CSE 477. VLSI Systems Design - GUC

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ELCT 503: Semiconductors German University in Cairo (GUC) ELCT503 Semiconductors Fall 2014 Lecture 09: BJT Circuit Analysis Dr. Hassan Mostafa د. حسن مصطفى[email protected]

Transcript of CSE 477. VLSI Systems Design - GUC

ELCT 503: Semiconductors German University in Cairo (GUC)

ELCT503

Semiconductors Fall 2014

Lecture 09: BJT Circuit Analysis

Dr. Hassan Mostafa

حسن مصطفى. د

[email protected]

ELCT 503: Semiconductors German University in Cairo (GUC)

Introduction

npn transistor

pnp transistor

ELCT 503: Semiconductors German University in Cairo (GUC)

Introduction

Symbol

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DC Models

npn

BEJ BCJ

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DC Models

iC – vCB characteristics for common-base configuration

ELCT 503: Semiconductors German University in Cairo (GUC)

DC Models

iC – vCE characteristics for common-emitter configuration

(VA = Early voltage)

ELCT 503: Semiconductors German University in Cairo (GUC)

DC Models (npn)

Transistor OFF

(VBE < 0.7V) and (VBC < 0.4V)

IB = 0 IC = 0 IE = 0

Transistor in Active region

(VBE >= 0.7V) and (VBC < 0.4V)

VCE >=0.3V

IC = IS exp (VBE/VT)

IC = β IB

IE = IC + IB = (β+1) IB

IC = IE

β = / (1- ) and β >> 1

= β / (β+1) and <= 1

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DC Models (npn)

Transistor in Saturation region

(VBE >= 0.7V) and (VBC >= 0.4V)

VCE =0.2V

ICsat = βforced IB

IE = IC + IB = (βforced+1) IB

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DC Models (pnp)

Same equations as before but after changing letters order: VBE VEB

VBC VCB

VCE VEC

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Example1

β = 100

VBE = 0.7V at IC = 1 mA

Design the circuit such that:

IC = 2 mA

VC = +5V

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Example1

Solution

Transistor works in the active region

kmA

RC 52

515

VVVVV

V

CBBC

BE

4.05

7.0

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Example1

Solution

kI

VR

mAmAII

V

VVVV

VmA

mAVV

E

EE

CE

E

EEBBE

TBE

07.7)15(

02.22*100

1100/

717.0

0717.0

717.0)1

2ln(*7.0

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Example2

β = 100

Find IC, IB, IE, VB, VC, and VE

Assume Active region VBE = 0.7V

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Example2

Check:

(VBE >= 0.7V)

(VBC = -1.3V < 0.4V)

Correct

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Example3

β = 100

Find IC, IB, IE,

VB, VC, and VE

Assume Active

region

Check:

(VBE >= 0.7V)

(VBC = 3.52V > 0.4V)

Incorrect

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Example3

Assume

Saturation

region

Check:

(VBE >= 0.7V)

(VBC = 0.5V > 0.4V)

Correct

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BJT in Logic Gates RTL (Resistor Transistor Logic)

Increasing vBE decrease vo

C

T

BESCCCEo

T

BESC

BE

CCoC

BE

CCCCCEo

RV

vIVvv

V

vIi

ActiveAssume

ONVvWhen

Vvi

OFFVvWhen

RiVvv

*)exp(*

)exp(*

5.0

0

5.0

*

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When vBE increases such that vBC becomes larger than 0.4V

Transistor enters Saturation Region

BJT in Logic Gates

Vvv CEo 2.0

Which Logic gate this Circuit implements?

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Think…

Which Logic gate this Circuit implements?

BJT in Logic Gates

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Small Signal Models

Derivation : BJT must be in active region to use the small signal model

TV

CI

bev

ci

mg

V

vIi

iIi

VvV

vIi

V

vI

V

v

V

VIi

V

vVIi

V

vIiIi

T

beCc

cCC

Tbe

T

beCC

T

beC

T

be

T

BESC

T

beBESC

T

BEScCC

)1(*

)exp()exp()exp(

)exp(

)exp(

B

T

m

C

T

b

be

T

beCcb

cCCbBB

I

V

gr

I

V

i

vr

V

vIii

iIiiIi

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Small Signal Models

-model

C

Ao

mC

T

B

T

T

Cm

I

Vr

gI

V

I

Vr

V

Ig

||

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Small Signal Models

T-model

C

Ao

mC

T

E

Te

T

Cm

I

Vr

gI

V

I

Vr

V

Ig

||

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Common Emitter Amplifier

Why it is called common emitter?

Bypass capacitors

Open circuit at DC

Short circuit at small signal

How to draw the small signal equivalent circuit?

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Common Emitter Amplifier

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Common Emitter Amplifier

Amplifier parameters:

Voltage gain (AV) = vo/vi <Units V/V>

Input resistance (Rin) = vi/ii <Units >

Output resistance (Rout) = vo/io|vs=0 <Units >

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Common Emitter Amplifier

Voltage gain (AV) = vo/vi = - gm*(ro//RC // RL) V/V

= - gm*(RC // RL) when ro >> RC and RL

= - gm*(RC ) when ro >> RC and RL =

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Common Base Amplifier

VA = ro =

Voltage gain (AV) = vo/vi = (RC //RL)/ re = +gm (RC //RL)

= (RC )/ re = +gm (RC) when RL =

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Common Collector Amplifier (Emitter follower)

VA = ro =

Voltage gain (AV) = vo/vi = RL / (RL + re)

= 1 when RL =

Why it is called Emitter follower?