An inverter converts DC input voltage into AC output voltage. There are various types of inverters including single-phase and three-phase inverters. Single-phase inverters include half-bridge and full-bridge configurations. Current source inverters directly control AC current instead of voltage. They use thyristors and commutating capacitors to generate quasi-square wave output current from a constant DC current source.

1. Power
Electronics
INVERTER

2. Introduction
Inverter is a device which convert a DC input supply
voltage into symmetric AC voltage of desired magnitude
and frequency at the output side. It is also know as DC-AC
converter.
Ideal and practical inverter have sinusoidal and no-
sinusoidal waveforms at output respectively.
If the input dc is a voltage source, the inverter is called a
Voltage Source Inverter (VSI). One can similarly think of a
Current Source Inverter (CSI), where the input to the circuit
is a current source. The VSI circuit has direct control over
‘output (ac) voltage’ whereas the CSI directly controls
‘output (ac) current.

3. Applications of inverter
For low and medium power applications, square-
wave or quasi-square wave voltages may be
acceptable
For high –power applications ,low distorted
sinusoidal waveforms are required.
Using high speed power semiconductor devices
the harmonic contents at output can be reduced
by PWM techniques.
Industrial applications:- variable ac motor,
induction heating , standby power supply ,UPS
(uninterrupted power supply)
Inputs are (battery, fuel cell, solar cell, other dc

4. Classifications of inverter
Inverter can be mainly classified into two types-
1) Single-phase inverter
2) Three-phase inverter
 Turn-ON and turn-OFF controlling devices are-
a. Bipolar junction transistor[BJTs]
b. Metal oxide semiconductor field-effect
transistor[MOSFETs]
c. Insulated-gate bipolar transistor[IGBTs]
d. Gate turn-OFF thyristor[GTOs]

5. Classifications of inverter (Cant’d)
 Voltage fed inverter[for constant input voltage]
 Current fed inverter[for constant input current]
 Variable DC-link inverter[for controllable input
voltage]
 Resonant-pulse inverter :- If the output voltage or
current of inverter is forced to pass through zero
by creating an LC resonant circuit.

6. Single phase inverter:-
Principle of operation:-The inverter
circuit consists of two choppers. when
only transistor Q1 is turned ON for a time
T0/2 then instantaneous voltage across
the load is (V0=Vs/2). If the transistor Q2
only is turned ON for a time T0/2,-VS/2
appears across the load. This type of
inverter is called Half bridge inverter.
24
2 2
0
S
T
SV (rmV s) output voltage

dt  

1
 2
 T0
Root-mean sq2 uare 


V0   
0

7. Single phase inverter:-
(Cant’d)

8. Single phase inverter (Cant’d)
Instantaneous output voltage in the form of Fourier
series is-
 Due to quarter wave symmetry along the X-axis, both
a0 and an are zero (or even Harmonic voltages are
 Output voltage-
 v0 =0 for n=2,4…..
0
0
2 n1
a
V 

an cosnt bn sinnt
n 2
absent ). then bn isb n

4 V S





n1,3,5,...
VS
V0   sinnt
n

9. Single phase inverter (Cant’d)
 For an inductive load ,the load current can not change
immediately with the output voltage. diode D1 and D2 are
known as feedback diodes .Transistor can be replaced by
any switching device.
 There must be a minimum delay time between the
outgoing device and triggering of the next incoming device,
otherwise there would occur a short-circuit between
devices.


 Maximum conduction time of a device would be 

on  td .
T0
2
t

10. Single phase inverter (Cant’d)
For an RL load instantaneous load current i0 is
Note-in most applications the output power due to the
fundamental current is generally the useful power ,and
the power due to harmonic current is dissipated as heat
and increase the load temperature.

0 ni  sinnt  2VS
R2
 nL2
n1,3,5.... n

11.  EXAMPALE -1. A single phase halfbridge inverter has a
resistive load of R=2.4 and the DC input voltage is
48V,Determine-
A. Rms value of output voltage at the fundamental frequency V01
B. output power
C. Average and peak current of each transisto
D. The peak reverse blocking voltage VBR.
E. The THD
F. The DF and
G.The HF and DF of the LOH.
Solution:- Vs=48V, R=2.4 .
A. V01 =
B.
 0.4548  21.6V
2
0

VS
V
2

48
 24V
S
2VS
 0.45V
 2

12. The output power is
C. The peak transistor current is Ip=24/2.4=10A. because each
transistor conducts for a 50% duty cycle ,the average
current of each transistor is IQ=0.5*10=5A.
D. The peak reverse blocking voltage VBR=2*24=48V.
E. V01 =
So THD =
F.
G. The LOH is the third ,V03 = V01/3
HF3=V03/V01 = 1/3=33.33% and
DF3=(V03/32)V01 =1/27=3.704%.
Because V03/V01 =33.33% which is greater than 3%,LOH=V03.
R
S
0
V 2
V  2 .4
24 2
 2 4 0 WV 0 
0.45VS
Sh
1
 v2
0  v2
01  0.2176V

 2

And the RMS harmonic voltage V   V 2
on 

n3,5,7...
 48.34%
S
S
0.45V 
h

V01
V 0.2176V 
5 73
1
22
2 2
2
2
03
2


    
    



1
2
 2

 

SS 01 n2,3...   0.45V
......... 
0.024VS
5.382%
   V05   V07 V
0.45Vn
 Von 

v
DF 
1


13. Single phase Bridge inverter
It consists of four choppers. Full bridge converter is also
basic circuit to convert dc to ac. An ac output issynthesized
from a dc input by closing and opening switches in an
appropriate sequence. There are also four different states
depending on which switches are closed.
When transistors Q1 and Q2 are turned on simultaneously,
the input voltage appears across the load. If transistor and
are turned on at the same time, the voltage across the is
reversed and is –Vs.
Transistor Q1 and Q2 acts as switches S1 and S2. ,
respectively.

