Everything You Need to Know About Inverters

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INVERTERS
11
11.1 INTRODUCTION
An inverter is a converter circuit which is us~d to conv~rt de pow~r into ac power at
desired output voltage and frequency. The block diagram representation of inv~rtc;;r is
shown in Fig. 11.1 . In an invertex. _in_Qu_t is_de 'Oilage. but output is ac at desir~d ' oilagc
with required f}·equency. The output voltage may be fixed or ,~ariable. Similarly. the
fregL~~cy of output voltage is also tixcd or variable. Usuall the output voltage can t c
controlled by pulse width modulation (PWM) techniqu~, gain control of inn~rt~r {ac
ouiput voltage/de input voltage) and controlling modulation indc;;x.
Input
DC supply ----1•~1 Inverter
Output AC with
1---~10 variable voltage
and variable frequt:ncy
Q¥111 Block diagram representation of inverter
lnver1ers are most commonly used in the follo·ing application
• Variable speed induction motor dri es
• Adjustable speed ac drives
• Induction heating
• Uninterruptible power supply (UPS)
• Standby power supply
• HVDC power transmission
• Variable voltage and variable frequency pOJ;er supply
• Battery operated vehicle drives
~he Phase controlled or line commutated full convet1ers can operate inverter mode at the
tne frequency only. These converters are called line commurared im·errers. But the line
cornrnutated inverters require an ac supply which is used for commutation of thvristors
at the · . • ·
80
output. Therefore. 1ine commutated mverters can not work as tsolated ac ,·oltaue
us~~ce With variable voltage and variable frequency. When a thyristor-based inverter~is
to Provide an isolated ac voltage source. the forced commutation technique mu, t
10
b)
0
,.

680 Power Electronics
be used to turn OFF SCR ·. Consequently. thyristor-based inverters are costly and .il!t_lky. But th
inverters are suitable for high po:ier applicatrons. In low power and moderate power appli£a!Jo1Js,
gate commutation switching devices such as GTOs;-BJis, MOSFETs·and IGBTs are used in invPrr.....;
In this chapter, the classification of inverters, principle of operation of single-phase and three-ph
inverters with R load and RL load, three-phase bridge inver1ers, performance parameters of inve
different methods of voltage control and harmonic reduction of inverters are discussed in detail.
operating principles of resonant converters such as series, parallel, series-parallel, quasi-resonant,
current and zero-voltage converters are also incorporated in this chapter.
11.2 CLASSIFICATION OF INVERTERS
Inverters can be classified depending upon the following factors:
1. Input source
2. Commutation
3. Circuit configuration
4. Wave shape ofoutput voltage
11.2.1 Input Source
Based on the nature of input source, inverters are classified as current source inverter (CSI) and vol
source inverter (VSI).
Current Source Inverter (CSI) In this type
Current
source
Input
Inverter
of inverter, a current source with high internal
impedance is used as input of inverter. Figure
11.2 shows the block diagram representation of
current source inverter. In CSI, the supply current
does not change very rapidly, but the load current
can be controlled by varying de input voltage of
CSI. This inver1er is commonly used in very high
power applications such as induction motor drives.
QWffj Block diagram representation ofcurrent
inverter (CSI}
Voltage Source Inverter (VSI) In volt-
age source inverter (VSI), a de voltage source
with very small internal impedance is used as
Input
Output
AC
Voltage vinput of inverter. Figure 11.3 shows the block de
source
diagram representation of voltage source in-
j_+ I
l- Inverter rLoadj
1
verter. The de side terminal voltage is constant,
but the ac side output voltage may be constant
or variable irrespective of load current. The VSI
QMIII Block diagram representation of voltage source
inverter (CSI}
can be classified as half-bridge VSI and full bridge VSI.
11.2.2 Commutation
According to commutation method, inverters may be classified as line commutated inverters and
forced commutated inverters.

Inverters 681
J,ifJe commutated inverters Single-phase or three-phase fully controlled converter acts as an
jJI.vcrter <vhen the firing angle a is greater than goo(; radian) . This is a voltage source inverter and
theused switching devices such as thyristors are naturally commutated.
Forced c_om"!uta_ted inverte~s In these type of inverters, additional circuits are require~ _for
mrnutauon ol thyrrstor · Dependrng upon the commutation technique, these inverters are classtfJed
~ . d .
liSanauxilrary commutate tnverters and complementary commutated inverters.
Auxiliary commutated inverters In case of auxiliary commutated inverters, an auxiliary thy-
ri tor must be used to turn OFF a conducting thyristor. For example, modified McMurray inverters or
McMurray inverters are auxiliary commutated inverters.
complementary commutated inverters In complementary comrnutated inverters, tightly
couple inductors can be used to turn OFF the other thyristor of a pair. For example, McMurray Bedford
inverters are complementary commutated inverters.
11.2.3 Circuit Configuration
According to circuit topology or connection of semiconductor switches, inverters can be classified as
series inverters. parallel inverters, half bridge inverters and full bridge in erters.
Series inverters In series inverters. inductor L and capacitor Care connected in series with the
load. In this inw rter L and C are used as commutating elements and the performance of inverter
depends on the alue or/, and c.
Parallel inverters In case of parallel inverters. commutating elements are connected in parallel
with the conducting thyristor.
H~lf·bridge inverters and full bridge inverters In half-bridge inverters, only one leg of
bndge ~xists. In case of full bridge inverters. c:itht:r t'Olegs or three legs are existing for single-phase
or thre~:-phase inverters respectively.
11.2.4 Wave Shape of Output Voltage
ht a_n ideal inverter. output voltage must be pure!_ sinu ·oidal. But due to swi_tching of semiconductor
~Vtces _as pt:r n:quirement of inverter opaation. output voltage is non-sinusoidal and it co-ntains
larnloni~:s. Depending upon the output 'Oitage wa'dorm this inverter can also be classified a~ square
wa · ~ . .
ve tnw ncrs and pulse width modulauon mverters.
Square wave inverters A square v•ave inverter generates a square wave ac output voltage of
eonstant · · · · I f · bin amplttude. The amplitude ot the output vo tage o mverter can e controlled by varying the
. ?Ut de VOita'ge.
P111 . .
eo s~ Wtdth modulation inverter In pulse width modulation inverters, the output volta,[e
~f~one or more pulses in each half cycl~. By varying th~ width of these p~: the amplitude
teehn;t t voltage can e contro~ougl~ t~ m.~ac.volt~~ !S c?nstant~ There are drfferent control -
q tes o pu se width modulation whrch are explamed m SectiOn 11 .7.

682 Power Elccrroni _
11.3 PERFORMANCE PARAMETERS OF INVERTERS f ·.~
ln an idc~l inverter. the output n)ltage nwst be purdy ~inusoidal. However, the output voltage of .·:
· 1 · · · · 1 1 1 · 1. · . 1' 111,1·rlll''rlt'rl '''>mponcnts as well as harmonicpract1ca lll'Crters rs non-slllllSLHC a ant 11 con ,11ns 1 '• '" • ...., . . ··.
c0mponrnts. sually the perfnrn.ancc ot' an inverter is measured by the followtng rcrformanc~
paramcten;:
Harmonic factor of nth harmonic {HFn) The llarmonic l:rctor is a mcasurc of ~he indiv.iduai
harmonic contribution in the output volta.!!<.: nf a·1 ill crter. This is defined by the ratto of the r~s
,·oltage of a particular harmonic compont:nt to i;1-: rms voltagr: or the fundamental component. It IS .-
represented by
where.
HF = VII
II 1
I
I~' is rms value of the 11
111
harmonic com1wncnt andII
V1 is rms value of the fundaml.!ntal component
Total Harmonic Distortion (THD) The total harmonic distortion is a measure of closeness in ·
a shape between the output voltage waveform and its fundamental component. This is defined by the
ratio of the rms value of the total harmonic component or the output voltage to the nns value of the ··
fundamental component. It is represented by
[_,t _v,;J
112
, , _ , , 112
THD = 11--..'.4.::.... [ nns I ]
vl vl
where, l'nllS is the rms value of output voltage.
Distortion Factor (DF) The distortion factor is used to measure the amount or harmonics that
remains in the output voltage waveform, after the waveform has been subjected to second order . ·
attenuation (divided by n The distortion factor is represented by
[ I. (v,~ )::!]"2
-1' '" 11 -DF = ~~- -..'.........
