chap15 - CHAPTER 15 ‘ 15.1(a Field Effect.modulation of...

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Unformatted text preview: CHAPTER 15 ‘ 15.1 (a) Field Effect...modulation of the semiconductor conductivity by an electric field applied normal to the surface of the semiconductor. (b) Channel...nondepleted 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, the drain current flows aubof the drain contact in a p-channel device. Holes are the channel carriers in a p—channel device and by definition these must flow along the channel into the drain. The current has the same direction as the hole flow—from source to drain and out of the drain contact. ((1) Gradual channel approximation...1n this approximation it is assumed the electrostatic variables in one direction (say the y—direction) change slowly compared to the rate of change of the electrostatic variables in a second direction (say the pdirection). The )2- direction dependence is then neglected and the electrostatic variables computed using a pseudo—one—dimensional analysis at each point y. (e) Pinch-off...complete depletion of the channel region; touching of the top and bottom depletion regions in the symmetrical J~FET. (i) As given by text Eqs. (15.18), BI . 8d = —D ...d:ra1n conductance BVD ngconstant a] 8m = —D ...transconductance aVG szconstant , (g) In a long channel J—FET, [D(VG held constant) E constant for VD > Vpsat. Thus gd E 0 and the gd conductance in Fig. 15.1963) can be neglected in drawing the equivalent circuit; (h) MESFET...metal semiconductor field effect transistor. D—...depletion mode; E-...enhancement mode. 15—1 (i) Once lgyl exceeds ~lO4 V/cm, the carrier drift velocity is no longer proportional to the magnitude of the electric field as assumed in the long-channel analysis. (i) In the two-region theory the carrier drift velocity is set equal to vsat at all points in the channel between yl and the drain. y1 is the point in the channel where iffy! has increased to t)SaJ(low—field mobility). 15.2 (a) If d << Lp, the two pn junctions will be interacting like in a BJT. Moreover, the biases are equivalent to active mode biasing in a BJT. Obviously, we are being asked for the common base output characteristics (Fig. 10.411 or Fig. 11.4d) of a bipolar junction transistor. (b) Since here d >> LP, the two pn junctions do not interact, and we simply have two diodes in parallel. E— 1 ———sirnpie diode (2X) characteristic DB (0) The biasing here is identical to that normally encountered in standard J~FET operation: The physical properties are also those of a J~FET.' The desired characteristics are clearly just the ID—VD characteristics of the LEI“ with VD —> VDB and VG —> VEB. 15—2 L52 (a). Following the Hint one obtains, y V0) I [Ddy' = [Dy = ZqanNDaf [1— W(V')/a]dV' 0 0 2 Z N ‘ ~_ 3/2 ._ 3/2 3 4 Jun Da {V n ; (Vbi_VP)HV+Vbi VG) m (Vb: VG) 3 ng~Vp Vbi-VP Note that, given the parallel development, setting VD—a V inside the Eq. (15.9) braces yields the foregoing integration result. Eliminating [9 using Eq. (159) then yields V+V -—v 3/2 V '—V 3/2 y = VbrVP Vbi—VP c: Answer L 2_ . VD+Vbi-VG 3/2 Vbi—VG 3’2 VD — (VbrVP) 3 Vbi—VP Vbi~VP (b) If VG = O, VD = 5V, Vbi 2 IV and VP = ~8V, y _ V _ (2/9) [(V+ n3/Z— 1] L 5 ~ (2/9)(63/2 — 1) and The above data was used in constructing Fig. 15.l1(c). lS~-3 l 5.4 Differentiating Eq.