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# A p-channel MOSFET with a heavily-doped p-type polysilicon gate ...

By Eric Boyd,2014-05-02 00:20
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A p-channel MOSFET with a heavily-doped p-type polysilicon gate ...with,Type,Gate

UNIVERSITY OF CALIFORNIA

College of Engineering

Department of Electrical Engineering and Computer Sciences

EE 130 Prof. Liu

Spring 2007

Homework Assignment #11 Solutions

Problem 1: MOS Fundamentals

a) The flat-band voltage (V) of this PMOS capacitor is determined as follows: FB

V(( FBMS

where (for n-type Si): 18D NkT10F(ln().026Vln()0.060V(8)0.48V10iqn10

EG ?((|(|)4.03V(0.56V0.48V)4.11VSSiF2q

?V5.15V4.11V1.04VFB

b) The sketches of the energy-band diagrams (for each condition) are shown below:

MOSMOSMOSMOSMOSMOS

| qV| qV| qV|||ox ox ox

EEEEEEEE ccccccccEEEE|qV|qV|qV|||FSFSFSFSG G G EEEEEEEE= = = = EEEEcccc FMFMFMFMvvvvEEEEEEEEvvvvFSFSFSFSEEEEcccc|qV|qV|qV|||G G G EEEEvvvvEEEE= = = = EEEE|q|q|q(((| is | is | is FMFMFMFMvvvvSSS small, small, small, ~~~000

| qV| qV| =0, | =0, |q|q((|=0|=0ox ox SS

i) Flat-band i) Flat-band ii) Accumulation ii) Accumulation

(V(V(V(V= = VV))> > VV))GGFBFBGGFBFB

MOSMOSMOSMOSMOSMOS

|||| qV| qV| qVox ox ox |||| qV| qV| qVox ox ox

EEEEEEEEccccccccEEEcccEEEE= = = = EEEEEEEE= = = = EEEEEEEFMFMFMFMvvvvEEEFMFMFMFMvvvvFSFSFS|qV|qV|qV|||cccG G G EEEEEEFSFSFSvvv|q|q|q(((|||SSSEEEvvv|q|q|q|||(((SSS

iv) Strong inversion iv) Strong inversion iii) Equilibriumiii) Equilibrium

(V(V= 0V)= 0V)(V(V< < VV))GGGGTT

c) For a MOS capacitor with n-type semiconductor:

qNDSiF22( VVTFBF2(Cox

qNDSiF22(VVVT ?1.042(0.48)Cox

where

14SiO3.98.8510F/cm6222 C1.7310F/cm1.73F/cmox7x210cmo

19183122?1.610C?10cm?110F/cm?0.96V ?V0.08VT621.7310F/cm

?V0.248VT

Thus at threshold (when V = V) the voltage dropped across the semiconductor is GT

2( = -0.96V, and the voltage dropped across the gate oxide is Q/C= -0.328V. F depox Note that the magnitude of the voltage dropped across the oxide is less than the

magnitude of the voltage dropped across the semiconductor.

d) Let’s rewrite the equation for V to see the impact of gate work function (() and TM

qNqNDSiFDSiF2(22(2semiconductor doping (hence (): FVVTFBFMSF(2(((2CCoxox

qNEDSiFG2(2??MSiFF? (((2 ?qCox2(

qNEDSiFG22(MSiF ((qCox2

++The work function of n poly-Si is 1.12eV lower than that of p poly-Si. Thus, if the gate

++material were to be changed from p poly-Si to n poly-Si, V would become more T

negative (shifted by -1.12V). To counteract this change (i.e. to keep V = -0.248V), the T

doping in the semiconductor must be changed such that

VVTnewTold,,

2qN22qN2((DSiFnewDSiFoldEE,,GG((((MnpolySiFnewMppolySiFold,,,,2qC2qCoxox

2qN22qN2((DSiFnewDSiFold,,1.12V1.12V0.48V0.328V0.312V((FnewFold,,CCoxox

Note that this equation is impossible to satisfy if ( is negative. Thus, ( must be F,newF,new

positive, i.e. the semiconductor must be changed to p-type.

