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Unformatted text preview: CHAPTER 8 ' a; (a) It is the time period (ts) during which the observed current associated with the turn—off
transient remains constant at a large negative value—See Fig. 8.1(b). The excess minority
carriers stored in the quasineutral region are being removed during this time period. (b) With reference to the turn—off transient pictured in Fig. 8.1(b), t, = In — ts, where tn is
the total time for the reverse current to decay to 10% of its maximum magnitude and ts is
the storage delay time. (c) . This is precisely what happens during the ts portion of the turn—off transient.
Even t ough there is an excess of carriers at the edges of the depletion region making UA >
0, the external circuitry enables a reverse current ﬂow that acts to eliminate the excess. (d) A delay in switching from the on— to the offstate arises because a ﬁnite amount of time
is required to remove the excess minority carriers stored in the quasineutral regions on the
two sides of the junction. (e) The excess carriers are removed by recombination and reverse injection. . Apn(x,t) > 0 implies Apn(xn,t) > 0, the necessary condition making vA > 0.
. Referring to Eq. (8.2), if i > 0, the slope of a pn(x,t) versus x plot must be
negative at x = xn. (h) The electrical response of the step—recovery diode is special in that the tr portion of the
transient is very short compared to the storage delay time. Physically, step recovery diodes
are actually a PIN type structure with very abrupt junctions. (i) . Both the approximate expression (Eq. 8.8) and the more exacting expression
(Eq. if: for Is vary only as the ratio of I}: and IR. Thus, increasing both [1: and [R by the
same amount will have no effect. (j) . During tumon recombination obviously dominates over generation because
there is a carrier excess. Thus, the inevitable loss of carriers via recombination will indeed
retard the buildup of stored carriers. (This fact is conﬁrmed mathematically by Eq. 8.10.) £1 (a) . Apn(xn,t) = pn(xn,t) ~ pm < 0. A carrier deficit at the edge of the
dep etlon region in teates the junction is reverse biased. , (b) Invoking the law of the junction,
momma“) = NDPnO/Z = "3/2 .—_~ ”12 eqUA/kT 0r vA = (kT/q) ln(1/2) = 410259 in? = “0.013 v (c) iReverse]. Since dApn/dxiﬁxn = dpn/dxipxn > 0, it follows from Eq. (8.2) that i < 0. 8—1 3; A comparison of the plot displayed below and Fig. 8.6 indicates the revised charge~control
expression is indeed a signiﬁcant improvement. The improvement is clearly greatest at the
largest IR/IF values. \2 Revised chdrééedntrol VVX ........ ts/taup lR/IF MATLAB program script... % Comparison of the ts/taup versus IR/[F computed
% using Eq.(8.9) and the revised charge control expression %Initialization
clear; close
%ts/taup calculation %Plotting results
Iratio=logspace(—2, l ); %Iratio=IR/[F loglog(Iratio,ts 1 ,'— g')
%Revised charge control expression axis([l .Oe—2,10,l 002,101); grid
x: l ./lrati0; hold on
ts l=log((l+x)."2./(l+2.*x)); loglog(Iratio,ts2)
%Equation (8.9) xlabelCIR/IF‘); ylabelCts/taup’)
ts2=erﬁnv(l./(l+1ratio))."2; text(0.22, 1.4,'Revised charge~control’);
‘ text(0.22,0.35,'Eq. (8.9)')
hold off 8—2 ‘ M (a) Because the diode is open circuited, i = 0. Thus, based on Eq. (8.2), the slope of all
the pn(x,t) versus )5 curves evaluated at x = xn should be zero. pn(x, t) The general solution is QPO) = QP(O+) 6"”?
