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As we explained in class, the power ratio (out to in) is the same as the voltage ratio. We start with the definitions of output and input power.   Then we realize that the output current and input current are effectively the same, with a high beta on the transistor.    Combining the relationships, &'(*+,j EHOJQJUjEHOJQJUjEHOJQJUjGEHOJQJUOJQJjEHOJQJU&()*,-./0&H$7$8$ Ir+D&H$7$8$ +D&H$7$8$ a+DJ&H$7$8$  +DJ&H$7$8$ U +DJ&()*,-./0 1h/ =!"#$%Oh+'0\   $ 0<DLTss elishavk lis Normal.dot elishavkt2isMicrosoft Word 8.0@@@vQC՜.+,D՜.+,< hp  University of Idaho1  Title 6> _PID_GUIDAN{5BEC4A23-B7DC-4F94-BDFC-2BC854114E0A}Ole :OlePres000.$Ole ;E0$ \&WordMicrosoft Word  g "System fj & -@Times New Roman؟ww w fj- .2 Xg Problem 6.1. Explain the relationship between energy efficiency and the output-to-input8!22,N22=32,22,!,,22'222,H,,2,2,"30,"!,,2.0,222,2222!2!222.  s.^2 s7g voltage ratio in the simplified linear power converter.22,1,!,222,'N2!,22,,!22H-!,223,!,!.Y.2 Y[g As we explained in class, the power ratio (out to in) is the same as the voltage ratio. WeeH'H,,32,2,22,,''2,22I,!!,2!2222!'2,',N,,'2,22-1,!,2_,..[2 5g start with the definitions of output and input power.',!H22,2,!222'2!2222,2222222H,!..2 Zg Then we realize that the output current and input current are effectively the same, with a=2,2H,",,-,2,2,2222,2!",2,22222,2!",2,",,"!,,2,02,',O,H2,.d.52 dg high beta on the transistor.2122,,223,!,2''2!..52 g Combining the relationships,C2N22212,!,,22'22'.-@"Arial 0؟ww w fj-g &M..  --g L  ^ ^--  Times New Roman؟ww w fj-.    !P !out     !V +!out3    !I I!outM  Symbol ؟ww w fj-.! E. - -%  &&'- -.  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The output power is 100W. a. Determine the efficiency of the converter of Fig 6.1 when it is used for this application. b. How much energy is lost in the transistor in one year? c. using the electric rate in your area, what is the cost of the energy loss for one year?  ChrPropMap6Y ParPropMap8Y RangeElem;Y ParPropData9 RangeData<@VEmbedMap0;LinkMap.Y;YLinkData/@NormalArial eqRegionA@TR*)tree? p? ?dV.in??t100?V@A@T*? p? ?dV.out ?!?t 30"? V#@A@TV)+$? p%? $&?d%P.out'?%(?t'100)?'W*@A@T,+? p,? +-?d,cents.?,1/@A@TI -0? p1? 02?d1dollars3?14?t31005?3cents6*@T.B-@La. Find the efficiency. First find the current from the power and voltage.6L8L7;L89@V09;.L:;L;/@NormalArial <@A@TV./=? p>? =??d>I.out@@?