THE UNIVERSITY OF NOTTINGHAM

Bipolar Transistor

-Biasing- Professor Dino Isa

BSEE (Hons), PhD

Professor of Intelligent System

Department of Electrical and Electronic

Engineering

2

Bipolar Transistor Normal Operating Conditions

junction forward biased junction reversed biased

is forward biased. The depletion region has been reduced in width due to the applied forward bias, resulting in a heavy flow of majority carriers from the p to the n type material.

3

is reversed biased. The depletion region has been increased in width. No majority carrier current flows across the reversed biased junction but all the majority carriers which

manage to get into the base through thermal excitation will cross the reversed biased CB

junction because in the base these majority carriers coming from the emitter are actually

minority carriers.

Majority carrier current from the emitter cross the forward bias eb junction and find

themselves in the n type base where they become minority carriers.

Since most minority carriers will cross a reversed biased p-n junction, most of the holes

generated in the emitter will find themselves in the collector, hence,

; Active region operation Or more accurately,

! Where range from 0.98 to 1 depending on device construction

4

Class Example

For the bipolar transistor below, indicate, using arrows of different sizes the direction of

majority and minority current flow, reflecting the magnitudes of those currents. Also

indicate the direction of the flow for currents in the external circuit.

5

Majority and minority carrier flow of a pnp transistor

6

(a) pnp transistor;

7

(b) npn transistor;

8

Common Base Configuration

Pnp transistor rigged to measure the output common base characteristics (next page)

For each hole that recombines with an electron in the base (base loses one electron), a hole

leaves the base (electron enters base).

1. For the circuit above, increase starting from 0 volts (going more negative) as increases, ie the CB junction becomes more reverse biased, also increases. When is big enough, 1 and ! . This is because only a very small portion of the emmiter current is able to enter the collector region at low values of reversed

biased voltages 0. 2. Once reaches a value large enough to ensure that almost all carriers enter the

collector, the curves more or less levels off. In the active region, therefore, is very close to 1 and is essentially constant.

3. When the EB junction is open and no emitter flows, no minority carrier is injected

into the base. Under these conditions only a very small "leakage" current flows, we

call this current $.

9

Origin of $ .

10

Output or collector characteristics for a common-base transistor amplifier

(npn)

11

Input or driving point characteristics for a common-base silicon transistor

amplifier (npn)

12

Common Base Input Characteristics

As gets more reversed biased, increases. This increase in "causes" a corresponding increase in ; so for a specific is higher for a which is more reversed biased.

13

Review Common Base Equations.

& ! ' Where

' ! ' $

(

! )*+,-./0 $).1,-./0

14

Typical values of voltage amplification for the common base configuration vary from 50 to

300. The current amplification ( is always less than 1 for the common base

configuration. This latter characteristics should be obvious since

! and is always less than 1 The base amplifying action was produced by transferring a current from a low to a high resistance circuit. The combination of the two terms results in the label transistor;

Transfer + Resistor 2 transistor

ACTIVE: FORWARD BIAS 2 3 REVERSE BIAS 2 43

15

Common Base Configuration

In the active region, the collector junction is reversed biased, while the emitter junction is

forward biased.

Current Convention

!

16

Transistor Amplifying Action

Input resistance looking from e to b is small due to the forward biased junction.

Output resistance looking from c to b is large due to the reverse biased junction.

! 565 ! 7889:78 ! 10;< Assume ! 1 = ! ! > > ! 10;<

And > ! >? ! 10;

17

Common Emitter Configuration

Previously;

! (1) ! H:I $ HJK

(Coming from Emitter) (Coming from Base. Akin to L (reverse saturation current in reverse biased P-N

junction) $ like L is temperature sensitive.

From JJ , or J)*+J where ! 0.9 2 0.99

= ! $ (2) Manipulating (1) and (2),

! 14 $14

At ! 0 , we define ! $ ! $14 NO PQRRSC TBAUVSW TRA; PABBPCAR

CA ;VCCR UVCO CO XYD AZS

18

[ ! [14 !1

14 [

= [ is much larger than [

Returning to the previous page,

! 14

Define ; \] ! ^

= In terms of ^, ! J\] J,\] 2 ^ J,\]

! ^ [

Neglegating [

^

19

Common Emitter Forward Current Amplification Factor

Common emitter transfer characteristics:

out current in current output voltage

^ JJ where and are collector and base current of a particular operating point in the linear region. ^ varies from 60 2 200 typically.