VI
Lowest Order Harmonics {LOH} It is the lowest ti·equency harmonic with a magnitud<: greater __.·
than or equal to three percent of the magnitude of the fundamental component of the output voltage. ::·
For higher the frequency of LOH, the distortion will be lower in the current waveform. ·
11.4 SINGLE-PHASE HALF-BRIDGE VOLTAGE SOURCE INVERTER,'.
Figure 11.4 shows the circuit confi gluration ofa sin!.!le-phase half-bridoe votanr: soup··· itl·~rt~r (VSl). .>:.~ - b ~ ' ~~ . .
Here switches§1 and S2 are pmver semiconductor switches such as IGBT B.IT. !'viOSfET. etc. When the ·~
~-witch is closed. cu'rren~I_ows tlu·ough dcvrcCS,'"yhll.~switd~cs arc o~'encd. ct~.,~,--il~,~·s through-(liOd~·:_·:.
In this section, the operating principle of half bridge ~;it!, ·/? ~lld /Uio;-d is e:x""jJlaim:."Zri,1detail.< ..:

~;,
+
I'
-2
.,.. v0
Load
+
v
2
o,
,. J--
l'o
+
'/)2 2
__.. '·'
GJWIII Single-phase half-bridge Inverter
--
I.~I
s,
+ -
f,,
11.4.1 Single-Phase Half-Bridge Voltage Source Inverter
with R load
111ll'IHM td_........
The operation of a single-phase half-bridge inverler with R loud ~nn be divided Into two dill' I' Ill
modes such as
I. Mode I ( 0 ,; t ,; ~) : Switch S1 conducts and S2• D1• and D ur·e in OF!'
2. Mode II ( ~,; t,; T): Switch S1 conducts and 51• D1• and D arc in OH
1. Mode I ( 0,; (,; ~) When the switch s, is closed for ·~ . i.e.. half or the lime pcl'iud. ~ INup·
plied across load and current flows through load is
2
VR . The switching diugrnm of mode I iHHhown In
T
fig. II.S(a) and the path of current tlow during 0 ~ I ~ 2 is also shown in Fig. 11 .5(u),
2. Mode 11 ( ~ ,;
1
,; T) At 1
~ ~ , switch S1 is opened and switch 52 iN dosed lbo· ~ duration. ·rh .11
again _ V is applied across load and -
2
VR current will tlow through loud. The swikhing dlu~rum
2 T
o~mode II is given in Fig. Il.5(b) and the path of current llow during 2 ~ I 1' is ulso d picl •d 111
Fig. ll.S(b).
th The gating signals of
51
and s2
, output voltage and .~urrent wa~vcfor.ms ar~.;; s.ho~n in 1:ig. II ,(1 , H i' .
.e o~tput voltage waveform is square wave. The cunent wa:etorm IS .also s1mllur to output. vollu~e
Wavetonn. During R load, diodes D, and D:. are not conductmg. Tht: lre~lu':ncy of' oul ut voHu c
~~~ . The output fiequency can be controlledb y ~~ying the ON lime nnd OFF oin1c ur •witch .•.
he operation of half-bridge inverter is represented by Table 11 .1.
l!
'•

+
v -~
2 -
.--1--
R .~.
lo 1
~ ~l___________~,-~__TD,
+
v -~
2 -
+
R
--
(~ (~
QWIJJ (a) Switching diagram of Fig. 11.4 in Mode 1(b) Switching diagram of Fig. 11.4 in Mode II
G.ating
(a) signal 1---------.
l gl 1
ofS1 ~-------L----~-J--------~--------L-------_k------~~
5T
2
.G.ating
(b) signal
T
2
T 3T
2
2T
fg2 r
ofS1 t---------....L..------___.~.________....L.._________.~.________________~... Time
3T
2
T
2
Output Vo
voltage v
2
(c)
T
v 2
2 --- - --- -
f o
Output
voltage L
2R
(d) T
v __ __ ___2_
2R
T
T 3T 2T
2
T 3T 2T
2
2T
5T
2
l5T
2
I
Time
Time
t
Q&lij (a) ig1, gating signals of switch 51 (b) ig2' gating signals of switch 52 (c) Output voltage V0
(d) Output current i0
T
- StST
2
2
v
2
2R
v
2R

lnvmrrs 685
Figure I I.6(c) shows the ou;ut voltage waveform. The average value ofoutput voltage is
I
v o(av ) =T Ivo(t). dt =0
0
me rrns value of output voltage is
Vo<nn•l =[T~l!v~(t) dr =[T~2[(~)'-dtr=~
Hence. the nns value of square wave is equal to the peak value.
The output voltage ~ can be expressed using Fourier series as
where,
V
0
(f) = L an cos n(J)( + f bn sin nmt
n=l n=l ·
I 21r
an=- I V0
(1)cos rUJJt · dmt and
7r 0
1 2tr
bn =- J V0
(t)sin nmt ·dmt
7r 0
Due to half-wave symmetry, only b11 components are prese~
21r
bn =- J v0
(t)sin nmt · dmt
no
Therefore,
2trV V
or b =-J-sinnmt-dmt asv0 (t)=-
n 7r 2 20
2V=- where, n = 1, 3, 5...
nn
The output voltage can be expressed as
00
2V .
v (1) = L -sm nmt
o nn"=1.2.3...
The rrns value of the n1
h component is
1 2Vv = ~- where, n == I, 3, 5
11
J2 nn
The rms value of fundamental component is V1 = 0.45 V
Example 11.1
120 V, determine
A single-phase half-bridge inverter feeds a resistive load of 5 n. When the voltage V 15
2
(a) nns value ofthe fundamental component of output voltage
(b) the output power
(c) the average and peak current of transis~ors which are used in inverter
(d) the peak inverse voltage (PIV) of tranststors
(e) the lowest order harmonics and the corresponding harmonic factor
(f) third harmonic distortion factor

Solution
' v( J/111'1/,'
2
1211 v lllld /( - ~ u
Th~rdi ,rc, V 2 i' 120 V IJO V
The output voltuK~o: of' liinKI · plut . · hull' hrid~ · inv ·rr ·r r ·xpr 'fl~cd l• J
¥ · 2V .
v)1) L '"' noll
n - 1./, 1 ll'lf
(II) Th~o: r'ntH vulue ol' f'undsu11~nt u l <.:lllllf'OIII.:III of' outf!Ul volta~c iii
v, h2
: 52
/~O vv 107.9()4 v
(h) The rrnHvalue of' output voll~tge il4 V111
,,,~, V 120 V
2
y 2 12( 2
,.h . I p orn,~J . J 2 "(J Wc outj')ut l'owcr rs C'Jllll In 11 • - - - ; - • -~- Watt • ~, ~tt
(c) The current and voltage wuvcfi.Jrrn of' inverter iHdcj')icted in 11ig. 11 .7.
v 240
The average l:Urrenl of' each transistor 0.~/,11~e,,k 1
a:: 0.5':/ - =0.5 x -
2 5
=12 A
21< /..
v 240
The peuk current of' each transistor i~o~ /,1,~:ukJ = 2/( =
2
x S= 24 A
Output v,
voltage
i -120 v 1--------,
r T
v 2
T - -t2o v ----.---- ~...---____.
Output Vo
current
i = 24A~------~
T T
- i = -24 A ---- - --- ~ ~------~
Current 1,,
3T
2
3T
2
2T
2T
5T
2
5T
2
through )
transi;~r 24 A 1------:::::-L------:--------::;-;:--------:=-----l____~
T T 3T 2T
Current 1
transistor
T 2
5T
2
through
41
)
Tz 24 A-
r---~----~T~--~3~T-----~2~T~------------~
2
Voltage and current waveforms of single-phase half-wave inverter

fhe peak inverse voltage (PIV) of each tra . . V
1
l nststor 1 equal to 2 x- =V =240 V
2
The towest order harmonics is v.,, =2V x _I__ J2v J2 x 240 998 V
tel -" nm l
31C r.:;- - - =35.
...;2 31C 31C
The corresponding hannonic factor is HF3
=V nm,= 35.998 =0.3333
v llrms ) 107.994
(O Third hannonic distortion factor
ly2 1,2]1/_ '
THD = nns - I _ [120- -107.994
2
]
112
VI - 107.994 =0
·
4844
where. Vrms is the rms nlue of output voltage.