(15.9) with respect to VD with VG held constant yields ._ 1/2 ZZqZHnNDal1_(VD+Vb1 VG) J m: 0 VG=constant L Vbi—VP . (VDsat+Vbi—VG)1"2 Vbi~VP 9.12 3V1) Solving we obtain =1 01‘ VDsat = VG — VP NOTE: The bottom depletion width is the same as at equilibrium; the top depletion width is greater than a. 1/2 1/2 (Vbr-Viyrl] ZKSEQ Vbi] , (IN D II normal situation top gate bottom gate depletion width depletion width Thus 2%1— VP)1/2 = (Va — VPT) 1/2 + viii/2 0f 2 VPT = Vbi ~l2(Vbi~VP)1/2 M Vlii/Zl Given Vbi = 1V, Vp = m8V, one obtains VPT =1—[2f91112 01‘ VpT : —24V The above answer is clearly consistent with part (a). The top depletion width needs to be wider than when the two gates are tied together, thereby necessitating a larger applied lVgl. : / Assumes Vpr < VGT < 0, NOTE: Although the bottom VGB = 0, the bottom depletion width still contributes to the constriction of the channel. - - -LARGERJ ’. 9 “_,-»~~ -“—Lv-‘ ~ -u‘w (d) When VD = V1333“, WT + W}; —) 2a and V(L) = VDSm. Also 1/2 WT = 580 (Vb;i + V— VGT)] ...top depletion width. 4N D 1/2 WE = [ZKZSEO (Vbi + V“ {43)} ..,bottom_ depletion width ' G D Since in the problem at hand VGB = 0, we obtain at pinch-off 2K 8 1/2 2K 8 “2 2a 2 S 0 (Vbi+VDsarVGT}] + S 0 (Vbi+VDsat) (IND qND But from part (13)"- 1/2 1/2 2a : 2K580 (Vbi—VPT) + ZKSEO Vbi qND . 4N D So finally, cancelling the ZKSEO/QND factor everywhere, 1 2 [(vbi—vml/Z + Vbi’ = (vm Vast—var?” + (vb-1+vDsaal/i] 15—5 (e) From the part (0) answer, one can tell by inspection that V1353; for VGB = 0 operation will be greater than VDSat for VGB = VGT operation. The top side depletion width needs to be wider, in turn necessitating more current flow and a higher V1333; at the pinch—off point. (Alternative) Using the parameters of part (b), if Vbi = 1V, VP 2 —8V and VpT = —24V, one concludes V1333: VG — Vp = 6V for VGB = VGT = —2V operation and V1333: 5 7V from the part ((1) result if VGT = —2V. Again VDSm (VGB = 0 operation) is greater than VDsat (VGB = VGT operation). Note that the two Vpsat's are equal if VGT = 0. (0 Since the top and bottom depletion widths are not equal, the symmetry of the structure is destroyed and one must start by revising Eq.(15.3). Za-WB (y) 20-478 0’) ID = 4] max = 2] (QHnNDd‘Y‘) dx = qzunNDgi;{2a_WB(y)—Wr<y>i d Wm) Wm) y or I = 2 Z d—V[1~WT+WB] D 61 #nNDa dy 20 Integrating next over the length of the channel yields, 2 z N V” [D = m] [1~WT+WB]dV ...revised Eq.(15.5) L 0 2a Using the WT, W3, and 2a expressions presented in part (d), one obtains wwwg z (Vbi+V“VGT)1/2 + (Vbi+V)1/2 VGB : 0 2 a (Vbi—VI’r)1/2 + Vt}? and VI) _ zqzynzvpa [1 (Vbi+V“VGT)1/2 + (Vbi+V)1/2 l 2 L 0 (VbH/PT)”2 + Vbi/ dV 1D Performing the integration gives the desired solution 3 21 ___ ZqzynNDa [VD 2 (Vbi+VD-VGT)3/2 + (Vbi+VD)3/2 — (VbrVGrP/Z — Vii _,J ID m 2 L 3 (Vbi—VPT) 1/2 + Viiil 15~6 1 5.6 (a) The general W—relationship for one—sided power—law profiles was noted to be (Eq. 7.6), 1/(m+2) W = [Q’H’iJfBKSfQ (Vbi_VA)] For a linearly graded junction m = 1 and b = No/a, or (b) It should be noted first of all that ...in the nondepleted left—hand : __ : L n _ ND NA No a side of the channel (W S x S a) giving 1 £11 ...in the left-hand portion _ _ L = J N ‘ JNY " WW0 agy ‘qfll‘NO a dy of the conducting channel Neglecting the ,th doping dependence, we can write 01‘ 2 ID = qZflnNoa (Ll/[1 — } "revised form of qu(15.3b) (The "2" appears in front of the first integral above because equal contributions are obtained from the left— and right—hand sides of the channel.) Integrating over the length of the channel then yields, qu Nga VB 2 [wwwng [14de But from part (a), 15—7 1/3 W = FWD" (Vbi + V~ Vd] where VA 2 VG " V 4N0 and 3K 8 1/3 a : [ S 0a(vb1”VP)] 4N0 so Wf’i (Vbi"VG)5/3] 1 Vbi—VP qanNoa 3 I -— .