Chapter 15 - iCHAPTEH 15 15.1(a Field Effect.modulation of...

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Unformatted text preview: iCHAPTEH 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 soarce and drain in a J—FET. (c) As viewed from the exterior of the device, the drain current flows out-0f the drain contact in a p—channel device. Holes are the channel carriers in a p-channel device and by definition these must flow aIOng 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. (d) Gradual channel approximation...ln 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 x—direction). The y- direction dependence is then neglected and the electrostatic variables computed using a pseudo-one—dimensional analysis at each pointy. (e) Pinch-off...complete depletion of the channel region; touching of the top and bottom depletion regions in the symmetrical JwFET. (0 As given by text Eqst (15.18), a! . 8 = ""9— ...drain conductance aVD Vg=constant 31 gm = ——D— ...transconductance aVG VD=constant (g) In a long channel J—FET, IDWG held constant) 5 constant for VD > V9531. Thus gd :—:( and the gd conductance in Fig. 15.19(b) can be neglected in drawing the equivalent circuit. i 4L Go—O- D + Ug 8,“th S S (h) MESFET...metal semiconductor field effect transistor. D-...depletion mode: E-...enhnncement mode. lS-l (i) Once I834 exceeds ~104 Want, the carrier drift velocity is no longer proportional to the magnitude of the electric field as assumed in the long-channel analysis. (j) In the two-region theory the carrier drift velocity is set equal to vsal at all points in the channel between yl and the drain. )2] is the point in the channel where {Eyi has increased to mat/(low-field mobility). 15.2 = (a) If a‘ << Lp, the two pn junctions will be interacting like in a BJT. Moreover, the biases are equivalent to active mode biasing in a BIT. bias D-B. bias Obviously, we are being asked for the common base output characteristics (Fig. 10.4a or Fig. 11.4d) of a bipolar junction transistor. (b) Since here a' >> Lp, the two pn junctions do not interact, and-we simply have two diodes in parallel. E43 1 ' - characteristic V D-B (c) 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 J-FET with VD ——) VDB and V3 m) VEB- 15-2 141.2 (a) Following the Hint one obtains, y V0) f Indy' = Ivy = ZqZunNnaf [1 - W(V')/a]dV’ 0 0 ' 2 2' V0) y = M] [1 —W/a]dV' ID 0 = ZqZHnNDa ID {V — %(Vbi-VP)[ V+Vbi-VG J3” __ (Vbi-VG F1} Vbi*VP Vbi—VP Note that, given the parallel development, setting VD—> Vinside the Eq. (15.9) braces yields the foregoing integration result. Eliminating ID using Eq. (15.9) then yields . 3 . 3 WH‘WVW‘H“) ”~ (W) ”l Z. __= M 4: Answer L V1) - 2-(l/’bi--VP)[(‘—--—VD+Vbi—VG)m- (VO‘FVGrnJ 3 Vbi-VP Vbi-VP (b) If VG = 0, V9 = 5v, vbi = 1V and Vp =-8V, _ v — (2/9)[(V+1)3’2— 1] 1 L 5 — (2/9x63f2— 1) and The above data was used in constructing Fig. 15.1 1(c). 15-3 15,4 Differentiating Eq.(15.9) with respect to VD with VG held constant yields Q12 = A2421” “ND“ 1 _. 