A single-phase half-bridge inverter has a resistive load of I 0 Q and the center tap de input
1
oitage 1s 100 V. Determine (a) rrns value of output voltage, (b) rrns value of fundamental component of output
1
oltage. (c) first three harmonics of the output voltage waveform. (d) fundamental power consumption in load
and (e) rms power consumed by load.
Example 11.2
Solution
vGn·en: - =100 V and R = I 0 Q
2
Therefore. V =2 x 100 =200 V
v
(a) The rms value of output voltage is Vo(rms l =2 =I00 V
(b) The rms alue of fundamental component of output voltage
__I_ 2V _ I 2 x 200 =90 07 V
VI -J2_1C -.fi lC .
(c) Since V =~ first three harmonics of the output voltage waveform is
II ' •
n . . v, 90.07
rms value of third harmomc ,·oltage 1s V3 = 3 = -
3- = 30.023 V
. VI 90.07
rms value of fifth harmonic voltage 1s V5 =- =--= 18.014 V
5 5
. V1 90.07
rms value of seventh harmonic voltage IS V7 =7 =- 7- =12.867 V
. V 2
90.07
2
(d) Fundamental power consumption in load IS ~ =--J?= 10 =811.26 Wan
2 2
(e) Volnn 1 I00 ()()()
nns power consumed by load Pnn = ~ = )o =I Watt
11·4.2 Single-Phase Half-Bridge Voltage Source Inverter
With RL Load
For RL I d . . ., h .
is ct·tro oa , the output voltage wave fonn IS s1m1 ar tot at With R load but the load c. l11ere fi . . ·
1
urrent waveform
lllust . nt rom the load current w1th resistive oad. The output voltage and curr t "-
. rated · r· · 1
h en wave,orrn are
ll'lto r, tn tg. 11.8. The operation of a smg e-p ase half-wave inverter with RL 1 d . .
our different d h oa can be d1v1ded
mo es sue as

(a)
f gl 1G_ating
signal r------,
ofSt ~_____j_______L______~------~~----~5(?T~----.
T T 3T 2T
22 2
G_ating
(b)
/12 1signal
of~ ~----~~--~~------~~----~~----------~
(c)
(d)
Output V0
voltage v
2
v
2
Output
current 1o
-fo
T '" T 3T 2T
2 2
T T 3T 2T 5T
2 2 2
--------
Time
~~--~~~---.~~--~~~--~.-~--~5~T------•
2
Conduction ~---r----:::---r-=-.---=--,--;:.-r---::::--,--;::--r---::;----.~r----::;----,
(e) o_f IDt I St ID2l S2 IDt I s. ID2l S2 I Dtl s.
devices
QMII:I (a) i91 gating signals af switch 51, (b) ~92 gating ~ignals of switch 52, (c) Output voltage VO'
(d) Output current i0
and (e) Conductton of devtees
1. Mode I (0 $; t $; t1): Diode D 1 conducts
2. Mode II (r1$; t $; ~): Switch S1 conducts
3. Mode Ill ( ~ $; t $; t2 ): Diode D2 conducts
4. Mode IV (12 $; t $; T): Switch S2 conducts
1. Mode 1(O ~ t ~ t1) At t = 0, the gating signal is removed from switch s2
and it becomes Of
At this instant the load current is i0 which is equal to its negative peak value (- / . Due to i
load, the load current can not be reversed instantly and then diode D 1
starts to at 1 =0. Su
·- ~ v I
sequently, the output voltage across load is 2 and the load current i0
increases form its negative
value (- /0 ) as the current cannot reverse instantaneously due to inductive load. Then the toad ~..rrr:...
flows_through_diode D 1• 111. the time interval 0 $ t $ t 1, the voltage across load is positive, but~ -
. --

~
Inverters 689
nt is negative. [fence the ner stor d . . fi d b k
cl.lrre Iy through D anathe I e Ill tnductance L during previous cycle must be e ac
~~· d. I . oad current ecreas s o = t t e foad current oecomes
zero-The swtt~ tng. taf_raml of mo .en1n :Fig. 1 ~a a~d the pat~ of current flow during 0 ~
0 1
is also s own tn tg. I.9(a). ·
,,
AAode II (t1 5: t 5: T) At the · · ON Thez, rw• 2 mstant t = t 1
, diode D1
becomes OFF but switch s, IS •
t tl
. . . T
J,t starts o ow 111 postttve direct· d · . · · k 1 I at t--curre ton an 1t reaches 1ts maximum postt1ve pea va ue o - 2
T
During the time interval 115: 1 5:2, both the output voltage as well as current is positive and energy
stored in inductance L. The switching diagram of mode 11 is shown in Fig. ll.9(b) and the path of
T
current flow during t 1 $ I $
2 is shown in Fig. 11.9(b).
v
2
R L __.. i, ~
I
v '
2
I 1 I
I_ - - - - - - - - - - - - - - - l - - - - _ I
v
2
v
2
+
R
+ I
I
~
- I s, 8J- I I
I I
'-- - - - -- - - - - -- - - - .L - - - - _,
(a) (b)
lififQ (a) Switching diagram of half bridge inverter with R-L Load in mode I (b) Switching diagram of half
bridge inverter with R-L Load in mode II
3. Mode 111 ( ~ $ I 5: 12
) 1t 1 =~, S.vitch 51 becomes OFF. At this instant, the load current i0
is
equal to its po. iti e peak va lue (/). Due to inductive load. the load current can not be reversed instantly
T
and then diode 0 '2 • t·trt 10 conduct at 1 = '2. After that the output voltage across load is - ~ and the
load current ;" decrea~cs form it po itive peak value /11 as the current cannot reverse instantaneously
due to inducti e load. This current flows through diode D2. During the time interval ~ ~ 1 $; 12 , the
output voltage is negative but the load current is positive. It decreases slowly and reaches zero at
~~ 12· The energy stored in inductance will be released an~ fed back to de supply during this period.
Jgurc 11 .10(a) show the load current t1ow path through diode D2 and load and the switching diagram
of mode III.
4
· Mode IV (t < t < n At instant 1 = 1'2. diode D2 becomes OFF and switch s is ON Th l d
Cllrr 2 - - . . d . h . . 2 . e oa
d
. ent starts to flow 1·n negative d1rect1on an 1t reac es maximum negative-/ at 1 = T Th . h.1a r . . o . e sw1tc mg
d. ~ arn of mode 1v is shown 111 F1g. 11 . 1O(b) and the current flow path through s and 1 d . 1t plct d . . f h If b .d . . 2 oa 1s a so
e 1n Fig. ll . lO(b). The operatiOn o a - n ge Inverter IS represented by Table 11 _2 _

Table 11.2
nme
0 ~I~ t 1
T
11~I~-
2
II
T
-~I~ I?
2 -
III
IV 12 ~ t ~ T
(a)
v
2
v--
2
v--
2
Negative
Positive
Positive
Negative
v ~-------
2
v
2
+
-----------
(b)
(a) Switching diagram ofhalfbridge inverter with RL Load in mode Ill (b) Switching diagram of
bridge inverter with RL Load in mode IV
The output voltage can be expressed as
V = i (t)R + L dio(t)
2 ° dt
T
forO< t :::;-
2
Assume that the initial condition is in(t = 0) =-lo . Then the output current can be expressed as
io(t) = 2~ (1-e-f)- foe-f
L
where, -r =-
R
Assume at t =~,i0 ( t =~)=10
Then (T) V( -L) _Li - =- 1- e 2
• - I e 2
• = I
o 2 2R . o o
or
V ( -L) ( _L)2R I - e 2r = Io 1+ e u

~---------------------------------------------------------~~~~~·~~~·691
I =-(1-t--:r
( 2R(1 -! )+t' -r
~tler substituting the vnlu~ of 1 in Eq (II 1)
' u . . • Wt' gt't
. ( ') ' (J -L)1
0
U)=- 1-e-7 _ - e :: - 1
,R - t ' r
- 2R (1 -·!:)+ t' ••
or (.(t = ~(1- - ('-t], R 1 _.r_
- + (: :r
. T
Dunng 2$ t $ T . the 'OltJge equation can ~ written as
v di (t')
-- = i0
(t')R + L___,_Cl_
2 dr
Assume i0 (r =~)=!" and i,,(r = T) =-I ,
{Theoutput current can be expressed as
h
, T
w ere. r = r- -..,
(
. io(T ) =- 2RI ( J- 2_.r_ e-~~-~ :'J T
for -::;; r::;; T
. 1+ e :, 2
The m1s output voltage is
'-- [ 1 T'-(v)2 ]"- vV = - f - dr =
mls T /2 0 2 2
Due to half-wave symmetry. only b,, components are present
2;rV
b =-J- sin nwt ·dmr
n 7r 20
2V 1 3 -= - where,n= , . .:> ...
