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Holes and electrons determine device characteristics. Three terminal device Control of two terminal currents. Bipolar Junction transistor. Amplification and switching through 3 rd contact. V. I. p. n. I. V. I 0. How can we make a BJT from a pn diode?.

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Presentation Transcript
bipolar junction transistor

Holes and electrons

determine device characteristics

Three terminal device

Control of two terminal currents

Bipolar Junction transistor

Amplification and switching through 3rd contact

how can we make a bjt from a pn diode

V

I

p

n

I

V

I0

How can we make a BJT from a pn diode?
  • Remember reverse bias characteristics
  • Take pn diode
  • Reverse saturation current: I0
test multiple choice
Test: Multiple choice

Why is the reverse bias current of a pn diode small?

  • Because the bias across the depletion region is small.
  • Because the current consist of minority carriers injected across the depletion region.
  • Because all the carriers recombine.
test multiple choice1
Test: Multiple choice

Why is the reverse bias current of a pn diode small?

  • Because the bias across the depletion region is small.
  • Because the current consist of minority carriers injected across the depletion region.
  • Because all the carriers recombine.
how can we make a bjt from a pn diode1

V

I

e-

p

n

h+

  • np and pn low
  • I0 small

I

V

I0

How can we make a BJT from a pn diode?
  • Remember reverse bias characteristics
  • Take pn diode
  • Reverse saturation current: I0
  • Caused by minority carriers swept across the junction
slide6

V

I

e-

p

n

h+

I

np and/or pn

V

I0

Test: Multiple choice

  • If minority carrier concentration

can be increased what will happen to I0?

  • Increase
  • Decrease
  • Remain the same
slide7

np and pn

V

I

e-

p

n

h+

  • If np and pn higher
  • |I0| larger

I

V

I0

  • If minority carrier concentration

can be increased near the depletion region edge, then I0 will increase.

slide8

V

I

e-

p

n

h+

I

V

I0

Test: True-False

pn

If we only increase

then |I0| will still increase.

how can we increase the minority carrier concentration near the depletion region edge

V

I

h+

p

n

e-

How can we increase the minority carrier concentration near the depletion region edge?
  • Take pn diode
  • Remember forward bias characteristics
  • How can we make a hole injector from a pn diode?
  • By increasing the applied bias, V.
  • By increasing the doping in the p region only
  • By applying a reverse bias.
slide10

V

I

h+

p

n

  • When using a p+n junction
  • diode current If ≈ hole current

e-

I

  • Ip pno (eeV/kT-1)
  • In npo (eeV/kT-1)
  • Since NA >> ND
  • np << pn
  • → Ip >> In

V

If

Hole injector

  • Take pn diode
  • Remember forward bias characteristics

p+

slide11

V

V

I0

I

h+

p+

n

e-

n

p

e-

h+

If W large, then?

W

Thus:

A forward biased p+n diode is a good hole injector

A reverse biased np diode is a good minority carrier collector

  • Recombination of excess holes will occur and excess will be 0 at end of layer
  • Recombination of excess holes will occur and excess will be large at end of layer
  • No recombination of excess holes will occur.
  • Recombination of excess electrons will occur and excess will be np0 at end of layer
slide12

V

V

I

I0

dpn

h+

p+

n

e-

n

p

e-

x

h+

Lp

If W large → holes

recombine

W

Thus:

A forward biased p+n diode is a good hole injector

A reverse biased np diode is a good minority carrier collector

Excess hole concentration reduces exponentially in W to some small value.

slide13

V

V

I

I0

dpn

h+

p+

n

e-

n

p

e-

x

h+

Lp

W

What is the magnitude of the hole diffusion current at the edge

x=W of the “green” region?

