Chap3 Utem (BJT Transistor Modeling)

Chapter 7: BJT Transistor ModelingChapter 7: BJT Transistor Modeling

2

Topic objectives

• At the end of the course you will be able to– Understand about the small signal analysis of circuit

network using re model and hybrid equivalent model

– Understand the relationship between those two available model for small signal analysis

3

INTRODUCTION:TRANSISTOR MODELING

• To begin analyze of small-signal AC response of BJT amplifier the knowledge of modeling the transistor is important.• The input signal will determine whether it’s a small

signal (AC) or large signal (DC) analysis.• The goal when modeling small-signal behavior is to make of a transistor that work for small-signal enough to “keep things linear” (i.e.: not distort too much) [3]• There are two models commonly used in the small signal analysis:

a) re modelb) hybrid equivalent model

4

How does the amplification be done?

• Conservation; output power of a system cannot be large than its input and the efficiency cannot be greater than 1

• The input dc plays the important role for the amplification to contribute its level to the ac domain where the conversion will become as η=Po(ac)/Pi(dc)

• Simply speaking…

5

Disadvantages

• Re model– Fails to account the output impedance level of device

and feedback effect from output to input

• Hybrid equivalent model– Limited to specified operating condition in order to

obtain accurate result

6

VS

VCC

C1

C2

C3

+

-

Vo

RS

Vi

+

-RE

RCR1

R2

VS

+

-

Vo

RS

Vi

+

-

RCR1

R2

•I/p coupling capacitor s/c• Large values• Block DC and pass AC signal • Bypass

capacitor s/c•Large values

DC supply “0” potential

Voltage-divider configuration under AC analysis

Redraw the voltage-divider configuration after removing dc

supply and insert s/c for the capacitors

• O/p coupling capacitor s/c• Large values• Block DC and pass AC signal

7

VS

RSR2 R1

Rc

Transistor small-signal ac

equivalent cct

Vo

Zi

Ii

Zo

Io

Vi

+ +

- -

B

E

C

Redrawn for small-signal AC analysis

Modeling of BJT begin

HERE!

VS

+

-

Vo

RS

Vi

+

-

RCR1

R2

8

AC bias analysis :

1) Kill all DC sources

2) Coupling and Bypass capacitors are short cct. The effect of there capacitors is to set a lower cut-off frequency for the cct.

3) Inspect the cct (replace BJTs with its small signal model:re or hybrid).

4) Solve for voltage and current transfer function, i/o and o/p impedances.

9

IMPORTANT PARAMETERS

• Input impedance, Zi

• Output impedance, Zo

• Voltage gain, Av

• Current gain, Ai

Input Impedance, Zi(few ohms M)

The input impedance of an amplifier is the value as a load when connecting a single source to the I/p of terminal of the amplifier.

10

VS Two-portsystem

Vi

Rsense

IiZi

+

-

Determining Zi

+

-

sense

isi

R

VVI

i

ii

I

VZ

Two port system-determining input impedance Zi

• The input impedance of transistor can be approximately determined using dc biasing because it doesn’t simply change when the magnitude of applied ac signal is change.

11

Demonstrating the impact of Zi

VS=10mVTwo-portsystem

Vi

Rsource

Zi

+

-

+

-1.2 kΩ

600Ω

mV6.6600k2.1

)m10(k2.1

RZ

VZV

Ω600R impedance, sourceWith

system the toapplied 10mV Full

0ΩR source, Ideal

sourcei

sii

source

source

12

Example 6.1: For the system of Fig. Below, determine the level of input impedance

VS=2mV Two-portsystem

Vi=1.2mV

RsenseZi

+

-

+

-

1 k Ω

A8.0k1

m8.0

k1

m2.1m2

R

VVI

sense

isi

:Solution

k5.18.0

m2.1

I

VZ

i

ii

13

Output Impedance, Zo (few ohms 2M)

The output impedance of an amplifier is determined at the output terminals looking back into the system with the applied signal set to zero.

Two-portsystem

Rsource

Vs=0V

Rsense

V

+

-

+

-

Io

Zo

Vo

Determining Zo

sense

oo

R

VVI

o

oo

I

VZ

cctopen become ZRZ oLo RLZo=Ro

Iamplifier

IRo

IL

RoL

Lo

II

RRFor

14

Example 6.2: For the system of Fig. below, determine the level of output impedance

Two-portsystem

Vs=0V

Rsense

V=1 V

+

-

+

-

Zo

Vo=680mV

20 kΩ

A16k20

m320

k20

m6801

R

VVI

sense

oo

:Solution

k5.4216

m680

I

VZ

o

oo

15

Example 6.3: For the system of Fig. below, determine Zo if V=600mV, Rsense=10k and Io=10A

Two-portsystem

Rsource

Vs=0V

Rsense

V

+

-

+

-

Io

Zo

Vo

mV500

k1010m600

RIVVR

VVI

senseoo

sense

oo

:Solution

k5010

m500

I

VZ

o

oo

16

Example 6.4: Using the Zo obtained in example 6.3, determine IL for the configuration of Fig below if RL=2.2 k and Iamplifier=6 mA.

