7.bjt transistor modeling.ppt

7/26/2019 7.BJT Transistor Modeling.ppt

1/49

BJT Transistor ModelingCHAPTER 5


2/49

2

Topic objectives

At the end of the course you will beable to

Understand about the small signalanalysis of circuit network using remodel and hybrid equivalent model

Understand the relationship between

those two available model for smallsignal analysis


3/49

3

To begin analyze of small-signal AC response of BJTamplifier the knowledge of modeling the transistor is

important.

The input signal will determine whether its 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) remodel

b) hybrid equivalent model

INTRODUCTION: TRANSISTOR

MODELING


4/49

4

How does the amplification be

done? Conservation; output

power of a system

cannot be large than its

input and the efficiencycannot be greater than 1

The input dc plays the

important role for the

amplification tocontribute its level to the

ac domain where the

conversion will become

as =Po(ac)/Pi(dc)


5/49

5

Disadvantages

Remodel

Fails to account the output impedancelevel of device and feedback effect from

output to input

Hybrid equivalent model

Limited to specified operating condition

in order to obtain accurate result


6/49

6

VS

VCC

C1

C2

C3

+

-

Vo

RS

Vi

+

-

RE

RC

R1

R2

VS

+

-

Vo

RS

Vi

+

-

RCR1

R2

I/p couplingcapacitors/c

Large values

Block DC and

pass AC signal Bypass

capacitors/c

Large values

DC supply

0 potential

Voltage-divider configurationunder AC analysis

Redraw the voltage-divider

configuration after removing dc

supply and insert s/c for the

capacitors

O/p coupling

capacitors/c

Large values

Block DC and

pass AC signal


7/49

7

VS

RS

R2R1 R

c

Transistor small-

signal ac

equivalent cct

Vo

Zi

Ii

Zo

Io

Vi

++

- -

B

E

C

Redrawn for small-signal AC analysis

Modeli ng of

BJT begin

HERE!

VS

+

-

Vo

RS

Vi

+

-

RC

R1

R2


8/49

8

1. Kill all DC sources

2. Coupling and Bypass capacitors are short circuit.

The effect of there capacitors is to set a lower cut-

off frequency for the circuit.

3. Inspect the cct (replace BJTs with its small signal

model:reor hybrid).

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

and o/p impedances.

AC bias analysis


9/49

9

Input impedance, ZiOutput impedance, ZoVoltage gain, AvCurrent gain, Ai

Input Impedance, Zi(few ohms M)

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

amplifier.

IMPORTANT PARAMETERS


10/49

10

VS Two-port

systemVi

Rsense

Ii

Zi

+

-

Determining Zi

+

-

sense

isi

R

VVI

i

ii

I

VZ

Two port system

-determining input impedance Zi

The input impedance of transistor can beapproximately determined using dc biasingbecause it doesnt simply change when the

magnitude of applied ac signal is change.


11/49

11

Demonstrating the impact of Zi

VS=10mV

Two-portsystem

Vi

Rsource

Zi

+

-

+

-

1.2 k

600

mV6.6600k2.1

)m10(k2.1

RZ

VZV

600Rimpedance,sourceWith

systemthetoapplied10mVFull

0Rsource,Ideal

sourcei

sii

source

source


12/49

12

Example 6.1: For the system of Fig. Below,

determine the level of input impedance

VS=2mV Two-port

systemV

i=1.2mV

Rsense

Zi

+

-

+

-

1 k

A8.0k1

m8.0

k1

m2.1m2

R

VVI

sense

isi

:So l u t i o n

k5.18.0

m2.1

I

VZ

i

ii


13/49

13

Output Impedance, Zo(few ohms2M)

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-port

system

Rsource

Vs=0V

Rsense

V

+

-

+

-

Io

Zo

Vo

Determining Zo

sense

oo

R

VVI

o

oo

I

VZ

cctopenbecomeZRZ oLo R

LZo=R

o

Iamplifier

IRo

IL

RoL

Lo

II

RRFor


14/49

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 k

A16k20

m320

k20

m6801

R

VVIsense

oo

:So l u t i o n

k5.4216

m680

I

VZ

o

oo


15/49

15

Example 6.3: For the system of Fig. below, determine Zoif

V=600mV, Rsense=10kand Io=10A

Two-port

system

Rsource

Vs=0V

Rsense

V

+

-

+

-

Io

Zo

Vo

mV500

k1010m600

RIVV

R

VVI

senseoo

sense

oo

:So l u t i o n

k5010

m500

I

VZ

o

o

o


16/49

16

Example 6.4: Using the Zoobtained in example 6.3,

determine ILfor the configuration of Fig below if

RL=2.2 kand Iamplifier=6 mA.