qNqNASiFASiF2(22(2Note that the equation for V is different for a MOS capacitor with p-type semiconductor: TVVTFBFMSF(2(((2CCoxox

qNEASiFG2(2??MSiFF? (((2 ?qCox2(

qNEASiFG22(MSiF ((qCox2

AAkTNN(ln().026ln()where FeV10

10iqn

+Noting that for an n poly-Si gate, and setting V = -0.248V: (4.03VTMSi

2qN2EASiFG(VTF(2qCox

2qN2ASiF(0.248V0.56VF(Cox

2qN2ASiF(0.312VF(Cox

19121021.610N10F/cm20.026ln[N/10]N??AA;；;；;；A 0.026ln?1062101.7310F/cm(

N??1010A;； 0.026ln4.6210N0.026ln[N/10]?AA1010(

15-3Solving by iteration, we obtain N~1.24x10 cm. A

Problem 2: MOS Charge Density +(a) For an NMOS capacitor with n poly-Si gate, 3nm-thick SiO gate dielectric, and 217-3substrate doping N=10 cm, A

14SiO3.98.8510F/cm6222C1.1510F/cm1.15F/cmox7x310cmo

17A NkT10F(ln().026Vln()0.060V(7)0.42V10iqn10

EG ((()4.03V(0.56V0.42V)5.01VSSiF2q

V((4.03V5.01V0.98VFBMS

ASiF2qN2( 3TFBFVV25.610V~6mV(oxC

i) When V < V, the semiconductor surface is accumulated. Thus, GFB

, Q=0, Q=0; depinv;；QCVVaccoxGFB

(1) ;；?QQQQQCVVTotaccdepinvaccoxGFB

ii) When V <V < V, the semiconductor surface is depleted. Thus, FBGT

, Q=0, Q=0; accinvQqNW2qN(depAAss

with qNASis2( VVGFBs(Cox

)?2qNCVV2ASioxGFB 2() ???(s11qNCASi??2ox2?

Therefore,

??2)? 2qNCVV?2()ASioxGFBQqN(qN??2211depAssAs?qN2C???ASi2ox?(

??2)? (2) qNCVV?22()ASioxGFBQQqN??211TotdepAs?qN?CASi2??2ox?(

iii) When V > V, the semiconductor surface is inverted. Thus, GT

Q=0, acc

(3) QQqNW2qN|2(|depdep,maxATAsF72 ?Q1.6710C/cmdep,max

; ;；QCVVinvoxGT

72 (4) ;；?QQQ1.6710C/cmCVVTotdepinvoxGT

The plot of the areal charge density is shown below:

Now, at V= V+1V, the semiconductor surface is inverted. From (3): G T

726262 ;；Q1.6710C/cm1.1510F/cm1V1.3210C/cmTot

(b) If the substrate doping N were to be increased, V would become more negative AFB

(since ( increases), V would increase, and Q would also increase in magnitude STdep,max

(since both N and ( increase, ref. equation (3)). AF

Consequently (since C is not affected), for V> V, Q increases (maintaining the oxG FBTot

same slope for V> V). For V < V, the Q curve is not affected. G TGFBTot

(c) If the oxide thickness x were to be decreased, C would increase. V and Q ooxFBdep,max

would not be affected. Consequently, V would decrease and the slopes of the Q TTot

curve in accumulation and inversion modes increase in magnitude.

Part (b)Part (b)Part (b)Part (c)Part (c)

Problem 3: MOS Capacitance

(a) As can be seen from the C-V curve above, C=C=C when V< 0V, which maxoxG

indicates that the substrate is p-type (i.e. holes accumulate at the surface of the p-type

substrate when V< 0V, as illustrated in figure (a) below). G

(b) The charge density diagrams are shown below:

(a) Point 1(a) Point 1(b) Point 2(b) Point 2

(c) From point 1 on the C-V curve,

143.98.8510F/cmSiOSiO4222CA?xA(1?10cm) oxcoc12xC10010Foox

7?x3.4510cm3.45nm o

(d) From point 2 on the C-V curve,

CCW1111180pFoxTmin?22(A?)CCA?CCCC20?10pFcSioxcSioxoxminminminWT

W 11T?0.04pFW0.04pF(A?)TcSiA?cSi1424 W0.04pF(10cm)(1pF/cm)0.04?10cm0.04mT

Therefore, since

2|2| sF(WTqNA

with NkTA F(ln()qni

17-3 Solving by iteration, we get: N~7.5x10 cm. A

(e) Since C = C in inversion (V > V), this is a high frequency measurement (as minGT

discussed in Slide 12 from Lecture 32).

(f) If the semiconductor doping concentration of were to be increased, V would FBdecrease while V would increase (since ( would increase). Additionally, C TSminwould increase slightly (since W decreases with increasing N). Therefore, the C-V TAcurve gets more ‘spread out’ as shown below (with C slightly higher), as discussed min

in Slide #4 of Lecture 33.

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