  • Magnitude of hole diffusion current at x=W is same as at x=0
  • Magnitude of hole diffusion current at x=W is almost 0
  • Magnitude of hole diffusion current cannot be derived from this layer.
slide14

V

V

I

I0

dpn

h+

p+

n

e-

n

p

e-

x

h+

Lp

if W large → holes

recombine

W

Reduce W

Thus:

A forward biased p+n diode is a good hole injector

A reverse biased np diode is a good minority carrier collector

Since gradient of dpn @ x=W is zero, hole diffusion current is also zero

bjt p np

p+

n

IC

B

V

V

IE

I

I

p

VBC

W < Lp

BJT p+np

E: emitter

VBC

EB

B: base

E

C

C: collector

C

E

Common base configuration

base short layer with recombination and no ohmic contacts at edges
Base: Short layer with recombination and no Ohmic contacts at edges.

Single junction

pno

pno

npo

npo

Double junction

npo

npo

pno

No Ohmic contact thus minority carrier concentration not

how will we calculate the minority carrier concentration in the base
How will we calculate the minority carrier concentration in the base?

Rate equation

Steady state

General solution of second order differential equation

With Ohmic contact C1=0

C2≠0

Without Ohmic contact C1≠0

C2≠0

planar bjt npn

n+-well for emitter

p-well for base

p+ Si

Ohmic contact

n+ Si

ohmic contact

device insulation

p-substrate

n-well for collector

B

C

E

n+ Si

p+ Si

p+ Si

p Si

n Si

p Si

Planar BJT - npn

For integrated circuits (ICs) all contacts have to be on the top

carrier flow in bjts

E

B

C

IE

IC

p+

n

p

holes

e- gain, reverse bias

holes

IE

IC

ICB0

I’B

I”B

Recombination

e- loss

e- loss, forward bias

IB

Carrier flow in BJTs

IB

IB = I’B + I”B – ICB0

control by base current ideal case

Electrostatically neutral

h+

e-

tp

recombine with

Wb << Lp

Control by base current : ideal case.

Based upon space charge neutrality

Base region

IE = Ip

tt transit time

tt < tp

Based on the given timescales, holes can pass through the narrow base before a supplied electron recombines with one hole: ic/ib = tp/tt

The electron supply from the base contact controls the forward bias to ensure charge neutrality!

how good is the transistor

E

IC

IEp

B

equilibrium

Injection of carriers

e-

VBE>0

Wb < Lp

x

h+

No amplification!

IEn

Amplification!

How good is the transistor?

C

  • IEp>>IEn
  • Wish list:

or g = IEp/(IEn + IEp) ≈ 1

g: emitter injection efficiency

  • IC≈ IEp

or B= IC/IEp≈ 1

B: base transport factor

or a= IC/IE≈ 1

a: current transfer ratio

+ (1-B) IEp

  • IB≈IEn

thus b= IC/IB= a/(1-a)

b: current amplification factor

ICB0 ignored

review 1 bjt basics

Forward biased p+n junction is a hole injector

Reverse biased np junction is a hole collector

B

E

W < Lp

Review 1 – BJT basics

IC

Forward active mode (ON)

IE

VBC

V

V

VBC

I

I

EB

E

C

p+

n

p

C

E

review 1 bjt basics1

Forward biased p+n junction is a hole injector

Reverse biased np junction is a hole collector

B

E

W < Lp

Review 1 – BJT basics

IC

Forward active mode (ON)

IE

VBC

V

V

VBC

IB=I’B

+I”B

I

I

EB

E

C

p+

n

p

C

E

review 2

IB = I’B + I”B – ICB0

Review 2

Amplification?

Recombination only case: I’B, ICB0negligible

ic/ib = tp/tt

Carriers supplied by the base current stay much longer in the base: tp than the carriers supplied by the emitter and travelling through the base: tt.

b = tp/tt

But in more realistic case: I’Bis not negligible

b = IC/IB

With IB electrons supplied by base = I’B = In

IC holes collected by the collector = Ip

currents
Currents?
  • In order to calculate currents in pn junctions, knowledge of the variation of the minority carrier concentration is required in each layer.
  • The current flowing through the base will be determined by the excess carrier distribution in the base region.
  • Simple to calculate when the short diode approximation is used: this means linear variations of the minority carrier distributions in all regions of the transistor. (recombination neglected)
  • Complex when recombination in the base is also taken into account: then exponential based minority carrier concentration in base.
minority carrier distribution

