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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 pchannel 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) Pinchoff...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 longchannel analysis. (i) In the tworegion 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 VbiVP 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+VbiVG 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
(VbrViyrl] 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 pinchoff 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?” + (vb1+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). ZaWB (y) 20478 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+VDVGT)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 lefthand 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) 159 O 0.5 1 1.5 2 2.5 3 35.
VD (volts) 1510 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) VbiVP Vb'rVP
and
ID VP 2 (VbiVPH 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 VbiVP.
g Gg[1_{vbi—VP/2)1/2]
Vbi—VP
G0 zw =2(1.6x1019)(1248)(1016)(5x105) =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 1511 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 1014x5 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 ]
,VbrVP
where
2 Z N
GO E 41 [:1 D11 The only parameter in G0 which is temperature dependent is ,un. Thus 1512 gmm _ unto 1—[Vbi(T)/(Vbi~VP)l1/2 gm(300K) un(300K) 1 [Vbi(300K)/(VbiVP)]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 powerlaw 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 (VbirVP)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(longchannel)
'L(1+i‘LKQ)0 “ “Mg
953: L “sat L 1515 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 longchannel
characteristics pictured in Fig. 15.16. (c) Per the deﬁnition in the problem statement, the longchannel 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.
 Spring '07
 HAMBLEN

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