microsoft word - ac analysis.pdf

4-1

BIPOLAR JUNCTION TRANSISTOR: AC ANALYSIS

INTRODUCTION

• Two techniques should be applied in doing the ac analysis to the amplifiers;

small–signal or large–signal techniques.

• The parameters of interest in the small signal ac analysis are the input

impedance, Zi, output impedance, Zo, voltage gain, AV and current gain, Ai.

TRANSISTOR MODELING

• A model is an equivalent circuit that represents the ac characteristics of the

transistor.

• The model uses circuit elements that approximate the behavior of the transistor

when ac input signal is applied to the transistor network.

• Three types of models: re model, hybrid π model and hybrid equivalent model.

• We discuss re model and hybrid equivalent model with emphasizing on the re

model.

• We discuss the modeling for common emitter configuration only.

1) re transistor model

Common Emitter Configuration

The common–emitter npn transistor configuration is shown in Figure 1:

B

E

C

E

Ic

Ib

Figure 1: common–emitter npn transistor configuration

EI

mV26=er

erβ

e

c

e

b

bβI

Ic

Ib

ro

Figure 2: new re model

bc βII =

4-2

Figure 3 shows the re equivalent circuit for the common emitter configuration if

Ω∞=or .

erβ

e

c

e

b

bβI

Ib

Ic

Figure 3: re equivalent model (ro = ∞∞∞∞)

2)The hybrid equivalent transistor model

• The structure is almost the same as re transistor model.

• Several quantities that are different from re transistor model are: hie, hre, hfe, and

hoe.

• These quantities are called the hybrid parameters are components of hybrid

equivalent model, where,

→ih input resistance

→rh reverse transfer voltage ratio

→fh forward transfer current ratio

→oh output conductance

Common Emitter Configuration

Figure 4 shows the hybrid equivalent model and the re The general structures of both

models are very similar. The relationship between the parameters are eie rh β= and

ββ ac ≅=feh .

e

c

e

b

bIfeh erβ

e

c

e

b

bβI

Ic

Ib

Ic

Ib

hie

Figure 4: (a) Hybrid (b)re model

(a) (b)

4-3

BJT SMALL SIGNAL ANALYSIS

• The ac equivalent circuit is obtained by:

i. Set dc sources to zero, replacing them by a short circuit equivalent.

ii. The external capacitors (CC, CB and CE) are also replaced by short

circuit equivalent.

• Three types of circuit involved: Fixed-Bias, Emitter-Bias and Voltage Divider

1) Fixed–Bias Circuit

• Figure 5(a) shows the complete circuit for fixed-bias configuration.

• For the ac analysis, the dc voltage is connected to ground and the capacitors

are replaced with short circuit equivalent as shown in Figure 5(b).

B

E

CIi

Io

Zo

Zi

VCC

RCR

B

CB

CCv

i

vo

(a)

b

e

cIi

Io

Zo

Zi

RCR

B

vi

vo

(b)

Figure 5: (a) complete circuit; (b) ac circuit

4-4

• The circuit in Figure 5(b) is redrawn as shown in Figure 6(a) to separate the

input section and the output section of the transistor.

• The re equivalent model is substituted into the circuit and the resulting ac

equivalent circuit is shown in Figure 6(b).

b

e

cIi

Io

Zo

Zi

RC

RB

vi

vo

(a)

bβIerβ

b

e

c

e

Ii

Io Z

oZi

RC

RB

vi

vo

IcI

b

(b)

Figure 6: (a) CE fixed–bias configuration; (b) ac equivalent circuit (ro = ∞∞∞∞)

Referring to Figure 4.11(b), the equation for input impedance is given by

erβ||RZ Bi = (4.2)

The output impedance can be determined as follows

Co RZ = (4.3)

The equation for the input is

erv βIbi = (4.4)

and the equation for the output is

Coo RI−=v (4.5)

Substituting bco βIII == ,

Cbo RβI−=v (4.6)

The definition for voltage gain,

i

oVA

v

v= (4.7)

Substituting equations (4.4) and (4.6) into equation (4.7),

erv

v

βI

RβIA

b

Cb

i

oV

−==

er

CV

RA −= (4.8)

4-5

The negative sign in equation (4.8) reveals a o180 phase shift between vo and vi.

Current gain is defined as

i

o

iI

IA = (4.9)

From Figure 4.11(b),

i

ii

ZI

v= (4.10)

Io is obtained from equation (4.5)

C

oo

RI

v−= (4.11)

Substituting equations (4.10) and (4.11) into equation (4.9), and solving the equation

for Ai,

C

i

i

o

i

i

C

o

i

i

C

o

i

oi

R

ZZ

RZRI

IA ×−=×−=÷−==

v

v

v

vvv

C

iVi

R

ZAA −= (4.12)

Effect of ro

Figure 7 shows an ac equivalent circuit using re model with effect of ro.

bβIerβ

b

e

c

e

Ii

Io

Zo

Zi

RC

RBv

iv

o

IcI

b

ro

Figure 4.12: ac equivalent circuit with effect of ro

Referring to the Figure 7, the output impedance become

oCo r||RZ = (4.13)