nn
The nns value of fundamental component is
v, =~=0.45Y 2lrms J2;r
The instantaneous load current can be exp essed as
oo 2V · 11WL
io(t) = L J ~ ., sm(nmt- ¢n) and tl>n =tan- 1-
n=I.3.5... mr R- +(nmLt R
Where, the impedance offered by the load to the n•h hamlonic is zn = JRl + (nmL )~
The fundamental load current is
2V . ( t~~ 1 wLi - sm rot - 'f't) and ¢1= tan- _
ol- nfR2+ (mL)'1 R

Example 11.3 A single-phase half-bridge inverter feeds an RL load with R = I0 nand L =0.1 H. When the
v
volta~e - is 120 V and the frequency of output voltage is 50 Hz, find
2
(a) The output current for the first two half cycles of output voltage.
(b) Third harmonic distortion factor (THD) of load current.
Solution
v
Given: - = 120 V, R = 10 Q and L = 0.1 H.
2
Therefore, V = 2 x 120 V = 240 V
(a) For the first or positive half cycles of output voltage ( 0 < 1 $ T), the output voltage can be expressed as
V - . ( )R L dio(t) 2--t I + - -
2 (J dl
V ( -L) -LAs i (1=0)=-1 i (!)=- 1-e r - / e r
o o• o
2R o where, r = ~ = .Q:_! = 0.01
R 10 ·
T T V l- e 2r
( )
( -L)
At 1=- i 1=- =I=-+-----+
2 ' o 2 ° 2R( -L)
1 I T
where, T =-=-=0.02 sand -=0.01 s
f 50 21+e 2r
Then i11 (t)=-V-(I-
2
_L e-})=-
2
-
4
-
0
-(t- 2
_!U_ e-otTJ= 12(1-1.9999e- 100')
2R 1+e 2r 2 X 10 1
+e-2 xo.o1
For the second or negative halfcycles ofoutput voltage ( ~ S 1ST),the output current can be expressed as
io(t) = __v (I- 2 L e_(1-:·nl)= __2_40_(1- ~-...ll.L e-•·~~:2'J
2R 1+e-2r 2xl0 l+e 2x oo1
= -12(1- 1.9999e- IOO<t-O.Oil)
(b) The instantaneous load current can be expressed as
. ( ) ~ 2V . 1 n(I)L
10 I= 41 J 2 2
Stn(nwt-q>n)andq>n=tan- - -
n =- 1.3.~... mr R + (nwL) R
The rms value of fundamental current
11 = 2v x 1 = !iv
1CJR
2
+(21CJL)
2
. J2 1CJR2 +(2TrjL)2
= J2 X 240 =1.6372 A
1rJ10
2
+ (2Tr X 50 X 0.1)2
Similarly, rrns value of third and fifth harmonics current are
J2v J2 x240
/3 = , = = 0.001349 A
3TrJR
2
+ (2Tr X 3}Lt 3TrJ102
+(21C X 3 X 50 X 0.1)2
J2v J2 x2401
s= 1 , = 1 =2.9I554x10-4 A
51Cv R
2
+(21r X 5jL}- 51rv102
+(21r X 5 X 50 X 0.1)2
THO of load current is
= ~rlj=--+~1;=-+~li:-+---:1~:-+-·-.. = Jo.OOt3492 + (2.91554 X 10-4)2 +...
/1 1.6372
=0.001178

~---~~~~--------------------------------------------~m~v~~n~e~~~69~
11.5 SINGLE-PHASE FULL .BRIDGE INVERTER
. res 11 .II and I1.I2 show single-phase full b . . .
f,gu (S S and S s ) d t . ndge mverters which consists oftwo paus ofcontrolled
r'tches 1' 2• 3• 4 an wo pairs of d. d · - ·- --:-- -
s~ of devic~~- conducts simuf - --·--- ·--_1.C? :es.(.0,~ L}2 a.n~ D3• D_4) .. Among these ~evt~~ly
~a·d is +V. Similari , when ta~eou~Jy...._When sw1tches 51 and 52 are ON, the output v~ltage
a~ross D D d D y SWitches 53 and 54 are ON, the output voltage across load IS -V.
owdes D,' . d 3
.' an Th 4 are used as feedback diodes The load of full bridge inverter will be either
resistive or m uctlve. e sequence of switching of s~itches and diodes for R load and RL load are
given below.
For RLoad
For RL Load
+ +
v v
litlll!l Single-phase full bridge inverter GJMIIfj Single-phasefull br~dge inverter using BJT
11.5.1 Single-Phase Full Bridge Inverter with R Load
Similar to single-phase half bridge inverter, the operation of a single-phase full bridge inverter with
Rload can be divided into two ditTerent modes such as
I. Mode 1 ( 0 ~ 1 ~ ~) : Switches 51 and 52 conduct
2. Mode 11 ( ~· ~ 1 ~ T): Switches S:. and 54 conduct
1. Mode 1( 0 ~ t ~ ~) In this mode, switches 51 and 52 are closed for ~ , i.e., half of the time
Period, and the voltage vis applied across load. Then current flows through load is V. The switch-
. R
Jng diagram of mode I is illustrated in Fig. 11.13(a) and the path of current flow during 0 $ t ~ T IS
also · . 2
gtven m Fig. 11.13(a).
t Mode 11 ( T < t < r) At t = T , switches 51 and S, are turned OFF and switches s a d 52 - · - 2 - 3 n · 4 are
turned ON c T · · V. 1· d I V
10r - duration Then agam- 1s app 1e across oad and -- current will flow through'2 . R
·~·

load. The switching diagram of mode II is shown in Fig. 11.13(b) and the path of current flow
~ $ t $ T is also depicted in Fig. 11 .13{b). At r= T, again S1
and S_ are turned ON and S~ and
tt.~.n OFF and cyclically witche are 0 and OFF repeatedly.
v
I
I
fio, s,[] fio,1 ~vo ~ , 1
· R
,8] ~o:·
'
'
( )
D~
s,BJ
R
s~
(b)
Mllll (a) Switching diagram of Fig. 11.11 in mode 1and (b) Switching diagram of Fig. 11.11 in mode
The gating ignal f ·witche 51, S2, S and -1· utput v ltage and current wavef m1 are . ho
in Fig. 11 .14. t thi time, the output ohage wave~ rm i a quare w ve. The urrent w ve~orm i
al o similar t
The frequency
ON time and
by Table II .
uLput' ltage wavt:foml. During R load, diod~ D1, D_, D3
n D are n t condu ·tin
f output v ltage i· f =~ .The output fre uency can be c ntr lied b · ar rrng t
FF time f witche . The peration f full-bridge invener ith R I
0 < I
T
I ' ~'
2 R 1 and ~
r
r v 1II - I
2 R , and 4
Figure 11 .14(c) ·how the output voltagt> wo t'~ rm. The :l eragt: value f utput ltage i
I T
VtJ(av ) = - JV0 (t) · dt = 0
To
The m1 alue of output oltage i·
v o( mlS ) =[-
1
-Jv~(r).dr ]l/_=[- 1
-Jv- .dr]
112
=v
T12 0 • T12 0
Therefore. the rms value of sq.uare wave is equal to the peak value.
The output voltage Vo can be expres ed using Fourier series as
00
V0 (t) =Lan cos nltX + Lbn sin ncm
n= l n=l

~-------~------------------------------------------------------~m~v~e~rt~e~~~~
(a)
(b)
(c)
(d)
fgl
Gating
signal 1g2 r------
ofS1 & S2
r---------Ir~------~--------_l_________L_________L________~Time
fg3
Gating 1g4
signal
ofs3& s4
Output
vo
voltage v
-Vo
fo
Output v
current
R
v
R
2 T 3T 2T 5T 1
2 2
r---------~--------~------__l_________L_________________~Time
T T 3T 2T 1
.. 2 2
T T 3T
2 2
------------
p.