RLZ

o=R

o

Iamplifier

IRo

IL

mA747.5 k2.2k50

)m6(k50

RZ)(IZI

:ruledividerCurrent

Lo

amplifieroL

:So l u t i o n


17/49

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.

Its the ratio of circuits output to circuits input.

The small-signal AC voltage gain can be determined by:

i

ov

V

VA


18/49

18

VS AvNL

Vi

Rsource

Zi

+

-

+

-

Vo

+

-

Determining the no load voltage gain

By referring the network below the analysis are:

cct)(openR

i

oLvNL

V

VA

loadno

vNLARZ

Z

V

VA

:resistancesourcewith

si

i

s

ovs


19/49

19

Example 6.5: For the BJT amplifier of fig. below,

determine: a)Vi b) Ii c) Zi d) Avs

VS=40mV

BJT amplifierA

vNL=320

Vi

Rs

Zi

+

-

+

-V

o=7.68V

+

-

1.2 k

mV24320

7.68

A

VV

V

VAa)

vNL

oi

i

ovNL

:Solution

sources

s

isi

RR

A33.13k2.1

m24m40

R

V-VIb)

k8.133.13m24

IVZc)

i

ii

192)320(k2.1k8.1

k8.1A

RZ

ZAd) vNL

si

ivs


20/49

20

Current Gain, Ai

This characteristic can be determined by:

i

oiIIA

BJT

amplifierV

i

Zi

+

-

Vo

+

-

Ii

RL

Determining the loaded current gain

Io

L

ivi

RZAA

Li

io

ii

Lo

RVZV

Z/VR/V

L

oo

R

VI


21/49

21

Employs a diode and controlled current source toduplicate the behavior of a transistor.

BJT amplifiers are referred to as current-controlled

devices.

Common-Base Configuration

Common-base BJT transistor

remodel

reequivalent circuit

reTRANSISTOR MODEL


22/49

22

E

BB

C

Common-base BJT transistor - pnp

Ic

Ie

e

b b

c

ec II

IcIe

remodel for the pnp common-base

configuration

e

b b

c

ec II

IcI

e

common-base reequivalent cct

re

currentemitter

oflevelDCtheisII

26mVr E

E(dc)

e

isolation

part,

Zi=re

e

b b

c

A0Ic

Ic

Ie=

0A

Determining Zofor common-base

re

Vs=0V

Zo

Therefore, the input impedance, Zi = re

that less than 50.

For the output impedance, it will be as

follows;


23/49

23

The common-base

characteristics


24/49

24

e

b b

c ec II Ie

re

Defining Av

=Vo

/Vi

for the common-base configuration

BJT common-base

transistor 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


25/49

25

1A

gain,Current

i

e

e

e

c

i

oi

I

I

I

I

I

IA

e

b b

c ec II Ie

re

Defining Ai=Io/Iifor the common-base configuration

BJT common-base


Vi Vo+

-

+

-

ZioZ RL

Io


26/49

26

Example 6.6: For a common-base configuration in figure

below with IE=4mA, =0.98 and AC signal of 2mV is

applied between the base and emitter terminal:

a) Determine the Zib) Calculate Avif RL=0.56kc) Find Zoand Ai

e

b b

c

ec II

IcIe

common-base reequivalent cct

re


27/49

27

Solution:

5.6

m4

m26

I

26mrZa)

E

ei

43.845.6

)k56.0(98.0

r

RAb)

e

Lv

98.0I

IA

Zc)

i

oi

o


28/49

28

e

b b

c

ec II

IcIe

common-base re equivalent cct

re

iI


29/49

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) Voif RL=1.2k c) Av d)Ai e) Ib

20m5.0

m10

I

VZa)

:Solution

e

ii

88mV5

(1.2k)0.98(0.5m)

RIRIVb) LeLco

8.58m10

m588

V

VAc)

i

ov

98.0Ad) i

A10

)98.01(m5.0

)1(m5.0

I-II-IIe)

ee

ceb


30/49

30

Common-emitter BJT transistorremodel

reequivalent cct.

