Transistorckk.com.tr/ders/analog/00 Transistor 02.pdfBJT Transistors • A transistor is a device which acts like a controlled valve. The current flow permitted can be controlled

Transistor

Dr. Cahit Karakuş

BJT Transistors

• A transistor is a device which acts like a controlled valve. The current flow permitted can be controlled.

• The bipolar junction transistor (BJT) is a three-terminal electronic valve - the output (collector) terminal current-voltage characteristics are controlled by the current injected into the input port (base). The BJT is a semiconductor device constructed from two pn junctions. There are two types of BJT: pnp and npn. Figures 1 and 2 show the circuit symbols and common current and voltage polarities during normal (active) operation.

DC Analiz

Vcc

VBB

Rc

Ic

C

E

BRB

VBE

VCE

IE

IB

CECCCC VIRV *

BEBBBB VIRV *

BC II *

B

BEBBB

R

VVI

C

CCSATC

R

VI

.0;0 OlurVVSaturasyonIIdayaVV CESATccCE

.0;0 OlurAIIKesmedeiseAI CBB

RC

VCE

VBE

Ic

IE

IB C

E

B

RE

RB

Vcc

EECECCCC IRVIRV **

EEBEBBCC IRVIRV **

BC II *

BBBBCE IIIIII *)1(*

CC

CE II

II *)1

(

RC

VCE

VBE

Ic

IE

IB C

E

B

RE

RB

VccEECEBCCCC IRVIIRV *)(*

EEBEBBBCCCC IRVIRIIRV **)(*

BC II *

BBBBCE IIIIII *)1(*

CC

CE II

II *)1

(

Vcc

RC

VCE

VBE

Ic

IE

IB C

E

B

RE

R1

R2

Vcc

VBB

Rc

Ic

C

E

BRB

VBE

VCEIB

IE

RE

21

2*RR

RVV CCBB

21

21 *

RR

RRRB

EECECCCC IRVIRV **

EEBEBBBB IRVIRV **

BC II *

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Sem I 0809/rosdiyana

Introduction to BJT Small Signal Analysis

CHAPTER 5

15

Introduction

•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 model

b) hybrid equivalent model

16

Disadvantage

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

Introduction

17

• The transistor can be employed as an amplifying device. That is, the output sinusoidal signal is greater than the input signal or the ac input power is greater than ac input power.

• How the ac power output can be greater than the input ac power?

Amplification in the AC domain

18


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)

19

• The superposition theorem is applicable for the analysis and design of the dc & ac components of a BJT network, permitting the separation of the analysis of the dc & ac responses of the system.

• In other words, one can make a complete dc analysis of a system before considering the ac response.

• Once the dc analysis is complete, the ac response can be determined using a completely ac analysis.


20

BJT Transistor Model • Use equivalent circuit

• Schematic symbol for the device can be replaced by this equivalent circuits.

• Basic methods of circuit analysis is applied.

• DC levels were important to determine the Q-point

• Once determined, the DC level can be ignored in the AC analysis of the network.

• Coupling capacitors & bypass capacitor were chosen to have a very small reactance at the frequency of applications.

21

The AC equivalent of a network is

obtained by:

1. Setting all DC sources to zero & replacing them by a short-circuit equivalent.

2. Replacing all capacitors by a short-circuit equivalent.

3. Removing all elements bypassed by short-circuit equivalent.

4. Redrawing the network.

BJT Transistor Model

24

Example

25

Example

26

Example

27

The re transistor model

• Common Base PNP Configuration

28

Common Base PNP Configuration

• Transistor is replaced by a single diode between E & B, and control current source between B & C

• Collector current Ic is controlled by the level of emitter current Ie.

• For the ac response the diode can be replaced by its equivalent ac resistance.

29


• The ac resistance of a diode can be determined by the equation;

Where ID is the dc current through the diode at the Q-point.

EI

mVre

26

30

• Input impedance is relatively small and output impedance quite high.

• range from a few Ω to max 50 Ω

• Typical values are in the M Ω

CBi reZ

CBZo


31

The common-base characteristics

32

Voltage Gain

re

RA

re

R

rI

RI

V

VA

rI

ZI

ZIV

RI

RI

RIV

LV

L

ee

Le

i

OV

ee

ie

iii

Le

LC

Loo

:gain voltage

: ageinput volt

)(

: tageoutput vol

33

Current Gain

1

i

e

e

e

C

i

oi

A

I

I

I

I

I

IA

• The fact that the polarity of the Vo as determined by the current IC is the same as defined by figure below.

• It reveals that Vo and Vi are in phase for the common-base configuration.

34


Approximate model for a common-base npn transistor configuration

35

Example 1: 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 Zi b) Calculate Av if RL=0.56k

c) Find Zo and Ai

e

b b

c

ec I αI

IcI

e

common-base re equivalent cct

re

36

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

37

Example 2: 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

Common Emitter NPN Configuration

• Base and emitter are input terminal

• Collector and emitter are output terminals

38

Common Emitter NPN Configuration

• Substitute re equivalent circuit

• Current through diode

39

bc II

bbe

bbbce

III

IIIII

)1(

40

• Input impedance

ei

ei

b

ebi

eb

eei

b

be

i

ii

rZ

rZ

I

rIZ

rI

rIV

I

V

I

VZ

; 1an greater thusually

)1(

)1( that so

)1(

:ageinput volt

:impedanceinput

41

The output graph

42

bI

c

e

bIi=I

b

re model for the C-E transistor configuration

re

ro

e

0AbI

c

e

bI

i=I

b

re

ro

e

Vs=0V

= 0A

oZ

impedance)high cct,(open ΩZ

the thusignored is r if

rZ

o

o

oo

Output impedance Zo

43

e

LV

eb

Lb

i

oV

eb

iii

Lb

Lco

Loo

r

RA

rI

RI

V

VA

rI

ZIV

RI

RIV

RIV

that so

:ageinput volt

: tageoutput vol

i

b

b

b

C

i

oi

A

I

I

I

I

I

IA

Voltage Gain Current Gain

re model for common-emitter

44

45

Example 3: 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

46

Example 4: 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=I

b

re model for the C-E transistor configuration

re

ro

e

RL

Io

47

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

6.8913

k40k2.1

r

rRvAc)

e

oL

48

Common Collector Configuration

• For the CC configuration, the model defined for the common-emitter configuration is normally applied rather than defining a model for the common-collector configuration.

Documents

Transistorckk.com.tr/ders/analog/00 Transistor 02.pdfBJT Transistors • A transistor is a device which acts like a controlled valve. The current flow permitted can be controlled