Power electronics...electrical engineering Inverters.pdf

Power electronics

1. Power Electronics
Inverters
1
Dr. Firas Obeidat

2. 2
Table of contents
1
•Single Phase Half Bridge Inverter – Resistive Load
2
•Single Phase Half Bridge Inverter – RL Load
5
•Single Phase Full Bridge Inverter
Dr. Firas Obeidat Faculty of Engineering Philadelphia University

3. 3
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Half Bridge Inverter – Resistive Load
Basic Operation
Consists of 2 choppers, 3-wire DC source.
Transistors switched ON and OFF alternately.
Each provides opposite polarity of V
s/2 across
the load.
When T
1 is ON through the period 0<t<T/2,
the output voltage equal to V
s/2.
When T
2 is ON through the period T/2<t<T,
the output voltage equal to -V
s/2. 2
s
V S
V R
Vo
C1
C2
io i1
i2
T1
D1
T2
D22
s
V
ao 2
s
V S
V R
Vo
C1
C2
io i1
i2
T1
D1
T2
D22
s
V
ao
T
1 ON, T
2 OFF, V
o = V
s/2 T
1 OFF, T
2 ON, V
o = -V
s/2 2
s
V S
V R
Vo
C1
C2
io i1
i2
T1
D1
T2
D22
s
V
ao

4. 4
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Half Bridge Inverter – Resistive Load
The frequency can be changed by controlling the conduction time of the
transistors.
The rms value for the output
voltage can be found as
??????
&#3627408476;,&#3627408479;&#3627408474;&#3627408480;=
2
??????

??????
&#3627408480;
2
2
&#3627408465;??????
??????2
0
=
??????
&#3627408480;
2

When T
1 is ON through the
period 0<t<T/2, the output
current equal to V
s/2R.
When T
2 is ON through the
period T/2<t<T, the output
current equal to -V
s/2R.
The output voltage frequency is
&#3627408467;
&#3627408476;=
1
??????
2
s
V 2
s
V
 R
V
s
2 R
V
s
2 t t t Vo
i2
i1R
V
s
2 t
ioR
V
s
2
 2
T T 2
T T 2
T T 2
T T

5. 5
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Half Bridge Inverter – RL Load
At t=0, the control signal is removed from T
2
and a control signal is applied to T
1. At this
moment the current is negative maximum.
The current can’t change the direction
directly to positive value due to the inductive
load. In this case, the current will flow from
the load through D
1 to the source and T
1 stay
disconnected in spite of existence the control
signal on it due to reverse biased. At t=t
1 the
current become zero and T
1 start to be
forward biased.
The operation of single phase half bridge
inverter with RL load can be divided into
four periods 2
s
V S
V R
Vo
C1
C2
io i1
i2
T1
D1
T2
D22
s
V
aoL
0<t<t
1 2
s
V R
Vo
C1
io i1
T1
D1
aoL

6. 6
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Half Bridge Inverter – RL Load
At t=t
1, the current change its direction to be
positive and T
1 start to conduct. The positive
current is increased until reach its positive
maximum value at t=T/2. At t=T/2 the control
signal is removed from T
1 and applied to T
2.
t
1<t<T/2
2
s
V R
C1
io i1
T1
D1
aoL
T/2<t<t
2 Vo
C2
i2
T2
D22
s
V
ao
At this moment the current is positive
maximum. The current can’t change the
direction directly to negative value due to the
inductive load. In this case, the current will
flow from the load through D
2 to the source
and T
2 stay disconnected in spite of existence
the control signal on it due to reverse biased.
At t=t
2 the current become zero and T
2 start
to be forward biased. At this period the
voltage become negative and the current
positive.

7. 7
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Half Bridge Inverter – RL Load
At t=t
2, the current change its direction to be
negative and T
2 start to conduct. The negative
current is increased until reach its negative
maximum value at t=T. At t=T the control
signal is removed from T
2 and applied to T
1.
t
2<t<T
Vo
C2
i2
T2
D22
s
V
ao
The rms value for the output
voltage can be found as
??????
&#3627408476;,&#3627408479;&#3627408474;&#3627408480;=
2
??????

??????
&#3627408480;
2
2
&#3627408465;??????
??????2
0
=
??????
&#3627408480;
2
2
s
V 2
s
V
 2
T T t t Vo
io
t1 t2
D1
ON
T1
ON
D2
ON
T2
ON
D1
ON

8. 8
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Half Bridge Inverter – RL Load
0<t<T/2

The output current can be found as
??????
&#3627408480;
2
=????????????
&#3627408476;??????+??????
&#3627408465;??????
&#3627408476;??????
&#3627408465;??????

??????
&#3627408476;??????=
??????
&#3627408480;
2??????
1−&#3627408466;
−
??????
??????
&#3627408481;
+??????
&#3627408476;&#3627408466;
−
??????
??????
&#3627408481;

At t=T/2, i
o(t)=I
o
??????
&#3627408476;=??????
&#3627408476;
??????
2
=
??????
&#3627408480;
2??????
1−&#3627408466;
−
??????
2??????
??????
−??????
&#3627408476;&#3627408466;
−
??????
2??????
??????
=
??????
&#3627408480;
2??????
1−&#3627408466;
−
??????
2??????
??????
1+&#3627408466;
−
??????
2??????
??????

Substitute the value of I
o in i
o(t) equation

??????
&#3627408476;??????=
??????
&#3627408480;
2??????
1−&#3627408466;
−
??????
??????
&#3627408481;
−
??????
&#3627408480;
2??????
1−&#3627408466;
−
??????
2??????
??????
1+&#3627408466;
−
??????
2??????
??????
&#3627408466;
−
??????
??????
&#3627408481;
=
??????
&#3627408480;
2??????
1−
2&#3627408466;
−
??????
??????
&#3627408481;
1+&#3627408466;
−
??????
2??????
??????