E

B

C

p(x)

B

DpE

Without recombination

pn0

With recombination

pn0

DpC

x

0

Wb

Minority carrier distribution
  • Emitter injects majority carriers into base.
  • dpn(0)=pno (exp(VEB/VT)-1)
  • Assume active mode: VEB>0 & VBC<0
  • Collector collects minority carriers from base.
  • dpn(Wb)=pno (exp(VBC/VT)-1)

dp(x)

0

currents simplified case

dp(x)

B

DpE

DpC

x

0

Wb

See expressions for diode current for short diode

Currents: simplified case
  • Then IE = total current crossing the base-emitter junction

Assume I”B=0 & IBC0= 0

  • Then IC = IEp gradient of excess hole concentration in the base
  • IB without recombination is the loss of electrons via the BE junction: I’B
  • Then IB = gradient of excess electron concentration in the emitter
narrow base no recombination i p

dp(x)

DpE

Linear variation of excess carrier concentration:

DpC

x

0

Wb

Narrow base: no recombination: Ip

→minority carrierdensity gradient in the base

DpE = pn0(e eVEB/kT – 1) ≈pn0 e eVEB/kT

DpC = pn0(e –e|VBC|/kT – 1) ≈ -pn0

Note: no recombination

collector current i p

Diffusion current:

Collector current: Ip

Hole current:

Collector current

No recombination, thus all injected holes across the BE junction are collected.

Base current??

look at emitter i n

dn(x)

Dnp

Linear variation of excess carrier concentration:

0

x

xe

0

Look at emitter: In

→ minority carrier density gradient in the emitter

Dnp = np0(e eVEB/kT – 1) ≈np0 e eVEB/kT

base current in

Diffusion current:

Base current: In

Base current:

The base contact has to re-supply only the electrons that are escaping from the base via the base-emitter junction since no recombination I”B=0 and no reverse bias electron injection into base ICB0=0.

slide32

Emitter current:

Current gain:

Emitter current

The emitter current is the total current flowing through the base emitter contact since IE=IC+IB (current continuity)

slide33

Short layer approach – summaryforward active mode

dc(x)

IE

=

IpEB

+

InEB

DpE

IC

=

IpBC

+

InBC

DnE

IC

IpBC

=

IpEB

IE

=

IB

+

IC

x

DpC

DnC

IB

=

IE

-

IC

-Xe

Wb

Xc

0

IB

=

InEB

slide34

General approach also taking recombination into account.forward active mode

dc(x)

DpE

DnE

x

-Xe

DpC

Xc

-LpE

LpC

DnC

0

Wb

< LnB

slide35

Which formulae do we use for the excess minority carrier concentration in each region?forward active mode

dc(x)

DpE

DnE

x

-Xe

DpC

Xc

-LpE

LpC

DnC

0

Wb

< LnB

Emitter

Collector

use LONG diode approximation

dnpE(x)=DnE exp(-(-x)/LpE)

dnpC(x)=DnC exp(-x/LpC)

in the base we must take recombination into account short diode approximation cannot be used

From:

In the base we must take recombination into account → short diode approximation cannot be used!

dp(x)

Excess hole concentration dp(x):

DpE

Exact solution of differential equation:

x

dp(x) = C1 ex/Lp + C2 e-x/Lp

DpC

Wb

Constants C1, C2:

DpE = dp(x=0)

DpC = dp(x=Wb)

in the base with recombination long diode approximation can also not be used
In the base with recombination → long diode approximation can also not be used!

dp(x)