14. Single phase Bridge inverter(cont’d)
The fundamental RMS output voltage obtained
from
= 0 for
n=2,4….
0For an RL load instantaneous load current i is

 220
n1,3,5....
n
S
i 
4V
sinnt  
n R nL
S
4VS
 0.90VV01(rms) 
 2
SRMS output voltagVe is 
 0
V 2
S d   V0 ( r m s )
2 



 
4VS
V0   sinnt
Fourier series of output voltage n1,3,5,... n

15. Single phase Bridge inverter(cont’d)

16. Single phase Bridge inverter(cont’d)
The output load voltage alternates between +Vs when Q1
and Q2 are on and -Vs when Q3 and Q4 are on, irrespective
of the direction of current flow. It is assumed that the load
current does not become discontinuous at any time. In the
following analysis we assume that the load current does not
become discontinuous at any time, same as for the half-
bridge circuit.
Bridge inverters are preferred over other arrangements in
higher power ratings.
With the same dc input voltage, output voltage is twice that
of the half-bridge inverter.

17. Current source inverter (CSI)
In the CSI, the current is nearly constant. The
voltage changes here, as the load is changed. In
an Induction motor, the developed torque
changes with the change in the load torque, the
speed being constant, with no
acceleration/deceleration. The input current in the
motor also changes, with the input voltage being
constant. So, the CSI, where current, but not the
voltage, is the main point of interest, is used to
drive such motors, with the load torque changing

18. Single-phase Current Source Inverter
The type of operation is termed as Auto-Sequential Commutated
Inverter (ASCI).
A constant current source is assumed here, which may be
realized by using an inductance of suitable value, which must be
high, in series with the current limited dc voltage source.
The thyristor pairs, Th1 & Th3, and Th2 & Th4, are alternatively
turned ON to obtain a nearly square wave current waveform.
Two commutating capacitors − C1 in the upper half, and C2 in
the lower half, are used. Four diodes, D1–D4 are connected in
series with each thyristor to prevent the commutating capacitors
from discharging into the load.
The output frequency of the inverter is controlled in the usual
way, i.e., by varying the half time period, (T/2).

19. Single-phase CSI
Circuit diagram

20. Single-phase CSI
Wave-form

21. Operation of Single-phase CSI
There are Two mode of operation:-
Mode1:-The Starting from the instant, t = 0− , the thyristor pair,
Th2 & Th4, is conducting (ON), and the current (I) flows through
the path, Th2, D2, load (L), D4, Th4, and source, I. The
commutating capacitors are initially charged equally with the
polarity as given, i.e., vC1 = vC 2 = −VC 0 .
At time, t = 0, thyristor pair, Th1 & Th3, is triggered by pulses at
the gates. The conducting thyristor pair, Th2 & Th4, is turned
OFF by application of reverse capacitor voltages.
The voltage, vD1
is obtained by going through the closed path,
abcda
as vD1 + Vco − (1 /(C / 2))⋅ ∫ I ⋅ dt = 0
It may be noted the voltage across load inductance, L is zero.

22. Operation of Single-phase CSI(cont’d)
The value of vC is
v = vC1 C 2= vC= −Vco + (2 / C) ⋅ ∫ I ⋅ dt , which, if
computed at t = t1 ,
vC1
= vC2= vC (t1 ) = −Vco + (( 2 ⋅ I ⋅ t1 ) / C) = −Vco +
((2 ⋅ I ) / C)⋅ (C /(2 ⋅ I ))⋅VC 0=0

23. Operation of Single-phase CSI(cont’d)
Mode2:- Diodes, D2 & D4, are already conducting, but
at t = t1 , diodes, D1 & D3, get forward biased, and start
conducting. Thus, at the end of time t1, all four diodes,
D1–D4 conduct. As a result, the commutating
capacitors now get connected in parallel with the load
(L).
natural frequency:-
f0= 1/((2 ⋅π ) ⋅(L ⋅ C) ), ω0 = (2 ⋅π ) ⋅ f0= 1/(L ⋅ C)
time period:-
T = 1/ f0 = (2 ⋅π ) /ω0 = (2 ⋅π ) ⋅ (L ⋅ C)

24. Operation of Single-phase CSI(cont’d)
The voltage across capacitor is
vC = vL = L ⋅ (di0 / dt) = (( 2 ⋅ I ) ⋅ (ω0 ⋅ L))⋅ sin (ω0 ⋅ t)
The total commutation interval is,
tc = t1 + t2 = (1+ (π / 2))/ω0 = (1 + (π / 2))⋅ (L ⋅ C) .
The procedure remains nearly same, if the load
consists of resistance, R only. The procedure in
mode I, is same, but in mode II, the load
resistance, R is connected in parallel with the two
commutating capacitors. The direction of the
current, I remains same, a part of which flows in
the two capacitors, charging them in the reverse
direction