3V1) ngconstam‘ L Solving we obtain [my/2] set: 0 Vbi-VP (VDsat+Vbi—VG)U2 2 1 Vbi—VP 01' VDsat = VG '” VP NOTE: The bottom depletion width is the same as at equilibn'um; the top depletion width is greater than a. (b) We can state 2K 5 ”2 2K 5 ”2 2K 8 ”2 2a = 2[ S o (Vbi—VP)] = S 0 (Vbi-VPT) + S 0 Vbi qND 4ND (IND It 11 fl normal situation - top gate bottom gate depletion width depletion width Thus 2(Vbi —— W)”2 = (Va — VPTWZ + v1}! 2 01" ‘ 2 VP'I‘ = Vbi-[2(Vb‘r'VP)I/2— V39] 15-4 Given Vb] = 1V, Vp = —8V, one obtains qu‘ = 142119—112 01' m = .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 Wet.“ / Assumes Vpr < VGT < 0. NOTE: Although the bottom V03: 0, the bottom depletion width still contributes to the constriction of the channel. (d) When VD: VDsat, WT + WB -——) 2a and V(L)= VDsat- Also U2 WT = [“530 (Vbi + v— Vgfi] ...t0p depletion width 4ND 1R WB = [z—flsrf—O—(Vbi + V- V034 ...bottorn depletion width 0 D Since in the problem at hand VGB = 0, we obtain at pinchvoff 2K 601/2 2K 8 1/2 20 = —S— glVbi+VDsarVoT)] +[ S 0(Vbi+VDsat)] (IND (IND But from part (b)... 1/2 U2 20 = [___QZK55 (Vbi—Vprl] + —QQKSE Vbi (IND 4ND So finally, cancelling the 2K580/qND factor everywhere, 2 (Vbi— vprfl” +v v1! — -(Vbi+ VDsat-VGT) ”2 + what/Dan”? , 15-5 (c) From the part (c) answer, one can tell by inspection that VDsat for V5}; = 0 operation will be greater than V1353; for V3}; = VGT operation. The top side depletion width needs to be wider, in turn necessitating more current flow and a higher V1353; at the pinch-off point. (Alternative) Using the parameters of pan (1)), if Vbi = 1V, Vp = —8V and Vpr = —24V, one concludes V951“ = VG — Vp = 6V for VGB = VGT = —2V operation and VDsatEZ- 7V from the pan (d) result if VGT =—2V. Again V1333; (V53 = 0 operation) is greater than Vpsm (V013 = VGT operation). Note that the two Vpsat's are equal if VGT = 0. (f) 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—wB 0’) In = a] new: = Z! (qpnNnglf) dx -—- qZunNDggtza-Waeywwn WTU’) WTU‘) 0r ___ g1[ _WT+WB] ID ZthtnNDa dy 71 __2a Integrating next over the length of the channel yields, 2 z N VD I I [D = M! [1 -M]dV ...revised Eq.(15.5) L 0 20 Using the WT, W3, and 2a expressions presented in part ((1), one obtains Wat/B = W ...VGB=o 20 (Vbi-VPT)”2 + V11? mfi VD ,D 2 2q2unNDa 1_(vbi+V—VGT)‘/2+(vbi+mm (W L 0 ' (Vbi—VPT)”2 + V61” Performing the integration gives the desired solution It) I 2%:an [v9 _ 2 WM] 3 fl 1 L (Vbi—Vpr)”2 +be 15-6 15,6 (21) The general W-reiationship for one-sided power-law profiles was noted to be (Eq. 7.6) ‘ ll(m+2) w = [_____(m+:);{580 (vbi—VA)] For a linearly graded junction m = 1 and b = Nola, or (b) It should be noted first of all that ...in the nondepieted left-hand = _ = 3: "~ND NA Noa sideofthechanneHWSxSa) giving = = L = 3; Q ...in the left-hand portion J” JN’ qnnNo as)” ”MN“ a dy of the conducting channel Neglecting the ,un depin g dependence, we can write a a 1D-=ZZI (qunNold—V)dx = quHnAJQiV-I xdx = qZMnN39Ma2-w2) we) or _ dV Hi 2 - ID — qanNoa dy [1 —(a )] ...reVISed form of Eq.(15.3b) (The "2" appears in front of the first integral above because equal contributions are obtainei from the left- and right-hand sides of the channel.) Integrating over the length of the channel then yields, VD ID = MI [1_(%)2]dv L 0 But from part (a), 15-7 1/3 W = [giivg-Ofl (Vbi + V» Vd] where VA = VG — V 1§.7 Noting V . 