Exact solution of differential equation:

dp(x) = C1 ex/Lp + C2 e-x/Lp

DpE

Long diode approximation:

dp(x) = C3 e-x/Lp

Boundary condition at BC junction cannot be guaranteed

x

LnB

DpC

Wb

slide38

http://www.ecse.rpi.edu/~schubert/Course-ECSE-2210-Microelectronics-Technology-2010/http://www.ecse.rpi.edu/~schubert/Course-ECSE-2210-Microelectronics-Technology-2010/

slide39

Extraction of currents in the general approach.forward active mode

dc(x)

IE

=

IpEB

+

InEB

IC

=

IpBC

+

InBC

DpE

IC

IpBC

DnE

IE

=

IB

+

IC

x

IB

=

IE

-

IC

-Xe

DpC

Xc

-LpE

LpC

DnC

0

Wb

< LnB

IB

=

InEB

+

IpEB

-

IpBC

Term due to recombination

slide40
Currents: Special case when only recombination in base current is taken into account: Approximation: IB’=0

dp(x)

  • Assume IE=IEp & IBC0= 0

B

DpE

Starting point:

  • Then IE = Ip(x=0)
  • and IC = Ip(x=Wb)

x

DpC

0

Wb

  • IB=IE - IC

=I”B

all currents are then determined by the minority carrier gradients in the base
All currents are then determined by the minority carrier gradients in the base.

Injection at emitter side: DpE = pn0(e eVEB/kT – 1)

Collection at collector side: DpC = pn0(e eVCB/kT – 1)

dp(x)

  • IE = Ip(x=0)

DpE

  • IC = Ip(x=Wb)

B

DpC

x

0

Wb

expression of the diffusion currents

Hyperbolic functions

Expression of the diffusion currents

Diffusion current: Ip (x) = -e A Dp ddp(x)/dx

Emitter current: IE ≈ Ip (x=0)

Collector current: IC ≈ Ip (x=Wb)

Base current: IB ≈ Ip (x=0) - Ip (x=Wb)

IE ≈ e A Dp/Lp (DpE ctnh(Wb/Lp) - DpC csch(Wb/Lp) )

IC ≈ e A Dp/Lp (DpE csch(Wb/Lp) - DpC ctnh(Wb/Lp) )

IB ≈ e A Dp/Lp ((DpE + DpC) tanh(Wb/2Lp) )

Superposition of the effects of injection/collection at

each junction!

Note: only influence of recombination

non ideal effects in bjts

VBC

BE

E

C

p+

n

p

C

E

B

Original base width

V

V

I

I

Depletion width changes with VBC

Effective base width

Metallurgic junction

Non-ideal effects in BJTs
  • Base width modulation
base width modulation

base width modulated

VA

Base width modulation
  • Early voltage: VA

iC

Wb

ideal

IB

-vCE

conclusions
Conclusions
  • Characteristics of bipolar transistors are based on diffusion of minority carriers in the base.
  • Diffusion is based on excess carrier concentrations:
    • dp(x)
  • The base of the BJT is very small:
    • dp(x) = C1 ex/Lp + C2 e-x/Lp
  • Base width modulation changes output impedance of BJT.
slide47

iC

RL

ib higher

iB

iC

ECC /RL

ECC

RS

ib

es

iE

es

-vCE

ECC

t

p-type material

n-type material

On

Off

slide48

iC

iC

ic=biB

-vCE

RL

iB

ECC

RS

es

Es

t

iE

-Es

slide49

iC

ic=biB

iC

-vCE

RL

ECC

RS

es

Es

t

iE

-Es

slide50

iC

ic≠biB

iC

-vCE

RL

ECC

RS

Ic= ECC /RL

es

Es

t

iE

-Es

switching cycle

iB

IB

IB≈Es/RS

iC

RL

iB

-IB

dp

ECC

RS

DpE

QB

t2

es

iE

ts

Qs

DpC

DpE

t1

Es

t1

ts

t2

t’s

0

t

-Es

x

-pn

DpE

tsd

t0

Wb

iC

IC≈ECC/RL

IC

Switching cycle

Switch to ON

Switch OFF

iC

ECC /RL

-vCE

ECC

charge in base linear
Cut-off

VEB<0 & VBC<0

DpE=-pn & DpC=-pn

Saturation

VEB>0 & VBC≥0

DpE = pn (eeVEB/kT – 1)