Still remain controlled-current source (conducted

between collector and base terminal)Diode conducted between base and emitter

terminal

Input OutputBase & Emitterterminal

Collector & Emitterterminal

Common-Emitter Configuration

c


31/49

31

common-emitter BJT transistor

EE

B

C

Ib

Ic

bc II

e e

b

Ic

Ib

remodel npn common-emitter configuration

bc II

c

e e

b

Ic

Ii=I

b

Determining Ziusing r

eequivalent model

re

Ie+

-

Vbe

+

-

Vi

(1)

Ii

ViZi

gives(1)intosubtitute

andIbreIereVbeVi

b

eb

b

be

I

rI

I

VZi

erZi

7k~6tohundredbetweenrangesi

Z


32/49

32

The output graph

Output impedance Z


33/49

33

bI

c

e

bIi=I

b

remodel for the C-E transistor configuration

re ro

e

0AbI

c

e

bIi=I

b

re

ro

e

Vs=0V

= 0A

oZ

impedance)highcct,(openZ

thethusignoredisrif

rZ

o

o

oo

Output impedance Zo


34/49

34

e

b b

c

bco III Ii=Ib

re

Determining voltage and current gain for the

common-emitter amplifier

BJT common-emitter


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

ovrIRI

VVA

gain,Voltage

iA


35/49

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 kis applied c) Ai with the 2 kload

975)125.8(120rZ

125.8m2.3

m26

I

26mra)

ei

Ee

:Solution

15.246125.8

k2

r

Rb)A

e

Lv

120I

IAc)

i

oi


36/49

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

26m

ra)

ei

Ee

:Solution

bI

cbIi=I

b

remodel for the C-E transistor configuration

re

ro

e

RL

Io


37/49

37

67.77

)80(

k2.1k40

k40

Rr

r

I

Rr

)I(r

A

Rr

)I(rI

I

I

I

IiAb)

(cont)Solution

Lo

o

b

Lo

bo

i

Lo

boL

b

L

i

o

6.89

13

k40k2.1

r

rRvAc)

e

oL


38/49

38

Hybrid Equivalent Model

re model is sensitive to the dc level of operationthat result input resistance vary with the dcoperating point

Hybrid model parameter are defined at anoperating point that may or may not reflect theactual operating point of the amplifier


39/49

39


40/49

40

Hybrid Equivalent Model

The hybrid parameters: hie, hre, hfe, hoeare developed andused to model the transistor. These parameters can be found

in a specification sheet for a transistor.


41/49

41

Determination of parameter

0VVo

i

12

0VVi

i

11

o12i11i

o

o

V

Vh

I

Vh

VhIhV

0AIo

o22

0VVo

i21

o

o22i21O

o

o

V

Ih

IIh

,0VVSolving

VhIhI

H22is a conductance!


42/49

42

General h-Parameters for any

Transistor Configuration

hi = input resistance

hr = reverse transfer voltage ratio (Vi/Vo)

hf = forward transfer current ratio (Io/Ii)

ho = output conductance

C itt h b id


43/49

43

Common emitter hybrid

equivalent circuit

C b h b id i l t


44/49

44

Common base hybrid equivalent

circuit


45/49

45

Simplified General h-Parameter ModelThe model can be simplified based on these approximations:

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

Simplified


46/49

46

Common-Emitter re vs. h-Parameter

Model

hie = re

hfe =

hoe = 1/ro


47/49

47

Common-Emitter h-Parameters

acfe

eie

h

rh


48/49

48

Common-Base re vs. h-Parameter

Model

hib = re

hfb = -

C


49/49

49

Common-Base h-Parameters

1

fb

eib

hrh

7.bjt transistor modeling.ppt

Documents