Section 5.6 Small Signal Model & Analysis

Section 5.6

Small Signal Model & Analysis

Quiz No 1 DE 27 (CE)

State the purpose and four steps (each) that are to be taken for carrying out

(a)DC Analysis

(b) Small signal Analysis

06-03-07

The operation of the transistor as an amplifier.

Conceptual circuit with the signal source eliminated .

(vbe =0)

DC Analysis Signal source eliminated

RI-VVV

1

II

II

II

II

0

CCCCCEC

EB

CB

CE

SC

T

BE

VV

be

e

v

Active Mode VerificationVC>VB-0.4 V

The collector Current & Trans-conductance

beBEBE vVv

T

be

T

be

T

BE

T

BE

Vv

CC

Vv

VV

SVv

SC

eIi

eeIeIi

ionApproximatSignalSmallmVv

VvIi

Vv

be

T

beCC

Tbe

10for Valid

1

The collector Current & Trans-conductance

CI toalProportionDirectly tanT

Cm

bemc

beT

Cc

VIcectranscondug

vgi

vVIi

Component Signal Bias DC

beT

CCC v

VIIi

For vbe<< VT, the transistor behaves as a voltage-controlled current device

The trans-conductance of the controlled source is gm

Output resistance is infinity

Linear operation of the transistor under the small-signal condition:

Base Current & Input Resistance at the Base

e

me

bee

beme

mb

be

bC

bemb

bemC

rrgi

vr

vgi

givr

ii

vgi

vgi

1

B

T

m

b

be

bem

T

beCb

CBbBB

beT

CCCB

beT

CCC

IV

gr

ivr

r

vgVvIi

IIiIi

vVIIii

vVIIi

base theinto LookingEmitter & Basebetween resistanceinput

Emitter Current & Input Resistance @ Emitter

eπ

eb

eeb

C

eeebbe

eebe

mmE

T

e

bee

e

beT

Ebe

T

CCe

eEE

CCCE

rrii

rriir

riririvriv

ggIV

ivr

r

vVIv

VIii

iIi

iIii

1 1

1

1

1resistanceemitter asKnown Emitter theinto looking

Emittter, & basebetween resistance Small

Voltage Gain

T

RC

T

CCCmv

CT

CCm

be

cv

beCmcCc

cCCC

cCCCCC

cCCCC

CCCCC

VV

VRIRgA

RVIRg

vvA

vRgiRviRVv

iRIRViIRV

iRVv

DC-AC Models

Large Signal Model Small Signal Model

The amplifier circuit

Figure 5.51 Two slightly different versions of the simplified hybrid- model for the small-signal operation of the BJT.

Figure 5.52 Two slightly different versions of what is known as the T model of the BJT.

Small Signal Analysis

• Coupling Capacitors– Couples the input signal vi to the emitter while

blocks the DC signals

– Don’t let dc biasing established by VCC &VEE be disturbed. when vi is connected

– Capacitor is of very large value –infinite, acts as short circuit at signal frequency of interest.

Application (Steps) : Small Signal Model• Suppress ac independent sources

– ac Voltage Sources be short circuited– ac Current Sources be open circuited– Capacitors be Open circuited

• Determine DC operating Point IC

• Suppress DC independent sources – DC Voltage Sources be short circuited– DC Current Sources be open circuited– Capacitors be short circuited

• Replace BJT with small signal Model

• Analyze the resulting circuit of find voltage gain & input/output resistance

E

T

me

B

T

m

ebbemc

T

Cm

IV

gr

IV

gr

i ivgiVIgCalculate

CE

CB

SC

III

II

II

B

T

BE

VV

eActive Mode Verification VBE > 0.7 VVC> VB-0.4 V


The Early Effect

• In real world

– (a) Collector current does show some dependence on collector voltage

– (b) Characteristics are not perfectly horizontal line

CEC vi

Figure 5.19 (a) Conceptual circuit for measuring the iC –vCE characteristics of the BJT. (b) The iC –vCE characteristics of a practical BJT.

T

BE

BE

T

BE

vv

SCC

A

CE

C

CEA

atConsvCE

Co

A

CEvv

SC

eIIIv

vI

vv

vir

vveIi

''o

C

o

1

tan

point operatingat values theare &I

as defined and

infinitenot is resistanceoutput that indicates slope Nonzero

1

The Early Effect

Figure 5.58 The hybrid-pi small-signal model, in its two versions, with the resistance ro included.

Coo

Cobeo

CCEBBE

C

A

C

CEAo

RrrRrvv

IVIV

IV

IVVr

10 if neglected becan reducedslightly isgain Thus ||-g

valuesbias DC are & and &

m

The hybrid- small-signal model, with the resistance ro included.

