Class E Resonant Inverter 116v6n

Class E Resonant Inverter

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Mode 1 Operation – Turn Q1 ON at t = 0 Turn Q1 OFF when vo = 0 volts

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For sinusoidal current,

Q7   0.072 The current through the transistor (switch)

iT  is  io The switch is turned OFF when the output voltage becomes = 0, and the current is “transferred” to the branch containing the capacitor. ECE 442 Power Electronics

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Mode 1

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Mode 2 Operation • Q1 is turned OFF • Diode D limits negative switch voltage

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Capacitor current becomes

iC  iS  io When the switch current falls to zero,

dvT iC  Ce dt ECE 442 Power Electronics

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Mode 2

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Waveform Summary

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Example 8.9 • A class E inverter operates at resonance and has VS = 12 Volts and R = 10 Ω. • The switching frequency is 25 kHz. – Determine the optimum values of L, C, Ce, and Le

• Use MultiSim to plot the output voltage v0 and the switch voltage vT for k = 0.304. Assume that Q = 7. ECE 442 Power Electronics

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Optimum Parameters Le  0.4001

R

s

2.165 Ce  Rs 1 s L   0.3533R s C ECE 442 Power Electronics

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Example 8.9 (continued) Le 

0.4001R

s

(0.4001)(10)   25.47  H 3 2 (25 10 )

2.165 2.165 Ce    1.38 F 3 Rs 10(2 )(25 10 ) QR

(7)(10) L   445.63 H 3 s 2 (25 10 )

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1 s L   0.3533R s C  QR  1 s   0.3533R   s  s C 1 C  0.0958 F s (QR  0.3533R)

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Check the damping factor and resonant frequency R  R C 2 L    1 0 2 L LC 10 0.0958 106  2 445.63 106   0.0733 f0 

1 2 LC



1 2 (445.63 106 )(0.0958 106 )

f 0  24.39kHz ECE 442 Power Electronics

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Example 8.9 XSC1 G T

A

Le 25.47uH

C

XFG1

L 445.63uH

0.0958uF

Q1 Vs 12 V

B

Ce 1.38uF

Rb

R 10 Ohm

500 Ohm BJT_NPN_VIRTUAL*

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Switch Voltage

Load Voltage

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Class E Resonant Rectifier

vs  Vm sin t

very large

for power factor correction ECE 442 Power Electronics

Vo  constant 17

Mode 1 Operation -- D1 OFF

vLC  Vs sin t  Vo

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Mode 2 Operation -- D1 ON

vL  Vs sin t  Vo iL  I m sin(t   )  I o ECE 442 Power Electronics

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D1 switches OFF at 0 volts (0 voltage switching)

• When the current iL falls to 0, the diode turns OFF. – When iL falls below Io, C discharges via D1 – At turn-off, iD=iL=0 and vD=vC=0. – The capacitor current, iC=C(dvC/dt)=0, or (dvC/dt) = 0. ECE 442 Power Electronics

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Waveform Summary

iL = i C + i D

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Example 8.10 • A Class E rectifier supplies a load power of PL=400mW at Vo=4V. The peak supply voltage is Vm=10V. The supply frequency is f=250kHz. The peak-to-peak ripple on the dc output voltage is ΔVo=40mV. • Determine the values of L, C, and Cf. • Determine the rms and dc currents of L and C. ECE 442 Power Electronics

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Example 8.10 (continued)

• Choose C=10nF. • The resonant frequency will be 250kHz. • Details on the following slide

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1

f0 

2 LC 1 1 L 2 2  2  40.6  H 3 2 9 4 f 0 C 4 (250 10 ) (10 10 ) V02 PL  R V02 42 R   40 3 PL 400 10 I0 

V0 4   100mA R 40

V0 

I0 2 fC f

I0 100 103 Cf    5 F 3 3 2 f V0 2(250 10 )(40 10 ) ECE 442 Power Electronics

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2 2 I rms  I dc2  I rms  I 1 rms 2  ...

V0 4 I dc  I 0    100mA R 40 Vm 10 Im    250mA R 40 2 250 I L ( rms )  1002   203.1mA 2 I L ( dc )  100mA

250 I C ( rms )   176.78mA 2 I C ( dc )  0 ECE 442 Power Electronics

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Example 8.10 (MultiSim) XSC1

A

B

XSC2 G

G

T

T

C

A

B

10nF L 40.5uH

Rsample 1 Ohm

D1 DIODE_VIRTUAL

Vs 7.07 V 250kHz 0Deg

Cf 5uF

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R 40 Ohm

+

3.812

U1 V DC 1MOhm

-

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Load Current 500 mAp-p

Load Voltage 50 mVp-p

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