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# Assignments - 3. Solutions to Selected Problems 3.1 dc-dc...

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EE8410 Power Electronics 12 3. Solutions to Selected Problems 3.1 dc-dc Switch-mode Converters Step-down Converters (Chapter 7, Moham) 7-1 V O =5V, V d =10V to 40V, P O 5W, f s =50kHz. Find the minimum inductance to keep the converter in the continuous conduction mode under all conditions. Solution: For a given load and output voltage, the likelihood that the inductor current will fall to zero is increased by lowering the duty ratio and thus increasing the OFF time. The duty ratio is lowest when V d =40V. P O /V O =5W/5V=I O =1A H 43.75 L µ = × × = = = = ] 5 40 [ 1 50000 2 125 . 0 ] [ 2f D L ; 125 . 0 40 5 ] [ 2 I 5, - 7 Eq. from conduction continuous For S O O d O O d S V V I D V V L f D 7-2 V o =5V, f s =20kHz, L=0.001H, C=470 µ F, V d =12.6V, I o =0.2A. Find V o . Solution: Is the circuit operating in the continuous mode? [] () 2.01mV = × × × × × = = = = = = 6 2 o 2 o OB 10 470 001 . 0 8 000 , 20 5 6 . 12 5 1 V 8 1 V 24 - Eq.7 from so mode, continuous in the is 397 . 0 12.6 5 D ere wh 0754 . 0 2 i 5, - Eq.7 From LC V D T It V V L f D o s o d s 7-3 Find the RMS ripple current through L. (optional) Solution: -0.0755A t T S DT S (1-D)T S i L , ripple 0.0755 A

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EE8410 Power Electronics 13 () s A s A s D s A dt di s A dt di dt di L V L L L µ 50 T : Note 151 . 0 7600 50 397 . 0 i is current ripple peak - to - peak the Therefore, . 397 . 0 6 . 12 5 ; 5000 001 . 0 5 , t During 7600 001 . 0 5 6 . 12 , t During ; s L off on = = = = = = = = = = 43.66mA current ripple RMS ] [ T 1 s 50 t s 19.84 for 5000t - 0.1747 84 . 19 DT : Note s 19.84 t 0 for 7600 0755 . 0 ) ( , 0 2 , s s , = = < < = = < < + = RMS i t i s t t i ripple L T ripple L ripple L s 7-5 Calculate the ripple voltage in problem 7-2 if the load current is reduced to I OB /2. (optional) Solution: 1.66mV = = = = × × = = = = + = = = O 2 1 1 1 o 1 1 d o V 4, - problem7 in derived equation the Using 0.281. D ; 52 . 1 1197 . 0 , 1197 . 0 ; 001 . 0 000 , 20 2 6 . 12 2 0377 . 0 I 14, - Eq.7 ing 52 . 1 ; 6 . 12 5 V V ; 0377 . 0 D D D L D T V Us D D D A I s d o One-quadrant chopper + V d _ G i D D T i T i O L R E a + V O _ Fig 7-a Circuit diagram of one-quadrant chopper 7-a In the one-quadrant chopper circuit of Fig. 7-a, V d =600V, E a =200V, L=4mH, R=1.5 , T s =4000 µ s, t on =2500 µ s. Show that the output current i O is continuous. Solution:
EE8410 Power Electronics 14 . continuous is i 0, I Since 3 . 133 3 . 178 5 . 1 200 1 1 1.5 600 I 5 . 1 2667 4000 937 . 0 2667 2500 2667 5 . 1 4 O min 5 . 1 937 . 0 min > = = = = = = = = = 45A e e T t s mH s on τ µ 7-b For the chopper circuit of problem 7-a, determine: (a) The average output voltage and current, V O and I O . (b) The output current at the instant of IGBT turn-off. Solution: 1.5 200 - e - 1 e - 1 1.5 600 I 1.5 10 4 5 . 1 10 4 T 9375 0. 10 4 5 . 1 10 2.5 t ) ( 1.5 200 - 375 R V 600 0.625 V T t V continous. is current a, - 7 Prob. From ) ( 1.5 - 0.9375 - max 3 - 3 - s 3 - 3 - on O on O 179.9A 116.7A 375V = = = × × × = = × × × = = = = = × = = b E I a a O 7-c In the chopper circuit of Fig.7-a, V d =600V, E a =200V, L=1mH, R=1.5 , T s =4000 µ s, and t on =2500 µ s. Show that the output current i O is discontinuous. Solution: ous. discontinu is i 0, I Since 133 41 5 . 1 200 1 1 5 . 1 600 I 0 . 6 667 4000 748 . 3 667 2500 667 5 . 1 1 O min 0 . 6 748 . 3 min < = = = = = = = = = -92A e e T t s mH s on 7-d For the chopper circuit of problem 7-c, determine: (a) The average output voltage and current, V O and I O .

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## This note was uploaded on 11/02/2009 for the course ECE ece464 taught by Professor Abc during the Spring '09 term at Jacksonville College.

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Assignments - 3. Solutions to Selected Problems 3.1 dc-dc...

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