Problem 5.130

• Find the common-emitter amplifier shown in Fig. P5.130, Let VCC =9V, R1 = 27kΩ, R2 = 15kΩ, RE = 1.2kΩ, and Rc = 2.2kΩ. The transistor has β = 100 and VA = 100 V. Calculate the dc bias current IE. If the amplifier operates between a source for which Rsig = 10 kΩ and a load of 2kΩ replace the transistor with its hybrid-Π model, and find the value of Rm, the voltage gain and the current gain i

o

ii sig

o

vv

Figure P5.130

Figure P5.130

DC AnalysisSuppress the AC (independent Sources)

Short Circuit Voltage Sources

Open Circuit the CapacitorsCalculate DC Node Voltages & Loop Currents

Figure P5.130

DC Analysis

BEvBBV

v9

CI k2.2RI

BR

EIk2.1

β = 100 , α = 0.99VA = 100VIE = ?, Rin = ?, overall gain vo/vsig, io/i1

kR

VkkVV

B

CCBB

64.927152715

21.3152715

Solution P5.130

• DC Values

1

B

E

BEBBE RR

VVI

mAI

mAI

C

E

92.194.199.0

94.1

10164.92.1

7.021.3

BEv

v9

CI k2.2RI

BR

EIk2.1

3.21 V

9.64 KΩ

1.94 mA

1.92 mA

• Check for ModeVVVV EBEB 33.22.194.17.0

Solution P5.130

BEv

v9

CI k2.2RI

BR

EIk2.1

3.21 V

9.64 KΩ

1.94 mA

1.92 mA

VVV BC 446.233.2776.44.0

VIRVV CCCCC 776.492.12.29

ACTIVE MODE VCB > - 0.4 V

• Small Signal Model

E

kRrRkrRR

Coo

Bi

11.2||15.1||

Solution P5.130sigv

k10

k27 k15r

bevbemvg

or CR LR

ov

IC = 1.92 mAVT = 25 mVβ = 100 , α = 0.99VA = 100 V

kIVr

kg

r

VmAVIg

C

Ao

m

T

Cm

1.52

3.18.76

100

/8.76025.0

94.199.0

E

sigvk10

k27 k15r

bevbemvg

or CR LR

ov

i

o

s

i

s

ov v

vvv

vvA

Solution P5.130

AAR

RRvv

iiA

RRvi

Rvi

L

isig

S

o

i

oi

insig

Si

L

oo /3.45

LCoi

o

isig

i

S

i RRrvv

RRR

vv ||||-g

15.11015.1 m

VVRRrgRR

RA LComisig

iv /13.8||||

E

sigvk10

k27 k15r

bevbemvg

or CR LR

ov

CE with pi Model

CE with ‘T’ Model

Comparison ‘pi’ Vs ‘T’ Model

E

sigvk10

k27 k15r

bevbemvg

or CR LR

ov

Single Stage BJT Amplifier

• Three Configurations

– Common Emitter (CE)• Common Emitter (CE) with Emitter Resistance

– Common Base (CB)– Common Collector (CC)

Figure 5.60 (a) A common-emitter amplifier using the structure of Fig. 5.59.

Amplifiers Configurations

Common Emitter

Amplifiers Configurations Common Emitter

DC AnalysisSuppress Independent ac Source Voltage source ----- Short Cct Current Sources --- Open

Capacitors ---- Open Cct

Redraw the Circuit

Analysis

IE=IIC=αIE

IB=(β+1)IE

VC=VCC-ICRC

VB=-IBRB

VE=VB-VBE

VC

VE

VB

gm=Ic/VT

rл=β/gm

re=α/gm

Amplifiers Configurations Common Emitter

Small Signal AnalysisSuppress Independent DC Source Voltage source ----- Short Cct Current Sources --- Open

Capacitors ---- Short Cct

Redraw the Circuit by replacing BJTWith pi Model

Analysisgm=Ic/VT

rл=β/gm

re=α/gm

Find Rin, Rout, Voltage Gain vo/vi

Common Emitter

Rin=RB||rл

Rout=RC|RL

RRRRRRRg

vv

RRgvv

RRRRR

vv

vv

vv

vvA

Bsig

BCLm

i

o

CLmsig

be

Bsig

B

sig

be

sig

be

be

o

i

ov

||||||

||

||||

vbe

+

-

Short Circuit Current Gain Ais

Ais = ios/iiios=-gmvbevbe=vi=iiRinAis=-gmRin

Summary : CE rrRR Bi ~||

rg

Rv

vgiiA m

in

be

bem

i

oi

ComLomisig

iv RrgRRg

RRRA ||~||

CCoo RRrR ~||

Input Resistance

Output Resistance

Open Circuit Voltage Gain

Short Circuit Current Gain

Low to moderate typically a few kilohms

Voltage gain of a few hundred

Current gain equal to β

Output Resistance is relatively low

Figure 5.61 (a) A common-emitter amplifier with an emitter resistance Re.

