计算机科学-浙江大学集成电路原理与设计Exercise19-19本习题答案4

浙江大学电气工程学院吴晓波，赵梦恋

模拟与数模混合集成电路

2018-2019学年复学期2018年5月

Table4.1

TypicalParameterValue

Parameter

SymbolParameterDescriptionnChannelp-ChannelUnits

%Thresholdvoltage(6-0)0.7x015-0.7±0.15V

K'Transconductanceparameter110.0±10%50.0士10%出A/V】

(insaturation)

yBulkthresholdparameter0.40.57

XChannellengthmodulation004(L=1Rm)005(L=1nm)v-«

parameter0.01(L=2H.m)0.01(L=2

2所|SuxfmjcpuleHialalix0.70.8V

inversion

4-1.AnactiveresistivevoltagedividerwasshowninFigure4.1.IfVDD=5V,VSS=0V,VOUT=2.5V.

I=15|iA,usingmodelparametersinTable4.1,determinetheW/LratioofMlandM2.

Assume入=0.

Figure4.1

Solution:

1D1

乙L

二生w9

1D2(7)2(%S2-VTPY

2LJ

w2/°i1

⑦1=

K^vGS1-vTNy11.88

W1

(E)2=

K^VGS2-VTPY5.4

4-2.DeterminethecurrentIinFigure4.2andcalculateitsTC(TemperatureCoefficient)using

CN20processshowninTable4.2.NeglectoxideencroachmentandassumethatM5is

operatinginthesaturatior.region.Allunlabclcdn-channelsare15/5andallunlabeled

p-channelsare70/5.AssumeResistorRispolysiliconandhasatemperaturecoefficientof

15(X)ppm/℃,thresholdvoltageVihasatemperaturecoefficientoftz=2.3x105V/°C.

vss

Figure4.2

Table4.2

frocewGV20NMOSC7V2OPMOS

InfoniuHionApp.AApp.A

VDD(V55=0)5V-5V

2.0pm

minimum4atl2.0pm

”-L--”0.6gm0.8pm

minimuiD卜匕13.0pm3.0pm

DW-"JW60.14pm0.16pm

S%0.83V0.91V

KP.pA/V55017

0.06for▲25Hm0.06forL>5jim

/^(gA/pm)

%800a17>un'800aF/Mm,

12g(〃W)36kQ(LW)

38ps114ps

Solution:

W.100

UsingCMOS20NMOSmodel,since—=----»1,

2

=2/4=2/2«2—=2•—=2〃4

R830x10,

Themoreaccuratesolutioncanbederivedas

I=2h=2h=

2倍+林糕十六

"*=3%R=J5°x83°xI°，=2.93X10,

整、耳*心膺等*(83。*哂=2.44x10。

252S

^50M/V100X(83OX1O)=8.61X1O

/

rJVn11[W7i1)

RP\R2R、伙RfkR?)

J0.83IiI2x0.831)

(830K2.93xlO483OxlO3V2.93x1042.44xlO10J

=2.018/zA

IdlId(Vn1If2WiI1IdVT1dR

2

〃1dT1dT{R小R?R、仇RR)VTdTRdT

where

\_dR=\500ppm/nC

~R~dT

器…=2„C

ihus

1dV1dR-a1dR2.3x10^

-,T——-=-———=—16

TCF=-15(X)x10=-4271ppmPC

VTdTRdTVTRdT0.83

4-3.Figure4.3illustratesvariouswaystoimplementthelayoutofaresistordivider.Choosethe

layoutthatbestachievesihegoalofa2:1ratio.Explainwhytheotherchoicesarenot

optimal.

B

-

v

W

2R-

7

W

-

A

(C)W)(*)

Figure4.3

Solution:

OptionAsuffersthefollowing:

-Orientationofthe2Rresistorispartlyorthogonal(othe1Rresistor.Matchedresistors

shouldhaveIhesameorientation.

-Resistorsdonothavetheappropriateetchcompensating(dummy)resistors.Dummy

stripesshouldsurroundallactiveresistors.

OptionBsuffersthefollowing:

-Resistorsdonothavetheappropriateetchcompensating(dummy)resistors.Dummy

stripesshouldsurrouncallactiveresistors.

-Resistorsdonotshareacommoncentroidastheyshould.

OptionCsuffersthefollowing:

-Resistorsdonotshareacommoncentroidastheyshould.

-Uncertaintyisintroducedwiththeadditionalnotchatthecontacthead.

OptionDsuffersthefollowing:

-Resistorsdonothavetheappropriateetchcompensating(dummy)resistors.Dummy

stripesshouldsurroundallactiveresistors.

OptionEsuffersthefollowing:

-Nothing

OptionFsuffersthefollowing:

-Violatestheunit-matchingprinciple

-Resistorsdonothavetheappropriateetchcompensating(dummy)resistors.Dummy

stripesshouldsurrouncallactiveresistors.

-Resistorsdonotshareacommoncentroidastheyshould.

UnitMatchingEtchComp.OrientationCommon

Centroid

(a)YssNoNoYes

(b)YesNoYesNo

(c)YesYesYesNo

(d)YssNoYesYes

(e)YssYesYesYes

(f)NoNoYesNo

Clearly,option(e)isthebestchoice.

4-4.DetermineVrcf(OutputVo.tagc)inFig4.4andtheconditionsunderwhichtheTCofVrefis

zero.AssumeK=10.

Outputvolume

vss

Figure4.4

解：

ForthecircuitinFig6.1,weget

V«fisdependentontemperatureandweget

dVr.fr、“a内dVdz

------二£•In4---------+------

dTdT

—0.085wV/JCand-=-2mV/℃

dT=dT

LetVrefhaszerotemperaturecoefficientandget

dVrefJ-d%dV3c

-------=£•