Since ^ ! ( and ! ( and using ! ,

We get ^ ! \] or ! ``a\

Common emitter characteristics can be derived from common base characteristics and vice

versa.

20

Characteristics of a silicon transistor in the common-emitter configuration: (a)

collector characteristics; (b) base characteristics. (npn)

21

Why is $ small? The eb depletion region is bigger then when the eb junction was forward biased. (See diode @ no bias)

22

Bipolar Junction Transistor

Common Collector Characteristics.

BUT

and

Looking at the common collector characteristic graph,

remains negative if remains above 0.7V.

To keep reversed biased, must be larger than 0.7v

Input characteristic graph : versus

Output characteristic graph : versus

23

Common collector characteristics (npn)

Input Characteristics

Output Characteristics

24

Common Collector (Input Characteristics)

What the

diagram

shows

,

If is allowed to increase to a point where it approaches that means,

(the B-E junction is no longer forward biased) and .

25

The Purpose of Bias

The purpose of biasing a transistor amplifier is to set a dc output level somewhere in the

middle of the total range of output voltages so that an ac waveform can be superimposed

on it and results in an undistorted and amplified ac output signal.

An ac input current causes the output voltage to vary and below the bias voltage.

26

Graphical Analysis of Small Signal Amplifiers

1. Assume is forward biased at 0.65V. The 0.03V variation of causes to vary between 0.62V and

0.68V.

Input Graph

2. The variation in causes a variation in . What is this variation?

3. The variation in causes a variation in which causes a variation in . What is this variation?

Output Graph

27

The base emitter voltage varies from 0.62 to 0.68V as L varies b 0.03V about the bias voltage of ! 0.65V. This input voltage variation causes to vary between 20c< and 40c

28

As the base current varies between 20 and 40c

29

Find the value of ?g so that > will have maximum symmetrical swing.

ANS: For maximum symmetrical swing, ! 2( ! 6

4 ! ? ! ^?

! 29.7c<

4 ?g 4 ! 0 = ?g ! 178.5i

30

TRANSISTOR MAXIMUM RATINGS;

COMMON EMMITER;

k )*l )*l )*l

k )*l !

31

ASSUMING 0 , FIND ,

! ?7?\ ?m7 ! 2 ! 4 ! 2 4 0.7 ! 1.3 ! ? !

1.31.5i ! 0.867;<

! 4 ? ! 13.33 ! 4 ! 13.33 4 1.3 4 12.03

32

USE ! 140 , CALCULATE ,

! ! ?7?\ ?7 !3.9i

39 3.9i 22 ! 2

? ! ?\?7?\ ?7 ! 3.55i

p ! 4 ? ^ 1? !2 4 0.7

3.55i 1411.5i ! 6.05c< ! ^ ! 1406.05c< ! 0.85;<

! 4 ? ? ! 12.2

0.7 ! ?:L

Almost the same because we picked ?7 small compared to ?:L qr therefore the assumption that 0 is valid

p 4 ? 4 4 ? ! 0 4 ? 4 4 ^ 1? ! 0 4^ 1? ! ? 4

! 4 ? ^ 1?

! ! ^ ! ^ 1

33

Exact Analysis Using Thevenin Equivalents

When ?7is not much smaller as compared to ?qr

1. Thevenin Equivalent Resistance,

? ! 11 ?\( 1 ?7(! ?\?7?\ ?7

2. Thevenin Equivalent Voltage,

! ?7?\ ?7

34

[ ! ?7?\ ?7

35

1. Short voltage supply, open current source.

2. Calculate ?Ns, ?Ns ! 2i 3i ! 5i

3. Remove (open current) load, Apply 4V source;

[ ! 4 Apply 2;< current source; (short 4 source)

[ ! 2;< t 2i ! 4 = [ /,/ ! 8

36

No Emitter Resistor! Anomaly!

Common emitter! Read value of to calculate ! Cannot assume 0

! 0.7 22 4 0.739i ! 0.546;<

1. Assume 0 77tu.vuvau.v ! 2 vollts at B But ! 0.7, where ! 0 = cannot be 2 = 0 not valid

37

2. 22 4 10000 4 ! 0 22 4 10000 4 ! 0

3. 22 ! 39000 w 4 0.7 ! 0

4. ! w 4 ww ww x 8.y]8u.vz ! 0.179;<

= J{8.u||9:J{`J ! ! 3.66;< Too big, assumption that ww ! 0 is not valid

= 22 4 10000 ! ! 22 4 36.6 ! 414.6 ! (Cannot be true)

is too big. Need emitter resistor to limit

38

! 77]8.yuv888 ! 0.5;<

! 0.7

! ^ ! 1000.54;< ! 5.4;<

22 4 ? ! ! 22 4 5.410000 ! 22 4 54 !

