Transcript Single-phase half

Single-phase half-bridge inverter
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Operational Details
3-wire DC source
•
•
•
•
Consists of 2 choppers, 3-wire DC source
Transistors switched on and off alternately
Need to isolate the gate signal for Q1 (upper device)
Each provides opposite polarity of Vs/2 across the load
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Q1 on, Q2 off, vo = Vs/2
Peak Reverse Voltage of Q2 = Vs
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Q1 off, Q2 on, vo = -Vs/2
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Waveforms with resistive load
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Look at the output voltage
rms value of the output voltage, Vo

2
Vo  
 To

1
2

V
 Vs
0 4 dt   2


To
2
2
s
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Fourier Series of the instantaneous output
voltage
ao 
vo     an cos(nt )  bn sin(nt ) 
2 n 1
ao , an  0
0


Vs
1  Vs
bn   
sin(nt )d (t )   sin(nt )d (t ) 
   2
2
0

2Vs
bn 
 n  1,3,5,...
n

2Vs
vo  
sin(nt )
n 1,3,5,.. n
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rms value of the fundamental component

2Vs
vo  
sin nt
n 1,3,5,.. n
1 2Vs
Vo1 
2 
Vo1  0.45Vs
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When the load is highly inductive
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Turn off Q1 at t = To/2
Current falls to 0 via D2, L, Vs/2 lower
+
Vs/2
-
+
Vs/2
-
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Turn off Q2 at t = To
Current falls to 0 via D1, L, Vs/2 upper
+
Vs/2
-
+
Vs/2
-
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Load Current for a highly inductive load
Transistors are only switched on for a quarter-cycle, or 90
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Fourier Series of the output current for an
RL load

vo
vo
2Vs
io  
 
sin(nt   n )
2
2
Z R  jn L n 1,3,5,... n R  (n L)
n L
 n  tan (
)
R
1
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Fundamental Output Power
In most cases, the useful power
Po1  Vo1 I o1 cos 1  I R
2
o1

2Vs
Po1  
2
2
 2 R  ( L)
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
 R

14
DC Supply Current
• If the inverter is lossless, average power
absorbed by the load equals the average
power supplied by the dc source.
T
T
 v (t )i (t )dt   v (t )i (t )dt
s
0
s
o
o
0
• For an inductive load, the current is
approximately sinusoidal and the fundamental
component of the output voltage supplies the
power to the load. Also, the dc supply voltage
remains essentially at Vs.
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DC Supply Current (continued)
T
1
0 is (t )dt  Vs
T

2Vo1 sin(t ) 2 I o sin(t  1 )dt  I s
0
Vo1
Is 
I o cos(1 )
Vs
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Performance Parameters
• Harmonic factor of the nth harmonic (HFn)
Von
HFn 
Vo1
for n>1
Von = rms value of the nth harmonic component
V01 = rms value of the fundamental component
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Performance Parameters (continued)
• Total Harmonic Distortion (THD)
• Measures the “closeness” in shape between a
waveform and its fundamental component

1
2
THD 
(  Von )
Vo1 n2,3,...
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2
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Performance Parameters (continued)
• Distortion Factor (DF)
• Indicates the amount of HD that remains in a
particular waveform after the harmonics have
been subjected to second-order attenuation.
2


1
 Von  
DF 
   2  
Vo1  n  2,3,...  n  
Von
DFn 
for n>1
Vo1n 2
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Performance Parameters (continued)
• Lowest order harmonic (LOH)
• The harmonic component whose
frequency is closest to the fundamental,
and its amplitude is greater than or equal
to 3% of the amplitude of the fundamental
component.
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Single-phase full-bridge inverter
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Operational Details
•
•
•
•
Consists of 4 choppers and a 3-wire DC source
Q1-Q2 and Q3-Q4 switched on and off alternately
Need to isolate the gate signal for Q1 and Q3 (upper)
Each pair provide opposite polarity of Vsacross the load
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Q1-Q2 on, Q3-Q4 off, vo = Vs
+ Vs -
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Q3-Q4 on, Q1-Q2 off, vo = -Vs
- Vs +
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When the load is highly inductive
Turn Q1-Q2 off – Q3-Q4 off
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Turn Q3-Q4 off – Q1-Q2 off
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Load current for a highly inductive load
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Example 6.3 – MultiSim7
C1
1000uF
Q1
1V0V
XFG1
C2
1000uF
Q4
1V0V
D1
DIODE_VIRTUAL
R
L
10 Ohm
31.5mH
D3
DIODE_VIRTUAL
Q3
1V0V
D2
DIODE_VIRTUAL
Q2
1V0V
C
112uF
D4
DIODE_VIRTUAL
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Example 6.3 using the scope
XSC1
G
T
C1
1000uF
Vs
220 V
Q1
1V0V
XFG1
D1
DIODE_VIRTUAL
Rs
1 Ohm
C2
1000uF
Q4
1V0V
R
A
B
L
D3
DIODE_VIRTUAL
Q3
1V0V
D2
DIODE_VIRTUAL
Q2
1V0V
C
9 Ohm 31.5mH 112uF
D4
DIODE_VIRTUAL
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Fourier Analysis of load current
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