Quiz No 2 (DE 27 CE)

Redraw the circuit for DC analysis Redraw the circuit for Small Signal pi model analysis

13-03-2007



Small Signal Analysis : CE with Emitter ResistanceInput Resistance

Multiplication by a factor (1+β) is known as the Resistance Reflection Rule.

Analysis

Small Signal Analysis : CE with Emitter ResistanceVoltage Gain

Voltage gain is lower than that of CE because of the additional term (1+β)Re

Small Signal Analysis : CE with Emitter ResistanceCurrent Gain

Small Signal Analysis : CE with Emitter Resistance

Summary

Re introduces negative feedback gives it the name emitter degenerative resistance

Comparison ‘T’ Vs ‘pi’ Model

A common-base amplifier

A common-base amplifier with its T model.

CB has low input resistance

CB is non-inverting amplifier

Small Signal Analysis : CB

• Very Low input resistanceRin=re

• Short Circuit Current Gain is nearly unity

• Open circuit Voltage Gain is equal to CE and is positivegm RC

• Relatively high output resistance (Rc) same as CE

• Excellent high frequency performance

• As short circuit current gain is unity Current Buffer, it accept an input signal current at a low input resistance and delivers equal current at a very high output resistance.

Summary : CB

An emitter-follower circuit : Common Collector

Non-unilateral Amplifier Input Resistance depends upon RL

Output Resistance depends upon Rsig

Common Collector An emitter-follower circuit : T model

An equivalent circuit of the Emitter Follower - CC

An equivalent circuit of the Emitter Follower - CC

Overall Voltage Gain is less than unity:RB>>Rsig, (β+1)(re+(ro||RL))>>(Rsig||RL)

The voltage at the emitter (vo) follows very closely the voltage at the input thus give the circuit the name Emitter Follower

The emitter follower : Reflecting resistance into emitter

For RB>> Rsig & ro >> RL Gain approaches Unity when Rsig/(1+β)<<RL89

Short Circuit Current Gain = 1+β

Common Collector : Output Resistance

Output Resistance is low

Summary : Common Collector

Non-unilateral Amplifier Input Resistance depends upon RL

Output Resistance depends upon Rsig

High Input Resistance

Low out Resistance

Voltage Gain ≈ unity

Relatively Large Current = 1+β

An equivalent circuit of the emitter follower

BJT Configurations

Common Emitter

Common Emitter with Emitter Resistance

Common Base : Current Buffer

Common Collector : Voltage Follower

Summary & Comparison

Comparison of Transistor Configurationsж

Quantity

Common Emitter (CE)

Common Collector (CC)

Common Base (CB)

AI Current Gain High (-50) High (50) Low (0.98)

AV Voltage Gain High (-136) Low (0.99) High (1.4)

Ri Input Resistance Medium (1 kΩ) High (154 kΩ) Low (21 Ω)

Ro

Output Resistance High (∞) Low (80 Ω) High (∞)

ж re = 1.1 kΩ, β = 50, RL= Rs = 3kΩ

Problem 5.135• The amplifier of Fig. P5.135 consists of two identical

common-emitter amplifier connected in cascade. Observe that the input resistance of the second stage, Rin2, constitutes the load resistance of the first stage.– For Vcc = 15V, R1 = 100kΩ , R2 = 47kΩ , RE = 3.9kΩ , Rc = 6.8kΩ ,

and β = 1000, determine the dc collector current and dc collector voltage of each transistor.

– Draw the small-signal equivalent circuit of the entire amplifier and give the values of all its components. Neglect ro1 and ro2

– Find Rin1 and vb1/vsig for Rsig = 5 kΩ– Find Rin2 and vb2/vb1.– For RL = 2kΩ , find vo /vb2– Find the overall voltage gain vo /vsig

Figure P5.135

Solution P5-135DC Analysis

Suppress the AC (independent Sources)Short Circuit Voltage Sources

Open Circuit the Capacitors

Calculate DC Node Voltages & Loop Currents

Solution P5-135DC Analysis

kR

VV

B

BB

32147

10047

8.41547100

47

VkIRVVmAII

mARR

VVI

CCCCC

EC

BE

BEBBE

5.88.696.01596.097.099.0

97.0

1

VmAVIg

T

Cm /8.3

•β=100, α=0.99

Small Signal Model

Suppress the DC (independent Sources)Short Circuit Voltage SourcesOpen Circuit Current Sources

Short Circuit the Capacitors

Draw the Small Signal Model

Small Signal Model

sigv

SR 1bv

1r

iv

C

1imvg1CR

1BR 2r2imvg

2CR LR

ov

2inR1inR

krRRRrr

kRR

kg

rr

kRR

Binin

oo

CC

m

BB

4.2||

8.6

6.2

32

121

21

21

21

21

sigv

SR 1bv

1r

iv

C

1imvg1CR

1BR 2r2imvg

2CR LR

ov

2inR1inR

VVvv

vv

vv

vv

VVvvRRvgv

VVvv

RRvgv

VVRR

Rvv

i

o

i

i

S

i

S

o

i

oLCimo

i

i

inCimi

Sin

in

sig

i

/1292..