T T 3T
2 2
-----------
21 5T
2
I21 5T
2
'
,.
Time
. I
Time
t
(e) Conduction ,---·-------------~-------------,------r--------T--------,
ofdevices l----~~--~~~____L___~2_~-~~--_j___~!...~i2___ j___!~_~_§~_j__~!..&-~~-J
ull§l (a) i91
, i92
gating signals ofswitches 51 and 52, (b) i93' i94 gating signals ofswitches 53 and 54
(c) Output voltage V"' (d) Output current i0 (e) Conduction of devices
} 2TC } 2TC •
Where a =_I v
0
(t)cos nOJt · dOJt and bn =- JV0 (t)sm nOJt · dOJt
n n Xo
0
Due to half-wave symmetry, only b11
components are present.
Therefore
'
or
2!C
b =-Iv0
(t)sin nOJt · dOJt
n n 0
2!C
b =-IVsin nOJt · dOJt
n n 0
4V
Th nn
e output voltage can be expressed as
oo 4V .
v (t) = L -sm nOJt
o nnn = l.2.3...
wheren= 1, 3, 5...

8
;:llll1.!
'.~:I
'.~.
St~n I
,fS, ...~ .·;
tn)
T
,
l g -( Iatin•'
(,-1='i,,11:11
~,r S: & s~
b)
T
2
l"(t
O utput
volttH.!.I: ,.
T
2
- V 1--------------
(d)
Output
CUIT.!1t
(e) Conduction D,
of devices D2
~ Tint'
r JT 27' 51' I
22
~
Tirno
T 3T 2T
2
limo
T 3T 2T 5T
2 2
Time
QWIIII (a) iu1, iu2 gating signals ofswitches 51 and 52, (b) ig31 ig4 gating signals of switches 5
3
and 541
(c) Output voltage VO' (d) Output current i0 and {e) Conduction of devices
D and D., starts to conduct at t = 0. Subsequently. the output voltage across load is +V and the loadI - .
current i increases fi·om its negative peak value (- /J as the current cannot reverse instantaneously()
dut: to inductive load. Then the load current flows through diode D 1 and D2
• During the time interval
0 $ 1 $ t 1• the voltage across load is positive. but load current is negative. Therefore, the energy stored
in inductance L during previous cycle must be fed back to de supply through feedback diodes D1
and D., and the load current decreases slowly. At I = I 1, the load current becomes zero. The switching
diagra;,, of mode I is shown in Fig. 11 .17(a) and the path of current flow during 0 s t s t
1
is also
depicted in Fig. 11 .17(a).
2. Mode II (r1S t S ~) At the instant t =t 1, feedback diodes D 1 and D2
are turned OFF but switches
S1 and S2 are turned ON. Then load current starts to flow in positive direction through switches S,
T Tand S2 and it reaches its maximum positive peak value 10 at t = 2. In the time interval t
1
S t ~ 2•

~---------------------------------------------------------------~~~~~~~699
th the output voltage as well as current . . . . .
bO . IS POSitive and energy stored in inductance L. The swrtchang
d. grarn of mode II ts shown in Fig 11 l?(b T
s~~wn in fig. 11.17(b). · · ) and the path of current flow during t1S t ~ 2 i.s also
+
v
I
I
s,BJ
I I
I I
s,8JtD,I
I
I
(a)
+
v I •
: _..lo
s.[J~D,
I
I
I
(b)
QWflfl (a) Switching diagram of Fig. 11.11 in mode 1, (b) Switching diagram of Fig. 11.11 in mode II
3. Mode Ill ( ~ ~ t ~ t2 ) At t = T , switches 51 and 52 are turned OFF. At this instan~ the load
2
current is i0 is equal to its positive peak value (10
). As the load is inductive, the load current cannot
T
be reversed instantly and then diodes D3 and D4 start to conduct at t =-. Then the output oltage
2
across load is -V and the load current i0
decreases form its positive peak value !0
as the current can not
reverse instantaneously due to inductive load. During the time interval T $ t $ t, , the output volta£e
2 - ~
is negative but the load current is positive. The load current decreases gradually and reaches zero at
t == t2• Hence, the energy stored in inductance will be released and fed back to de source during this
period. Fig. 11.1 8(a) shows the load current flow path through diodes D3 and D4 and load and the
switching diagram of mode III.
4. Mode IV (t2
~ t ~ n At instant 1 = t2, dio~es D3 ~nd lJ_4 ar~ turned OFF and switches s3
and S~
are turned ON. Then load current starts to flow m negattve d1rect10n through switches s3
and 5~ and it
reaches maximum negative - Io at I = T. The switc~ing diagra~ of ~ode IV is shown in Fig. 11. 18 b)
and the current flow path through 53, 54 and load IS also depicted m Fig. ll.l8(b). The operation of
full-bridge inverter is represented by Table 11 .4.
II
III
Operation offull bridge inverter
T
t < t <-, _ -2
v
- V
Positive
Positive
: IV · 1 < t < T - V Negative
..___ : 2- - ____.,:___________________ ________~---------- - ----------
----- _____j__~ -- - -- - -------------- --------- ... _--
s, and S .,
S- and 5~

I - I
I I
s,8J Ao, sEJ1 1 '+- Vo ~ 1
I I - R L + I
I I I +
s,8J
I
~ - I
I lo I
s,[] D4 s,[] to,I
I
v I
s,8JAD,I I
I I
I I
(a) (b)
Q¥111:1 (a) Switching diagram of Fig. 11.11 in mode 111, (b) Switching diagram of Fig. 11.11 in mode IV
The nns value of output voltage is
Vo(nns) =[-
1
-JV~(t)· dt]
112
=[-
1
-LJ
2
V
2
· dt]
112
= V )
T12 0 T12 0
· ·
I Hence, the nns value of square wave is equal to the peak value.
The load current can be expressed as
and
where,
i0 (t)=- 1- _Le r forO:=;t:=;-V ( 2 _LJ T
R 1+ e 2r 2
i0
(t)=-- 1- _Le rv( 2 ~J
R 1+ e 2r
L
't'=-
R
T
for-::; t:=; T
2
The output voltage ~ can be expressed using Fourier series as
00
4V
V0
(t) = L, -sin nmt
n=l,2,3... ntr
The nns value of the nth component is
1 4V
vn = r;;- where, n = 1, 3, 5...
...;2 ntr
The rms value of fundamental component is
4V
"'nns= r;; = 0.9 V
v27r )The instantaneous load current i0 can be expressed as
oo 4V
io(t)= L ~ sin(nmt-l/J )andl/J =tan-1 nwL
2 2 n "n=t,3,5... ntr R +(nmL) R
where, Z11
= ~,-R-2_+_(_n_m_L_)_2 is the impedance offered by the load to the nth harmonic.
The fundamental load current is
. 4V . ( A. _ 1 mL
l01 =
1
sm mt- 'I'J) and 4J1=tan -
trvR2
+(mL)2
R
-------------·--

~:::::~=============-----------------------------------------~~~n~ve~n~e~~~701, • A full bridge inverter h .
· 0 H as a de sour . . .
10 n L == .1 and c == 4.7 J.LF Th . ce voltage of 240 v. The inverter supplies a RLC load With
R:;:; ' · e operatm fi
nt at fundamental frequency (b) the g requency of inverter is 500 Hz Determint: (a) rms load
urre . ' rms value f 1 ·
:d (e) THD m load current.
0
oad current, (c) power output, (d) average supply current
solution
, Giren: v==.240 V, R = 10 Q , L = 0.1 H, C== 4.7 F -
of'l.e inductive reactance at fundamental fire J.l. and I- 500 Hz
111 quency 1s
.. xL =2nfl. =2n x soox 0.1 n=314.2857 n
The capacitive reactance at fundamental fire .
quency 1s
1
Xc=-= 1
2nfC 2n x 500 x 4.