The equation for vo is,

)r||(RβI oCbo −=v (4.14)

Substituting equations (4.4) and (4.14) into equation (4.7), the voltage gain is given as

erv

v

βI

)r||(RβIA

b

oCb

i

oV

−==

er

oCV

r||RA −= (4.15)

Since ro does not appear at the input section, ro does not affect the input impedance and

the current gain. Therefore the input impedance and the current gain can be determined

from equation (4.2) and (4.12) respectively.

erβ||RZ Bi = (4.2)

4-6

C

iVi

R

ZAA −= (4.12)

Example 1 Referring to Figure E1 and assuming Ω∞=or , sketch the ac

equivalent circuit using re model. Then, determine;

a. re

b. Zi and Zo

c. AV and Ai

d. vo if mVt5sin10i =v

kΩ2.2kΩ420

+15 V

031β =

Ii

Io

Zo

Zi

CB

CC

vi

vo

Figure E1

Solution

DC analysis:

µA34.05kΩ420

V0.7V15

R

VVI

B

BECCB =

−=

−=

Using the approximation, CE II ≅ ;

mA4.43µA)5(130)(34.0II CE ==≅

AC analysis:

bβIerβ

b

e

c

e

kΩ420 kΩ2.2

Ii

Io

ZoZ

i

vi

vo

IcI

b

Figure E1(a): ac equivalent circuit

a. Using equation (4.1); Ω=== 869.5Am43.4

mV26

I

mV26

E

er

b. Using equation (4.2); Ω59.761)5.869(0)3(1||k420β||RZ Bi =ΩΩ== er

4-7

Using equation (4.3); kΩ2.2RZ Co ==

c. Using equation (4.8); 374.855.869

k2.2RA C

V −=Ω

Ω−=−=

er

Using equation (4.12); 9.821k2.2

59.761374.85)(

R

ZAA

C

iVi =

Ω

Ω−−=−=

d. Using equation i

oVA

v

v= , we can calculate vo as;

Vt5sin3.75V)(0.01374.85vAv iVo −=−==

Example 2 Repeat Example 2 with kΩ10ro = .

Solution

The DC analysis is same as in Example 1. Therefore

µA34.05IB = and mA4.43IE =

AC analysis:

bβIerβ

b

e

c

e

kΩ420 kΩ2.2

Ii

Io

Zo

Zi

vi

vo

IcI

b

ro

Figure E.2: ac equivalent circuit

a. Using equation (4.1); Ω=== 869.5Am43.4

mV26

I

mV26

E

er

b. Using equation (4.2); Ω59.761)(5.8690)3(1||k420β||RZ Bi =ΩΩ== er

Using equation (4.13); Ω=ΩΩ== k1.8k10||k2.2r||RZ oCo

c. Using equation (4.15); 307.255.869

k1.8r||RA oC

V −=Ω

Ω−=−=

er


59.7615)2.073(

R

ZAA

C

iVi =

Ω

Ω−−=−=

d. Using equation i

oVA

v


Vt5sin703.V)10.0(527.03A iVo −=−== vv

4-8

2)Emitter–bias Circuit

Circuit with CE (Bypassed)

• Figure 7 shows the emitter-bias circuit with bypass capacitor, CE.

• For the ac analysis, CE is replaced with a short circuit equivalent. Therefore the

emitter resistor, ER will not appear in the ac equivalent circuit (RE is bypassed

by short circuit).

• Hence the ac equivalent circuit for emitter-bias circuit is the same as the ac

equivalent circuit for fixed bias circuit.

• Therefore equations (4.2) to (4.12) can be used to determine the required

parameters.

erβ||RZ Bi = (4.2)

Co RZ = (4.3)

er

CV

RA −= (4.8)

C

iVi

R

ZAA −= (4.12)

B

E

CIi

Io

Zo

Zi

RCR

B

vi

vo

RE

VCC

CB

CC

CE

Figure 7: CE emitter–bias configuration

Effect of ro The equations for determining Zi, Zo, AV and Ai for circuit with ro are:

erβ||RZ Bi = (4.2)

oCo r||RZ = (4.13)

er

oCV

r||RA −= (4.15)

C

iVi

R

ZAA −= (4.12)

4-9

Example 3 Referring to Figure E3 and assuming Ω∞=or , sketch the ac equivalent

circuit. Then, determine;

e. re

f. Zi and Zo

g. AV and Ai

h. vo if mVt5sin15i =v

kΩ420

kΩ8.0

kΩ2.1

+18 V

021β =

Ii

Io

Zo

Zi

vi

vo

CB

CC

CE

Figure E3

Solution

DC analysis:

µA46.33k)8)(0.121(k420

0.718

R1)(βR

VVI

EB

BECCB =

+

−=

++

−=

Using the approximation, CE II ≅ ;

mA024.µ)46.30)(32(1II CE ==≅

AC analysis:

bβIerβ

b

e

c

e

kΩ420 kΩ2.1

Ii

Io

Zo

Ziv

iv

o

IcI

b


a. Using equation (4.1); Ω=== 468.6mA02.4

mV26

I

mV26

E

er

b. Using equation (4.2); Ω73.774)468.6(0)2(1||k420β||RZ Bi === er

Using equation (4.3); kΩ.21RZ Co ==

4-10

c. Using equation (4.8); 53.185468.6

k.21RA C

V −=−=−=

er

Using equation (4.12); 9.811k.21

73.774)53.185(

R

ZAA

C

iVi =−−=−=

d. Using equation i

oVA

v


Vt5sin8.72)510.0(53.185A iVo −=−== vv


The DC analysis is the same as in Example 3. Therefore;

µA46.33IB = and mA024.IE =

AC analysis:

bβIerβ

b

e

c

e

kΩ420 kΩ2.1

Ii

Io

Zo

Ziv

iv

o

Ic

Ib

ro

Figure E4: ac equivalent circuit


mV26

I

mV26

E

er

b. Using equation (4.2); Ω73.774)468.6(0)2(1||k420β||RZ Bi === er

Using equation (4.13); Ω=== k071.k10||k.21r||RZ oCo

c. Using equation (4.15); 65.165468.6

k071.r||RA oC

V −=−=−=

er

Using equation (4.12); 95.061k.21

73.774)65.165(

R

ZAA

C

iVi =−−=−=

d. Using equation i

oVA

v


Vt5sin48.2)510.0(65.165A iVo −=−== vv

4-11

3)Voltage-divider Circuit

Circuit with CE (Bypassed) Figure 8(a) shows a voltage divider bypassed configuration. The ac equivalent circuit

using re model is shown in Figure 8(b).

B

E

CIi

Io

Zo

Zi

RC

R1

vi

vo

RE

VCC

CB

CC

CE

R2

(a)

bβIerβ

b

e

c

e

Ii

Io

Zo

Zi

RC

vi

vo

IcI

b

R1

R2

(b)

Figure 8: (a) CE voltage–divider configuration;

(b) ac equivalent circuit (ro = ∞∞∞∞)

Referring to the input side of Figure 8(b), the input impedance can be determined as

follows

erβ||R||RZ 21i = (4.26)

The output section of the voltage divider circuit is the same as the output section of

emitter bias circuit. Therefore, the same equations can be used to calculate Zo, AV and

Ai.

Co RZ = (4.3)

er

CV

RA −= (4.8)

C

iVi

R

ZAA −= (4.12)

4-12

Effect of or

The equations for determining Zi, Zo, AV and Ai for circuit with ro are:

erβ||R||RZ 21i = (4.26)

oCo r||RZ = (4.13)

erv

v oC

i

oV

r||RA −== (4.15)

C

iVi

R

ZAA −= (4.12)

Example 7 Referring to Figure E7 and assuming Ω∞=or , sketch the ac equivalent

circuit. Then, determine;

a. re

b. Zi and Zo

c. AV and Ai

d. vo if mVt5sin15i =v

kΩ2.2

µF10kΩ54

µF1

kΩ0.8 µF22

kΩ8.8

24 V

110β =

Ii

Io

vi

vo

Figure E7

Solution DC analysis:

Test whether the condition 2E R10βR ≥ is satisfied.

EβR = (110)(0.8 k) = kΩ88

2R10 = 110(8.8 k) = kΩ88

Therefore the condition 2E R10βR ≥ is satisfied.

V3.36(24)k8.8k54

k8.8)(V

RR

RV CC

21

2B =

+=

+=

V2.660.73.36VVV BEBE =−=−=

Therefore, mA3.33k0.8

2.66

R

VI

E

EE ===

4-13

AC analysis:

bβIerβ

b

e

c

e

kΩ8.8 kΩ2.2kΩ45

Ii

Io Z

oZ

i

vi

vo

IcI

b



mV26

I

mV26

E

er

b. Using equation (4.26); Ω33.771)808.7(0)1(1||k8.8||k54Zi ==

Using equation (4.3); Ω== k2.2RZ Co

c. Using equation (4.8); 76.281808.7

k2.2

r

RA

e

CV −=−=−=


33.771)76.281(

R

ZAA

C

iVi =−−=−=

d. Using equation i

oVA

v


Vt5sin23.4)015.0(76.281A iVo −=−== vv


Solution

The DC analysis is similar to Example 7. Therefore IE = 3.33 mA

AC analysis:

c

e

kΩ2.2bβI

erβ

b

e

kΩ8.8kΩ45

Ii

Io

Zo

Zi

vi

vo

Ic

Ib

ro

Figure E8: ac equivalent circuit


mV26

I

mV26

E

er

4-14

b. Using equation (4.26); Ω33.771)808.7(0)1(1||k8.8||k54Zi ==

Using equation (4.13); Ω=== k1.8k10||k2.2r||RZ oCo

c. Using equation (4.15); 53.230808.7

k1.8r||RA oC

V −=−=−=

er


33.771)53.230(

R

ZAA

C

iVi =−−=−=

d. Using equation i

oVA

v


Vt5sin46.3)510.0(3.5230A iVo −=−== vv

microsoft word - ac analysis.pdf

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