39

Example (dc load line)

The silicon transistor in the CE bias circuit below has a of 100. Find the bias point

algebraically. Find it graphically.

Repeat when is changed to

GRAPHICALLY

1)

2)

3) Calculate , : know : can

determine value of

ALGEBRATICLY

Changing does not have any effect on the slope of the load line

40

Dc bias with voltage feedback

4 \? 4 ? 4 4 ? ! 0 But \ ! ! ! ^ 1

= U WC 4 ^ 1? 4 ? 4 4 ^ 1? ! 0

! 4 ? ^ 1?? 4 \? 4 4 ? ! 0 Assume \

! 4 ? ?

41

Find ,

10 4 2i 4 ! 0 But ! ^

! 10020c< ! 2;<

10 4 20002;< 4 ! 0 4 ! 410 4 ! 6 }~N

42

Find [ when ^ ! 50, 100 and 200 for q ! 0, q ! 5. @q ! 0 [ ! 5

@q ! 5

! F]\88z ! F]8.y\88888 ! .u\88888<

! 43c<

! ^ ! 5043c< ! 2.15;<

5 4 2.151000 4 ! 0 ! 5 4 2.15

! 2.85 ! [

43

^ ! 100 ! 4.3;< ! 5 4 4.3

! 0.7 ! [ CP CP

44

= 5 4 1000100 4 100000 4 0.7 ! 0 5 4 100000 4 100000 4 0.7 ! 0 5 4 200000 4 0.7 ! }

! 4.3200000 ! 21.5c< , ! 10021.5c< ! 2.15;<

= ! 5 4 10002.15;< ! 2.85

45

Find [, Assume :

5 4 1000 4 4 1000 ! 0

5 4 :20000 4 :10000 ! 0

5 4 :20000 4 4 1000 ! 0

Calculate Find , then calculate

46

Calculate the dc bias current and voltage for the circuit above.

! 4 ? ^ 1? ?

! ^ 1 !

! 4 ? ?

47

Find and for the circuit above

! ? ^ 1? ?

! ^

! 4 ?

48

Find

?w?w ? !43i

43i 10i 10 ! 8.11

! ! 8.11 0.7 ! 8.81

! 4 ? !10 4 8.812i ! 0.595;<

! ? ! 0.595;

At node \ , V 4 V> ! 0 ; V ! V> 4 2 And V>?> ! > Also ~?> ! > 4 3

Substitute (2) into (1) and using (3),

O&B 4 B> 1 O$&( ! > O&O$& 4

V>O$& ! V>?>; O&O$& V ! V> ?> 1 O$&( 4 4

By definition,

VJK !V>V

Where from 4

V> !1 O$&(1 OA( ?>

O&V

!1 O$&( O&V?O$& 1 O$&(

= V(

! O&1 O$&?> ! 1001 12 t 10]|t 2 t 10u

! 97.7

79

Compared to

If we had neglected O$& (open circuit)

V>?> ! > ! ~?> ! O&~?> V V( !

O&?>?> ! O&

! 97.7

Inclusion of O$& reduces current gain

Previously

found

80

Voltage Divider Bias (equivalent circuit)

DC network

AC network

Step (1)

81

Step (2) Hybrid Equivalent Circuit

(3) (4)

But

And

Where

82

Review current Division

And

But

And

Also, (current division)

Need

Current gain of voltage

divider bias circuit

83

CE unbypassed Emitter Resistor

Step (1) Ground the DC Power Supply;

Step (2);

Calculation;

Step (1)

Step (2)

But

And

Step (3)

If then

(Current division)

84

Common Emitter (1 Bias Resistor Base)

q ! ?||Oq& [ ! ?

85

Case 1 (most common)

? Oq& = q

Since $ ! 4[? And [ ! ! O& And ! q ,

= [ ! 4O&q? , but q ! q Oq&(

! 4O&q?Oq&

=

86

Case 2

Cannot assume q [ ! 4O&? q ! Oq&

= [\ !4O&Oq& ?

Same as before

(logical because ? is in parallel with Oq&)

The difference is in the current gain expression derived below;

87

Common Emitter unbypassed Emitter Resistor

Find formulas for q , [ ,

88

[ ! 4[? ! 4? ! 4O&? ! ]5 ? since ! 5

= [q !

89

Summary notes for hybrid equivalent circuits.

