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chapter 15 - CHAPTER 1 5 lS_1(a Field Effect.tnodulation of...

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Unformatted text preview: CHAPTER 1 5 lS-_1 (a) Field Effect..tnodulation of the semiconductor conductivity by an electric field applied normal to the surface of the semiconductor. (b) Channel...nondepieted current carrying portion of the semiconductor “bar" between the source and drain in a J—FET. (c) As viewed from the exterior of the device, die drain current flows onrvof the drain contact in a tit—channel device Holes are the channel carriers in a prchannel device and by definition these must flow along the channel into the drain. The current has the same direction as the hole flowmfrotn source to drain and out of the draih oontact. (d) Gradual channel approximation...ln this approximation it is assumed the electrostatic variables in one direction {say the gar-direction} change slowly compared to the rate of change of the electrostatic variables in a second direction (say the x—direction). The y- direction dependence is then neglected and the electrostatic variables computed using a pseudo—one—dirnensional analysis at each pointy. (e) Pinch—off...complete depletion of the channel region; touching of the top and bottom depletion regions in the symmetrical J—FET. (f) As given by text Eqs. (15.13). s = Li“ mdrairt conductance EVD Vg=constaml 3m = Lin ...transconductance 3V0 VDr-ocatstant (g) In a long channel .l—-FET. IDWG held constant} 5 constant for V1) 2:- VDsat- Thus 3.: E El and the gs conductance in Fig. 15.19{b) can be neglected in drawing the equivalent circuit. (h) MESFET...matal semiconductor field effect transistor. D-...dep1etion mode; Easenhancement mode. 15-1 [i] Dace IEyl exceeds ~lfl4 Wen-t. the carrier drift velocity is no longer proportional to the tnagnitude of the electric field as assumed in the long—channel analysis. (j) In the two—region theory the carrierdrift velocity is set equal to tan at all points in the channel between 3:1 and the drain. yl is the point in the channel where IEYI has increased to thafllow-field mobility}. 15.2 (a) If a' is: Lp, the two pa junctions 1will he interacting like in a BJT. Moreover, the biases are equivalent to active mode biasing in a BIT. bias D—H bias Obviously. we are being asked for the common base output characteristics (Fig. lfl.4a or Fig. ll.4d} of a bipolar junction transistor. {h} Sinee here d b} LP, the two pr: junctions do not interact, and we simply have two diodes in parallel. E—C f characteristic V [LB (e) The biasng here is identical to that normally encountered in standard J-FET operation. The physical properties are also those of a l—FET. The desired characteristics are clearly just the tin—VD characteristics of the J—FET with VD —:t VD}; and V5 —lr V53. 15-2 111.1 (a) Following the Hint ant: nbtains, Jr Vi?) I [My = ID? = quflnfiflaf [l- W(V')t’a]d'tv" 'D t] 2' VB} y = w] WW D (I _ thzflnNDfl * z , V+Vbi-VG 3’1 fits-Vern} _ ID {V 3 (vhf-VP)“ Vhi—VP ) [Vbi-VP Note that, givan the parallcl dcvclopmcnt, setting VDF} Pit-[side the. Eq. (15.9) braccs flclds that foregoing integration result. Eiiminating In using Eq. {15.9} than yields v v- v 4'1 v- v H l V — %(Vtt—Vp)[(—-—-—-; F"; Gr — [Vb:VG]3 ] = —______hr' P t“ P = Answer L . f2 . H I VD — 2'{Vhi—VP) [(V_D+Vb.—VG)3 - [El—:Er ] r i {h} we =0, VD =5v. vb} = w and VP Lav, v ~ (mfltw 113”- 1] 1 L 5 — (N'Wtfim — I) and The above data was used in mnstructjng Fig. 15.1103}. i 5.3 Differentiating Eq.