1

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

2

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

4

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

6

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

Draw band diagrams for different emitter

7

Preparing to compute IPreparing to compute ICC

• Why do we show asymmetrical hemi-Maxwellians?

8

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?

9

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 ?

11

Collector current: boundary conditionsCollector current: boundary conditions

12

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?

13

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?

16

Base current: componentsBase current: components

• Which IB components do we need to consider?

(iv)

17

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

18

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

24

y

Base-spreading resistanceBase-spreading resistance

25

Capacitance Capacitance

V

QC

V

+ +

- -

Generally:

Specifically:

1

2

26

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

°

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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?

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HF MOS HF MOS

What is this?