5 main basic types of circuits in dc analysis;

1. Common Emitter Fixed Bias or base biased circuit. 2. Common Emitter Voltage Divider or beta immunity bias.. 3. Common Emitter unbypassed Emitter resistor. 4. Emitter Follower. 5. Common Base.

For ac analysis, transistor models are used to simplify the network so that mesh, node,

Thevenin's and Norton's equivalents can be used. The transistor models replace the symbol

of individual transistor when placed in a circuit diagram.

In ac analysis, assume frequency is high enough so that capacitors can be shorted and

voltage power supplies can be shorted to ground.

When using the principle of superposition in a circuit which has both current and

voltage sources, what do you do to the voltage source when you want to calculate

only the contribution of the current source? YOU SHORT THE VOLTAGE SUPPLY.

Same principle is used for small signal ac analysis of transistor circuits.

The reason for using h parameters is convenience with which they can be measured at

audio frequencies, typically 1 KHz. They provide an accurate model if the frequency is low

enough that the inter electrode capacitances may be neglected.

The steps to performing the small signal low frequency ac analysis using the hybrid

parameter model is as follows;

1. Short all dc voltage sources to ground.

2. Assume 1/jwC to be small at the operating frequency (short capacitors).

3. Redraw circuit implementing steps 1 and 2 above.

4. Draw hybrid equivalent circuit replacing the transistor with its model.

5. Neglect appropriate parameters.

Common Emitter Hybrid Parameter Model

90

Progress Test 3

1) q 2) $ 3)

91

The model

Remember the approximate CB output

characteristic,

Input characteristics

For ac conditions, the input impedance at

the emitter of the CB transistor can be

determined by;

Differentiate the diode current equation

(The small signal input impedance of

the transistor in its common base

configuration is defined as;

)

Resistance similar to that of a

forward biased diode

output resistance looking from

collector to base and is given by;

92

' is usually big, so we can approximate with an open circuit

93

Substitute & for the Diode between E and B . . . and get the & model

94

(1)

(2)

Comparing (1) and (2),

if current power points upwards!

is determined by setting

(open circuit current source) for both cases.

95

96

Remembering the approximate CE characteristics;

As increases,

! is more forward biased

and more majority carriers

cross into the base resulting in

a larger base currents.

The finite slope of the

plot effectively acts as an

output resistance. It results

from base width modulation.

Increasing causes a greater

reverse bias across the

junction increasing the

depletion region and reducing

base width. This has 2 affects

both leading to an increase in

the collector current.

1) Reduce chance of recombination within the

effectively smaller base.

2) The charge gradient is increased across the base as

the flow of electrons

increases

(looking into the base)

(sometimes called in

datasheets)

Assume is a constant, which is

not exactly true since decreases

as the transistor reaches

saturation. Since we are in the

small signal region we can

assume is a constant for all

values of

Comparing the and hybrid parameter models,

( very big)

97

Common Base Example

Find & , q , $ ,

98

Common base pnp ac equivalent circuit & model

From the dc circuit on the previous page,

! 2 4 0.71i ! 1.3;<

= & ! 26;1.3;< ! 20 q ! ? |& ! 1 | 20 ! 19.61 [ ! ? ! 5i

99

Example: common base npn, & model

Draw the & ac equivalent circuit of the CB amplifier above, and find

100

101

Example

Common emitter npn

& ac equivalent circuit problem

Draw the & equivalent circuit. Find & , q , [ ,

102

To find ^& use dc analysis, (circuit on previous page)

! ?7?\ ?7 !8.2i2256i 8.2i ! 2.81

! 4 ! 2.81 4 0.7 ! 2.11

! ?( !2.111.5i ! 1.41;<

& ! 26; ! 18.44

103

[ 2

q ! ?\|?7|^& ! 1.35i [ ! ? ! 6.8

104

[ ! 50i

q ! ?\|?7|^& ! 1.35i [ ! ?||[ ! 5.98i

105

Example Sheet 4a

1. What is the reactance of a 10F capacitor at a frequency of 1KHz? For networks in

which the resistor levels are typically in the in the kilohm range, is it a good

assumption to use the short circuit equivalence for the conditions just disabled?

How about at 100KHz?

2.

Find q, q, $, $,

106

4. Given Oq& ! 1i, O& ! 2 t 10],

107

5.

Draw the small signal ac equivalent circuit and determine expressions for;

a) q b) $ c)

108

6.