where paralleling the analysis in the text QP(O+) = 11:13)
Thus Qp(t) = [Wipe[NP 8~3 (c) If the charge is assumed to decay uasistatically, then, referring to Eq. (8.15),
q
tht) = IorpleqvA/kT— 1) E torpeqvA/kT
Equating the part (b) and (c) expressions for Qp(t) gives [Ftp e'mp = 101p eqvA/kT
01‘
eqvA/kT = (IF/10) et/rp But from the statement of the problem
[12/10 a e WON/k?" Therefore eqvA/kT = quONIkTe—t/tp
or :(d) The part (0) result suggests a very simple way of determining 1p (known as the Open—
Circuit Decay Method). After forward biasing the diode, one openccircuits the device and
monitors the voltage drop across the diode as a function of time. Provided the decay
follows the ideal fonm, 7p is readily determined from the slope of the data 8—4 a; Since under steady state conditions [F = 10(e qVON/kT._ 1) E 10 quON/kT 3 '
V s £11451) = 0.0259 MEL) = 0.716 V
ON ‘1 [0 1015 Next, solving Eq. (8.16) for i, one obtains in general f z = —rp1n[1—(10/1F)(eqvA/kT_1)] 5 41,141 _ eqUA/kT/quON/kT]
or t : ~11, 1,,[1 _ eq(vA—V0N)/kT]
Corresponding to DA = 0.9V0N, {90% = _(10—6)1n[1_e«(o.1)(0.716)/(0.0259)] : 65.1 “sec
To reach DA = 0.95 VON,
[95% = “(10—6) 1,,[1 _ e40.05)(0.716)/(0.0259)] = 289 “sec Note that the 9070—9596 portion of the transient takes much longer than the 0—90% portion
of the transient this property of the turn—on transient was noted at the end of Section 82. 8—5 M (a)
P1105, 0
same slopes
greater than
the t < 0 slope
(b) Here for t > 0,
de , Qp
__ = 1 _ _
dt F2 1p
{QM} dQP : t
9pm [F2 — QP/Tp
Q90) H ‘
t = «1:? 141132 — 23) = ‘19 ln{I———~—~——F2 QP/Tp)
Tp Qp(0*)=1mp [F2 — [F1
Thus
1'52 —@ = (1m " 11:1) 6"”?
79
and QPU) = 11127;) — (IFZ—[FDTP e’1/Tp 86 Invoking the quasistatic assumption (Eq. 8.15), we can also write Q90) = 10%(eqvA/kT~1) Therefore, equating the Qp(t) relationships, eqvA/kT_1 : LE;_(1F2IF1) e'mp 01‘ Note as a check that the foregoing expression reduces to Eq. (8.16) if [F1 —+> 0 and
I F2 —9 I F. a1 (a) We note using Fig. 8.6 that tS/rp E 0.22 when IR/Ip = 1. Thus rs = 0.22psec, or the
diode becomes reverse biased before it is pulsed back to the ON condition. Based on the
above information, we conclude i(t) (b) Since pulsing is occurring from reverse bias, we can assume by analogy with the text
turnon development that the pulsing effectively occurs from i : 0 with Qpﬂtésee) r: 0.
The situation here is completely analogousall but identical—to the turnuon situation
considered in Section 8.2, except I is replaced by t— l/Jsec. Consequently, the required
expression is just Eq. (8.16) with the tin exp(—t/1'p) replaced by t — lusec 87 §._8 (a) In the CPG program
yon(l) =Apn(0,I)/Apnmax
Thus
quy%T__
yonm =£______L
quONkT__1
and
._DA = L774 1+ onl quON/kT_1
VON (VON ml 3/ ( >1 ll The desired onN values can be obtained by inserting the following five lines into the
last segment of the CPG program. Place before for i=1: j, Place after yon= (A—B) / 2; VON=O . 5; vj= (kT/VON) *log (l+yon (1) * (exp (VON/k‘I') 1) );
vrel= ; %Vrel=vA/VON vrel=€vrel,vj] ; kT=0.0259; V After the program is run, vrel is read out from the Command window. (b) Appropriately modifying Eq. (8.16), $ = (Eli) lull +(1 ~ e"/TpXe ‘IVON/kT~ 1)]
VON The computational results based on the above relationship are recorded along with the exact
results in the table below. Note in all cases that vA/V0N(exact) > vA/VON(quasistatic). exact quasistatic exact quasistatic
1’59 vA/V ON ”AN ON {(3}; vA/V ON DA” ON
0.1 0.9449 0.8782 1.1 0.9923 0.9790
0.2 0.9612 0.9115 1.2 0.9933 0.9814
0.3 0.9701 0.9301 1.3 0.9941 0.9835
0.4 0.9760 0.9425 1.4 0.9949 0.9853
0.5 0.9802 0.9517 1.5 0.9955 0.9869
0.6 0.9835 0.9588 1.6 0.9960 0.9883
0.7 0.9860 0.9644 1.7 0.9965 0.9896
0.8 0.9881 0.9691 1.8 0.9969 0.9906
0.9 0.9897 0.9730 1.9 0.9973 0.9916
1.0 0.9911 0.9762 2.0 0.9976 ' 0.9925 88 ...
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 Spring '07
 HAMBLEN

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