From the course textbook, Power Electronics Circuits, Devices, and Applications, Fourth Edition, by M.S. Rashid, do the following problems

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1 ECE 427 Homework #2 From he course exbook, Power Elecronics Circuis, Devices, and Applicaions, Fourh Ediion, by M.S. Rashid, do he following problems 1. Problem.1 on page 1. Draw he oupu volage and inpu curren waveforms. Compue he average oupu volage. Change he resisance o a DC curren source wih he same average curren value. Then draw he oupu volage and inpu curren waveforms and compue he average oupu volage..1 A single phase bridge recier of Figure.a has a purely resisive load R=5 Ohms, he peak supply volage Vm=17V, and he supply frequency of f=6hz. Deermine he average oupu volage of he recier he source inducance is negligible. R 5. Ω V m 17 V f 6Hz ω 2f rad sec Compue he average value of he oupu volage. 1 v o_avg V ω m sin d v o_avg V Plo he oupu volage and he inpu curren for his resisive load. v o V m sin i i V m sin R 2 15 v o 1 i i ω ω

2 Calculae he average oupu curren. Then se he oupu curren o his value as requesed. v o_avg i o_avg i R o_avg A Le he oupu curren be i o_avg Plo he oupu volage and he inpu curren for his consan curren load. v o V m sin i i ω ω v o 1 i i ω ω 2. Problem.15 on page 11. A single phase bridge recier of Figure.a is supplied from a 12V, 6Hz source. The load resisance RL=14 Ohms. a. Design a C-filer so ha he ripple facor of he oupu volage is less han 5%. b. Wih he value of capaciior Ce in par (a), calculae he average load volage Vdc. R L 14Ω f 6Hz RF 5% V m 12 2V f 6Hz RF = 1 4R L f C e 1 1 RF C e C RF4R L f e 625μF The exbook answer is from a previous ediion where he ripple was saed as 7%. V m 1 V o_avg 2 2 R L 2f C V o_avg V e

3 . Problem.17 on page 12. Also plo he waveform for vo. The singel phase half wave recier in figure P.16 has a purely resisive load of R. Deermine: a. he efficiency b. he form facor FF c. he ripple facor RF d. he Transformer Uiliizaion Facor TUF e. he Peak Inverse Volage PIV f. he Cres Facor CF of he inpu curren g. he inpu power facor PF. Assume Vm=1V. V m 1V ω v o V m sin Le R 1Ω ω v o ω a. he efficiency 2 1 V dc V dc v ω 2 o d V dc 1.81 V I dc I R dc 1.81 A 2 1 V rms v ω 2 o 2 d V rms 5 V V rms I rms I R rms 5 A P dc V dc I dc P dc 1.1kW P ac V rms I rms P ac 2.5kW P dc efficiency efficiency % P ac

4 b. he form facor FF Recall V dc 1.81 V V rms 5 V V rms FF FF % V dc c. he ripple facor RF Recall V dc 1.81 V V rms 5 V 2 2 V ac V rms V dc V ac V V ac RF RF % V dc d. he ranformer uiliizaion facor TUF Recall P dc 1.1kW V rms 5 V P dc TUF TUF.45 V rms I rms This procedure is he same as Example.5 on page 96 of he exbook. The answer is no he same as he back of he book. e. Peak Inverse Volage PIV PIV V m sin PIV 1 V 2 f. Cres Facor CF v o_pk V m v o_pk v o_pk 1 V I pk R I pk 1 A I rms 5 A I pk CF CF 2 I rms g. Inpu Power Facor PF P dc 1.1kW V rms 5 V I rms 5 A lagging 1 P dc PF PF.45lagging V rms I rms

5 4. Problem.28 on page 1. For par b, ploing he FFT ampliude is sufficien. The hree phase bridge recier of Fig..11 supplies a ripple-free load curen of Ia. The primary and secondary of he ransformer are conneced in Y. Assume a ransformer urns raio of uniy. a. Draw he waveforms for currens in D1, D, D5, and he secondary phase curren of he ransformer b. Express he secondary phase curen in Fourier series. c. Deermine he inpu PF and HF of he inpu curren. Le V m v ab V m sin v bc V m sin ω 2 v ca V m sin ω vbc vab -vca vbc -vab vca 1 v bc v ab.5 v ca v bc v ab v ca ω The curren flows in diode D1 when vab and -vca is he greaes volages. The diode is he consan Ia.

6 i D1 ω 2 ω ω ω 4 ω ω 5 2 i D ω 2 ω 2 ω ω ω ω 5 2 i D5 ω 2 ω ω ω 4 ω 5 ω 2 1 i D1 i D i D ω Seconday phase curren is he sum of id1 and id4

7 i D4 ω 2 ω ω ω 4 ω ω 5 2 i a ω 2 ω ω ω 4 ω ω i a ω

8 NN 1 ps 2 NN k k 1 ps 1 i aak i a ps 2 1 iaff o ps ff i aa.6.4 iaff ok k Find he harmonic facor. I rms_1 2 iaff o1 I rms_ I rms i ω 2 a 2 d I rms.816 HF I rms_1.954 I rms Power facor is power in divided by he produc of rms volage in and rms curren in. For an ideal recier, power ou equals power in. Using symmery on he dc side, we ge, for 1/6 of a cycle,