___.___.._ V _— V ‘——V D ( D 5( bl P) be—Vp L 15.7 Noting 1 El =G'-V—;V"”V1 bl» D0 DsathG=0 andRs=RD=0 0{ P 3 ( b1 mi AiWrWz) and introducing vref = #vp—Z (Vbi*VP)[1 L V“ m 3 Vbi~VP gives [130 = GO Vref Using the results from Exercise 15.3 we can then write: 15—8 oFor VD S VDsat 1D VD 1D G —- = ——— 0(Rs+RD) 1130 me 1D0 [D _ 3/2 [D 3/2 ‘ - VD— —GoRDVref+Vbi—-VG —GoRs Vref+VbinG 2 (Vbi—VP) I Do _ I DO 3 me , Vbi - VP Vbi — VP 3/2 11) “GS” oRsV t+Vb'—VG IDsat __ VG‘VP_IDsatGORS 2(Vbi_VP) 1" [DO m l [DO ' m 190 3 Va Vbi—VP The foregoing relationships can be iterated using the fzero function in MATLAB to determine I D/IDQ or I Dsat/I Do as a function of VD with VG held constant at preselected values. Running the P_()5_07.m file on the Instructor’s disk yields the results reproduced below and on the next page. With GoRs = GORD = 0, one obtains the same characteristics as those displayed in Fig. 15.16. Although the characteristics retain their same general shape when GoRs = GORD > 0, an increase in the series resistances causes a significant decrease in [mat and a slight increase in V1332“. l 0 0‘5 1 1 ‘5 2 2.5 3 3.5 4 4.5 5 VD (volts) 15-9 O 0.5 1 1.5 2 2.5 3 35. VD (volts) 15-10 15.8 (a) Since the gate is shorted to the source, VG = 0. Also, I = ID and V: VD. Thus, refening to Eqs.(15.9), (15.12), and (15.13), ._ . 3/2 . 3/2 G = :51?— =Go{ _Z(———Vb* VPHWVDJ’VW —(———Vb1 ) ] ...0 SVDSVpsaF —Vp V1) 3 V1) Vbi-VP Vb'r-VP and ID -VP 2 (Vbi-VPH Vb' 3/2] _ G = sat :0 ————~— 1— ‘ ...V 2V —-—V sat VD 0 VD 3 VD Vbi‘VP D Dsat P Likewise (utilizing Table 15.1), (11}; VD+Vbi)1/2] _ =-———==r _=G1—--——~,——‘ ...OSV SV ——V 8’ WI) JVG_() 4 ( Vbi_VP D Dsat P and d1 1 z gsat = Dsa‘VG = 0 ...V9 2 V1353; = —Vp dVD (b) With VD = VDSm/Z = —Vp/2 R = 1 z 1 G Gall -ilvb“VP)l(—-V""V"”)m-{ )3’21} 3 —VP Vbi—Vp Vbi-VP. g Gg[1_{vbi—VP/2)1/2] Vbi—VP G0 zw =2(1.6x10-19)(1248)(1016)(5x10-5) =2.00><10"4 s R 2‘ 1 3/2 3/2 = 16.9 kg 4 {i 3.12 .1. (2X10 ){113X2 (3) 3) r : ——-~—-—1—-—~—1«/-é—— = 4 Z w m 1 15-11 15.9 (a) The same development as presented in Section 17.3.2 can be followed with the replacement of CO with Cg. (b) At maximum (whether one considers below or above pinch—off biasing), one can write gmgangaufgtqg L CG 2 2] K5802 dy 0 W Also, in general, Since a 2 W(y) L CG 2 2f K5802 dy = 2K380ZL 0 a a If gm is replaced by something greater than or equal to itself, and CG is replaced by something less than or equal to itself, then it follows that L, gm < ZqZunNDa . a = qunNDaZ max ZKCG ” 27CL ZKSEOZL ZKKS‘EOLQ (c) . . qunNDaZ (1.6x10‘19)(1248)(1016)(5X1()'5)2 fmaxalmlt) = = \ 2nKseoL2 2n (11.8)(8.85 x 10-14x5 x 1044)2 = 3.04 GHz 15.10 (a)/(b) With the device saturation biased and VG : O, we conclude from Table 15‘} that . 1/2 gm = GO[1“( Vbl ] ,Vbr-VP where 2 Z N GO E 41 [:1 D11 The only parameter in G0 which is temperature dependent is ,un. Thus 15-12 gmm _ unto 1—[Vbi(T)/(Vbi~VP)l1/2 gm(300K) un(300K) 1- [Vbi(300K)/(Vbi-VP)]U2 with vbi = (kT/q) ln(NAND/ni2) and 2K “2 a =[ 380 (Vbi—VP) (IND 0f (Vbi—VP)lT = (Vbi~VP)l300K = (qNDa2)/(2Ks€0) The required computations for both part (a) and part (b) are performed by file P_15_10.m on the Instructor's disk. The ,th vs. T dependence was established employing the empirical—fit relationships found in Exercise 3.1 and programmed in file P"03_03.m. The ni vst Tdependence was computed following the procedure outlined in Exercise 2.4(a). The resultant gm(T)/gm(300K) and ,un(T)/,un(300K) plots reproduced in the follow: ing figure clearly exhibit a power-law type dependence, with a least squares fit yielding 8m(T)/gm(300K) = (77300)“1‘647. The variation of the transconductance with temperature is seen to arise primarily from the variation of the carrier mobility with temperature. 