Power Electronics
Inverters
Dr. Firas Obeidat
1
Table of contents
1
2
5
• Single Phase Half Bridge Inverter – Resistive Load
• Single Phase Half Bridge Inverter – RL Load
• Single Phase Full Bridge Inverter
2
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 Vs/2 across
the load.
 When T1 is ON through the period 0<t<T/2,
the output voltage equal to Vs/2.
 When T2 is ON through the period T/2<t<T,
the output voltage equal to -Vs/2.
Vs
2
VS
Vs
2
C1
R
o
D1
io a
Vo
C2
T1
i1
i2
D2
T2
T1 ON, T2 OFF, Vo = Vs/2
Vs
2
C1
VS
Vo
Vs
2
Vs
2
VS
Vs
2
R
o
D1
io a
R
o
T2
D1
io a
Vo
C2
i1
i2
D2
C2
C1
T1
T1
i1
i2
D2
T2
T1 OFF, T2 ON, Vo = -Vs/2
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Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Single Phase Half Bridge Inverter – Resistive Load
The rms value for the output
voltage can be found as
𝑉𝑜,𝑟𝑚𝑠 =
2
𝑇
𝑇 2
0
𝑉𝑠
2
2
𝑑𝑡 =
𝑉𝑠
2
Vs
2
T
T

Vs
2
Vs
2R
 When T1 is ON through the
period 0<t<T/2, the output
current equal to Vs/2R.
 When T2 is ON through the
period T/2<t<T, the output
current equal to -Vs/2R.
The output voltage frequency is
1
𝑓𝑜 =
𝑇
Vo
t
2
i1
T
2
Vs
2R
T
i2
T
T
t
t
2
Vs
2R

Vs
2R
io
T
T
t
2
The frequency can be changed by controlling the conduction time of the
transistors.
4
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Single Phase Half Bridge Inverter – RL Load
The operation of single phase half bridge
inverter with RL load can be divided into
four periods
Vs
2
VS
D1
R io a
C1
o L
Vs
2
T1
i1
i2
Vo
C2
D2
T2
0<t<t1
At t=0, the control signal is removed from T2
and a control signal is applied to T1. At this
moment the current is negative maximum.
The current can’t change the direction
V
directly to positive value due to the inductive s
2
load. In this case, the current will flow from
the load through D1 to the source and T1 stay
disconnected in spite of existence the control
signal on it due to reverse biased. At t=t1 the
current become zero and T1 start to be
forward biased.
D1
R io a
C1
o L
T1
i1
Vo
5
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Single Phase Half Bridge Inverter – RL Load
t1<t<T/2
At t=t1, the current change its direction to be
positive and T1 start to conduct. The positive V s
current is increased until reach its positive 2
maximum value at t=T/2. At t=T/2 the control
signal is removed from T1 and applied to T2.
D1
R io a
C1
o L
T1
i1
T/2<t<t2
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 D2 to the source
and T2 stay disconnected in spite of existence V s
the control signal on it due to reverse biased. 2
At t=t2 the current become zero and T2 start
to be forward biased. At this period the
voltage become negative and the current
positive.
o
a
i2
Vo
C2
D2
T2
6
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Single Phase Half Bridge Inverter – RL Load
t2<t<T
o
At t=t2, the current change its direction to be
negative and T2 start to conduct. The negative
current is increased until reach its negative V s
maximum value at t=T. At t=T the control 2
signal is removed from T2 and applied to T1.
Vs
2
The rms value for the output
voltage can be found as
𝑉𝑜,𝑟𝑚𝑠 =
2
𝑇
𝑇 2
0
𝑉𝑠
2
2
a
i2
Vo
D2
C2
T2
Vo
t1
V
 s
2
t2
T
t
T
2
io
𝑉𝑠
𝑑𝑡 =
2
t
D1
ON
T1
ON
D2
ON
T2
ON
D1
ON
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Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Single Phase Half Bridge Inverter – RL Load
The output current can be found as
0<t<T/2
𝑉𝑠
𝑑𝑖𝑜 𝑡
= 𝑅𝑖𝑜 𝑡 + 𝐿
2
𝑑𝑡
𝑅
𝑅
𝑉𝑠
−𝐿 𝑡
−𝐿 𝑡
𝑖𝑜 𝑡 =
1−𝑒
+ 𝐼𝑜 𝑒
2𝑅
At t=T/2, io(t)=Io
𝑇
𝑉𝑠
=
1−
2
2𝑅
𝑅
− 𝑇
𝑒 2𝐿
𝑉𝑠 1 −
𝐼𝑜 = 𝑖𝑜
−
=
𝑅
2𝑅
−2𝐿𝑇
1+𝑒
Substitute the value of Io in io(t) equation
𝑅
−2𝐿𝑇
𝑒
𝑅
−2𝐿𝑇
𝐼𝑜 𝑒
𝑅
−2𝐿𝑇
𝑒
𝑅
−𝐿 𝑡
2𝑒
𝑅
𝑅
𝑉𝑠
𝑉𝑠 1 −
𝑉𝑠
−𝐿 𝑡
−𝐿 𝑡
𝑖𝑜 𝑡 =
1−𝑒
−
𝑒
=
1−
𝑅
𝑅
2𝑅
2𝑅
2𝑅
−2𝐿𝑇
−2𝐿𝑇
1+𝑒
1+𝑒
8