{15.9) with respect to V3. with W; held constant yieids are = Eq2un~na11_[vu+vtewsrfl] m: n BVD Vg=emtsmnt L Vet-VP Solving we obtain [Vm+Vbi—VG )lfl _ 1 ! Verve | er NOTE: The bottom depletion widdt is the same as at equilibrium; the top depletion width is gteater than t1. I .- 1 1K 11"? i [Vht—VFTi‘] + ——1155 Vet] i QND i Di! it ii . nonnal situation top gate bottom gate depiefien width depietitm width Thus PI 2{Vhi Vii”: = {Vbi -VP1‘)m + Vial-fl 01' vie: Vei- —[2(vh-VP1”1— v if] 15—4 Given Vb; = 1V, V];- = —BV, one obtains vpr = 1—[21I'ET—t]2 CIT VPT = —24V The above ammo is clearly.Ir consistent with part (a). The top depletion width needs to be wider than when the two gates ale tied together, thereby necessitating a larger applied W51. / Assumes WT‘CVGT": fl. ND'I‘E: Although the boom V63: 1'], the bottom depletion width still connibutes to the mnao'iction of the channel. (a) When VD: rpm, WT + W3 —: 2a and m: Vow Also 1.31 WT = [flfl [Vet + V, VET}] mtep depletion width (J D W]; = [H—Seflmfi V— VGBJF ...bntton1 depletion width Since in the problem at hand V53 = I], we obtain at pinch-off mlfl—Ssfl [Vbi+VDsat—VGT]]”2 +[2K—SE‘1 [Vbiwnsadrfl But from part {In}... in In =[—‘12K;E {Vet—VH1] +[2K—HE Vet] ‘3" D So finally, cancelling the HSWqND factor everywhere. tan—WW + V1152 = We VDmt—VGT}IR + Newman“ 15-5 {c} From the part (c) answer, one can tell by inspection that VD“ for VGE = fl operation will be greater than FM for VGB = VGT operation. The tap side depletion width needs to be wider, in turn necessitating more current flow and a higher V3331 at the pinch-off point. {Alternative} Using the parameters of part (b), if Vb]- = 1V, VP: 4V and Vpr = 44‘s", one concludes VD“, = VG — Vp = 6‘! for V5]; = VGT = —2V operation and V133”; TV from the part (d) result if VET = -21.,F_ Again FM [V53 = fl operation} is greater than FM (V53 = V51- operation). Note that the two Unsafs are equal if VGT = i}. {f} Since the top and bottom depletion widths are not equal, the symmetry of the structure is destroyed and one must start by revising 31115.3). Eel-WED] Iii-W30} ID = 4:]- Jnydx = z] ‘qpnNDfidi'i'Fx = qznnNng-Efla—Wam—Wnyll Wrol Wro} GIT r = 2 2 fl[ cwfiWB] D It ilnNn-fl d}, 1 2a Integrating next over the length of the channel yields, VD a a n in = flit—E] [1 _fl+gfli’fl]dv ...teviSed Eq.{15.5) 0 Using the WT, W3, and 2a expressions presented in part {:1}, one obtains wT+wB = {VeiJrV—Vo‘rlm + new” ---VGB = fl 2“ (Vbi-Vnrlm + Vii: and VB = iszflnNod {I _ (Vai+V~VoTlm+ {1%in ”‘1 a L in (W (Venom + Vet Jrti Pectforndng the integration gives the desired solution a: 2 2:32me [in g ; (an+VD—VGT)3H + (VaiWnlm — (tn—Vania — Vii] L 3 (venom + Vii: lS-fi lite (a) The general IW'—t'neflatiotnship for one-sided power-law profiles was noted to he (liq. 16), [m+2]K3£0 Jlf{m+2} W = --—--_- V ' V qt: { hr A} For a linearly graded junction m = l and s = Note, or 113 W = [MWH— 1%)] We (b) It should be noted firstof all that ...in the nond loted left—hand E ND - NA = Nui- ED side of the channel {W S x S a) giving = = a; = Lfl ...in the left—hand portion J” I”? qun as? fipnfln a fly of the conducting channel Neglecting the 11.3 doping dependence, we can write a In- — 22! (eunfiuld—V)dx= Zeno—M?! xdx = annN—‘ld—Vlal— W2) WU} w or ID = qzunnfla fidfll "(lgfl -..revised fonu of Eq.