1 10 gmratio or mobility ratio 8 TiK) 15—13 1 5. 1 l The device subject to analysis is pictured below -r- In the two region model the long—channel theory can be employed for drain biases below saturation. Paralleling the solution to Problem 15.5(f), let WT(y) be the top gate (MS) depletion width and W307) the bottom gate (p+—n) depletion width. In general Za—WBU) Za~WB ID = J] JNydx = Z I (qflnNDZ—V‘) dx = (IZHnNDi‘Y‘lza"WB“WTl WTO’) WT y y 01' ID = quHnNDafl(1—WT+WB) dy 261 Integrating next over the length of the channel yields, VD ZthtnNDa I ( ID : ——~— 1.1 Now ."top depletion width 1/2 J “bottom depletion width WB = [ZKSEO (VbiB + V— VGB) (IND and, given total depletion of the channel occurs when VGT = VP and VD : VGB 2: 0, 1/2 1/2 20 = [ZKSEO (VanVpll + [ZKSEO VbiB] CIND qND 15—14 Thus WT+WB _ (VbiT+V—VGT)1/2 +(Vbi13+V—VGB)”2 2a _ 1 2 (VbirVP) 1’2 + Vbi’B Substituting the depletion width relationship into the ID expression and performing the integration finally yields the desired computational relationship. [D 2 G0 VD _ ; (VbiT+VD“VGT)3/2 + (VbiB+VD—VGB)3/2 ~ (Vbi'I‘-Verr)3/2 - (VbiB“VGB)3/2 1/2 3 (Vbir-VP)1’2 + VbiB 15.12 Setting ,uo —> —~,un and 8 ——> a, = ~dV/dy in Eq. (15.21), and replacing u“ in Eq. (15.2) with the resulting M8) expression, one obtains JNy = —q Jim- ND—dK (15.2') 1+s‘u_nd_V dy Dsat dy and #11 W l ID = 2qZ —— NDa— —— (15.3b) 1 £41 dy( a) Usat dy or #n W W W ID1+-——=ZZN -——~— . vsat y qfln Dady‘ a) Integrating over the length of the channel and remembering ID is independent of y, we obann L V1) VD ID dy + if: W = ZthtnNDa (1 "ZW—Vv 0 “33‘ 0 0 Of V [D : ZqanNDa I D(1__W_)dv : ID(long-channel) 'L(1+i‘LKQ)0 “ “Mg 953: L “sat L 15-15 15. 1 3 Since dV/dy = ~8y, differentiating both sides of the Problem 15.3(a) result with respect to y yields V+Vbi—VG )1/2 _8 + 8 _ L : y y Vbi—VP L V1) 2 Wm VP)[(VD+Vbi—VG)3/2 (Vbi—VGr/z] 3 Vb‘r—VP Vbi—VP Next solving for 8), gives V +V- V 3/2 V-—V 3/2 VD gwbi VP)“ D br- G) _( bl G) , Vbi—VP . Vb’r'VP (Vteri—Vgrl2 _ VbiflvP eyL = 1 8y = 833; when V(L) = VD : V1332“. Thus, substituting into the preceding equation 'V +V -—V 3/2 V -*V 3/2 VDSat 2 (Vbi VP) Dsat in G) w( b1 G) 3 Vbi—Vp Vbiva (Vnsat+Vbi—VG)“2 _ 1 Vbi—VP EsatL = (15.26) 1514 (a) The MATLAB m—file P_15_14.m found on the Instructor's disk was constructed to calculate and plot the FET I D—VD 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 file was employed in constructing Fig. 15.23. (b) An FET with a channel length of L = lOOym qualifies as a long—channel device. With L z 100,um the computed characteristics are indeed identical to those of the long-channel characteristics pictured in Fig. 15.16. (c) Per the definition in the problem statement, the long-channel theory begins to “fail” when EsatL = -—5.575V. Although there are a number of approaches that could be employed, the author obtained this result'by simply monitoring the command window output of [mat/1130 (VG20) as a function of L with 83m held constant at "104 V/cm. 15~l6 ...
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This note was uploaded on 04/05/2009 for the course ECE 3040 taught by Professor Hamblen during the Spring '07 term at Georgia Tech.

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chap15 - CHAPTER 15 ‘ 15.1(a Field Effect.modulation of...

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