Chapter 8 - CHAPTER 8 El (a) It is the time period (:5)...

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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 turn-off 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 off-state 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 step-recovery 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 P-I-N 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 turn-on recombination obviously dominates over generation because there ts a earner excess. Thus, the inevitable loss of carriers via recombination will indeed retard the build-up 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 8-1 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 .0e-2,10,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 charge-control'); 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 8-3 (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 open-circuits 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 10-15 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% =._(10-6)}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 turn-on transient was noted at the end of Section 8.2. 8-5 (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 8-6 132'“ 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 8-8 ...
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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.

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Chapter 8 - CHAPTER 8 El (a) It is the time period (:5)...

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