7
x
10
_6 n =67.6982 n
The impedance offered to the nth harmonic component is
Therefore,
z.=~R'+xL<c)'
Zl=Jii2;(.XL -Xc)
2
=Jl02+(314.2857-67.698)2 =246.79.0
z3= R
2
+ ( 3XL- ~J= 10
2
+ ( 3 X314.2857-
67
·:
98
r= 920.3254 n
Z5 = R
2
+(5XL --¥r= l0
2
+(5x314.2857-
67
·:
98
r=1557.92.0
Z7 = R2
+(7XL -~r= 10
2
+(7x314.2857-
67
·~98
r =2190.3516.0
z9= R2
+(9xL- ~cJ= 10
2
+(9x314.2857-
67
·:
98
J=2821.067n
The rms value of nth harmonic component of output voltage is
0.9V 0.9 X 240 216
v =--= =-
II 11 11 n .
The rms value of of nth harmonic component of output current 1s
v 216
I=-"=-
11 Z nZn n .
(a) The rms .value of load current at fundamental frequency 1s
V1 216_ 216 A=0.875A asV1=216V
11=-2 =z-- 246.79
I I
216 A I _ 216 _ 216 _
Similarly, I =216 = = 0.078233 , 5 -
52 - 5 x 1557 92
- 0.027729 A,
3 3Z 3 x 920.3254 5 •
3
I 216_ 216 =0.014087A,I9= ~~
6
= 9x 2~~~.067 =0.00850A
1= 7z7
-7x2190.3516 9
(b) The rms value of load current is equal to
, , /2+/2+/2+ ...
I= lj + 13 + 5 7 9
= 0.8752+ 0.07822
+ 0.0277
2
+ 0.0140
2
+ 0.0085
2
+ ··· = 0.8790 A

(c) The po"''l'r output i,
~ .,
P,,= 1-R = 0.879- x 10 = 7.72 Watt
(d ) The a cragc supply current is
I .=P,, =7
·
72
=0.032 A
U v 240
(c) T/ //) in load current is
THD
J1
2
- ~~ Jo.8792
- 0.8752
= = =0.095
II 0.875
Example 11.6 A single-phase full bridge inverter has a resistive load of I0 Q and it is operated from a 120
uc input voltage. Determine (a) rms value of output voltage, (b) rms value of fundamental component of
voltage. (c) first three harmonics of the output voltage waveform, (d) fundamental power consumption in I
(c) nns power consumed by load and (f) transistor rating.
Solution
Gil·en: V = 120 V and R = 10 Q
(a) The rms value of output voltage is V:,cnnsl =V =120 V
(b) The rms value of fundamental component of output voltage
I 4V I 4 x 120
V1= r;::- = r;:: = 108.09 V
/2 7r /2 7r
(c) As V:,=~, first three harmonics ofthe output voltage waveform is
n
rms value of third harmonic voltage is V3
=~ =108
·
09
=36.03 V
3 3
' h h . I . VI 108·09 21 618rms value of tift armomc vo tage IS V5 =- = = . V
5 5
nns value of seventh harmonic voltage is V7 = V1 =
108
·
09
= 15.441 V
7 7
. . . Vj2
108.092
(d) Fundamental power consumptiOn tn load IS ~ =- = =1168.34 Watt
R 10
vo~nns) 1202
(e) rms power consumed by load prms =R =w=1440 Watt
(I) Transistor rating: Voltage rating VeE;?: V;?: 120 V and
v v 12
Current rating /Tpcak;?: R;?: 12 A and /Tnns;?: J2R;?: J2 =8.485 A
Example 11.7 A single-phase full bridge inverter is connected to a RL load where R = 10 Q and L = 0.2 J-1.
1rthe inverter is supplied from 220 V de source and its output voltage frequency is 50 Hz, determine the expressions
f()r steady state current for the first two half cycles.
Solution
Given: V =220 V,.f= 50 Hz, R = 10 Q and L =0.2 H
During the tirst half cycle. 0 ~ t ~ T , the KVL equation for RL load is
2
di
V=Ri +L__!L
I) dt

Inverters 703
uming ': ' 0 0 . th~ solution orabo . • . .
c cquutton 1s
,~,(1) = !:.(, - t·-f')- -20( _IQ')
R - - 1- e ~~ ~ - "?(I - 5or)
I0 - "'- - t'
;I 1 C. the.· ' tluc.· of current i~ ; ( 1 = T)_V( -~' )" -- 1- (' _,
2 R
r()urillg th' St'COild holf C clc. _
T or 0 ' ..- T T 1 I I .I ~- wh~:n.: 1' =1 - - the KVL equation for N - oa< IS
2 2
= Hi L!5:_
t• tit'
·:;uming initial l:Ondition i ( , =T)- v(l -~,r) h11 _ - R - e- .tc olution of ab vc equation i
. ( ') v v( /1.1) J( •
', f =- R + R 1 - (' ~ I (' - ,,
'tcaJ stult' l'Urrcnt li.1r the first half cydc ( 0 S t s ~) 1s
i, (I) = 22( I - e-~or) - 5.385t'-~or
T
where. 0 S t' S -
2
Stc.·ad_ lute current for tht• second hulf cyde ( 0 S t' S ~) 1s
i (I') = - 22 + 22(2 - t'-
0
.
5
)t'- 50
'' + 5.385e- 50''
"
h
, T
w ere./ =1 - -
2
! ~.unplt• ll.H A ingk-phusl' transistoriscd full bridg~ in~erter. has a resistive load of 5 n ancJ it is uper:Jtccl
frurn a 9(1 v ck input voltage. Dett:rminc (a~ total harm~HIIl: d•s.tortJon. (b) distortion factor. (c) harmonic l~u.:tor
and distortion factor at loWl'SI order hannontl', (d) 1mr1srstor ratmgs.
Solution
<.ih 'II: v 9(1 v and R 5 n
(a} Tht• nns ,·aluc.· of output voltage is v...tm,.1= V = 96 V
(b)
The rms 'ltiUt' of fundmn.:nlal cumponl'nl of outpul vullagl'
~ =0.9 v = 0.9 96 =86.4 vlm"
~ 2 J9 2 '
rms ham10nic voltag~ is V,,= ~,(nn'l - V,m•~ = 6 - 86.4~ = 41.845 V
. v,, 41.845
Total hannonic disiOrtion JS TDH = y;-= 86 4
V =0.4843
Inn' ·
..t(:J']"'Distonion factor is DF = ~
I
~ V, 86.4 ?8 8 V ~ v, 86.4 I V - I; 86.4
As Vn=-;. ~=3=-3-=- . . s=5=s= 7.28 V and 7-7=7= 12.34 V

Th refore 2. (v; )2=(v~ )~+ ( v~)2+(vi)2=(28~8)2+ ( 17}8)' + C2;~4r10.7811
n=.U .7 n 3 5 7 3 5
[ -f (~;)']1/2
Then OF= "-:~.~.?... = JJ0.7811 = 0.03800 = 3.8%
v, 86.4
(c) Harmonic factor and distortion factor at lowest order harmonic
rms valtH.: of third harmonic voltage is V1
= ~ =
86
·
4
= 28.8 V
. 3 3
v 28.8
Harmonic factor HF3 = 3
rm~ = - - = 0.3333 = 33.33%
vlnm. 86.4
v :lpns 28.8
Distortion factor OF1 = ~ = l = 0.03703 = 3.703%
· v,...11
, 86.4
(I) Transistor rating: Voltage rating VeE~ V ~ 96 V and
C . V 2 V 19.2 3 57 Aurrent ratmg l r....uk ~- ~ I~. A and IT .~ ~ ~ ~ =I .
•- R mu ~2R ~2
. . lrpcak 13.57
Avt:ragt: transrstor current rs /Tav =- - =-- =6.785 A
2 2
Example 11.9 A single-phase full bridge irWI.:rh.:r is connected to a RC load where R = I0 n and C = 50 J.!F. If
the inverter is supplied ti·om 220 V de source and its output voltage frequency is 50 Hz. determine the expressions
for steady state current for the first two half cycle .
Solution
Given: V = 220 V.f= 50 Hz. R = 10 n and C = 50 J..lF.
During the first half cyck. 0 :S t :S T , the KVL equation for RC load is
2
V =Ri + _!_Ji ·dt(I c 0
. dq h b . h .As 1 : : - • t e a ove equatron can c wntten as
u dt
V =Rdq + q
df C I ( )
Solution of the above equation is q(t) = CV(l - e-~~c) and Vc(t) =!I!_= V(l- e- Rc)
c
dV (t) V -
j (f)= C C =-e RC
o dt R
Then
Therefore. Vc(t) = V(l- e- i <' ) = 220(1- e lOxSOxlo-ll )= 220(1- e- 2000') and
j (f)= ve--k- =220 e IOx 5Ux iU- o =22e-2000t
o R lO
r . . ( r) ( __I_)At r = '2.the value ot current ts Vc t =
2
= V l - e zRc
. T
Durmg the second half cycle, - :S t :S T
2
or 0 ~ t' ~ T where t' t T th KVL · d ·2 = - 2 e equation for RL loa ts

Inverters 705
dq h b .
