BJT Transistor Modeling - I Electronic Circuits First Term Second Year (11CS Batch) 1

1

BJT Transistor Modeling - I

Electronic CircuitsFirst Term Second Year (11CS Batch)

2

Model

A model is the combination of circuit elements, properly chosen, that best approximates the actual behaviour of a semiconductor device under specific operating conditions.

Small Signal BJT Model

re Model Hybrid Parameter Model

3

Small Signal AC Analysis

1. Set all dc sources to zero and replace them by a short-circuit equivalent circuit.

2. Replace all capacitors by a short circuit equivalent.

3. Remove all elements bypassed by the short circuit equivalents introduced by steps 1 and 2.

4. Redraw the network in a more convenient and logical form.

4

AC Analysis

Consider the following BJT circuit

5

AC Analysis

Step 1,2,3:

Step 4: Transistor Small Signal ac

Equivalent Circuit

6

The Important Parameters: Zi, Z0, Av , Ai

Input Impedance (Zi):

Two-PortSystem

Ii

ZiVi

+-

I0

Z0 V0

+-

i

ii I

VZ

7

Input Impedance (Cont.)

Two-PortSystem

RSense

VS

Vi Zi

Ii

+

-

sense

isi R

VVI

i

ii I

VZ

8

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

Two-PortSystem

RSense

VS Vi = 1.2 mv Zi

Ii

+

-

2mv

1k +

-

Solution: A8.0101

102.1102

R

VVI 3

33

sense

isi

k5.1A8.0

mv2.1

I

VZ

i

ii

9

Output Impedance (Z0)

sense

00 R

VVI

0

00 I

VZ

The output impedance is determined at the output terminals looking back into the System with the applied signal set to zero.

The output impedance of a BJT amplifier is resistive in nature and,depending on the configuration and placement of the resistive elements, Z0 can vary from a few ohms to a level that can exceed2 M.

10

Example: For the system of the following Figure, determine the level of output impedance.

Solution:

A16100020

106801

R

VVI

3

sense

00

k5.421016

10680

I

VZ 6

3

0

00

11

Amplifier Gain• Gain– A multiplier that exists between the input and

output of a circuit.– For example, if the gain of an amplifier is 100,

then the output signal is 100 times as great as the input signal under normal operating conditions.

• Types of Gain:– Voltage gain, AV

– Current gain, Ai

– Power gain, Ap

12

Voltage Gain

The small signal ac voltage gain is defined as

For the system of the above Fig., a load has not been connected to the output terminal, the gain is therefore referred to as no load voltage gain (AvNL).

For the transistor amplifiers, the no load gain is greater than the loaded voltage gain.

i

ov V

VA

13

Using the Voltage Divider Rule

AvNL VS Vi

Zi

Ii

+

-

Rs +

-

+

-

V0

si

sii RZ

VZV

si

i

s

i

RZ

Z

V

V

i

0

s

i

s

0v V

V

V

V

V

VA

s

vNLsi

i

s

0v A

RZ

Z

V

VA

s

14

Example: For the BJT amplifier of the following Fig., determine Vi, Ii, Zi and Avs.

Solution:

BJT amplifierAvNL = 320

Vs = 40mv Vi

Zi

+ Rs

+

-

+

-

V0 = 7.68v

-

1.2k

mV24320

68.7

A

VV

V

VA

vNL

0i

i

0vNL

A33.13k2.1

mV24mV40

R

VVI

s

isi

,k8.1A33.13

mV24

I

VZ

i

ii

192)320(k2.1k8.1

k8.1

ARZ

ZA vNL

si

ivs

15

Current Gain (Ai)

BJTamplifier RL V0

+

-

I0

Vi Zi

+

-i

0i I

IA

From the circuit

i

ii Z

VI and

L

00 R

VI

Li

i0

ii

L0

i

0i RV

ZV

ZV

RV

I

IA

L

ivi R

ZAA

16

Example: For the given BJT circuit, determine:(a) Ii

(b) Zi

(c) V0

(d) I0

(e) Ai.

