ECE 442 Power Electronics 1
Class E Resonant Inverter
ECE 442 Power Electronics 2
Mode 1 Operation – Turn Q1 ON at t = 0 Turn Q1 OFF when vo = 0 volts
ECE 442 Power Electronics 3
The current through the transistor (switch)
T s oi i i
7
0.072
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.
ECE 442 Power Electronics 4
Mode 1
ECE 442 Power Electronics 5
ECE 442 Power Electronics 6
Mode 2 Operation
• Q1 is turned OFF
• Diode D limits negative switch voltage
ECE 442 Power Electronics 7
Capacitor current becomes
C S o
TC e
i i i
dvi C
dt
When the switch current falls to zero,
ECE 442 Power Electronics 8
Mode 2
ECE 442 Power Electronics 9
Waveform Summary
ECE 442 Power Electronics 10
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 11
0.4001
2.165
10.3533
es
es
ss
RL
CR
L RC
Optimum Parameters
ECE 442 Power Electronics 12
Example 8.9 (continued)
3
3
3
0.4001 (0.4001)(10)25.47
2 (25 10 )
2.165 2.1651.38
10(2 )(25 10 )
(7)(10)445.63
2 (25 10 )
es
es
s
RL H
C FR
QRL H
ECE 442 Power Electronics 13
10.3533
10.3533
10.0958
( 0.3533 )
ss
ss s
s
L RC
QRR
C
C FQR R
ECE 442 Power Electronics 14
Check the damping factor and resonant frequency
0
6
6
0 6 6
0
21 2
10 0.0958 10
2 445.63 100.0733
1 1
2 2 (445.63 10 )(0.0958 10 )
24.39
RR CL
LLC
fLC
f kHz
ECE 442 Power Electronics 15
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
ECE 442 Power Electronics 16
Load Voltage
Switch Voltage
ECE 442 Power Electronics 17
Class E Resonant Rectifier
sins mv V t
oV constantfor power factor correction
very large
ECE 442 Power Electronics 18
Mode 1 Operation -- D1 OFF
sinLC s ov V t V
ECE 442 Power Electronics 19
Mode 2 Operation -- D1 ON
sin
sin( )L s o
L m o
v V t V
i I t I
ECE 442 Power Electronics 20
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 21
Waveform Summary
iL = iC + iD
ECE 442 Power Electronics 22
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 23
Example 8.10 (continued)
• Choose C=10nF.• The resonant frequency will be 250kHz.• Details on the following slide
ECE 442 Power Electronics 24
0
2 2 2 3 2 90
20
2 20
3
00
00
30
3 30
1
21 1
40.64 4 (250 10 ) (10 10 )
440
400 10
4100
40
2
100 105
2 2(250 10 )(40 10 )
L
L
f
f
fLC
L Hf C
VP
R
VR
P
VI mA
R
IV
fC
IC F
f V
ECE 442 Power Electronics 25
2 2 21 2
00
22
( )
( )
( )
( )
...
4100
4010
25040
250100 203.1
2100
250176.78
20
rms dc rms rms
dc
mm
L rms
L dc
C rms
C dc
I I I I
VI I mA
RV
I mAR
I mA
I mA
I mA
I
ECE 442 Power Electronics 26
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
+
-
ECE 442 Power Electronics 27
ECE 442 Power Electronics 28
Load Current
500 mAp-p
Load Voltage
50 mVp-p