>@A?d@@P.out@B?@@V.out@C@A@T 0@D? p@E?@D@F?d@EI.out@G?@E@H?+@@GSerial_DisplayNodeW@I?@G@J*@T#s301Bcc-@BFrom the figure we see that the circuit is just a series circuit. 6B8B@K;B@L9@V0@M;.B@N;B@O/@NormalArial @P@A@TLFb&X2@Q? p@R? @Q@S?d@RI.in@T?@RI.out@U@A@TLbX3@V? p@W? @V@X?d@WP.in@Y?@W@Z?d@YV.in@[?@YI.in@\*@T{*4B-#Efficiency is power out / power in:6#8#@];#@^9@V0@_;.#@`;#@a/@NormalArial @b@A@TH!5@c? p@d? @c@e?d@d\h@f?@d@g?d@fP.out@h?@fP.in@i@A@T6@j? p@k?@j@l?d@k\h@m?@k@n?+@@m@W@o?@m@p*@T7B-#b. Energy loss in a year is found:6#8#@q;#@r9@V0@s;.#@t;#@u/@NormalArial @v@A@T O08@w? p@x? @w@y?d@xT@z?@x@{?t@z1@|?@zyr@}@A@T9@~? p@?@~@?d@T@?@@?+@@@W@?@hr@J0`@@A@T@X:@? p@?@@?d@P.in@?@@?+@@@W@?@@@A@T2;@? p@?@@?d@P.out@?@@?+@@@W@?@@@A@T4J;@<@? p@? @@?d@W.loss@?@@?p@@@?@@?d@P.in@?@P.out@?@T@@A@T,>J@=@? p@?@@?d@W.loss@?@@?+@@@W@?@@@A@Tx, J@>@? p@?@@?d@W.loss@?@@?+@@@W@?@@?d@kW@?@hr@*@Tk{x?:-"c. Cost here is 6 cents per kwhr.6"8"@;"@9@V0@;."@;"@/@NormalArial @@A@T?@@? p@? @@?d@C.cost@?@@?d@W.loss@?p@@?@@?t@6@?@@?d@cents@?@@?d@kW@?@hr@@A@T6A@? p@?@@?d@C.cost@?@@?+@@@W@?@cents@@A@T FB@? p@?@@?d@C.cost@?@@?+@@@W@?@dollars@*@T[@-hB(p(p-ASProblem 6.7 A b uck converter has an input of 60V and an output of 25V. The load resistor is 9 ohms, the switching frequency is 20kHz, L=1mH, and C=200mF. a. Determine the duty ratio. b. Determine the average, peak, and rms inductor current. c. etermine the average source current. d. Determine the peak and average diode current. 6S@;@ ChrPropData7@@;@7@oSymbol@@;@7@@@@8S@;S@9@V0@;.S@;S@/@NormalArial @@A@TT2@? p@? @@?d@V.in@?@@?t@60@?@V@@A@T@? p@? @@?d@V.out@?@@?t@25@?@V@@A@TH@? p@? @@?d@f.s@?@@?t@20@?@kHz@@A@T@? p@? @@?d@T@?@@?t@1@?@f.s@@A@T1@? p@? @@?d@\ms@?@@?@@@?t@10@?K@@?@6@?@secA@A@TM 'A? pA? AA?dALA?AA?tA1A?AmHA@A@T A? pA ? AA ?dA CA ?A A ?tA 200A ?A \mFA@A@T? A? pA? AA?dARA?AA?tA9A?AohmA@A@T  A? pA?AA?dATA?AA?+@A@WA?A\msA*@T+;8Btt-@Da. Find the duty ratio. Duty ratio in a buck converter is found by6D8DA;DA9@V0A;.DA ;DA!/@NormalArial A"@A@T(>am9XA#? pA$? A#A%?dA$DA&?A$A'?dA&V.outA(?A&V.inA)@A@TLL] XA*? pA+?A*A,?dA+DA-?A+A.?+@A-@WA/?A-A0*@THB0 0 -@{b. Find the average, peak, and rms inductor current. The average inductor current is the same as the average load current.6{8{A1;{A29@V0A3;.{A4;{A5/@NormalArial A6@A@T(~VA7? pA8? A7A9?dA8I.L_avgA:?A8A;?dA:V.outA? pA??A>A@?dA?I.L_avgAA?A?AB?