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ECE 442 Power Electronics1 Class E Resonant Inverter.

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Presentation on theme: "ECE 442 Power Electronics1 Class E Resonant Inverter."— Presentation transcript:

1 ECE 442 Power Electronics1 Class E Resonant Inverter

2 ECE 442 Power Electronics2 Mode 1 Operation – Turn Q1 ON at t = 0 Turn Q1 OFF when v o = 0 volts

3 ECE 442 Power Electronics3 The current through the transistor (switch) For sinusoidal current, The switch is turned OFF when the output voltage becomes = 0, and the current is “transferred” to the branch containing the capacitor.

4 ECE 442 Power Electronics4 Mode 1

5 ECE 442 Power Electronics5

6 6 Mode 2 Operation Q 1 is turned OFF Diode D limits negative switch voltage

7 ECE 442 Power Electronics7 Capacitor current becomes When the switch current falls to zero,

8 ECE 442 Power Electronics8 Mode 2

9 ECE 442 Power Electronics9 Waveform Summary

10 ECE 442 Power Electronics10 Example 8.9 A class E inverter operates at resonance and has V S = 12 Volts and R = 10 Ω. The switching frequency is 25 kHz. –Determine the optimum values of L, C, C e, and L e Use MultiSim to plot the output voltage v 0 and the switch voltage v T for k = 0.304. Assume that Q = 7.

11 ECE 442 Power Electronics11 Optimum Parameters

12 ECE 442 Power Electronics12 Example 8.9 (continued)

13 ECE 442 Power Electronics13

14 ECE 442 Power Electronics14 Check the damping factor and resonant frequency

15 ECE 442 Power Electronics15 Example 8.9

16 ECE 442 Power Electronics16 Load Voltage Switch Voltage

17 ECE 442 Power Electronics17 Class E Resonant Rectifier for power factor correction very large

18 ECE 442 Power Electronics18 Mode 1 Operation -- D1 OFF

19 ECE 442 Power Electronics19 Mode 2 Operation -- D1 ON

20 ECE 442 Power Electronics20 D1 switches OFF at 0 volts (0 voltage switching) When the current i L falls to 0, the diode turns OFF. –When i L falls below I o, C discharges via D 1 –At turn-off, i D =i L =0 and v D =v C =0. –The capacitor current, i C =C(dv C /dt)=0, or (dv C /dt) = 0.

21 ECE 442 Power Electronics21 Waveform Summary i L = i C + i D

22 ECE 442 Power Electronics22 Example 8.10 A Class E rectifier supplies a load power of P L =400mW at V o =4V. The peak supply voltage is V m =10V. The supply frequency is f=250kHz. The peak-to-peak ripple on the dc output voltage is ΔV o =40mV. Determine the values of L, C, and C f. Determine the rms and dc currents of L and C.

23 ECE 442 Power Electronics23 Example 8.10 (continued) Choose C=10nF. The resonant frequency will be 250kHz. Details on the following slide

24 ECE 442 Power Electronics24

25 ECE 442 Power Electronics25

26 ECE 442 Power Electronics26 Example 8.10 (MultiSim)

27 ECE 442 Power Electronics27

28 ECE 442 Power Electronics28 Load Current 500 mA p-p Load Voltage 50 mV p-p

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