DpC = 0 (VBC=0)

dp

dp

DpE

DpE

-pn

DpC

x

Wb

x

Wb

Charge in base (linear)

VBC>0

slide53

Currents - review.forward active mode

dc(x)

IE

=

IpEB

+

InEB

IC

=

IpBC

+

InBC

DpE

IC

IpBC

DnE

IE

=

IB

+

IC

x

IB

=

IE

-

IC

-Xe

DpC

Xc

-LpE

LpC

DnC

0

Wb

< LnB

IB

=

InEB

+

IpEB

-

IpBC

Term due to recombination

switching cycle review

iC

RL

iB

dp

ECC

RS

DpE

t2

es

iE

ts

DpC

Es

0

t

-Es

x

DpE

Wb

Switching cycle - review

iB

Switch to ON

Common emitter cicuit

IB

IB≈Es/RS

With IB>ICmax/b

Over-saturation

-IB

QB

Qs

DpE

t1

Load line technique

t1

ts

t2

-pno

t0

iC

iC

ECC /RL

ICmax≈ECC/RL

IC

<< DpE

pno

-vCE

ECC

switching cycle review1
Switching cycle - review

iB

Switch OFF

Common emitter cicuit

IB

iC

RL

iB

-IB

≈-Es/RS

dp

ECC

RS

DpE

QB

t2

DpE

es

iE

t’s

Es

Qs

t

-Es

DpC

t3

-pno

Load line technique

t2

t3

t4

t’s

0

t4

x

tsd

iC

Wb

iC

ECC /RL

IC≈ECC/RL

IC

-vCE

ECC

calculating the delays
Calculating the delays
  • Since the currents and minority carrier charge storage are determined by the pn diodes, the delays are calculated as in the pn diode.
    • Knowledge of current immediately before and after switch
    • Stored minority carrier charge Qp(t) cannot change immediately → delay.
  • The additional parameter is the restriction on the maximum collector current imposed by the load.
slide57

RL

C

p

RS

vbc

e(t)

ECC

B

n

veb

p

t

E

iC

ICsat

dpnB(x)

E

B

C

QB

IBtp

Qsat

IB

IB

IB

IB

IB

tsat

t

0

WB

x

tsat

t

t<0

t<tsat

veb

= 0→ON≈0.7V

E - p

B - n

t≥tsat

RS

+E>>0.7V

ON switching

OFF=0→ON

t≥0

driving off

ib

IB

IB

t

-IB

t

Qb

IBtp

IBtp

Qs

t

Qs = ICtt

iC

iC

ib

-IBtp

t

tsd

tsd

IC

IC

t

Driving off

Time to turn the BJT OFF is determined by:

  • The degree of over-saturation (BC junction)

2) The off-switching of the emitter-base diode

CASE 2: OFF=-IB

0N (saturation)→OFF

CASE 1: OFF=IB=0

0N (saturation)→OFF

Qb

t

slide59

IB=0

tsd

Qsat

IB

tsd

tsd

OFF switching

0N (saturation)→OFF - CASE 1: OFF=IB=0

RL

C

p

RS

vbc

e(t)

ECC

B

n

veb

p

t

E

iC

tsd

dpnB(x)

ICsat

t<0

E

B

C

QB

t≥0

IBtp

tsd

t

0

WB

x

t

t<0

t<tsd

veb

= 0.7V (ON)→0V

E - p

B - n

t≥tsd

RS

E=0V

slide60

tsd

t≥0

-IB

tsd

tsd

Qsat

IB

tsd

tsd

0N (saturation)→OFF - CASE 2: OFF=-IB

RL

C

p

RS

vbc

e(t)

ECC

B

n

veb

p

t

E

iC

dpnB(x)