3/2 lDo=1Dsanlvg=03ndRS=RD=0 = Go{—VP-%(Vber)[l {TS/P} J} and introducing Vb 3/2] v r = —Vp——(vb- WM 1 H 1'8 I— Vbi—VP gives [Do = Go Vrcf Using the results from Exercise 15.3 we can then write: 15-8 0For VD S V1352“ .51.). — h—IAGOGQfiRD) [DO _ Vref 1130 ' 3f2 1 3/2 . VD— "LD'GORDVref‘l'Vbi-VG AGORS me-l-be-VG _;(Vbr-VP) [Do _ 100 3 Vbi - VP Vbi — VP 3/2 J"D9 Vbi — VP ) (L—GDsm oRsVref+Vbi—VG l.— The foregoing relationships can be iterated using the fzero function in MATLAB to determine 11311130 or lbw/[Do as a function of VD with VG held constant at preselected values. Running the P_05__07.m file on the Instructofs disk yields the results reproduced below and on the next page. With GoRs = CORD = 0, one obtains the same characteristics as those displayed in Fig. 15.16. Although the characteristics retain their same general ' shape when GoRs = GORE > 0, an increase in the series resistances causes a significant decrease in 113m and a slight increase in Vpsat. o' 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 VI) (veils) 15-9 O 0.5 1 1 .5 2 2.5 3 3.5 4 4.5 5 VD (volts) O 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 VD (volts) 15-10 L535 ‘ (a) Since the gate is shorted to the source, V0: 0. Also, I: ID_a_nd V: VD. Thus, refern'ng to Eqs.(15.9), (15.12), and (15.13), G = I__D =G0‘l 2[_V__biV—VP)[___ 3/2 3/2 ‘ VD+Vbi) ______( Vbi ) [2]} ...0 s VD s V0531: —-VP _ Vbr‘VP VI) 3 VD Vbi VP and 1 ID 413-2 (Vbi"VP) (* Vb J) G = _s_at =G l— 1 > V -— —V sat VD 0"— VD -3 VD .Vbi-'VP f- VD Dsat“ P Likewise (utilizing Table 15.1), . 1/2 ' M) J ...0 5 V1) S V1352“: -—Vp dVD Vbi—VP and (1'1 353; ~ DMWCFO ~ 0 V0 2 VDsat - —VP dVD R = L =————-——*1—-—____ G 60 1-1(Vbi‘VP) (Vbi—VP/z')3n_ A)” 3 —Vp Vbi—VP Vbi‘VP ,. =1 = 1 , 8 G [I _{Vbr-VP/2)U2J (i Vbi-VP Go _242#£ND“ w.2(16x1019x1248x1016x3 x 105) —200x104s R _ 123/2 =.16 9 k9 (2x104){li§-I%i((—) %) J} r = ______l_.__ = 27.3 kg (2 x 10-4)[1—(§-)”2] 15-1]- 15,2 (3) The same development as presented in Section 17.3.2 can be followed with the replacement of Co with C (3. (b) At maximum (whether one considers below or above pinch-off biasing), one can write 2 Z a 8m 5 GO = —"—'—q HEN!) Also, in general, 1.. CG = 2! may 0 W Since a 2 W(y) L CG 2 2! K3202 dy =2WKSZOZL 0 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 gm < ZqanNoa, a = qunNo02 fmax = 27:63 ' ZnL 2Ks£OZL ZEKSEOLQ (C) fmaxflimit) = M =Wfl 2xK580L2 21c (11.8)(8_35 x 10-1er5 x 10.4); = 3-04 GHz 15.10 (a)/(b) With the device saturation biased and V0 2 O, we conclude from Table 15.1 that - Vbi ”2] = G 1— .—— gm 0[ (Vin—VP) where E quunNDa G 0 L The only parameter in Go which is temperature dependent is IJn. Thus 15-12 ngT) _ ( MD 1— [Vbi(T)/(ng-Vp)] 1’2 p gm(300K) _ 1160010 1 — [Vbi(300K)/(Vbi—VP)] U2 with Vbi = (161751) 111(NAND/n12 ) and 1/2 a = [2K3 80 (Va—1’9] " (We 01' (Vbi-'VP)]T = (Vbi—VP)|300K = (qNDa2)/(2Ks€o) The required computations for both part (a) and part (b) are performed by file P"15_10.m on the Instructor's disk. The ,un vs. Tdependence was established employing the empirical-fit relationships found in Exercise 3.1 and programmed in file P_O3_03.m. The n; vs. T dependence was computed following the procedure outlined in Exercise 2.4(a). The resultant gm(T)/gm(300K) and un(T)/ttn(300K) plots reproduced in the follow- ing figure clearly exhibit a power-law type dependence, with a least squares fit yielding gm(T)/gm(300K) = (TBOO)*l-547. The variation of the transconductance with temperature is seen to arise primarily from the variation of the carrier mobility with temperature. 