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Single Phase Half Bridge Inverter – RL Load
T/2<t<T
𝑡′
𝑇
=𝑡−
2
′
𝑉𝑠
𝑑𝑖
𝑡
𝑜
− = 𝑅𝑖𝑜 𝑡 ′ + 𝐿
2
𝑑𝑡
𝑖𝑜
𝑡′
𝑅
𝑉𝑠
−𝐿
=−
1−𝑒
2𝑅
𝑡′
+
𝑅
− 𝐿 (𝑡 ′ )
𝐼𝑜 𝑒
At t=T/2, io(t)=Io
𝐼𝑜 = 𝑖𝑜
𝑅
−2𝐿𝑇
𝑒
𝑇
𝑉𝑠 1 −
=−
𝑅
2
2𝑅
−2𝐿𝑇
1+𝑒
Substitute the value of Io in io(t) equation
𝑖𝑜 𝑡 ′ = −
𝑉𝑠
1−
2𝑅
𝑅
− 𝐿 (𝑡 ′ )
2𝑒
1+
𝑅
−2𝐿𝑇
𝑒
9
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Single Phase Full Bridge Inverter
is
The output voltage Vo in single phase full
bridge inverter can be Vdc, -Vdc, or zero,
depending on which switches are closed.
i3
i1
D1
T1
Load
a
VS
Vo
T4
D4
Switched Closed
Output Voltage Vo
T1 and T2
+Vdc
T3 and T4
-Vdc
T1 and T3
0
T2 and T4
0
is
is
VS
T3
a
Load
io
Vo=Vs
b
a
VS
T4
T2
T1 and T2 Closed
Load
io
b
i2
Vo=-Vs
is
T1
VS
T3
a
Load
b
a
VS
Load
T1 and T3 Closed
b
Vo=0
Vo=0
T4
T3 and T4 Closed
T3
T2
D2
i4
is
T1
D3
io b
T2
T2 and T4 Closed
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Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Single Phase Full Bridge Inverter
T1 and T4 should not be closed at the same
time, nor should T2 and T3. Otherwise, a
short circuit would exist across the dc source
The switches connect the load to +Vdc when
T1 and T2 are closed or to -Vdc when T3 and
T4 are closed. The periodic switching of the
load voltage between +Vdc
and -Vdc
produces a square wave voltage across the
load. Although this alternating output is
nonsinusoidal, it may be an adequate ac
waveform for some applications.
is
i3
i1
D1
T1
a
VS
Load
D4
i2
Vs
2
Vo
T
T
V
 s
2
Vs
2R
Philadelphia University
t
2
i1
2
Vs
2R
T
i2
T
T
t
t
2
Vs
2R
io
T
T
t
2
The waveforms when resistive load
Faculty of Engineering
T2
D2
i4
V
 s
2R
Dr. Firas Obeidat
T3
Vo
T4
T
 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.
D3
io b
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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.
Switches T1 and T2 close at t=0. The
voltage across the load is +Vs, and
current begins to increase in the load
and in T1 and T2. The current is
expressed as the sum of the forced and
natural responses.
The waveforms when RL load
where A is a constant evaluated from the initial condition and τ=L/R. at t=0,
i(0)=Imin.
→
12
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Single Phase Full Bridge Inverter
At t=T/2, T1 and T2 open, and T3 and T4 close. The voltage across the RL load
becomes -Vs, 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)=Imax.
→
In steady state, the current waveforms for RL load can be described by
𝑉𝑑𝑐
𝑉𝑑𝑐 −𝑡 𝜏
+ 𝐼𝑚𝑖𝑛 −
𝑒
𝑅
𝑅
𝑖𝑜 𝑡 =
−𝑉𝑑𝑐
𝑉𝑑𝑐 −(𝑡−𝑇 2)
+ 𝐼𝑚𝑎𝑥 +
𝑒
𝑅
𝑅
0<𝑡<
𝜏
𝑇
2
𝑇
<𝑡<𝑇
2
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Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
Single Phase Full Bridge Inverter
An expression is obtained for Imax by evaluating the first part of io(t)
equation at t=T/2.
And by symmetry
Substituting –Imax for Imin in i(T/2) equation yields
The power absorbed by the load can be determined from (Pac=Irms2R)
The rms load current is determined by
The power supplied by the source must be the same as absorbed by the load.
Power from a dc source is determined from (Pdc=VdcIs)
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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, Vs=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)
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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, Vs=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)
The power absorbed by the load is
(c) Average source current can also be computed by equating source and load power
16
Dr. Firas Obeidat
Faculty of Engineering
Philadelphia University
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