ICsat

t<0

E

B

C

QB

IBtp

t

0

WB

x

t

t<0

veb

= 0.7V (ON)→-E

E - p

B - n

t<tsd

RS

-E

t≥tsd

slide61

tsd

STORAGE DELAY TIME: tsd

tsd

tsd

shorter delay

0N (saturation)→OFF - CASE 1: OFF=IB=0

0N (saturation)→OFF - CASE 1: OFF=-IB

iC

t<tsd

iC

t<tsd

tsd

ICsat

ICsat

t≥tsd

t≥tsd

t

t

transients

t

ts

iC

IC≈ECC/RL

IC

Transients

Turn-on: off to saturation

slide63

RL

C

p

RS

vbc

e(t)

ECC

B

n

veb

p

t

E

iC

ICsat

QB

IBtp

Qsat

tsat

t

tsat

t

Time to saturation

ON switching

OFF=0→ON

t≥tsat

t<tsat

t=tsat

transients1

t

ts

iC

IC≈ECC/RL

IC

oversaturation

Transients

Turn-on: off to saturation

ts = tp ln(1/( 1 – IC/b IB))

ts small when:

tp small

IC small compared to b IB

transients2

toff

t’s

iC

IC ≈ ECC/RL

IC

Transients

Turn-off: saturation to off

Storage delay time: tsd

slide66

tsd

Time from saturation

0N (saturation)→OFF - CASE 1: OFF=IB=0

t<tsd

iC

tsd

ICsat

t≥tsd

t

transients3

Determined by

EB diode

toff

t’s

iC

IC ≈ ECC/RL

IC

NO oversaturation

Transients

Turn-off: saturation to off

Storage delay time: tsd

tsd = tp ln(b IB /IC)

tsd small when:

tp small

BUT

tsd large when:

IC small compared to b IB

transients4

Determined by

EB diode

t

ts

toff

t’s

iC

iC

IC ≈ ECC/RL

IC≈ECC/RL

IC

IC

oversaturation

NO oversaturation

Transients

Turn-off: saturation to off

Turn-on: off to saturation

Storage delay time: tsd

ts = tp ln(1/( 1 – IC/b IB))

tsd = tp ln(b IB /IC)

ts small when:

tp small

IC small compared to b IB

tsd small when:

tp small

BUT

tsd large when:

IC small compared to b IB

solution to dilemma the schottky diode clamp
Solution to dilemmaThe Schottky diode clamp

B

B

metal

C

C

C

B

B

E

E

I

V

0.3

0.7

Schottky diode

pn diode

large signal equivalent circuit
Large signal equivalent circuit
  • Switching of BJTs
    • LARGE SIGNAL

iC

RL

iB

ECC

RS

es

iE

iC

t

ebers moll large signal circuit model for large signal analysis in spice
Ebers-Moll large signal circuit model for large signal analysis in SPICE

Not examinable

Is valid for all bias conditions.

The excess at the BC is taken into account what is essential for saturation operation and off-currents.

superposition eb bc influence

dp

dp

dp

DpE

DpE

DpC

DpC

IEN

ICN

IEI

ICI

Wb

x

Wb

x

Wb

x

SuperpositionEB & BC influence

Take EB & BC forward biased.

Charge in base:

=

+

negative

IE = IEN + IEI

Where IEN, ICI are pn diode currents of EB and BC respectively.

IC = ICN + ICI

ebers moll equations

Diode currents

Ebers-Moll equations

IE = IEN + IEI

IC = ICN + ICI

IE = IES (eeVEB/kT –1) – aI ICS (eeVCB/kT –1)

IC = aN IES (eeVEB/kT –1) – ICS (eeVCB/kT –1)

ebers moll equations1

Collected currents

IEI = aI ICI

ICN = aN IEN

IE = IES (eeVEB/kT –1) – aI ICS (eeVCB/kT –1)

IC = aN IES (eeVEB/kT –1) – ICS (eeVCB/kT –1)