I
. :::;; - , t e a ove equatiOn can be wrJ.tte dq q
AS o dt nas -V=R-+-
dt' c
rning initial condition V(r=T)_v(1 -,f-) . .ASSU c 2 - - e - c , the solution of above equatiOn IS
Therefore,
At
q(t') =-cv(t-e--k)+CVc (T
2
)e-;~ T
2
where, 0 ~ t' ~-
Vc (t') =q~') =- V (1- e-!If)+ Vc ( ~)e-~~ =- v+V (2 - e-rfc)e- ~~
. ( ') _ dVc(t') V ( __I_) __t___
10 1 -C =-- 2-e 2NC e RC
dt R
Vc (t') =-V + v(2-e-rlc )e-fc- =-220 + 220(2- e
=-220 + 220(2 _ e-2c )e-20001'
_O x I _ _ _, __
6
, 0- J ) ·
2x iOx 50x iU-6 e 10 x 5U x l0
V ( _ ) __(__ 220 ( 20x i0- J )
1
• ,
i,(t') =- R 2 - e ~RC e I<C =-10 2- e 2x i Ox 50x i 0-
6
e IOx 5U x i0-
6
= -22(2 - e - 20)e-2000t
_ _I_
(
T) l-e2NC l-e2 x l0x 50x lo-
6
I - 20
Vc I =- =V =V = 220 =220 - e , "" 220 V
2
u _I_ '0 IU- 1 I - 0
l+e-2RC - x . +e-
I+ e 2 x 1ox su x 1o- o
Steady state voltage for the tirst half cycle is
Vc(t) =220(1- e-2000
')- 220e-
20001
and
Steady state voltage for the second half cycle is
Vc (t') =-220 + 220(2- e-
20
)e-
2000
'' + 220e-
2000
t'
Example 11.10 Fundamental component
of voltage
Fundamental r-.....,.:....-'--_-_,/....:.'..----
component Output voltage
v
ofcurrent
A single-phase bridge inverter delivers
power to RLC series load with R = 5 Q and wL = I0 Q.
The time period of output voltage is 0.2 ms. What is the
value of C to obtain load commutation? Assume thyristor
tum-OFF time is 16 J..lS and circuit turn-off time is 1.5/q.
Assume that load current contains fundamental component
only. ~1----~t------~~Time
Solution
The value of c should be such that the circuit operates
in underdamped condition with RLC load. Figure 11 .19
shows the waveforms of fundamental component voltage
and fundamental component current. The current leads
Voltage by an angle ¢.
Th Xc-XL
en tan ¢ =_;--=-
Since current leads voltage by :0angle ¢, Xc > XL
' /
' ... _..."'
I
I
I
I
I
I
I
I
t
Waveforms offundamental com1 ponent
vo tage and fundamental component
current
~ must be equal to the circuit tum-OFF time 1.5!q =1.5 X 16 X 1o-6
=24 J..lS =tc
Frequency _I_ I =5kHz
f - T - 0.2 X 10-3

706,~P~o~w~er~E~Ie~c~tr~on~i~a~-----------------------------------------------------------
¢ =Wl(' =2Trftr=2Tr X 5 X IOJ X 24 X Io - 6 = 0.7536 rad = 43.160
X -XL
tan ¢ =___,.c"-~
R
Xc-10
or, tan 43.16 =
5
or
or
0.9367 =Xc -
10
5
- - - - = 14.6835
2Tr X 50 XC
or, XC=lO + 5 X 0.9367 = 14.6835
c- 1 =216.89J.tF
or, - 21r x 50 x 14.6835
11.6 THREE-PHASE INVERTER,.------r----.----r-~----r-l
The output of a single-phase inverter is a
non-sinusoidal waveform and it consists of
harmonics component. This is suitable only
for low power industrial applications. When
the more number ofswitching elements are
used in an inverter, the output waveform
will be non-sinusoidal, but its harmonics
content will be reduced. Hence the output is
nearer to sinusoidal. Therefore, three-phase
inverters are used for high power industrial
applications.
A three-phase inverter can be formed
after combining three half-bridge single
phase inverters in parallel. It consists of
six switching devices (SCRs, power BJTs,
MOSFETs, IGBTs, etc.) and six diodes.
The circuit configuration of a three-phase
inverter is shown in Fig. 11.20(a) and (b).
The load may be connected either in delta
or star. The delta connected load and star
connected load are depicted in Fig. 11 .21 (a)
and (b) respectively.
The circuit arrangement of inverter as
shown in Fig. 11 .20 may be operated with
star connected or delta connected load and
the switching devices can be triggered in
two different modes of operation such as
I. I80° conduction mode
2. 120° conduction mode
In this section, the operating principle
of above two modes of an inverter are
discussed elaborately.
+
v
+
v
A B c
Three phase load
(a)
A B c
Three phase load
(b)
(a) Three-phase inverter using switches and diode~
and (b) Three-phase inverter using transistors and
diodes ·
z B-----
c-------.J C--------------(a) (b)
•G•M•Ii~!.l~~~~l• (a) Delta connected load and (b) Star connected
load

''
'
--- Inverters 107
11.6.1 180o Conduction Mode
In this mod~, each switch conducts for 180oduration in each cycle of the output voltage. Each leg of
three-phase J~verter const~ts of two switches, one is a part of positive group switches and other _JS a
part of negatJ~e group _switches. When a positive group switch of a leg conducts for Igoo duration,
its correspondmg negattve group switch of same leg conducts for next Igoo duration as one compl_ete
cycle is equal to 360o duration. For example, if the switch 5, of leg I is ON for Igoo duration dunng
0° ~OX :S l goo, then the switch 54 of leg I is ON for 1goo duration during Igoo ~ (JJ{ ~ 360o. The
sequence of switching of semiconductor switches s,, 5
2
, s3
, 5
4
, 5
5
, 5
6
are depicted in Fig. I I.22. In
· h· h h · d t· e Hence twothis sw1tc mg sc erne, t ree SWitches from three different legs are conducte at a 1m · . ' .
switches from the same leg are not switched on simultaneously. The one complete cycle of swJtch~ng
can be operated into six modes and each mode operates only for 60° duration. The different operatmg
modes are given in Table I I.5. The switching sequence is
Mode-l
Mode-ll
Mode-III
Mode-IV
Mode-V
Mode-VI
QMIIJI Sequence ofswitching ofsem~conductor switches
1!
0 5: {t)f 5:-
3
1! 27!
-<{t)t 5:-
3- 3
47!
1! 5: (l)f 5: -
3
41! < {t)/ 5: 57!
3 - 3
' ', ___... _..._
c- --·------------------------- h t during each interval, only three switches conducts and two from the
--- - - -------- ----- ----- -- --.L... ----- -- --- - --- -- --- ------ - ------------ -- _:
It is clear from Table li .S ~ a negative group or vice versa. Assume that the load is connected in
P<>sitive group and other one roVm V and VcN· The line to line voltages are VA
8
V and v . The
sta It es are 4N• BN . ' Be c 4·
rand the phase vo ag c. · ix different modes are g1ven below: ·
equivalent circuits of inverter ,or s
1
( . rr) 0 ring mode I, switches S1, 56, and S5 conduct for 0 s wr s 1C and the· Mode I 0 s; wr S: 3 u 3
equivalent circuit is shown in Fig. 11 '23.
...

c
St v R
3
+ +
v v
s4
B
QWIIJI Equivalent circuit of Fig. 11.20 during 0 $;rot$; 1C when 51, 56 and 55 are closed
3
The current i0
is given by
. v v 2 vl - - =--
0- RIIR+R- R +R 3 R
2
The lines to neutral voltages are
i v
R
asRIIR=-
2
. 2V
V -V _ oR-AN- CN-- - -
2 3
as z =--
o 3R
2
then V8N=-VN8 =--V
3
(
TC 2TC)2. Mode II 3 :5 rot :5 3 In mode II, switches S1, S6, and S2 conduct for 1C :5 rot :5 2n and the
3 3
equivalent circuit is shown in Fig. 11.24.