+@AA@WAC?AAAD*@T *:-The peak inductor current is68AE;AF9@V0AG;.AH;AI/@NormalArial AJ@A@T(EO0AK? pAL? AKAM?dALI.L_pkAN?ALAO?dANI.L_avgAP?ANAQ?@APAR?tAQ1AS?AQ2AT?pAPAU?ATAV?@AUAW?@AVAX?dAWV.outAY?AWLAZ?pAVA[?AZA\?tA[1A]?A[DA^?AUA_?tA^1A`?A^f.sAa@A@TP$:v0Ab? pAc?AbAd?dAcI.L_pkAe?AcAf?+@Ae@WAg?AeAh*@T Sc4`:-!Finding the rms inductor current,6!8!Ai;!Aj9@V0Ak;.!Al;!Am/@NormalArial An@A@T(vUAo? pAp? AoAq?dApI.L_minAr?ApAs?dArI.L_avgAt?ArAu?@AtAv?tAu1Aw?Au2Ax?pAtAy?AxAz?@AyA{?@AzA|?dA{V.outA}?A{LA~?pAzA?A~A?tA1A?ADA?AyA?tA1A?Af.sA@A@TXA? pA?AA?dAI.L_minA?AA?+@A@WA?AA*@T8B"-&and offsetting the rest of the cycle, 6&8&A;&A9@V0A;.&A;&A/@NormalArial A*@T^!BFF- For 0?dB= I.pk_diodeB??B=B@?+@B?@WBA?B?BB*@T + "( B-)The average value of the diode current is6)8)BC;)BD9@V0BE;.)BF;)BG/@NormalArial BH@A@T2 w ]X BI? pBJ? BIBK?dBJ I.avg_diodeBL?BJBM?@BLBN?tBM1BO?BMTBP?%BLBQ?@BPBR?@BQBS?dBRDBT?BRTBU?BQTBV?BPBW?dBVtBX?BVBY?dBX I.pk_diodeBZ?BXB[?@BZB\?p@B[B]?B\B^?dB]I.L_minB_?B]I.L_pkB`?pB[Ba?B`Bb?dBatBc?BaBd?dBcDBe?BcTBf?BZBg?dBfTBh?BfBi?dBhDBj?BhTBk@A@TL Fb X Bl? pBm?BlBn?dBm I.avg_diodeBo?BmBp?+@Bo@WBq?BoBr*@T ; % B+@+@-@Problem 6.8 The buck converter of Fig 6.3a has Vs=30V, V0=20V, and a switching frequency of 40kHz. The output power is 25W. Determine the size of the inductor such that the minimum inductor current is 25% of the average inductor current. 68Bs;Bt9@V0Bu;.Bv;Bw/@NormalArial Bx@A@T$ H: &0 By? pBz? ByB{?dBzV.sB|?BzB}?tB|30B~?B|VB@A@T`$ : w0 B? pB? BB?dBV.oB?BB?tB20B?BVB@A@T$ : 0 B? pB? BB?dBP.oB?BB?tB25B?BWB@A@T($ h: :0 B? pB? BB?dBf.sB?BB?tB40B?BkHzB*@TK )k X B  -@rThe minimum value of inductor current is positive and non-zero, so our continuous conduction approach will work. 6r8rB;rB9@V0B;.rB;rB/@NormalArial B@A@Tv > ! B? pB? BB?dBDB?BB?dBV.oB?BV.sB@A@T B? pB?BB?dBDB?BB?+@B@WB?BB@A@T [ > B? pB? BB?dBI.L_avgB?BB?dBP.oB?BV.oB@A@T B? pB?BB?dBI.L_avgB?BB?+@B@WB?BB@A@T   M B? pB? BB?dBI.L_minB?BB?tB0.25B?BI.L_avgB@A@T   B? pB?BB?dBI.L_minB?BB?+@B@WB?BB*@T h ! :PP- We know that6 8 B; B9@V0B;. B; B/@NormalArial B@A@T& U D@ B? pB?,BB?dBI.L_minB?BB?dBI.L_avgB?BB?@BB?tB1B?B2B?pBB?BB?@BB?@BB?dBV.oB?BL.minB?pBB?BB?tB1B?BDB?BB?tB1B?Bf.sB*@Ts ~ 8 :ff-Rearranging this, 68B;B9@V0B;.B;B/@NormalArial B@A@T 2 B? pB? BB?dBL.minB?BB?@BB?dBV.oB?pBB?BB?tB1B?BDB?BB?@BB?p@BB?BB?dBI.L_avgB?BI.L_minB?B2B?Bf.sB@A@T  ! B? pB?BB?dBL.minB?BB?+@B@WC?B