1 Heterojunction Bipolar Transistors Heterojunction Bipolar Transistorsfor High-Frequency Operation...

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Heterojunction Bipolar TransistorsHeterojunction Bipolar Transistors

forfor

High-Frequency OperationHigh-Frequency Operation

D.L. Pulfrey

Department of Electrical and Computer EngineeringUniversity of British ColumbiaVancouver, B.C. V6T1Z4, Canada

pulfrey@ece.ubc.ca

http://nano.ece.ubc.ca

Day 3A, May 29, 2008, Pisa

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OutlineOutline

• What are the important features of HBTs?

• What are the useful attributes of HBTs?

• What are the determining factors for IC and IB?

• Why are HBTs suited to high-frequency operation?

• How are the capacitances reduced?

3

Schematic of InGaP/GaAs HBTSchematic of InGaP/GaAs HBT

• Epitaxial structure

• Dissimilar emitter and base materials

• Highly doped base

• Dual B and C contacts

• Identify WB and RB

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HETEROJUNCTION BIPOLAR TRANSISTORS

• The major development in bipolar transistors (since 1990)

• HBTs break the link between NB and

• Do this by making different barrier heights for electrons and holes

• NB can reach 1E20cm-3

e-

h+• Key feature is the wide-bandgap emitter

- this improves fT and fmax

- this allows reduction of both WB and RB SHBT

An example of Bandgap Engineering

5

2.5

2

1.5

1

0.5

05.4 5.5 5.6 5.7 5.8 5.9 6 6.1

Lattice Constant, A

Ban

dg

ap, e

V

Si

Ge

GaAs

GaP

AlP

AlAs

InAs

InP90

8070

6050

4030

2010

a = 5.6533 Amatched to GaAs

°

°

a = 5.8688 Amatched to InP

°

Selecting an emitter for a GaAs baseSelecting an emitter for a GaAs base

AlGaAs / GaAs

InGaP / GaAs

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InGaP/GaAs and AlGaAs/GaAsInGaP/GaAs and AlGaAs/GaAs

Draw band diagrams for different emitter

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Preparing to compute IPreparing to compute ICC

• Why do we show asymmetrical hemi-Maxwellians?

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Current in a hemi-MaxwellianCurrent in a hemi-Maxwellian

Full Maxwellian distribution

Counter-propagating hemi-M's for n0=1E19/cm3

/1E20What is the current?

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Density of statesDensity of states

Recall:

In 1-D, a state occupies how much k-space?What is the volume in 3-D?

If kx and ky (and kz in 3-D) are large enough, k-space is approximately spherical

Divide by V (volume) to get states/m3

Use parabolic E-k (involves m*) to get dE/dk

Divide by dE to get states/m3/eV

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VelocitiesVelocities

Turn n(E) from previous slide into n(v) dv using

vR = 1E7 cm/s for

GaAs

Currents associated with hemi-M's and M's

= 1E7 A/cm2 for n0=6E18 /cm3

*

What is Je,total ?

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Collector current: boundary conditionsCollector current: boundary conditions

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Reduce our equation-set for the electron Reduce our equation-set for the electron current in the basecurrent in the base

What about the recombination term?

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Diffusion and Recombination in the baseDiffusion and Recombination in the base

In modern HBTs WB/Le << 1

and is constant

Here, we need:

1016

1017

1018

1019

1020

10-6

10-5

10-4

10-3

10-2

10-1

Doping (cm-3)

Diff

usio

n le

ngth

(cm

)

Le

Lh

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Collector current: controlling velocitiesCollector current: controlling velocities

Diffusion (and no recombination) in the base:

-1

Note:

- the reciprocal velocities

- inclusion of vR necessary in modern HBTs

*

* Gives limit to validity of SLJ10

110

210

30

1

2

3

4

5

6

7

8

9

10x 10

4

WB (nm)

J C

(Acm

-2)

vR

=infinityv

R =1e7 cm/s

15

Comparing resultsComparing results

• What are the reasons for the difference?

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Base current: componentsBase current: components

• Which IB components do we need to consider?

(iv)

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Base current components and Gummel Base current components and Gummel plotplot

• What is the DC gain?