10' TdewndenceofmeJFmamnducm gmratio or mobility ratio 7:3 I I x f 3 inobiiiryfrauo : T00 15-13 15.11 The device subject to analysis is pictured below In the two region model the long-channel theory can be employed for drain biases below saturation. Paralleling the solution to Problem 15.5(0, let WTOI) be the top gate (MS) depletion width and W30) the bottom gate (p+-n) depletion width. In general Za-WBU) 2a-WB ID =4] JNydx = ZJ (amp? dx = qZunaniKtza-WB—Wn _ Wm) WT 3' '. y In = 242unNDa§§(1—E%Vi] . Integrating next over the length of the channel yields, 01' VD ID z W! (bmdv L 0 2a Now In WT = [235,80 (VbiT + V" VGT)] ...top depletion width D ...bottom depletion width 1/2 WB = [K380 (VbiB + V- Veal] (IND and, given total depletion of the channel occurs when VGT = Vp and VD = V53 = 0, 2 In 20 = l K580 (VbiT-Vpl] + [—m—MSEO VbiB] (IN D L (IN!) 15-14 Thus WT+WB _ (VbiT+V'VGT)]/2 + (vbigw—VGBWZ 2 _ a (VbiPVP)ln + Vii/Ii Substituting the depletion width relationship into the ID expression and performing the integration finally yields the desired computational relationship. [D g Go [VD _ ; (VbiT+VD‘-VGT)3/2 + (VbiB+VD—VGB)3Q — (Vbi'ILVGTfin — (VbiB—VGB)3/2;i 1 3 (VbiT‘VPflfz + Vin/1% L112. - - Setting #0 —) —pn and 8 —> 8), = —dV/dy in Eq. (15.21), and replacing [in in Eq. (15.2) with the resulting MS) expression, one obtains my = —q ——“'3-—— NDQK (152') ' ”flag! dy ”sandy and . ID = 2:12 -#—H\Npafl[1 “ll-C) (15.3b') 1+ tin iV—j dy ‘ a Dsat dy 01' #11 W W W ID(1+—-—~— = 2 z Npa— 1—— Integrating over the length of the channel and remembering I D is independent of y, we obtain L VD VD 1D dy + L" W = ZqanNDa (1 —!Z)dv o ”33‘ o 0 01' V ZquDNDg 0(1_l'_V I _ [Daong-Channel) 0 a flu VD 1D=__.__ L (1.11m) ”sat L 15—15 15.1 5 ' Since dV/dy = —8y, differentiating both sides of the Problem 15.3(3) result with respect to y yields V+V - V 1/2 .8), + 8y(__12£_§.) Vb‘r-‘VP =_____,______________]. VD+Vbi-VG)3’2 _ (Vbi—VGF’Q t L 2 VD — §(Vbi—VP) ( Vbi-Vp Vbi—VP J Next solving for 8y gives V +V 1/ 3/2 V 1/ 3/2 VD _ %(Vbl_VP)[( D .bl' G) _( br‘ G) V+Vbi—VG)1"'2 _ 1 Vbi-VP ' 8), = 85m when V(L) = VD = V1332“. Thus, substituting into the preceding equation V +V '-V 3/? V -—v 3/? VDsat _. 2 (Vat-VP) “MD“ t” G) — ( b' G) .5 Vbi-Vp Vb‘r-VP (VDsat+Vbi-VG )1’2 _ 1 Vbi—Vp SsatL = (15.26) 15.14 (a) The MATLAB m-file P_15_14.m found on the Instructor's disk was constructed to calculate and plot the FET ID—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 = 100nm quaiifies as a long—channel device. With L = lOOttm 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 SsatL : —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 [1353me (V520) as a function of L with 85;“ held constant at #104 V/cm. 15-16 ...
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