7.bjt Transistor Modeling.ppt 6k71m

BJT Transistor Modeling CHAPTER 5

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

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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 model b) hybrid equivalent model 3

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)

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Disadvantages 

Re model • Fails to the output impedance level of device and effect from output to input



Hybrid equivalent model • Limited to specified operating condition in order to obtain accurate result

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VCC

DC supply  “0” potential •I/p coupling capacitor  s/c • Large values • Block DC and AC signal

R1

RC

C1 C2 RS

+ Vi

Vo R2

RE

C3

-

VS

+

-

Voltage-divider configuration under AC analysis

R1

+ Vi

VS

• By capacitor  s/c •Large values

RC

+ RS

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

Vo R2

-

Redraw the voltage-divider configuration after removing dc supply and insert s/c for the capacitors

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Modeling of BJT begin HERE!

Ii B R1

+

RC

Zi

+ RS

+ Vi

VS

Vo R2

-

RS

VS

Vi

R1 R2

Transistor smallsignal ac equivalent cct

C

Io

+

E Rc Zo

Vo

-

-

-

Redrawn for small-signal AC analysis

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AC bias analysis 1. Kill all DC sources 2. Coupling and By capacitors are short circuit. The effect of there capacitors is to set a lower cutoff frequency for the circuit. 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. 8

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. 9

Two port system -determining input impedance Zi Rsense +

VS

+

Ii Zi

-

Vi -

Two-port system

Vi Zi  Ii Vs  Vi Ii  Rsense

Determining 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.

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Demonstrating the impact of Zi Rsource +

VS=10mV

600 Ω

Zi

1.2 k Ω

-

+ Vi -

Two-port system

Ideal source, Rsource  0Ω Full 10mV applied to the system With source impedance, Rsource  600Ω ZiVs 1.2k (10m) Vi    6.6mV Zi  Rsource 1.2k  600

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Example 6.1: For the system of Fig. Below, determine the level of input impedance 1k Ω

+

VS=2mV -

Rsense

+

Zi Vi=1.2mV

Two-port system

-

Solution : Vs  Vi 2m  1.2m 0.8m Ii     0.8A Rsense 1k 1k

Vi 1.2m Zi    1.5k Ii 0.8 12

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. Rsense

Rsource + Two-port system

Vs=0V

Vo

Io Zo

-

Determining Zo Iamplifier

+

V

V  Vo Io  Rsense

-

Vo Zo  Io

IL

For Ro  RL

IRo Zo=Ro

RL

IL  IRo

Zo  RL  Zo become open cct 13

Example 6.2: For the system of Fig. below, determine the level of output impedance Rsense Two-port system Vs=0V

+

20 k Ω

Zo Vo=680mV -

+

V=1 V -

Solution : V  Vo 1  680m 320m Io     16A Rsense 20k 20k

Zo 

Vo 680m   42.5k Io 16

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

Rsource + Two-port system

Vs=0V

Io

Vo

Zo

-

+

V -

Solution : V  Vo Io  Rsense Vo  V  IoRsense

 600m  10 10k  500mV

Zo 

Vo 500m   50k Io 10

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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.

Iamplifier

IL IRo

Zo=Ro

RL

Solution : Current divider rule : Zo(Iam plifier) IL  Zo  RL 50k (6m)  50k  2.2k  5.747mA

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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:

Vo Av  Vi

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By referring the network below the analysis are:

Rsource +

VS -

Zi

+

+ Vi

AvNL

-

Determining the no load voltage gain

Vo

no load Vo AvNL  Vi

RL  Ω (open cct)

-

with source resistance : Vo Zi Avs   AvNL Vs Zi  Rs

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

Rs +

VS=40mV

1.2 kΩ

+

Zi

Vi

-

BJT amplifier AvNL=320

-

b) Ii 

+

a) AvNL 

Vo=7.68V -

Vi 

Vo Vi

Vo 7.68   24mV AvNL 320

Vs - Vi 40m  24m   13.33A Rs 1.2k

Rs  Rsource

c) Zi 

Vi 24m   1.8k Ii 13.33

d) Avs 

Zi 1.8k AvNL  (320)  192 Zi  Rs 1.8k  1.2k 19

Current Gain, Ai • This characteristic can be determined by: Io

Ii + Vi

+ Zi

BJT amplifier

RL

Vo -

-

Vo Io   RL

Determining the loaded current gain

Io Ai  Ii

Vo / RL VoZi   Vi / Zi ViRL

Zi Ai   Av RL 20

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 transistor re model re equivalent circuit 21

Ic

C

Ie

E

re 

B

B

Common-base BJT transistor - pnp

26mV  IE is the DC level of IE( dc)

emitter current Therefore, the input impedance, Zi = re

e

that less than 50Ω.