I C (

A/c

m2 )

VBE (V)

IB (injection)

IB (recombination)

IC

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Preparing for the high-frequency analysisPreparing for the high-frequency analysis

• Make all these functions of time and solve!

• Or, use the quasi-static approximation

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The Quasi-Static ApproximationThe Quasi-Static Approximation

q(x, y, z, t' ) = f( VTerminals, t')

q(x, y, z, t' ) f( VTerminals, t < t')

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Small-signal circuit components

gm = transconductance

go = output conductance

cebeb vgvgi 12

g = input conductance

g12 = reverse feedback conductance

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Recallg12=dIb/dVce

next

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Small-signal hybrid-Small-signal hybrid- equivalent equivalent circuitcircuit

What are the parasitics?

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HBT ParasiticsHBT Parasitics

• CEB and RB2 need explanation

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y

Base-spreading resistanceBase-spreading resistance

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Capacitance Capacitance

V

QC

V

+ +

- -

Generally:

Specifically:

1

2

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E B C

QNE

QNC

QNB

VBE+

BE

jEjEB V

QC

,

,

• QE,j is the change in charge entering the device through the emitter and creating the new width of the depletion layer (narrowing it in this example),

• in response to a change in VBE (with E & C at AC ground).

• It can be regarded as a parallel-plate cap.B

jEB W

AC

,

WB2

WB1

What is the voltage dependence of this cap?

Emitter-base junction-storage capacitanceEmitter-base junction-storage capacitance

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E B C

QNE

QNC

QNB

VBE+

BE

bEbEB V

QC

,,

• QE,b is the change in charge entering the device through the emitter and resting in the base (the black electrons),

• in response to a change in VBE (with E & C at AC ground).

• It’s not a parallel-plate cap, and we only count one carrier.

Emitter-base base-storage capacitance: Emitter-base base-storage capacitance: conceptconcept

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B

QNB

n(x)

xn(WB)

WB

)/exp(),0(

)/exp(),0(

202,

101

thBEpBE

thBEpBE

VVnVn

VVnVn

bEB

BE

bE

BBBth

BEpBBEbE

C

dV

dQ

WAnqWWnV

VnAWqVQ

,

,

BE

bE,

0,

Hence

V

Q Take

)()()exp(2

1)(

For the case of no recombination in the base:

What is the voltage dependence of CEB,b ?

Emitter-base base-storage capacitance: Emitter-base base-storage capacitance: evaluationevaluation

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Base-emitter transit capacitance: evaluation Base-emitter transit capacitance: evaluation

Q = 3q

qe = -2q

• What are q0 and qd ?

• Where do they come from ?

30fT from hybrid-pi equivalent circuit

• g0 and g set to 0

• fT is measured under AC short-circuit conditions.

• We seek a solution for |ic/ib|2 that has a single-pole roll-off with frequency.

• Why?

• Because we wish to extrapolate at -20 dB/decade to unity gain.

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Extrapolated fExtrapolated fTT

• Assumption:

• Conditions:

• Current gain:

• Extrapolated fT:

32 Improving fImproving fTT

• III-V for high gm

• Implant isolation to reduce C

• Highly doped sub-collector and supra-emitter to reduce Rec

• Dual contacts to reduce Rc

• Lateral shrinking to reduce C's

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Designing for high fDesigning for high fTT values values

Why do collector delays dominate ?

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How does Si get-in on the act?How does Si get-in on the act?2.5

2

1.5

1

0.5

05.4 5.5 5.6 5.7 5.8 5.9 6 6.1

Lattice Constant, A

Ban

dg

ap, e

V

Si

Ge

GaAs

GaP

AlP

AlAs

InAs

InP90

8070

6050

4030

2010

a = 5.6533 Amatched to GaAs

°

°

a = 5.8688 Amatched to InP

°

35

Developing an expression for Developing an expression for ffmaxmax

Assumption and conditions:

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Improving fImproving fmaxmax

• Pay even more attention to Rb and C

Final HF question:

How far behind are Si MOSFETs?

37

HF MOS HF MOS

What is this?

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