17

The re Transistor Model

(a) Common Base (PNP) Configuration Ie

Ic =Iere

Ie Ic

Ee I

mV26r

ei rZ

0Z

18

Voltage Gain:fV0 = -I0RL

= -(-Ic)RL

V0 = IeRL

and Vi = IeZi = Iere

Therefore,

BJTCommon-base

Transistoramplifier

RL V0

+

-

Vi Zi

+

-

Ic = IeIe

Z0 = I0

ee

Le

i

0v rI

RI

V

VA

e

Lv r

RA

For the current gain

e

e

e

c

i

0i I

I

I

I

I

IA

iA

19

Example: For a common-base configuration with IE = 4mA, = 0.98, and an ac signal of 2mV applied between the base and ammeter terminals:

(a) Determine the input impedance(b) Calculate the voltage gain if a load of 0.56 k is

connected to the input terminals.(c) Find the output impedance and current gain.Solution:(d) Input Impedance = Zi = re = 2610-3/(410-3 ) = 6.5

(b) 43.845.6

k56.098.0

r

RA

e

Lv

,Z0 (c) 98.0A i

20

The re Transistor Model

Common Emitter Configuration:

Ic = Ib

IcIc

Ib Ib

(a) CE (NPN) Configuration (b) re Model

bc II bbbce IIIII

bbe II)1(I

Ie

21

Ic = IbIi = Ib

vbe re vi

Ie

b

eb

b

ee

b

be

i

ii I

rI

I

rI

I

V

I

VZ

ei rZ

BJTCommon-

emitterTransistoramplifier

RL V0

+

-

Vi Zi

+

-

I0 = Ic = IbIi =Ib

Z0 =

Lb

LcL00

RI

RIRIV

ebiii rIZIV

eb

Lb

i

0v rI

RI

V

VA

e

Lv r

RA

b

b

b

c

i

0i I

I

I

I

I

IA

iA

22

Example: Given = 120 and IE = 3.2 mA for a common-emitter configuration with r0 = , determine:

(a) Zi

(b) Av if a load of 2 k is applied.

(c) Ai with 2 k load.

Solution: (a)

(b) Av = -RL/re = -2000/8.125 = -246.15

(c) Ai = = 120

,125.8mA2.3

mV26re 975125.8120rZ ei

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re Model for the Common-Emitter Configuration

Note: r0 may be computed from the BJT output characteristics.

24

Common Emitter Fixed Biased Configuration

(a) CE Fixed Bias Configuration(b) All capacitors and dc power supply removed

(c) Circuit with the substitution of the re model

25

From Fig. (c) of the previous slide, the input impedance may be computed as

Zi = RB ||re ohms

or

eB

Bei rR

RrZ

In the majority of situations, RB 10re. In such cases

B

Bei R

RrZ

ei rZ

The output impedance is: Z0 = r0 ||RC

or

26

C0

C00 Rr

RrZ

If r0 10RC

C0 RZ

Voltage Gain:

C0

0Cb0Cb0 Rr

rRIr||RIV

Bute

ib r

VI

C0

0C

e

i0 Rr

rR

r

VV

or

27

C0e

0C

i

0v Rrr

rR

V

VA

If r0 10RC, then

0e

0Cv rr

rRA

or

e

Cv r

RA

Current Gain:

C0

b00 Rr

IrI

eB

iB

C0

00 rR

IR

Rr

rI

ButeB

iBb rR

)I)(R(I

eBC0

0B

i

0i rRRr

rR

I

IA

If r0 10RC and RB 10re , then

iA

B0

0Bi Rr

rRA

C

ivi R

ZAA

Also, note that

28

Example: For the given network(a) Determine re.

(b) Find Zi (with r0 = ).

(c) Calculate Z0 (with r = ).

(d) Determine Av (with r = ).

(e) Find Ai (with r = ).

(f) Repeat parts (c) through (e)including r = 50 k in all calculations and compare results.

Solution:(a) A04.24

k470

7.012

R

VVI

B

BECCB

mA428.2A04.241100I1I BE

29

71.10mA428.2

mV26

I

mV26r

Ee

(b)

k069.1

71.10100k470

k47071.10100

rR

RrZ

eB

Bei

(c) k3RZ C0

(d) 11.28071.10

k3

r

RA

e

Cv

(e) 100A i

(f)

k83.2Rr

RrZ

C0

C00

24.264Rrr

rR

V

VA

C0e

0C

i

0v

13.94rRRr

rR

I

IA

eBC0

0B

i

0i

30

Voltage Divider Bias

31

Input Impedance:

e21i r

1

R

1

R

1

Z

1

e21

21ei rRR

RRrZ

Output Impedance:

