h

(min) (max)fe fe feh h h

Give examples Later


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Summary

AC concepts

Roles of capacitors in amplifiers

Common-emitter ac equivalent

circuit Amplifier gain

Gain and impedance calculations

Swamped amplifiersh-parameters

4 Common Emitter Amplifiers CH5

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