Ebers-Moll equations

IE = IEN + IEI

IC = ICN + ICI

a: current transfer factor

ebers moll equations2

IEO

IE = aI IC + IEO (eeVEB/kT –1)

IC = aN IE - ICO (eeVCB/kT –1)

ICO

Ebers-Moll equations

IE = IES (eeVEB/kT –1) – aI ICS (eeVCB/kT –1)

IC = aN IES (eeVEB/kT –1) – ICS (eeVCB/kT –1)

Where: aN IES = aI ICS

Or:

IE = aI IC+ (1- aN aI) IES (eeVEB/kT –1)

IC = aN IE - (1- aN aI) ICS (eeVCB/kT –1)

General equivalent circuit based on diode circuit

equivalent circuit

E

C

IE

IC

IB

B

Equivalent circuit

IE = aI IC + IEO (eeVEB/kT –1)

IC = aN IE - ICO (eeVCB/kT –1)

IE = aI IC + IEO (eeVEB/kT –1)

IC = aN IE - ICO (eeVCB/kT –1)

IE = aI IC + IEO (eeVEB/kT –1)

IC = aN IE- ICO (eeVCB/kT –1)

IE = aI IC + IEO (eeVEB/kT –1)

IC = aN IE - ICO (eeVCB/kT –1)

IE = aI IC + IEO (eeVEB/kT –1)

IC = aN IE - ICO (eeVCB/kT –1)

Valid for all biasing modes

description of different transistor regimes
Cut-off

VBE<0 & VCB<0

Active

VBE>0 & VCB<0

E

E

C

C

IE

IE

IC

IC

IB

IB

B

B

IC

IE

-VCB

Small!

IC0, IE=0

Description of different transistor regimes

IC = IC0 + aN IE

IE = -(1-aN) IES

IC = (1-aI) ICS

slide79
Now
  • Amplification and maximum operation frequency
    • SMALL SIGNAL equivalent circuit

Cj,BC

C

B

Cj,BE

Cd,BE

npn

Rp

gmvbe

Ro

vbe

E

definition of circuit elements
Definition of circuit elements
  • Transconductance

Cj,BC

C

B

Cj,BE

Cd,BE

Rp

gmvbe

Ro

E

slide81
Base input resistance

Cj,BC

C

B

Cj,BE

Cd,BE

Rp

gmvbe

Ro

E

slide82
Base-emitter input capacitances

Cj,BE

Depletion capacitance

Cd,BE

Diffusion capacitance

See SG on pn-diode

Cj,BC

C

B

Cj,BE

Cd,BE

Rp

gmvbe

Ro

E

slide83
Base-collector capacitance

Cj,BC

Depletion capacitance

Miller capacitance: feedback between B & C

Cj,BC

C

B

Cj,BE

Cd,BE

Rp

gmvbe

Ro

E

slide84

iC

IB

ideal

VA

-vCE

  • Output resistance

Cj,BC

C

B

Cj,BE

Cd,BE

Rp

gmvbe

Ro

E

current gain frequency
Current gain - frequency
  • Small signal current gain

Circuit analysis

Max gain

ib

Cj,BC

C

B

Cj,BE

Cd,BE

vbe

Rp

gmvbe

Ro

E

transit frequency f t
Transit frequency fT
  • Small signal current gain=1

t total transit time

Base transit time

Base-Emitter charging time

transit frequency f t1
Transit frequency fT
  • Base transit time

for p+n

Note: this approach ignores delay caused by BC junction (see 3rd year)

simplified small signal equivalent circuit
Simplified small signal equivalent circuit

Common-emitter connection

Active mode:

BE: forward, BC: reverse.

small signal equivalent circuit when other biasing connection is made

ie

ic

E

C

i’e

re

rc

ai’e

Cdif

CjE

CjC

B

Small signal equivalent circuit when other biasing connection is made

Common-base connection

Active mode:

BE: forward, BC: reverse.

conclusion
Conclusion
  • Delays in BJTs are a result of the storage of minority carriers.
  • Main delay in common BJTs is due to the base transit time tt.