C00 R

1

r

1

Z

1

C0

C00 Rr

RrZ

If r0 10RC

C0 RZ

Voltage Gain:

C0

C0b0Cb0 Rr

RrIr||RIV

Bute

ib r

VI

C0

C0

e

i0 Rr

Rr

r

VV

0Ce

C0

i

0v rRr

Rr

V

VA

If r0 10RC

e

Cv r

RA

32

Current Gain:

eC0

0

i

0i r'RRr

r'R

I

IA

If r0 10RC

e0

0

i

0i r'Rr

r'R

I

IA

Where R’ = R1||R2

ei r'R

'RA

Also note that

C

ivi R

ZAA

33

Example: For the network shown in the Fig., determine:

(a) re (b) Zi (c) Z0 (r0 = ) (d) Av (r0 = )

(e) Ai (r0 = ) (f) The parameters of part (b) throughif r0 = 50 k.

Solution: (a)v81.2)22(

k2.8k56

k2.8V

vRR

RV CC

21

2B

v11.27.081.2VVV BEBE

mA41.1k5.1

11.2

R

VI

E

EE

44.18mA41.1

mV26

I

mV26r

Ee

34

k35.1

rRR

RRrZ

e21

21ei(b)

(c) k8.6RZ C0

(d) 76.36844.18

k8.6

r

RA

e

Cv

(e) ,k15.7k56k2.8

k56k2.8

RR

RR'R

21

21

(f)

04.73r'R

'RA

ei

,35.1Z i

98.5Rr

RrZ

C0

C00

3.324rRr

RrA

0Ce

C0v

3.64r'RRr

r'RA

eC0

0i

35

CE Emitter Bias Configuration

Fig. (a)

Fig. (b)

Applying KVL to the input Side of Fig (b) we getVi = Ibre + IeRE

orVi = Ibre + ( + 1)IbRE

and the impedance Zb

is: b

Eeb

b

ib I

R1rI

I

VZ

Eeb R1rZ or

36

Since >> 1, therefore + 1 . \ Zb re + RE

or Eeb RrZ

Since RE is often much greater than re, the above equationis further reduced to

Eb RZ

Returning to Fig. (b), we havebBi Z||RZ

The output impedance Z0 is determined as:C0 RZ

37

CE Emitter Bias Configuration• Voltage Gain:

b

ib Z

VI

and CbC00 RIRIV

Cb

i RZ

V

b

C

i

0v Z

R

V

VA

Substituting Zb = (re + RE)

Ee

Cv Rr

RA

and for the approximation Zb RE

E

C

i

0v R

R

V

VA

38

CE Emitter Bias Configuration

Current Gain:

bB

iBb ZR

IRI

bB

B

i

b

ZR

R

I

I

Since

In addition, I0 = Ib

b

0

I

I

so thatbB

B

i

0i ZR

R

I

IA

bB

Bi ZR

RA

C

iii R

ZAA or

39

Example: For the network of the given Fig., determine (using appropriate approximations):(a) re (b) Zi (c) Z0 (d) Av and (e) Ai.

Solution:(a) Testing RE > 10R2

(210)(0.68k) > 10(10k) 142.8k > 100k satisfied. V6.1

k10k90

k10V

RR

RV CC

21

2B

V9.07.06.1VVV BEBE

mA324.1k68.0

9.0

R

VI

E

EE

64.19

324.1

mV26

I

mV26r

Ee

40

(b) The ac equivalent circuit is shown here.From this circuit

The testing conditions r0 10(RC +RE ) and r0 10RC are both satisfied. Using the

appropriate approximations yieldsZb RE = 142.8 k

Zi = RB||Zb = 9k||142.8k = 8.47k

(c) Z0 = RC = 2.2k

(d) Av = -RC/RE = -3.24

(e) Ai = -AvZi/RL = 144.1

k9RR

RRR||R'RR

21

2121B

41

Emitter Follower Configuration

Emitter Follower Configuration re Model for EF Configuration

42

Emitter Follower Configuration

Input Impedance (Zi):

Zi = RB||Zb

With ZB = re + ( + 1)RE.

orZb (re + RE)

and Zb RE.