[15.3h} (The "2" appears in front of the first integral above because equal contributions are obtained fi‘om the left- and right—hand sides of the channel.) Integrating over the length of the channel then yields, L But from part {a}, V In = M] ”ll-[slaw t] 15—1Ir U3 w = [MC—3:031!“ + v— HQ} where VA = VG — V e u and axe-egg ”3 r1 = [ [Vt-i - VP} quI SO E = [Vbi + V— Var” ‘3 Vm — VP Substituting into the In equation, I VD ID = inENDaJ [1" Phi + vi var/3] (W Vbi - VP fl and after integrating 153:-I Neting . I =: | =60 —V -;[Vb' V)[l I ”hi lam] ! DUI - Deal VEFUMRSJH P 3 l_ P Vbi—‘VF I and introducing 5 _ BIZ m: = #VP-%{Vbi-VP}|I FEL) i iii—VF fee = Ge er gives ._ Using the results {mm Exercise 15.3 we can then write: | | | 15-8 iFflt VD E V1353; ;_D = Eeflfldfls+flm foo Vtef Inn 332 if? . VD— iD—GORDVEI-ivm—VG lQGo-‘isi’mnVn—Vo _a[VbrVP] _¢o—_ — in; 3 Vrct Vet-VP Vat-VP 'Ffll' VI) :3 V3153; {12mm 1’ I" 1’ so {DEL=VG—VP_IJEELGGRS_Z phi—VP} EmiliLh—w—m—Jr 5 EH hr G) foo Vret fun 3 th VhifivP The foregoing relationships can be iterated using the {Zero function in MATLAB to detennine IDHDQ or Imaging as a function of V1} 1with VG held constant at preselected values. Running the P_{}S_fl‘i'.rn file on the Insu‘uctot's disk yields the results reproduced below and on the next page. With GOES = CORD = fl, one obtains the same characteristics as those diaplayed in Fig. 15.16. Although the characteristics retain their same general shape when GoRs = GDRD :- fl. an increase in the series resistances causes a significant decrease in Insal 311d a slight increase in VDM. VD {will} VI] [1' 011.5] 15—“) iii (:1) Since the gate is Shflflfid to the sourcc, VG = I] Also, I= In and V: VI). Thus, referring to Eqs {15. 9), (15 12), and {15.13}, I V v _m_3fl G=—D—=Gg {1—Z(v hi— DPHFDLVEL ] ...nsv 5v =4? VD Vbi—Vle (Vb:VP D “m P and _ In VP _2[ [be‘VF)I_ G -— __sa_1 =5 =— sax. VD fill“- VI} 3 FE”— V1312” VD 3 VDsat VP Likcwisn (utilizing Table 15.1}, _ din _ _ v +1," 1:2] 3 — — — l — I] <3 V 5 V — —V WI) £Jvn=a “Fwy? D Dsal P and (U gig gm = diff“? = fl “VD a VDsat = -VP ('3) With VD = VDsauQ = —VP!2 % Gob éfgflflfllm-(ifl _ 2041 _ [Li—F2 112] Vhi VP GU = Elf—W = 2 (1.5 x10—19}(1243)(mlfi){5 x m—SJ = 2m] x1434 s R=—-——l———=lfi.9kfl lifi-Elarwrfln = 213 Ml r: _..J._,_ {2 x Mlgjm] 15-11 {a} The same development as presented in Section 113.2 can be followed with the l , replacement qu0 with Ca. (It) At maximum {whether one mansiders below or abeve pinch—off biasing}, One can “Till: Z gm 5 GD = W A150, in general, I. t} W Since a a “2113!) L CG 2 4 M: a}: git—5:92!" [2' If gm is replaced by something greater than or equal to itself. and (3.5 is replaced by something less than or equal to itself, then it fellows that flgm._¢_=m f““" = 2an ML ZKSEOZL gflssobs {c} - #1.le (1.6x1n-19){1243)(m16}(s><tit-51.1 fmfltnut) = —-— =—_....__.—_ EEKSEOLE 21: (names x 10-14)(s x 1:514)2 = 304 CH: 15 15,] where The only parameter in GD which is temperature dependent is .11“. Thus 15—12 gmm _ Liam 1— [VhiWWhi-VPHM gmt3fltiK) _ (WK) 1_[Vbi{3m[Vhi—VP)]IH with Vhi = we lukewarm?) and 1H a :[Hfltvheei] GND 01' (Vet-Veils = (Vet—Veilsoett = (QNMEHOKSED) The required emaputatiuns for both part {a} and part (b) are performed by file P_15_1fl.trt on the Instructor's disk. The 31“ vs. Tdependettee was established employing the empirical—fit telationships found in Exercise 3.! and programmed in file P_fl3_fl3.m. The Hi vs. Tdependenee was eompubed following the procedure outlined in. Exercise 2.4[a]. The resultant gmmfgmiflflflK) and pnfljfllnflflflK} plots reproduced in the follow- ing figure