Silicon VLSI Technology Solutions Manual – Plummer, Deal & Griffin – Instructor's Guide

Solutions.Manual


SILICON VLSI TECHNOLOGY
. .



Fundamentals, Practice andModels S
. . . .




olutions Manual for Instructors
. . .




JamesD.Plummer M
. . .


ichael D. Deal Peter B.
. . . .


Griffin .




SILICON.VLSI.TECHNOLOGY 1 ©.2000.by.Prentice.Hall.
Fundamentals,.Practice.and.Modeling. Upper.Saddle.River,.NJ.
By.Plummer,.Deal.and.Griffin

, Solutions.Manual


Chapter 1 Problems
. .




1.1. Plot.the.NRTS.roadmap.data.from.Table.1.1.(feature.size.vs..time).on.an.expanded.s
cale.version.of.Fig..1.2..Do.all.the.points.lie.exactly.on.a.straight. line?.If.not.what.reas
ons.can.you.suggest.for.any.deviations.you.observe?

Answer:

250

200

150

100

50

0
1997 2002 2007 2012
Year

Interestingly,.the.actual.data.seems.to.consist.of.two.slopes,.with.a.steeper.slope.for.the.fi
rst.2.years.of.the.roadmap..Apparently.the.writers.of.the.roadmap.are.more.confident.of.t
he.industry's.ability.to.make.progress.in.the.short.term.as.opposed.to.the.long.term.

1.2. Assuming.dopant.atoms.are.uniformly.distributed.in.a.silicon.crystal,.how.far.apart.
are.these.atoms.when.the.doping.concentration.is.a)..1015.cm-3,.b)..1018.cm-
3,.c)..5x1020.cm-3.


Answer:

The.average.distance.between.the.dopant.atoms.would.just.be.one.over.the.cube.root.of.th
e.dopant.concentration:

x.=. N A .−1/ .3

(
a) x.=. 1x1015.cm−3. ) −.1./.3.
=.1x10−5.cm.=.0.1m.=.100nm

b) x = (1x10 ) −.1/3.
. .
18
. cm−3. =.1x10−6.cm.=.0.01m. =.10nm


SILICON.VLSI.TECHNOLOGY 2 ©.2000.by.Prentice.Hall.
Fundamentals,.Practice.and.Modeling. Upper.Saddle.River,.NJ.
By.Plummer,.Deal.and.Griffin

, Solutions.Manual


(
c) x.=. 5x1020.cm−3. ) −1/.3.
=.1.3x10−7.cm.=.0.0013m.=.1.3nm

1.3. Consider.a.piece.of.pure.silicon.100.µm.long.with.a.cross-
sectional.area.of.1.µm2..How.much.current.would.flow.through.this.“resistor”.at.roo
m.temperature.in.response.to.an.applied.voltage.of.1.volt?

Answer:

If.the.silicon.is.pure,.then.the.carrier.concentration.will.be.simply.ni..At.room.temperature,
10 -3
.ni.≈.1.45.x.10 .cm ..Under.an.applied.field,.the.current.will.be.due.to.drift.and.hence,




( )
I.=.In . +.Ip . =.qAn.i. n. +. p . 

(
=. 1.6x10−19.coul )(10 −8.cm2.
)(1.45x10 10
. carrierscm−3. )(2000cm volt
2
. ) 1volt
−1.sec−1.
. 
 10 cm
.
..

−2.


=. 4.64x10−12.amps.or. 4.64pA

1.4. Estimate.the.resistivity.of.pure.silicon.in.. ohm.cm.at.a).room.temperature,.b).77K,.
and.c).1000.˚C..You.may.neglect.the.temperature.dependence.of.the.carrier.mobility
.in.making.this.estimate.




Answer:

The.resistivity.of.pure.silicon.is.given.by.Eqn..1.1.as
1 1
=.
.
=.
( ) (
q .n n .+.pp . qni. .n. +.p )
Thus.the.temperature.dependence.arises.because.of.the.change.in.ni.with.T..Using.Eqn..1.
4.in.the.text,.we.can.calculate.values.for.ni.at.each.of.the.temperatires.of.interest..Thus
. .0.603eV
n i . =.3.1x1016.T 3/ .2 −.
exp
kT 

which.gives.values.of.≈.1.45.x.1010.cm-3.at.room.T,.7.34.x.10-21.cm-
3 18 -
.at.77K.and.5.8.x.10 .cm
3 2 -1 -
.at.1000.˚C..Taking.room.temperature.values.for.the.mobilities.,.µn.=. 1500.cm .volt .sec
1 2 -1 -1
.and.,.µp.=.500.cm .volt .sec ,.we.have,




SILICON.VLSI.TECHNOLOGY 3 ©.2000.by.Prentice.Hall.
Fundamentals,.Practice.and.Modeling. Upper.Saddle.River,.NJ.
By.Plummer,.Deal.and.Griffin

, Solutions.Manual


 = .2.15x105.cm.at.room.T
=.4.26x1035.cm.at.77K
=.5.39x10−4.cm.at. 1000.ÞC

Note.that.the.actual.resistivity.at.77K.would.be.much.lower.than.this.value.because.trace.a
mounts.of.donors.or.acceptors.in.the.silicon.would.produce.carrier.concentrations.much.h
igher.than.the.ni.value.calculated.above.

1.5. a).. Show. that. the. minimum. conductivity. of. a. semiconductor. sample. occurs
p
when. n. =. ni .
n
b). What.is.the.expression.for.the.minimum.conductivity?
c). Is.this.value.greatly.different.than.the.value.calculated.in.problem.1.2.for.the.intri
nsic.conductivity?

Answer:

a).
1.
 .=. =.q .n n .+.pp

( )
To.find.the.minimum.we.set.the.derivative.equal.to.zero.
 . . . .  n2.i..  n2 . 
.

=.
.
( )
q .nn.+.pp =. q nn.+.p
.
. ..=. q. n . +. p . . i. . =.0
n n n..  n. .  n2 . 

p
n 2 . =. n 2 . or n. =. n p
i. i n
n

b). Using.the.value.for.n.derived.above,.we.have:

 
 n2i 
min. =.q   n
. n i
p +. = .q n n i
p
+ pni n.  =. 2qni
p .  n p
n p.  n  p. 
 ni 
 n. 




c). The.intrinsic.conductivity.is.given.by

(
i. =.qni. n. +.p )
SILICON.VLSI.TECHNOLOGY 4 ©.2000.by.Prentice.Hall.
Fundamentals,.Practice.and.Modeling. Upper.Saddle.River,.NJ.
By.Plummer,.Deal.and.Griffin