RLZo=Ro

Iamplifier

IRo

IL

mA747.5k2.2k50

)m6(k50RZ

)(IZI

:ruledivider Current

Lo

amplifieroL

:Solution

17

Voltage Gain, AV

• DC biasing operate the transistor as an amplifier. Amplifier is a system that having the gain behavior. • The amplifier can amplify current, voltage and power.• It’s the ratio of circuit’s output to circuit’s input.• The small-signal AC voltage gain can be determined by:

i

ov

V

VA

18

VS AvNLVi

Rsource

Zi

+

-

+

-Vo

+

-

Determining the no load voltage gain

By referring the network below the analysis are:

cct)(open ΩRi

oLvNL

V

VA

load no

vNLARZ

Z

V

VA

:resistance sourcewith

si

i

s

ovs

19

Example 6.5: For the BJT amplifier of fig. below, determine: a)Vi b) Ii c) Zi d) Avs

VS=40mVBJT amplifier

AvNL=320Vi

Rs

Zi

+

-

+

-Vo=7.68V

+

-

1.2 kΩ

mV24320

7.68

A

VV

V

VA a)

vNL

oi

i

ovNL

:Solution

sources

s

isi

RR

A33.13k2.1

m24m40

R

V-VI b)

k8.133.13

m24

I

VZ c)

i

ii

192)320(k2.1k8.1

k8.1A

RZ

ZA d) vNL

si

ivs

20

Current Gain, Ai

• This characteristic can be determined by:

i

oi

I

IA

BJTamplifier

Vi

Zi

+

-

Vo

+

-

Ii

RL

Determining the loaded current gain

Io

L

ivi

R

ZAA

Li

io

ii

Lo

RV

ZV

Z/V

R/V

L

oo

RV

I

21

re TRANSISTOR MODEL

• employs a diode and controlled current source to duplicate the behavior of a transistor.• BJT amplifiers are referred to as current-controlled devices.

Common-Base Configuration

Common-base BJT transistorre modelre equivalent cct.

22

E

BB

C

Common-base BJT transistor - pnp

Ic Ie

e

b b

c

ec I αI

IcIe

re model for the pnp common-baseconfiguration

e

b b

c

ec I αI

IcIe

common-base re equivalent cct

re

current emitter

of level DC the isII

26mVr E

E(dc)

e

isolation part,Zi=re

e

b b

c

A0Ic

IcIe=0A

Determining Zo for common-base

reVs=0V

Zo

Therefore, the input impedance, Zi = re

that less than 50Ω.

For the output impedance, it will be as follows;

23

The common-base characteristics

24

e

b b

c ec I αI Ie

re

Defining Av=Vo/Vi for the common-base configuration

BJT common-basetransistor amplifier

Vi Vo

+

-

+

-

Zi

oZ RL

Io

LeLcLoo RIRIRIV

e

L

e

Lv

r

R

r

RA

gain, Voltage

eeiei rIZIV

ee

Lev

rI

RI

Vi

VoA

25

1A

gain,Current

i

e

e

e

c

i

oi

I

I

I

I

I

IA

e

b b

c ec I αI Ie

re

Defining Ai=Io/Ii for the common-base configuration

BJT common-basetransistor amplifier

Vi Vo

+

-

+

-

ZioZ RL

Io

26

Example 6.6: For a common-base configuration in figurebelow with IE=4mA, =0.98 and AC signal of 2mV isapplied between the base and emitter terminal:a) Determine the Zi b) Calculate Av if RL=0.56kc) Find Zo and Ai

e

b b

c

ec I αI

IcIe

common-base re equivalent cct

re

27

Solution:

5.6m4

m26

I

26mr Za)

Eei

43.845.6

)k56.0(98.0

r

RA b)

e

Lv

98.0I

IA

Ω Zc)

i

oi

o

28

e

b b

c

ec I αI

IcIe

common-base re equivalent cct

re

iI

29

Example 6.7: For a common-base configuration in previous example with Ie=0.5mA, =0.98 and AC signal of 10mV is applied, determine:a) Zi b) Vo if RL=1.2k c) Av d)Ai e) Ib