43

Emitter Follower Configuration

Output Impedance (Z0):

b

ib Z

VI

and

b

ibe Z

V1I1I

Substituting for Zb gives

Ee

ie R1r

V1I

or

Ee

ie

R1

rV

I

44

Emitter Follower Configurationbut

1

and

eee rr

1

r

so that

Ee

ie Rr

VI

If we now constructthe network defined By the above equation:

Ee

EeEe0 Rr

RrR||rZ

Since RE >> re, so

e0 rZ

45

Emitter Follower ConfigurationVoltage Gain (AV):

From the circuit given belowWe have

eE

iE0 rR

VRV

eE

E

i

0V rR

R

V

VA

Since RE >> re

1AV

46

Emitter Follower Configuration

From the Fig.

bB

iBb ZR

IRI

orbB

B

i

b

ZR

R

I

I

and be0 I1II

1I

I

b

0 or

so that

47

Emitter Follower Circuit

i

b

b

0

i

0i I

I

I

I

I

IA

bB

B

ZR

R1

and since ( + 1) ,

bB

Bi ZR

RA

or

E

iVi R

ZAA

48

Example: For the emitter follower network of the Fig., determine (a) re (b) Zi (c) Z0 (d) AV (e) Ai.

Solution: (a)

A42.20R1R

VVI

EB

BECCB

mA062.2I1I BE

61.12mA062.2

mV26

I

mV26r

Ee

(b) Zb = re + ( + 1)R = 334.56 kZi = RB||ZB = 132.72k

49

(c) Z0 = RE||re = 12.56

(d) 1996.0rR

RA

eE

EV

(e) 67.39ZR

RA

bB

Bi

50

Collector Feedback Configuration

51

Collector Feedback Configuration

Input Impedance (Zi):

F

i0

R

VV'I

With V0 = -I0RC and I0 = Ib + I’

Since Ib is generally much largerthan I’,I0 Ib andV0 = -(Ib)(RC) = -IbRC

Bute

ib r

VI

and

52

Collector Feedback Configuration

ie

CC

e

i0 V

r

RR

r

VV

Therefore

ie

C

FF

i

Fe

iC

F

i

F

0

F

i0 Vr

R1

R

1

R

V

Rr

VR

R

V

R

V

R

VV'I

The result is eeieiebi r'IrIr'IIrIV

iee

C

Feiebi Vr

r

R1

R

1rIrIV

53

Collector Feedback Configuration

eie

C

F

ei rI

r

R1

R

r1V

e

C

F

e

e

i

ii

r

R1

R

r1

r

I

VZ

But RC is usually much greater than re and 1 + RC/re RC/re

So that

F

C

ei

R

R1

rZ

F

C

ei

RR1

rZ

54

Collector Feedback Configuration

Output Impedance (Z0):

If we set Vi to zero as required to define Z0, the network will appear as shown below and

Z0 = RC||RF

55

Collector Feedback Configuration

Voltage Gain (AV):

At node C:'III b0

For typical values, Ib >> I’ and I0 Ib.V0 = -I0RC = -(Ib)RC

Substituting Ib = Vi/re gives usC

e

i0 R

r

VV

and

e

C

i

0V r

R

V

VA

56

Collector Feedback ConfigurationCurrent Gain (Ai):

Applying Kirchhoff’s law

Vi + VRF - V0 = 0

and Ibre + (Ib – Ii)RF + I0RC = 0

Using I0 Ib, we have

Ibre + IbRF – IiRF + IbRC = 0

Ib(re + RF + RC) = IiRF

Substituting Ib = I0/ from I0 = Ib yields

57

Collector Feedback Configuration

FiCFe0 RIRRr

I

and

CFe

iF0 RRr

IRI

In general RF + RC >> re

CF

iF0 RR

IRI

CF

F

i

0i RR

R

I

IA

For RC >> RF,

C

F

i

0i R

R

I

IA

or

C

F

i

0i R

R

I

IA

58

Example: For the given network, determine (a) re (b) Zi (c) Z0 (d) AV (e) Ai.

Solution: (a)

(b)

(c) Z0 = RC||RF = 2.66k

A53.11RR

VVI

CF

BECCB

mA32.2I1I BE

21.11I

mV26r

Ee

5.560

RR1

rZ

F

C

ei

(d) AV = -RC/re = -240.86

(e) 50RR

RA

CF

Fi