clearly exhibit a power—law type dependenee, with a least squares fit yielding gflhfgmflflflli} = (T flflfl)-1-‘54?. The valiation of the Uauseouduetanee with temperature is seen to arise pritttariljtr from the variation of the carrier Elfihflity' with temperature. in ............. fimrelio er mobility ratio Illa 15—13 The device subject to analysis is pictured below l lifl ! | l In the two region model the long-channel dreary can be employed for drain biases below saturation. Paralleling the solution to Problem 15.5(1'), let WTU) be the too gate (MS) depletion widtlt and W30?) the bentorn gate {pita} depletion width. In general ill-Welt} 2ill-lilti- fD=—ZJ Jflydr e 3] (4MB?)JI = qzflnNfliifllza—WBFWT] “nee we F i 01' In It if E E i l; E 1e Integrating next over the length of the channel yields, _ NEH-amt: VD _W]:W ' “L h 23 Elev I D L New HZ ww=lgflflhmn+v—nfifl “fimmmhmmwmm {We 11"?- W3 = zfiflj [VbiB + V— 13:53)] ...bottom depletion width 1? I) and, given total depletion of the channel occurs when VET = VP and Up = 1’03 = l}. in 1r: 2:: = [mll’bi'IL‘VPl] + [E52Q Vein] gift: eNn ! 15—14 Thus wwwL {VbiT'l‘V-VGTJIE‘F(VbiB+V-VGEl1‘Q 2a whiThVfimV + Vhia Substiniting the depietion width relationship into the In expression and performing the integlation {mall},F yields the desired computational relationship. 3 1.” in = 00 [VIII _ g {VbtrWD-Vorlm + '{l’h“shite—Von)Bug — {Vbi‘I‘VGTlafl - (VhiB-VGBJM Wen-VP)!” + Vain 1.1.12 Setting Jtin —} jun and E —} 8),: WM} in Eq. (15. 21} and replacing it“ in Eq [15 2} with the resulting ME} expression, one obtains INF = —q L Nnfl (152*) ”sat 0‘}? and flu fl W . ID = mg Nun [ “?I (15.31:) (1 +flfl if]; ”set if? that it Integrating over the length of the channel and remembering ID is independent of y, we obtain L Jun VD ID I dy+~1gL av =2qanNDnJ-H} [Ir—d1? i} Dr 1 GI Inl+£1fli}=2q2nnwpn%(i—%l lr’n 10- — ML (LEW = M L[1+“"VD 1+flV—D wait. was 15-15 1L1} Since dey 2—8” differentiating both sides of the Problem 153(3) result with respect to 3: yields . no _£1|. _+ £1!r V'i‘vbt—VG) l = $ L F +1! -—V V- V 3!? vs _ 2(abgyp)[‘_n_a_fir”_‘ hr G] 3 Vha—Vp Vol—VP Next solving for 83. gives v V; v E V‘ v 0- v. — twin—e 'r “f -[ ._. “1’ J EyL : Vat—VP Var-VP {V+Vm—Vglla _ 1 Vat-VP 8} = seat when VIILJ = Up = V3531. Thus, substituting into the preceding equation F +V._V 3:“? v-v 3”? Vnm _. %{Vnt-VP1[(M] _( hr" G] J Vat—VP Vat—VP [VD-331+Vbi—Vfirfl _ 1 Vat-VP «9th = (was) 15.14 (a) The MNILAB m-file P_15_14.m found on the Instructor’s disk was constructed to calculate and plot the PET 1'er characteristics predicted by the two region model. Characteristics numerically identical to those in Fig. 15.23 are obtained when the short channel parameters noted in the figure caption are input into the program. This is not too surprising since a version of the tile was employed in constructing Fig- 15.23. (b) An FET with a channel length of L = lfltltun qualifies as a long—channel device. With L = lmtun the computed characteristics are indeed identical to those of the long-channel characteristics pictured in Fig. 15.16. {c} Per the defutition in the problem statement. the long-channel dreary begins to "fail" when EfitL = -5.5T5V. Although there are a number of approaches that could be employed. the author obtained this result’by simply monitoring the command window output of landing {Vail} as a function ofL with 8531 held constant at —1'[_‘t4 ‘Wcttt. 15wlfi ...
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