9. 9
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Half Bridge Inverter – RL Load
T/2<t<T

−
??????
&#3627408480;
2
=????????????
&#3627408476;??????
′
+??????
&#3627408465;??????
&#3627408476;??????
′
&#3627408465;??????

??????
&#3627408476;??????
′
=−
??????
&#3627408480;
2??????
1−&#3627408466;
−
??????
??????
&#3627408481;
′
+??????
&#3627408476;&#3627408466;
−
??????
??????
(&#3627408481;
′
)

At t=T/2, i
o(t)=I
o
??????
&#3627408476;=??????
&#3627408476;
??????
2
=−
??????
&#3627408480;
2??????
1−&#3627408466;
−
??????
2??????
??????
1+&#3627408466;
−
??????
2??????
??????

Substitute the value of I
o in i
o(t) equation

??????
&#3627408476;??????
′
=−
??????
&#3627408480;
2??????
1−
2&#3627408466;
−
??????
??????
(&#3627408481;
′
)
1+&#3627408466;
−
??????
2??????
??????

??????
′
=??????−
??????
2

10. 10
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Full Bridge Inverter
The output voltage V
o in single phase full
bridge inverter can be V
dc, -V
dc, or zero,
depending on which switches are closed.
S
V Load
Vo
io
T3
D3
T2
D2
ba
T1
T4
D1
D4
i3
i2
i1
i4
is
Switched Closed Output Voltage V
o
T
1 and T
2 +V
dc
T
3 and T
4 -V
dc
T
1 and T
3 0
T
2 and T
4 0 S
V Load
Vo=Vs
ioba
is
T1
T2 S
V Load
Vo=0
ba
is
T1 T3 S
V Load
ba
is
T2T4
Vo=0 S
V Loadioba
is
Vo=-Vs
T4
T3
T
1 and T
2 Closed

T
3 and T
4 Closed
T
1 and T
3 Closed
T
2 and T
4 Closed

11. 11
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Full Bridge Inverter S
V Load
Vo
io
T3
D3
T2
D2
ba
T1
T4
D1
D4
i3
i2
i1
i4
is
The switches connect the load to +V
dc when
T
1 and T
2 are closed or to -V
dc when T
3 and
T
4 are closed. The periodic switching of the
load voltage between +V
dc and -V
dc
produces a square wave voltage across the
load. Although this alternating output is
nonsinusoidal, it may be an adequate ac
waveform for some applications.

T
1 and T
4 should not be closed at the same
time, nor should T
2 and T
3. Otherwise, a
short circuit would exist across the dc source

The current waveform in the load
depends on the load components.
For the resistive load, the current
waveform matches the shape of the
output voltage.

The waveforms when resistive load 2
s
V 2
s
V
 R
V
s
2 R
V
s
2 t t t Vo
i2
i1R
V
s
2 t
ioR
V
s
2
 2
T T 2
T T 2
T T 2
T T

12. 12
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Full Bridge Inverter
An inductive load will have a
current that has more of a
sinusoidal quality than the voltage
because of the filtering property of
the inductance.

The waveforms when RL load
Switches T
1 and T
2 close at t=0. The
voltage across the load is +V
s, and
current begins to increase in the load
and in T
1 and T
2. The current is
expressed as the sum of the forced and
natural responses.
where A is a constant evaluated from the initial condition and τ=L/R. at t=0,
i(0)=I
min.
→

13. 13
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Full Bridge Inverter
In steady state, the current waveforms for RL load can be described by
At t=T/2, T
1 and T
2 open, and T
3 and T4 close. The voltage across the RL load
becomes -V
s, and the current has the form
where B is a constant evaluated from the initial condition and τ=L/R. at
t=T/2, i(T/2)=I
max.
→
??????
&#3627408476;??????=
??????
&#3627408465;&#3627408464;
??????
+??????
&#3627408474;??????&#3627408475;−
??????
&#3627408465;&#3627408464;
??????
&#3627408466;
−&#3627408481;??????
0<??????<
??????
2
−??????
&#3627408465;&#3627408464;
??????
+??????
&#3627408474;????????????+
??????
&#3627408465;&#3627408464;
??????
&#3627408466;
−(&#3627408481;−??????2 )??????
??????
2
<??????<??????

14. 14
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Full Bridge Inverter
An expression is obtained for I
max by evaluating the first part of i
o(t)
equation at t=T/2.
And by symmetry
Substituting –I
max for I
min in i(T/2) equation yields
The power absorbed by the load can be determined from (P
ac=I
rms
2
R)
The power supplied by the source must be the same as absorbed by the load.
Power from a dc source is determined from (P
dc=V
dcI
s)
The rms load current is determined by

15. 15
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Full Bridge Inverter
Example: The full-bridge inverter has a switching sequence that produces a
square wave voltage across a series RL load. The switching frequency is 60
Hz, V
s=100 V, R=10 Ω, and L=25 mH. Determine (a) an expression for load
current, (b) the power absorbed by the load, and (c) the average current in
the dc source.
(a)

16. 16
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Single Phase Full Bridge Inverter
Example: The full-bridge inverter has a switching sequence that produces a
square wave voltage across a series RL load. The switching frequency is 60
Hz, V
s=100 V, R=10 Ω, and L=25 mH. Determine (a) an expression for load
current, (b) the power absorbed by the load, and (c) the average current in
the dc source.
(b)
(c)
The power absorbed by the load is
Average source current can also be computed by equating source and load power

17. 17