9 1 6 P dc V m cos dω P dc V m V rms cos dω V 2 rms.48 Line o neural volage rms 2 1 I rms i ω 2 a 2 d I rms.816 P dc PF PF.955lagging V rms I rms checking, I rms_1 PF PF.954lagging I rms 5. Problem 1. on page 547. A single phase half wave converer in figure P1.1 is operaed from a 12V, 6 Hz supply and he resisive load is R=5 Ohms. If he average oupu volage is 25% of he maximum possible average oupu volage,calculae a. he delay angle, b. he rms and average oupu cuirrens, c. he rms and average hyrisor currens, d. he inpu power facor. 1 v omax V ω 2 m sin d V V m 12 2V R 5Ω

10 For and, 2 1 V O V ω 2 m sin d 1.55 V V O.25 v omax V O _ave I R o_ave 2.71 A 2 1 V m _rms sin ω ω 2 R ( ) d I o_rms 7.54 A The hyrisors have he same currens as he oupu. The average and rms values are he same. V i_rms 12V I i_rms _rms I i_rms 7.54 A 1 V m P o V ω 2 m sin sin ω R ( ) d P o W P o PF PF.1lagging V i_rms I i_rms 1.22 Semiconrolled Reciier problem 1.22 abc A hree phase semiconverer in Figure P1.22 is operaed from a hree phase Y conneced 28V 6Hz supply and he load resisance R=1 Ohms. If i is required o obain an average oupu volage of 5% of he maximum possible oupu volage, calculae a. he delay angle b. he rms and average oupu currens, c. he aveage and rm hyrisor currens, d. plo one of he hyrisor currens over a plo of v ab. Ideny which hyrisor you choose. V m 28 2V f s 6Hz 2f s R load 1Ω Le us sar wih a diode recier o undersand he waveforms. v ab V m sin

11 v o V m sin V m sin 2 V m sin V m sin V m sin V m sin Plo he volage on he dc side cb ab ac bc ba ca v o 2 v ab Average volage is 2 Average curren, assuming an inducance-dominaed load, is ω s V o V o v 2 o d V 28.9 A R load This is he diode case, a siuaion ha yields he

12 i a maximum possible oupu volage. 5% of his is 2 V x 5% V o V Plo he curren in phase a cb ab ac bc ba ca 4 2 i a Now le us consider a six-hyrisor recier o undersand he waveforms. Le 2deg

13 v o V m sin V m sin 2 V m sin 2 V m sin V m sin V m sin V m sin ca cb ab ac bc ba ca v o 2 v ab

14 2 ω s V o V o v 2 o d V R load A i a ca cb ab ac bc ba ca 4 2 i a

15 Now consider he semiconrolled recier. Commuaions beween even-numbered devices, he diodes, occur a =. Commuaions beween odd-numbered devices, he hyrisors, occur a a specied value of. v o V m sin V m sin 2 2deg V m sin V m sin V m sin V m sin cb ab ac bc ba ca v o 2 v ab ca cb ab ac bc ba ca

16 2 ω s V o V o v 2 o d V R load A i a 2 This value of 2 degrees for a semiconrolled recier does no mee he 5% volage reques cb ab ac bc ba ca i a v ab

17 Now ake he semiconrolled bridge o 6 degrees. This is he delay angle limi of he diodes overaking he hyrisors. v o V m sin V m sin 2 6deg V m sin V m sin V m sin V m sin cb ac ba v o 2 v ab

18 2 ω s V o V o v 2 o d V R load A i a cb ac ba i a Take he angle pas 6 degrees. We ge an inabiliy o achieve he negaive par of he waveform. The hyrisor say on during he inerval pas 6 degrees; he diode commuaes.

19 Le 8deg v o V m sin 2 2 V m sin 4 4 V m sin cb ac ba v o 2 v ab ω s V o V o v 2 o d V R load A

20 8 deg i a cb ac ba 2 1 i a

21 A =9 degrees, he volage is half of he maximum volage. 9deg V x V v o V m sin 2 2 V m sin 4 4 V m sin cb ac ba v o 2 v ab A =9 degrees, he volage is half of he maximum volage. 2 ω s V o V o v 2 o d V R load A

22 i a cb ac ba 4 i a v ab

23 Now answer he problem's quesions: a. he delay angle 9deg b. he rms and average oupu currens, A The rms value is he same; he oupu curren is DC. This mus be he answer because he maximum volage is 28.9V and he resisance is 1 Ohms. Therefore, o conserve energy in he inducive par of he load, he load curren is always Vo/Rload. c. he average and rms hyrisor currens; plo one of he hyrisor currens over a plo of v ab. Ideny which hyrisor you choose. I choose Thyrisor T1. Is curren is he posiive curren labelled 61 and 12 in he diagram below. Ploing he hyrisor T1 curren from his, we have only he (12) par of he waveform. i T I kep rack of he phase relaionships by ploing vab() on each of my volage plos and now here on he las curren plo. I divided i by 1 o make he plo more readable; Phase angle is imporan for his plo and i shows up beer his way.

24 i T1 v ab ω ω s s I Ta i 2 T1 d.511 A i T1_rms 2 i T1 2 d 7.22 A This makes inuiive sense because he curren in phase a riggers when =9 degrees. I remains ON for 9 degrees before expiring. Tha is only ON for 1/4 of a cycle. Therefore, he average value is 1/4 of he Io value of 14.A or.5a. The RMS value is sqr(1/4) of he rms value of Io or 7.A.

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