1. Diode with an RLC Load v L (t) v C (t) V Co

2. Close the switch at t = 0 V Co

3. KVL around the loop

4. Characteristic Equation

5. 3 Cases Case 1 –  = ω 0 “critically damped” –s 1 = s 2 = -  –roots are equal –i(t) = (A 1 + A 2 t)e s1t

6. 3 Cases (continued) Case 2 –  > ω 0 “overdamped” –roots are real and distinct –i(t) = A 1 e s2t + A 2 e s2t

7. 3 Cases (continued) Case 3 –  < ω 0 “underdamped” –s 1,2 = -  +/- jω r –ω r = the “ringing” frequency, or the damped resonant frequency –ω r = √ω o 2 – α 2 –i(t) = e -  t (A 1 cosω r t + A 2 sinω r t) –exponentially damped sinusoid

8. Example 2.6

9. Determine an expression for the current

11. Determine the conduction time of the diode The conduction time will occur when the current goes through zero.

12. Conduction Time

13. Freewheeling Diodes Freewheeling Diode

14. Freewheeling Diodes D2 is reverse biased when the switch is closed When the switch opens, current in the inductor continues. D2 becomes forward biased, “discharging” the inductor.

15. Analyzing the circuit Consider 2 “Modes” of operation. Mode 1 is when the switch is closed. Mode 2 is when the switch is opened.

16. Circuit in Mode 1 i 1 (t)

17. Mode 1 (continued)

18. Circuit in Mode 2 I1I1 i2i2

19. Mode 2 (continued)

20. Example 2.7

21. Inductor Current

22. Recovery of Trapped Energy Return Stored Energy to the Source

23. Add a second winding and a diode “Feedback” winding The inductor and feedback winding look like a transformer

24. Equivalent Circuit L m = Magnetizing Inductance v 2 /v 1 = N 2 /N 1 = i 1 /i 2

25. Refer Secondary to Primary Side

26. Operational Mode 1 Switch closed @ t = 0 Diode D 1 is reverse biased, ai 2 = 0

27. V s = v D /a – V s /a v D = V s (1+a) = reverse diode voltage primary current i 1 = i s V s = L m (di 1 /dt) i 1 (t) = (V s /L m )t for 0<=t<=t 1

28. Operational Mode 2 Begins @ t = t 1 when switch is opened i 1 (t = t 1 ) = (V s /L m )t 1 = initial current I 0 L m (di 1 /dt) + V s /a = 0 i 1 (t) = -(V s /aL m )t + I 0 for 0 <= t <= t 2

29. Find the conduction time t 2 Solve -(V s /aL m )t 2 + I 0 = 0 yields t 2 = (aL m I 0 )/V s I 0 = (V s t 1 )/L m t 1 = (L m I 0 )V s t 2 = at 1

30. Waveform Summary

32. Example 2.8 L m = 250μHN 1 = 10N 2 = 100V S = 220V There is no initial current. Switch is closed for a time t 1 = 50μs, then opened. Leakage inductances and resistances of the transformer = 0.

33. Determine the reverse voltage of D 1 The turns ratio is a = N 2 /N 1 = 100/10 = 10 v D = V S (1+a) = (220V)(1+10) = 2420 Volts

34. Calculate the peak value of the primary and secondary currents From above, I 0 = (V s /L m )t 1 I 0 = (220V/250μH)(50μs) = 44 Amperes I’ 0 =I 0 /a = 44A/10 = 4.4 Amperes

35. Determine the conduction time of the diode t 2 = (aL m I 0 )/V s t 2 = (10)(250μH)(44A)/220V t 2 = 500μs or, t 2 = at 1 t 2 = (10)(50μs) t 2 = 500μs

36. Determine the energy supplied by the Source W = 0.5L m I 0 2 = (0.5)(250x10 -6 )(44A) 2 W = 0.242J = 242mJ W = (1/2)((220V) 2 /(250μH))(50μs) 2 W = 0.242J = 242mJ