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ECE 442 Power Electronics 1
Class E Resonant Inverter
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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
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.
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
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Capacitor current becomes
C S o
TC e
i i i
dvi Cdt
When the switch current falls to zero,
ECE 442 Power Electronics 8
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 11
0.4001
2.165
1 0.3533
es
es
ss
RL
CR
L RC
Optimum Parameters
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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
ECE 442 Power Electronics 13
1 0.3533
1 0.3533
1 0.0958( 0.3533 )
ss
ss s
s
L RC
QR RC
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 102 445.63 100.0733
1 12 2 (445.63 10 )(0.0958 10 )24.39
RR CL
LLC
fLC
f kHz
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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
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Load Voltage
Switch Voltage
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Class E Resonant Rectifier
sins mv V t
oV constantfor power factor correction
very large
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Mode 1 Operation -- D1 OFF
sinLC s ov V t V
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Mode 2 Operation -- D1 ON
sinsin( )
L s o
L m o
v V t Vi I t I
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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.
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Waveform Summary
iL = iC + iD
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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.
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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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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
ECE 442 Power Electronics 25
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
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
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