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chap08 - CHAPTER 8 a(a It is the time period(ts during...

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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 off-state 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 P-I-N 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 tum-on recombination obviously dominates over generation because there is a carrier 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.) £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 V-VX ........ 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) e-t/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 10-15 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 8-6 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;_(1F2-IF1) 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 turn-on development that the pulsing effectively occurs from i : 0 with Qpfltésee) r: 0. The situation here is completely analogous-all 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 8-7 §._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 8-8 ...
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