Ic

Ie

c Ic  α Ie

b

For the output impedance, it will be as follows;

Ie=0A

b

re model for the pnp common-base configuration

Ic

e Vs=0V

c re

Ic  0A

b e

Ie

re

b

Ic c Ic  α Ie

b common-base re equivalent cct

isolation part, b Zi=re

Determining Zo for common-base

Zo  

22

The common-base characteristics

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BJT common-base transistor amplifier

Ie e

+ Vi -

c

Zo  

re

Zi

Ic  α Ie

b

Io RL

b

+ Vo -

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

Vo  IoRL   Ic RL  IeRL

Vi  IeZi  Iere Vo IeRL Av   Vi Iere

Voltage gain,  RL RL Av   re re 24

BJT common-base transistor amplifier

Ie e

+ Vi -

c

Zo  

re

Zi

Ic  α Ie

b

Io RL

b

+ Vo -

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

Io  Ic Ie Ai    Ii Ie Ie Current gain, Ai    1 25

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 Zi b) Calculate Av if RL=0.56k c) Find Zo and Ai

e

Ie

re

Ic c Ic  α Ie

b

b common-base re equivalent cct

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Solution: 26m 26m a) Zi  re    6.5 IE 4m

RL 0.98(0.56k) b) Av    84.43 re 6.5 c) Zo  Ω Io Ai      0.98 Ii

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Ii  Ie e re

Ic c Ic  α Ie

b

b common-base re equivalent cct 28

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 Solution: Vi 10m a) Zi    20 Ie 0.5m b) Vo  IcRL  IeRL  0.98(0.5m) (1.2k)  588mV

c) Av 

d) Ai    0.98

e) Ib  Ie - Ic  Ie - Ie  0.5m(1   )  0.5m(1  0.98)  10A

Vo 588m   58.8 Vi 10m

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Common-Emitter Configuration Common-emitter BJT transistor re model re equivalent cct. Still remain controlled-current source (conducted between collector and base terminal) Diode conducted between base and emitter terminal Input Base & Emitter terminal

Output Collector & Emitter terminal 30

c

C

Ic

Ic b

B

Ib

E

E common-emitter BJT transistor

Zi 

Ic   Ib

Ib

Vi Ii

e

e

re model npn common-emitter configuration

(1) c

Vi  Vbe  Iere  Ibre and

Ic subtitute into (1)gives

Zi 

Vbe  Ibre  Ib Ib

Zi  re Z i rangesbetweenhundred to 6 ~ 7k

b + Vi e

Ic   Ib

Ii=Ib + Vbe -

Ie re e

Determining Zi using re equivalent model 31

The output graph

32

Output impedance Zo b

Ii=Ib c  Ib

re

ro

e

e

re model for the C-E transistor configuration

b

Vs=0V

Ii=Ib = 0A c

re

Ib  0A

e

Zo

ro e

Zo  ro if ro is ignored thus the Zo  Ω (open cct, high impedance) 33

Ii=Ib e

+ Vi

BJT common-emitter transistor amplifier

c Zo  

re

Zi  re

b

-

Io  Ic   Ib

b

RL

Io + Vo -

Determining voltage and current gain for the common-emitter amplifier

Vo  IoRL  Ic RL  IbRL

Vi  IiZi  Ibre Voltage gain, Vo IbRL Av   Vi Ibre

RL Av   re

Current gain, Io Ic Ib Ai    Ii Ib Ib Ai  

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Example 6.8: Given =120 and IE(dc)=3.2mA for a commonemitter configuration with ro=  , determine: a) Zi b)Av if a load of 2 k is applied c) Ai with the 2 k load Solution : 26m 26m a) re    8.125 IE 3.2m Zi  re  120(8.125)  975

RL 2k b)Av      246.15 re 8.125 c) Ai 

Io    120 Ii 35

Example 6.9: Using the npn common-emitter configuration, determine the following if =80, IE(dc)=2 mA and ro=40 k b) Ai if RL =1.2k 

a) Zi b

Ii=Ib

c) Av if RL=1.2k 

c Io re

 Ib

ro

RL

e

Solution : 26m 26m   13 IE 2m Zi  re  80(13)  1.04k

a) re 

re model for the C-E transistor configuration

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Solution (cont) Io IL  Ii Ib ro( Ib) IL  ro  RL ro( Ib) ro 40k r o  RL Ai     (80) Ib ro  RL 40k  1.2k  77.67 b) Ai 

c)Av  

RL ro re



1.2k 40k 13

  89.6

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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

38

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.

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Determination of parameter Vi  h11Ii  h12 Vo h11  h12 

Vi Ii

Vo  0V

Vi Vo

Vo  0V

IO  h21Ii  h22 Vo Solving Vo  0V , h21  h22 

Ii Io

Vo  0V

Io Vo

Io  0A

H22 is a conductance!

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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

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 = re hfe =  hoe = 1/ro 46

Common-Emitter h-Parameters

hie  re h fe   ac

47

Common-Base re vs. h-Parameter Model

hib = re hfb = - 48

Common-Base h-Parameters

hib  re h fb    1

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