/3.59||

/1.68

||

/32.0

21

21

222

1

2

2112

1

11

Small Signal Model

Figure P5.141 Common Base

For the circuit shown, Assume β=100

(a) Find the input resistance Rin

(b) Find the voltage gain vo/vsig

Figure P5.141 (Common Base)

DC Analysis





DC Analysis


IC

IB

IE

I = IB +IC=IE=0.33 mA

7533.0

25

E

T

E

T

T

Cme I

VI

V

VIg

r

β =100






e

o

e

e

sig

e

sig

o

iv

vi

vv

vv

VVRr

rvv

sige

e

sig

e /5.0

7511

ee

erv

i

VVvv

sig

o /9


75ein rR

)5.1||100(99.0)||( kkRRiv

LBe

o

ic=αie

vo

B

C

Eve

ie

Rin

Figure P5.143 Common Collector ( Emitter follower)

For the circuit shown, Assume β=40

(a) Find IE,VE,& VB

(b) Find the input resistance Rin

(c) Find the voltage gain vo/vsig







mARR

VVIB

E

BECCE 41.2

1

VRIV EEE 41.2

VVVV BEEB 11.3

β=40

37.10E

Te I

Vr


Small Signal AnalysisSuppress the DC (independent Sources)

Short Circuit Voltage SourcesOpen Circuit Current Sources



Small Signal Model P5-143

3.17)||)(1(|| LEeBi RRrRR

b

o

sig

b

sig

o

vv

vv

vv

isig

i

sig

b

RRR

vv

)||(||

lce

Lc

b

o

RRrRR

vv

VVRRr

RRRR

Rvv

lce

Lc

isig

i

sig

o / 621.0)||(

||

Small Signal Model P5-143

Ri

(β+1)ib

vb

vo

Figure P5.144

Problem

Problem

Problem 5.147

• For the circuit in Fig P5.147, called a boot-strapped follower:– Find the dc emitter current and gm, re, and rΠ Use β =

100. – Replace the BJT with its T model (neglecting ro), and

analyze the circuit to determine the input resistance Rin and the voltage gain vo/vsig.

– Repeat (b) for the case when capacitor CB is open –circuited. Compare the results with those obtained in (b) to find the advantages of bootstrapping.

Boot-Strapped Follower

P5-147

Figure P5.147

DC Analysis




Figure P5.147

DC AnalysisCalculate DC Node Voltages & Loop Currents

V9

V5.4

k10 k10

k2

EV

DC Analysis

mAII

mAk

I

RR

VVI

kRVV

EC

E

BC

BEBBE

B

BB

71.1

73.1

10110102

7.05.41

105.4

V9

V5.4

k10 k10

k2

EV

Solution 5-147

kg

r

gIVr

VmAVI

VIg

m

mC

Te

T

E

T

Cm

46.1

5.145.68

99.073.1

25

/5.68

Figure P5.147

Small Signal Model




B

E

C

αie

Figure P5.147

Small Signal Model

re

B

E

C

αie

B

E

C

αieie

i

Rin

Solution 5-147

B

E

C

αieie

i

Rin

krRrRi

vR

vii

vvi

iii

vR

eEeEb

in

be

bbe

e

bin

148.168))(1(1

)168148()99.01(

)48.1681()166748.14(

)1(

VVRR

RrR

Rvv

vv

vv

insig

in

eE

E

sig

b

bsig

/93.000

Solution 5-147

vo

kkRkR

R

krRrR

iv

R

ib

ibin

eEeEb

ib

12.18)20()20(

203))(1(1

Rib

VVRR

RrR

Rvv

vv

vv

insig

in

eE

E

sig

b

bsig

/64.000

Rin

The value of overall voltage gain and Rin obtained by using Bootstrap capacitor is higher than cct ,without BootstrappingBootstrapping is used to avoid loading of the input cct and to have higher gain.

Solution 5-147Without Boot-Strap Capacitor

Comparison of Transistor Configurations

Quantity

Common Emitter (CE)

Common Collector (CC)

Common Base (CB)

AI Current Gain High High Low

AV Voltage Gain High Low High

Ri Input Resistance Medium High Low

Ro Out Resistance High Low High

Date post:	08-Feb-2016
Category:	Documents
Upload:	pilar
View:	173 times
Download:	7 times

Section 5.6 Small Signal Model & Analysis

Documents