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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 out0f the drain contact in a p—channel device. Holes are the channel carriers in a pchannel 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
pseudoone—dimensional analysis at each pointy. (e) Pinchoff...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. lSl (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 longchannel analysis. (j) In the tworegion 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/(lowﬁeld 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 DB. 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, andwe simply have two
diodes in parallel. E43 1
'  characteristic
V
DB (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 IDVD characteristics of the JFET with VD ——) VDB and V3 m) VEB 152 141.2 (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 — %(VbiVP)[ V+VbiVG J3” __ (VbiVG 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/’biVP)[(‘——VD+Vbi—VG)m (VO‘FVGrnJ
3 VbiVP VbiVP (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). 153 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
VbiVP (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 (VbiVPT) + 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] 154 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+ VDsatVGT) ”2 + what/Dan”? , 155 (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 pinchoff 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 — qZunNDggtzaWaeywwn 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 (VbiVPT)”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 156 15,6
(21) The general Wreiationship for onesided powerlaw 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 lefthand = _ = 3:
"~ND NA Noa sideofthechanneHWSxSa) giving = = L = 3; Q ...in the lefthand 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 = quHnAJQiVI xdx = qZMnN39Ma2w2)
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 righthand 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), 157 1/3
W = [giivgOﬂ (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: 158 0For VD S V1352“ .51.). — h—IAGOGQﬁRD) [DO _ Vref 1130 ' 3f2 1 3/2
. VD— "LD'GORDVref‘l'VbiVG AGORS melbeVG
_;(VbrVP) [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) 159 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) 1510 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 4132 (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 11(Vbi‘VP) (Vbi—VP/z')3n_ A)” 3 —Vp Vbi—VP Vbi‘VP ,. =1 = 1 , 8 G [I _{VbrVP/2)U2J (i VbiVP 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 104)[1—(§)”2] 151] 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 pinchoff 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 101er5 x 10.4);
= 304 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 1512 ngT) _ ( MD 1— [Vbi(T)/(ngVp)] 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 empiricalﬁt 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 powerlaw type dependence, with a least squares ﬁt yielding
gm(T)/gm(300K) = (TBOO)*l547. 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 1513 15.11 The device subject to analysis is pictured below In the two region model the longchannel 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 ZaWBU) 2aWB
ID =4] JNydx = ZJ (amp? dx = qZunaniKtzaWB—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 (VbiTVpl] + [—m—MSEO VbiB]
(IN D L (IN!) 1514 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 “llC) (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 _ [DaongChannel)
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+VbiVG)3’2 _ (Vbi—VGF’Q t
L 2
VD — §(Vbi—VP) ( VbiVp 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
VbiVP ' 8), = 85m when V(L) = VD = V1332“. Thus, substituting into the preceding equation
V +V 'V 3/? V —v 3/? VDsat _. 2 (VatVP) “MD“ t” G) — ( b' G) .5 VbiVp Vb‘rVP (VDsat+VbiVG )1’2 _ 1
Vbi—Vp SsatL = (15.26) 15.14 (a) The MATLAB mﬁle 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. 1516 ...
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 Spring '08
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