Oq& ! 1, h ! 10], h ! 100, h ! 12cD, ?>2

a) Draw the O parameter equivalent circuit b) Find expressions for

109

b) By current division,

V> !O&V 1 O$&(1 O$&( ?>

V> V( ! 97.7

Mesh around the input,

43 4 VOq& 4 O&'& ! 0

But > ! ! O&V 1 O$&( ||?>

! O&?>V1 O$&?>

= V ! 1 O$&?>O&?>

( 1 )

( 2 )

Replace ( 2 ) into ( 1 )

4L 4 '&1 O$&?>O&?> Oq& 4 O&'& ! 0

4L 4 1 O$&?>O&?> Oq& 4 O& ! 0

! 4

110

7.

^ ! 100 $ ! 50

a) Determine & b) Calculate q and $ c) Find

111

3.

a) & ! qq !48;3.2;< ! 15

b) q ! & ! 15

c) ! ! 0.993.2;< ! 3.168;<

d) $ ! ?> ! 3.168; ! [

c)

q !

?? ?> ?>q

q ! Oq&

= $q !??>? ?>Oq& !

O&??>? ?>Oq&

= ? ?>

113

H61 IIC Introduction to Electronics

School of Electrical and Electronics Engineering

Example Sheet 4(b): Transistor Amplifiers

1. A single stage NPN transistor amplifier hfe = 100, hie = 1k has a collector load

resistor of 15 k. The ac input to the amplifier is obtained from a source having an

internal resistance Rs = 1 k. Using the h parameter equivalent circuit for the

transistor, calculate:

(i) The current gain

(ii) The overall voltage gain

(iii) The voltage gain from the transistor base to the output

[Ans: 100, -750, -1500]

2. In the CE circuit below, RE = 1k, RC = 8k, R1 = 10k, R2 = 85k, Rs = 0k, RL = 10k.

For the transistor hfe = 60, hie = 10k .

(i) The current gain

(ii) The transistor voltage gain

(iii) The source generator to output terminal voltage gain across RL.

The impedance of all capacitors and the dc voltage supply may be assumed to be

negligibly small at the signal frequency.

114

3. In the circuit below RS =0, RE = 1k, RL = 10k, and for the transistor hfe = 60, hie=

1k. Determine the current gain, transistor input resistance and transistor terminal

voltage gain.

SAME AS CLASS EXAMPLE JUST DIFFERENT NUMBERS

4. A silicon n-p-n transistor is connected so as to form a simple a.c. coupled common

emitter amplifier as shown in the circuit below. If VCC = 10V, RL = 1k and the current

is biased to give a collector voltage of 5V, find RB and then calculate (hie = 10k

hfe=100):

(i) The input impedance

(ii) The small signal voltage gain vo/vi.

SAME AS CLASS EXAMPLE JUST DIFFERENT NUMBERS

115

H61IIC Solutions to example sheet 4b

1.

(i) O& ! ! O&

116

2.

( i )

= [ [ 1 !O&[

YS 66E 1 !JJ, _____ ( 1 )

! J56w||66w|6|5 ______ ( 2 ) current division

Plug ( 2 ) into ( 1 ),

??> 1 !O&q?\||?7?\|?7|Oq&[

= [q ! Oq& !1001000010000

! 100

117

(ii) Transfer voltage gain

! J5]J, 6 ______ ( 1 ) ! [[ [?> !

[?

= [ ! ?>? [

! [ ?>? [

! [1 ?> ? = [ ! J\a6E 6 ! J66a6E ______ ( 2 )

Plugging ( 2 ) into ( 1 ),

= '( !Oq&4? ? ?>

! Oq&? ?>4O&?

= ' !4Oq&O&

? ?>?

! 1000060 8000 10000

8000 ! 4375

(iii) , ! ]J,6E ?? ?> ! [

= [ ! 4??>? ?> !4O&??>? ?>

118

q85||1010 85||10 ! Or in symbols,

q?7||?\?\||?7 Oq& !

= q ! ?\||?7 Oq&?7 ?\ L 4 q?L ! Oq& L ! Oq& q? ! Oq& ?\||?7 Oq&?7||?\

! Oq& ?\||?7 Oq&?7||?\

[L !4O&??>? ?>

Oq& ?\||?7 Oq&?7||?\

!4O&??>?7||?\? ?>?7||?\Oq& ?\||?7 Oq&

!4O&??>? ?> ?7||?\?7||?\Oq& ?\||?7 Oq&

But ?7|| ! \\ 6( a\ 6w( !\\ F888( a\ \8888(

! 8952 [L !

41008000100008921800089210000 10000 892

! 1008000108928953895218 ! 444.43