HWSolution8

HWSolution8 - ELEN E4301- Introduction to Semiconductor...

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Unformatted text preview: ELEN E4301- Introduction to Semiconductor Devices Spring 2005 Homework Solution Eight For NMOSFET, ow as _o_5s_ Em[_4_] 4 For PMOSFET, ow g—o.55+.’.‘.31n[_x] q n. with xm=20nm, Vgg=0V, we have N (6mg) n-channel MOSF’ET p—Channel MOSFET mi- .9- Iii- enhancement The Charis L's I ad = XI. "' gN‘XJm "Fm E31414!) dexg 4WHV’5] “ a4 = 1&1?” ‘ 26*NWH'4’HMJ'32‘W and Vrzv -+\/+l Icfi, '+_Q_l._ -‘ F5 s 43?! + fluiafi M m4» 53mm) 1 +3 (3) foersv0V,g=-EynQ,:>Q’-'—: 31' =4.l'r'xlllmcm”2 L 4} qW/uu (b) Q. =~Cs(VGs—V})=>VarVr =~§" =3.82V OX 2,51 (a) current is given by ID = -_WQ,.v. Using the long channel model, the device is in the saturation region W ,Cm, V —V In =p,C,,, 504, —V,.)’ auw at the source, Q..=Cm(Vg-Vr) :> v = Evin—m = 1.26 x lO’crn/s as Q” is assumed to be close to zero at the drain in the long channel model, the carrier velocity have to approach infinity. 0)) ID has to be calculated first. Using Equation 9.2.3, 8.fl=0.6MWcm Using Equation 9.2.4 with value given in Table 9.3, w3é4cm2fv~s Taking vm=8xl 0‘5ch and with Equation 9.2.7, we have 8m=4.39x10‘V/cm Therefore, using Equation 9.2.11, VDm—*l.62V and the device'is also in saturation At the source, WC“ V -V -V m, 15 = WCJVG ‘Vr Wash... :>v= W: 3.68x106cm/s At the drain, the carrier is moving at vaSX 106011115 ...
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This note was uploaded on 02/28/2012 for the course EE 4301 taught by Professor Laibowitz during the Spring '10 term at Columbia.

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HWSolution8 - ELEN E4301- Introduction to Semiconductor...

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