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Unformatted text preview: CHAPTER 8 El (a) It is the time period (:5) during which the observed current associated with the tumeoff
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 tum—off transient pictured in Fig. 3.1(b), r, = Inw t‘s, where In is the total time for the reverse current to decay to 10% of its maximum magnitude and I; is
the storage delay time. (c) Yes . This is precisely what happens during the ts portion of the turnoff transient.
Even t ough there is an excess of carriers at the edges of the depletion region making vA >
0, the external circrritry 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 quasineunal regions on the
two srdes of the Junction. ' (e) The excess carriers are removed by recombination and reverse injection.
(1') . Apn(x,t) > 0 implies Apn(xn,t) > 0, the necessary condition making DA > 0. (gt . Referring to Eq. (8.2), if i > O, the slope of a pn(x,t) versus 1: plot must be
negattve atx =xn. (h) The electrical re3ponse of the steprecovery diode is special in that the t, 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. . ' for ts vary only as the ratio of I}: and 13. Thus, increasing both If: and IR by the
same amount will have no effect. (j) . During turnon recombination obviously dominates over generation because
there ts a earner 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.) 8.2 (T). spud“) = pn(xn,t) — pno < 0. A carrier deficit at the edge of the dep etion region in tcates the junction is reverse biased.
(b) Invoking the law of the junction, "(In)P(xn) = NDPDDIZ = ni2/2 = :1? eqvA/kT
DA = OCT/(I) 100/2) = 0.0259 ln2 = 43.018 V (c) . Since dam/4:45,“ = dpn/diden > 0, it follows from Eq. (8.2) that t' < 0. Or 81 3.3. 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/Ip values. ts/taup lFi/lF MATLAB program script... % Comparison of the tsltaup versus IR/IF computed
% using Eq.(8.9) and the revised charge control expression %Initializalion
clear; close
%ts/taup calculation %Plottin g results
Iratio=logspace(2, l); %Iratio=IR/IF loglog(]ratio,ts l ,‘ g')
%Revised charge control expression axis([1 .0e2,10,I .Oe—2,10]); grid
x=1 JIratio; hold on
tsl=log((1+x)."2./(1+2.*x)); loglog(lratio,t32)
%Equation (8.9) xlabel('IR/IF'); ylabelCts/taup')
tsZwrﬁnvﬂ .l( l +Iratio))."2; text(0.22, i .4,'Revised chargecontrol');
text(0.22,0.35,'Eq. (8.9)')
hold off 8~2 3A (a) Because the diode is open circuited, i = 0. Thus, based on Eq. (8.2), the slope of all
the pn(x,t) versus 2: curves evaluated at x = xn should be zero. Pn(xs I) The general solution is thz) = Qp(0+)e"”P where paralleling the analysis in the text QP(0+) = 11%
Thus QpU) = Iprpe'wp 83 (c) If the charge is assumed to decay quasistatically, then, referring to Eq. (8.15),
Qp(t) = [orp(e‘1”A/"T— 1} a 1071, eqvA/kT
Equating the part (b) and (c) expressions for Qp(t) gives [FTP e"/Tp I: for? equkT
01'
eqvA/kT = (IF/10) e‘I'Tp But from the statement of the problem
1pm) 2 WON/k?" Therefore e qvA/kT = e qVON/kT e 4/1};
0r (d) The part (c) result suggests a very simple way of determining 1;, (known as the Open
Circuit Decay Method). After forward biasing the diode, one opencircuits the device and
monitors the voltage drop across the diode as a function of time. Provided the decay
follows the ideal form, 7;, is readily determined from the slope of the data. 8—4 3:2 Since under steady state conditions I]: = [0(qu0N/k7— 1) 5 10 e‘IVON’H VON a = 0.02591 = 0.716V
lo 1015 Next, solving Eq. (8.16) fort, one obtains in general = ~Tp1nil ~(10/1F)(89”A”‘T— 1)] a «p 1n[1 _ quA/kT/quON/kT]
01'
x = —rp1n[1 _ eﬂDA‘VOchT] Corresponding to DA = 0.9Von. {90% 2 _(10.5)1,{1*e—(o.t)(o.116)/(o.0259)] = 65.1 “sec
To reach DA = 0.95V0N, 195% =._(106)}n[1 _ e—(o.osxo.7t6)/(o.0259)] = 239 “sec Note that the 90%95% portion of the transient takes much longer than the 0—90% portion
of the transient. this property of the turnon transient was noted at the end of Section 8.2. 85 (a)
same slopes
greater than
the t < 0 slope
(b) Here for t > 0,
HQ? Q?
_. = 1 _ _
d: F2 1",
I290) 09? a t
me‘) 1F? ‘ QP/Tp
Q90)
t = ~1p1n(.’1=2 — = —1.’p lr{1
7p QP(0+)=IFlTp
Thus
11:2 93 = (IF2 ~1F1) 8""13
7p
and QPU) = 1F2Tp— (In—Imrp e'Wp 86 132'“ QP/Tp [Fz—IFI ) Invoking the quasistatic assumption (Eq. 8.15), we can also write
one) = Imp (WA/*1 1)
Therefore, equating the (212(1) relationships, eqvA/kT_1 z {ELUFz—Im) em?
10 lo
01' DAG) = 1%: 1.1[1 + ‘31— (IF2_IF1—) e '1”? I0 10 Note as a check that the foregoing expression reduces to Eq. (8.16) if [1:1 —> 0 and
1122 ~—) [12. 3.1 (a) We note using Fig. 8.6 that mp a 0.22 when IR/IF = 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
turn—on development that the pulsing effectively occurs from i = 0 with Qp(l}.£SCC) = 0.
The situation here is completely analogous—all but identical—to the turn—on situation considered in Section 8.2, except I is replaced by t— lttsec. Consequently, the required
expression isjust Eq. (8.16) with the tin exp(—t/1'p) replaced by t— lysec £3
(a) In the CPG program yon(1) = Apn(0.t)/Apnmax
Thus
eqvA/kT_ 1 on(1 = —————
y ) quON/kT __ 1 and A = (Eli) 1n[1 + yon(1)(e WONM" 1)]
VON The desired vA/VON 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=0. 5: vj= (kT/VON) *log (1+yon (1) * (exp (VON/kT) «1) ) ,
vrel= ; %vrel=vA/VON vrel= [vrel, vj] ; kT=0 . 0259; ' After the program is run, vrel is read out from the Command window. (b) Appropriately modifying Eq. (816), 55/1 1n[1 + (1 — e‘lTpXe WON/k?“w 1)]
VON t)A" 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/V0N(quasistatic). exact quasistatic exact quasistatic
5/32 vA/VON vA/VON {ER vA/VON vAfV 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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This note was uploaded on 01/31/2011 for the course ECE 3085 taught by Professor Taylor during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Taylor

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