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ECE 442 Power Electronics 1

Class E Resonant Inverter


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


The current through the transistor (switch)

T s oi i i

70.072

Q

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.


Mode 1


Mode 2 Operation

• Q1 is turned OFF• Diode D limits negative switch voltage


Capacitor current becomes

C S o

TC e

i i i

dvi Cdt

When the switch current falls to zero,


Mode 2


Waveform Summary


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.


0.4001

2.165

1 0.3533

es

es

ss

RL

CR

L RC

Optimum Parameters


Example 8.9 (continued)

3

3

3

0.4001 (0.4001)(10) 25.472 (25 10 )

2.165 2.165 1.3810(2 )(25 10 )(7)(10) 445.63

2 (25 10 )

es

es

s

RL H

C FRQRL H


1 0.3533

1 0.3533

1 0.0958( 0.3533 )

ss

ss s

s

L RC

QR RC

C FQR R


Check the damping factor and resonant frequency

0

6

6

0 6 6

0

21 2

10 0.0958 102 445.63 100.0733

1 12 2 (445.63 10 )(0.0958 10 )24.39

RR CL

LLC

fLC

f kHz


Example 8.9

Q1

BJT_NPN_VIRTUAL*

Le25.47uH

Vs12 V

C

0.0958uF

L445.63uH

R10 Ohm

Ce1.38uF

Rb

500 Ohm

XFG1

XSC1

A B

G

T


Load Voltage

Switch Voltage


Class E Resonant Rectifier

sins mv V t

oV constantfor power factor correction

very large


Mode 1 Operation -- D1 OFF

sinLC s ov V t V


Mode 2 Operation -- D1 ON

sinsin( )

L s o

L m o

v V t Vi I t I


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.


Waveform Summary

iL = iC + iD


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.


Example 8.10 (continued)

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


0

2 2 2 3 2 90

20

2 20

3

00

00

30

3 30

121 1 40.6

4 4 (250 10 ) (10 10 )

4 40400 10

4 10040

2

100 10 52 2(250 10 )(40 10 )

L

L

f

f

fLC

L Hf C

VP

RVRP

VI mAR

IVfC

IC F

f V


2 2 21 2

00

22

( )

( )

( )

( )

...

4 10040

10 25040

250100 203.12

100

250 176.782

0

rms dc rms rms

dc

mm

L rms

L dc

C rms

C dc

I I I I

VI I mAR

VI mAR

I mA

I mA

I mA

I


Example 8.10 (MultiSim)

Vs

7.07 V 250kHz 0Deg

L40.5uH D1

DIODE_VIRTUAL

C

10nF

Cf5uF

R40 Ohm

XSC1

A B

G

T

Rsample

1 Ohm

XSC2

A B

G

T

U1DC 1MOhm 3.812 V

+

-


Load Current

500 mAp-p

Load Voltage

50 mVp-p

class e resonant inverter - welcome - faculty pages - …€¦ · ppt file · web view ·...

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