chapter 11 bjt static characteristics

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Lecture 10

Semiconductor Device Physics

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com

2 0 1 3

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Chapter 11BJT Static Characteristics

Semiconductor Device Physics

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Deviations, due to model limitations

Deviations from the IdealChapter 11 BJT Static Characteristics

Common base

Common emitter

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dc 2

E B

B E E B

B E Edc

E B

1

12

D N W WD N L LD N L

D N W

Wx

D pB(x)

EBB0 ( 1)qV kTp e (VCB=0)

0

W

P+ N P

+ VEB

IE IC

Increasing –VCB

C dc B CE0I β I I

If –VCB increases→ W decreases→ bdc increases→ IC increases

Base-Width ModulationChapter 11 BJT Static Characteristics

Common-Emitter ConfigurationActive Mode Operation

Recalling two formulas,

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WB = xnEB + xnCB

Punch-ThroughChapter 11 BJT Static Characteristics

Punch-Through: E-B and C-B depletion regions in the base touch each other, so that W = 0.

As –VCB increases beyond the punch-through point, the E-B potential hill decreases and therefore increases the carrier injections and IC.

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Punch-through Avalanche

Increasing reverse bias of C-B junction

EC EB CB=V V V

Breakdown MechanismsChapter 11 BJT Static Characteristics

In the common-emitter configuration, for high output voltage VCE, the output current IC will increase rapidly due to the two mechanisms: punch-through and avalanche.

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pnp BJT

dc CB0C B

dc dc

α

1 α 1 α

M II I

M M

M : multiplication factor

Avalanche MultiplicationChapter 11 BJT Static Characteristics

Holes [0] are injected into the base [1], then collected by the C-B junction.

Some holes in the C-B depletion region have enough energy do impact ionization [2].

The generated electrons are swept into the base [3], then injected into the emitter [4].

Each injected electron results in the injection of IEp/IEn holes from the emitter into the base [5].

For each pair created in the C-B depletion region by impact ionization, (IEp/IEn) + 1 > bdc additional holes flow into the collector.

This means that carrier multiplication in C-B depletion region is internally amplified.

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Geometrical EffectsChapter 11 BJT Static Characteristics

Emitter area is not equal to collector area.Current does not flow in one

direction only.Series resistance.

Voltage drop occurs not only across the junction.

Current crowding.Due to lateral flow, current is

larger around emitter periphery than the collector periphery.

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diff

kT q

xE

xdiff : exponential decay constant

Graded BaseChapter 11 BJT Static Characteristics

Dopants are injected through diffusion.More or less falling exponential

distribution with distance into beneath of the semiconductor.

The doping within the base is not constant as assumed in ideal analysis.A function of position, having

maximum at E-B junction and minimum at C-B junction.

Creating a built-in electric field.The electric field enhances the

transport of minority carrier across the quasineutral width of the base. Increase of IE and IC.

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Due to recombination in emitter depletion region

Due to high level injection in base, base series resistance,

and current crowding

Gummel Plot

Figures of MeritChapter 11 BJT Static Characteristics

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E1 E2E1 E2 d p d p

qD qDdx dx

D D

Continuity of hole current in emitter

(1 polysilicon; 2 Si) Shallower slope less JP

higher g, b

E1 E2 E2 E2 E2

E1 E1

d p D d p d p

dx D dx dx

D D D

Polysilicon Emitter BJTChapter 11 BJT Static Characteristics

bdc is larger for a poly-Si emitter BJT as compared with an all-crystalline emitter BJT.

This is due to reduced dpE(x)/dx at the edge of the emitter depletion region.

Lower mp

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Summary on BJT PerformanceChapter 11 BJT Static Characteristics

High gain (bdc >> 1)One-sided emitter junction, so that emitter efficiency g 1Emitter doped much more heavily than base (NE >> NB).

Narrow base, so base transport factor aT 1.Quasi-neutral base width << minority-carrier diffusion length

(W << LB). IC determined only by IB (IC function of VCE or VCB)

One-sided collector junction, so that quasineutral base width W does not change drastically with changes in VCE or VCB.

Base doped more heavily than collector (NB > NC), W = WB – xnEB – xnCB for pnp BJT.

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Chapter 12BJT Dynamic Response Modeling

Semiconductor Device Physics

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Cut-off(OFF)

Idealized switching circuit

Saturation(ON)

Load line

Qualitative Transient ResponseChapter 12 BJT Dynamic Response Modeling

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cutoff

active saturation

saturation active

1

5

2 3

4

1

2 3 4

5

Qualitative Transient ResponseChapter 12 BJT Dynamic Response Modeling

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B BB

B

,dQ Q

idt

B B( , ) (0, ) 1x

p x t p tW

D D

B BB

B

0dQ Q

idt

B B B

0

( , ) (0, )2

W qAWQ qA p x t dx p t D D

Small

Interpreted as average lifetime of an excess

minority carrier

Charge Control RelationshipsChapter 12 BJT Dynamic Response Modeling

A pnp BJT biased in the active mode has excess minority-carrier charge QB stored in the quasineutral base.

