1. Diode with an RLC Load
vL(t)
vC(t)
VCo

2. Close the switch at t = 0
VCo

3. KVL around the loop

4. Characteristic Equation

5. 3 Cases Case 1  = ω0 “critically damped” s1 = s2 = - roots are equal
i(t) = (A1 + A2t)es1t

6. 3 Cases (continued) Case 2  > ω0 “overdamped”
roots are real and distinct
i(t) = A1es2t + A2es2t

7. 3 Cases (continued) Case 3  < ω0 “underdamped” s1,2 = - +/- jωr
ωr = the “ringing” frequency, or the damped resonant frequency
ωr = √ωo2 – α2
i(t) = e-t(A1cosωrt + A2sinωrt)
exponentially damped sinusoid

8. Example 2.6

9. Determine an expression for the current

10. 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
i1(t)

17. Mode 1 (continued)

18. Circuit in Mode 2 I1 i2

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
Lm = Magnetizing Inductance
v2/v1 = N2/N1 = i1/i2

25. Refer Secondary to Primary Side

26. Operational Mode 1 Switch closed @ t = 0
Diode D1 is reverse biased, ai2 = 0

27. vD = Vs(1+a) = reverse diode voltage primary current i1 = is
Vs = vD/a – Vs/a
vD = Vs(1+a) = reverse diode voltage
primary current i1 = is
Vs = Lm(di1/dt)
i1(t) = (Vs/Lm)t for 0<=t<=t1

28. ai2 = 0, D1 is reverse biased
is = i1
v1 = Vs

29. v2 = av1 = aVs
-v1 + vD/a – Vs/a = 0  vD = Vs(1+a)

30. Operational Mode 2 Begins @ t = t1 when switch is opened
i1(t = t1) = (Vs/Lm)t1 = initial current I0
Lm(di1/dt) + Vs/a = 0
i1(t) = -(Vs/aLm)t + I0 for 0 <= t <= t2

31. i1 = 0 (is = 0)
i1 becomes ai2
is = -ai2 (into source Vs)

32. vD = 0, D1 is forward biased

33. Waveform Summary

34. Find the conduction time t2
Solve
-(Vs/aLm)t2 + I0 = 0
yields
t2 = (aLmI0)/Vs
I0 = (Vst1)/Lm
t1 = (LmI0)/Vs
t2 = at1

35. Example 2.8 Lm = 250μH N1 = 10 N2 = 100 VS= 220V
There is no initial current.
Switch is closed for a time t1 = 50μs, then opened.
Leakage inductances and resistances of the transformer = 0.

36. Determine the reverse voltage of D1
The turns ratio is a = N2/N1 = 100/10 = 10
vD = VS(1+a) = (220V)(1+10) = 2420 Volts

37. Calculate the peak value of the primary and secondary currents
From above, I0 = (Vs/Lm)t1
I0 = (220V/250μH)(50μs) = 44 Amperes
I’0 =I0/a = 44A/10 = 4.4 Amperes

38. Determine the conduction time of the diode
t2 = (aLmI0)/Vs
t2 = (10)(250μH)(44A)/220V
t2 = 500μs
or, t2 = at1
t2 = (10)(50μs)

39. Determine the energy supplied by the Source
W = (1/2)((220V)2/(250μH))(50μs)2
W = 0.242J = 242mJ
W = 0.5LmI02 = (0.5)(250x10-6)(44A)2
W = 0.242J = 242mJ