20m5.0

m10

I

V Za)

:Solution

e

ii

88mV5(1.2k)0.98(0.5m)

RIRIV b) LeLco

8.58m10

m588

V

VA c)

i

ov

98.0A d) i

A10

)98.01(m5.0

)1(m5.0

I-I

I-II e)

ee

ceb

30

Common-Emitter Configuration

Common-emitter BJT transistorre modelre equivalent cct.Still remain controlled-current source (conducted between collector and base terminal)Diode conducted between base and emitter terminal

Input Output

Base & Emitter terminal Collector & Emitter terminal

31

common-emitter BJT transistor

EE

B

C

Ib

Icbc I I

c

e e

b

Ic

Ib

re model npn common-emitter configuration

bc I I

c

e e

b

Ic

Ii=Ib

Determining Zi using re equivalent model

re

Ie+

-

Vbe

+

-

Vi

(1) Ii

ViZi

gives (1) into subtitute

and IbreIereVbeVi

b

eb

b

be

I

rI

I

VZi

erZi 7k~6 to hundred between ranges iZ

32

The output graph

33

bI

c

e

bIi=Ib

re model for the C-E transistor configuration

rero

e

0AbI

c

e

bIi=Ib

rero

e

Vs=0V

= 0A

oZ

impedance)high cct,(open ΩZ

the thusignored is r if

rZ

o

o

oo

Output impedance Zo

34

e

b b

cbco I II Ii=Ib

re

Determining voltage and current gain for the common-emitter amplifier

BJT common-emittertransistor amplifier

Vi Vo

+

-

+

-

oZ RL

Io

ei rZ

e

Lv

r

RA

Ib

Ib

Ib

Ic

Ii

IoA

gain,Current

i

LbLcLoo RIRIRIV

ebiii rIZIV

eb

Lb

i

ov

rI

RI

V

VA

gain, Voltage

iA

35

Example 6.8: Given =120 and IE(dc)=3.2mA for a common-emitter configuration with ro= , determine:

a) Zi b)Av if a load of 2 k is applied c) Ai with the 2 k load

975)125.8(120rZ

125.8m2.3

m26

I

26mr a)

ei

Ee

:Solution

15.246125.8

k2

r

Rb)A

e

Lv

120I

IA c)

i

oi

36

Example 6.9: Using the npn common-emitter configuration, determine the following if =80, IE(dc)=2 mA and ro=40 k

a) Zi b) Ai if RL =1.2k c) Av if RL=1.2k

k04.1)13(80rZ

13m2

m26

I

26mr a)

ei

Ee

:Solution

bI

cbIi=Ib

re model for the C-E transistor configuration

rero

e

RL

Io

37

67.77

)80(k2.1k40

k40

Rr

r

IRr

)I(r

A

Rr

)I(rI

I

I

I

IiAb)

(cont)Solution

Lo

o

b

Lo

bo

i

Lo

boL

b

L

i

o

6.8913

k40k2.1

r

rRvAc)

e

oL

38

Hybrid Equivalent Model

• re model is sensitive to the dc level of operation that result input resistance vary with the dc operating point

• Hybrid model parameter are defined at an operating point that may or may not reflect the actual operating point of the amplifier

39

Hybrid Equivalent Model

The hybrid parameters: hie, hre, hfe, hoe are developed and used to model the transistor. These parameters can be found in a specification sheet for a transistor.

40

Determination of parameter

0VVo

i12

0VVi

i11

o12i11i

o

o

VV

h

IV

h

VhIhV

0AIo

o22

0VVo

i21

o

o22i21O

o

o

VI

h

II

h

, 0VV Solving

VhIhI

H22 is a conductance!

41

General h-Parameters for any Transistor Configuration

hi = input resistancehr = reverse transfer voltage ratio (Vi/Vo)hf = forward transfer current ratio (Io/Ii)ho = output conductance

42

Common emitter hybrid equivalent circuit

43

Common base hybrid equivalent circuit

44

Simplified General h-Parameter Model

The model can be simplified based on these approximations:

hr 0 therefore hrVo = 0 and ho (high resistance on the output)

Simplified

45

Common-Emitter re vs. h-Parameter Model

hie = rehfe = hoe = 1/ro

46

Common-Emitter h-Parameters

[Formula 7.28]

[Formula 7.29]

acfe

eie

h

rh

47

Common-Base re vs. h-Parameter Model

hib = rehfb = -

48

Common-Base h-Parameters

[Formula 7.30]

[Formula 7.31]

1

fb

eib

h

rh