While

In steady state,

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2

tB2

W

D

Interpreted as average time taken by minority carriers to diffuse across

the quasineutral base

BC B

( , )

x W

p x ti qAD

x

D

BC

t

Qi

(Active mode)

Base Transit Time tt Chapter 12 BJT Dynamic Response Modeling

B BB 2

B

(0, )( / 2 )

qAD Qp t

W W D D

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C Bdc

B t

I

I

B

Ct

Qi

BB

B

Q

i

The lifetime of a minority carrier before it recombines in the base is much longer than the time it requires to

cross the quasineutral base region

Relationship between tB and ttChapter 12 BJT Dynamic Response Modeling

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2

tB

,2

W

D B N n

kTD D

q

2

t

n2

W

kTq

From Figure 3.5, mn 801 cm2/(Vs)

ExampleChapter 12 BJT Dynamic Response Modeling

Given an npn BJT with W = 0.1 μm and NB = 1017cm-3. Find tt.

5 2

2

(10 cm)2.414 ps

2(25.86mV)(801cm /V s)

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B BBB

B

dQ QI

dt

BB BB B( ) tQ t I Ae

BB BB B( ) (1 )tQ t I e

BB BB Br

t tC

CCCC r

L

( )(1 ) for 0

( ) for

tQ t Ie t t

i tV

I t tR

SB BB

S

,V

i IR

tr : rise time, period of active mode

Turn-On TransientChapter 12 BJT Dynamic Response Modeling

During the turn-on transient:

The general solution is:

Initial condition: QB(0)=0, since transistor is in cutoff:

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CC tr B

BB B

1ln

1ItI

IBBtB > ICCtt

bdc IBB > ICC

bdc > ICC/IBB

bdc > bdc(saturation)

Turn-On TransientChapter 12 BJT Dynamic Response Modeling

In saturation mode:

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During the turn-off transient:

B BBB

B

dQ QI

dt

B/B BB B( ) tQ t I Ae

B/B BB B( ) (1 ) tQ t I e

B

CC sd/

BB BC Bsd

t t

for 0

(1 )( ) ( ) for

t

I t t

I ei t Q tt t

SB BB

S

,V

i IR

tsd : storage delay time

Turn-Off TransientChapter 12 BJT Dynamic Response Modeling

The general solution is:

Initial condition: QB(0)=IBBtB:

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CC tsd B

BB B

1ln It

I

Turn-Off TransientChapter 12 BJT Dynamic Response Modeling

The transient speed of a BJT depends on the amount of excess minority-carrier charge stored in the base and also the recombination lifetime tB.

By reducing tB, the carrier removal rate is increased, for example by adding recombination centers (Au atoms) in the base.

Tradeoff: bdc= tB/tt will decrease.

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Collector current (actual form)

Collector current (mathematical model)

Practical ConsiderationsChapter 12 BJT Dynamic Response Modeling

The foregoing analysis was highly simplified to avoid excessive amount of mathematics.

More realistic iC transient response is shown below.Added delay time td (due to charging of junction capacitance)

and fall time tf.

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Homework 8

Due: Monday, 02.12.2013.

1. (E11.36-B)The base of a silicon npn bipolar transistor is WB = 0.8 μm wide. The doping concentrations are NB = 2×1016 cm–3 and NC = 1015 cm–3. Determine the punch-through voltage.

Chapter 12 BJT Dynamic Response Modeling

3.Next slide.

2. (E11.37-B)The base impurity doping level is NB = 3×1016 cm–3 and the base width is WB = 0.7 μm. If the required minimum punch-through voltage is determined to be Vpt = 70 V, calculate the highest level of collector doping.

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Homework 8

Further data are given as follows: NE = 1019 cm–3, NB = 1017 cm–3, NC = 1015 cm–3, DE = 2 cm2/s, DB = 20 cm2/s, DC = 12 cm2/s, τE= 10–7 s, τB= τC= 10–6

3.(EIE130.AF00F)Consider a pnp silicon BJT of area A = 10–6 cm2 at 300 K. It operates in active region with VEB = 0.6 V and VCB = –1 V, so that WB = 0.6 μm. Assume that the emitter and collector regions are much longer than the respective minority carrier diffusion length.

Chapter 12 BJT Dynamic Response Modeling

a. What is the common-emitter d.c current gain of this transistor?

b. What are the excess minority-carrier concentrations at the edges of the depletion regions (at A, B, C, and D in the diagram above).

c. Calculate the collector current.