+ +
v v
R
Equivalent circuit of Fig. 11.20 during Tr $;rot$; 2Tr when 51
, 561
and 5
2
are closed
3 3
d

..--------------
i'
fhc lin 'S ftl flt:Ulruf '( ftu~s U:
·"'" = ~~~R =.:.
J

'
• ~ ~,!
IS# :::--
' 3R
us RIIR = R
'
NH ::: , . . ::: ~ R - h .
' - -; l n ~s = - ·,., = -- mli ·:' = - · ~ =--
.J
3. Mode Ill cor _ 1r .
2/r )
J unng modt' Ill.
tht• l.aquivalent circuil is ~hown in Fi 1 1
·wirrh ~ 1• -~· und , · mJu"·r f ) f -=--_~
Th
. ' . g. 1._5.
t: n1rrent 111 ts gl en by
1,= =-----
RIIR+R R
. +R JR
The line t neutrnl voltage · ure -
R
as RII R =-
. -
as 1 = - -v .., R
. 2
C=IR= -( 3 lhen s =- .v = _..=. '
3
Equivalent circuit of Fig. 11.20 during 2: :5: CtJt $Jr when Sv S.! and S
Similarly. during mode IV ( JC 5 aJt5 ~K). the line to neutral voltages are
V 2V
v _ v. =--and VBN= - -
AN- C 3 3
In mode v (~ 5 aJt5
5
J). the line to neutral voltages are
2V V
VAN =-) and VsN = VcN=]

During mode VI (
5
; $rot$ 2tr), the line to neutral voltages are
V 2V
VAN=VoN=--and VcN--
3 3
When the lines to neutral voltages are known for a complete cycle, the line-to-line voltages are
computed by using following expressions:
VAB = VAN- VBN
VBc = VBN - VcN
VcA = VcN - VAN
The output phase voltages ~-1N• V8N, VcN and line voltages VAB• V80 VcA are depicted in Fig. 11 .26.
The instantaneous line voltages can be expressed in Fourier series as they are periodic non-sinusoidal
waveform as given below.
oo 4V ntr . ( ntr)
vAn= L -cos-sm nwt+-
n=l ,3,5... ntr 6 6
VBc = f 4
V cos ntr sin(nrot- ntr)
n=l,3,5... ntr 6 2
oo 4V ntr . ( 5ntr)
vcA = L -cos-sm nwt + --
n=t.3,5... ntr 6 6
If the triple harmonics (n = 3, 9, 15, ...) are zero, cos ntr = 0.
The rms line to line voltage is expressed by
6
VL(rm•l=[~1V'dmtr=[~vt;-0)]"'
fi=..fj v =0.8165 v
The rms value of nth component of the line voltage is given by
4V ntr
VLn(nns) = r;:; cos-
....;2ntr 6
The rms value of fundamental component of the line voltage is
4V tr
VLI(mls) = r;:; cos-= 0.7797 V
....;2tr 6
The rms value of phase voltages is
VL(nns) fi
Vph(nns>= r:; = - V=0.4714 V
....;3 3
The power output of inverter with resistive load is equal to
v2 2 v2p = 3 ph(nns) = __
R 3 R
The rms value of switch current is given by
1 _ Iph(nns) _ Vph(nns> _ V
Switch(nns)- .J2 - fiR -
3
R

Garing
signal
of~
Gating
signal
ofS3
Garing
signal
of54
Gating
signal
of5s
Gating
signal
of56
0 1C 4
lg• 3 3
lg4
0
0
k4JT
3
v....a
VI------;
Conduction
mode
0
3
3Jr
•
Time
.... rot
Time
.... OJI
Time
OJI
Time
OJ(
Time
WI
Time
(i)(
Time
(i)(
Time
(i)(
Time
(i)f
Time
(i)f
Time
(i)(
Time
(J)f
Ill 5 s S 5 5 line voltages V V v 0
d h
~+ilj Gating signals ofS1, 1r ~ •
9
& A& so CA n P ose voltages VA,., vBNo VCN
nt

A thrc~-phase hridgc inwrta is fed from 400 V de supply. If the semiconducto .
(tranststors) w 11c ar~ used in in' erter conducts for 180° duration and the uwerter is supplying a star~swnchcs
resistive load l1f 10 n.determine - onnected
(a) nns v:tlue of per phase ,·oltage and line voltage
(b) m1s value of load ..:urrcnt
(c) nns value of current flows throul!h transistors
(dl Power dt'livered to load -
(cl Average source current
Solution
Giren: r ; 400 V. R ; 10 !.1
(a) The nns value of per phase load voltage is
VLtrmsl J2Vphrrm,
1
= .Jj =3V;0.4714Y=0.4714x400Y=l88.56V
The nns line to line voltage is
J2VLtrm•l= .Jj V =0.8165 V = 0.8165 X 400 V = 326.6 V
(b) nns value of load cumnt per phase
I ph= Vphtrm•l = 188.56 = 18.856 A
R 10
(c) nns value of current flows through transistors
v 400
lrmsttnn"''"' ' =JR=3xJo= 13.3333 A
(d) Power delivered to load
P.
vJ(rm>) 188.56
2
o = 3-R- = 3 x -
10
-Watt = 10.6664 kW
(e) Average source current is
I = ~_10666.4 _., V -
400
- 26.666 A
• A three-phase bridge inverter operates · 180o d · 0V
de battery and it is connected with a R load
10
con uctmnmode and it is operated by a 12
output voltage and (c) distortion factor ando:a~~nDet;nnmef(a) nns ltne to Ime output voltage, (b) fundamental
tc actor o output voltage wavefonn.
Solution
Given: V ~ 120 V, conduction of transistor is 180° and R ; 5 Q
(a} nns line-to-line output voltage
..[2
VLtnn•i=JJV =0.8165 V =0.8165 X 120= 97.98 y
(b) Fundamental output voltage
V _ 4V tr
Lt(rms,-Jj;cos6= 0.7797 V = 0.7797 X 120=93.564 V
(c) Distortion factor is
. DF=~'<-1 =93.564_
VLtrm>l 97.98 -0.9545
Harmonic factor ofoutput voltage waveform
HF=J I -1- ~DF2 - 'Jo:9S4s'2-I = 0.31256
lm1ftfm 7U
• - . A three-phase bridge inverter
. iSOoconducuon mode. Find (a) rrns value li operates is operated by 400 V de . .
tOd (cl fundamental output voltage in te neflo line output voltage (b) I mlput voltage and tt operatesall nns o line volta d · nns va ue me to phase output voltage
. ge an phase voltage.
solutton
Giren: V ~ 400 V. and conduction of transistor is 180,
(a) rnts hne-to-ltne output voltage is
.[2
VLrms! = J3V =0.8165 V=0.8165 X 400= 326.6 V
(b) rms value line-to-phase output voltage is
~ = 11Ltrm>t _ 0.81651' 0.8165x400
Pha.«(rm<t fj -~=~=188.568V
(c) Fundamental output voltage in tenns of line 1·oltagc
4V 1r
VLitrmst= Ji1rcos6=0.7797 V =0.7797 x400 =311.88 V
Fundamental output voltage in tenns of phase voltage
VLinnsl 311.88
Vphas<t(rms) = fj =h =180.069 V
• - • A three-phase bridge inverter operates is operated by 240 V de supply and it operates in
180° conduction mode. When the inverter is connected 10 astar connected R load with R = 5 n and L = mH and
inverter output frequency/,, = 50 Hz. compute (a) instantaneous line-to-line vol!age and line current in Fourier
series, (b) rms value line-to-line output vol!age, (c) rms value lin~ to phase output voltage and (d) fundamental
output voltage in terms of line voltage and phase voltage
Solution
Given: V; 240 V conduction of transistor is 180' . R=5Q. L =
10
mH and /.;
50
Hz
(a) lnstantaneo,us line-to-line voltage in Fourier series
~ 4V 111! ( IIIC )
v = L - cos- sin nw1 +6
AB ,= 1,).5 ."1! 6 Ill!
1
- ) are zero as cos-=0 ·
where, the triple hannonics (n = 3, 9. '··· · 6
w= 2nf ~21f x50= 314 and~=30'
4V ( TC ) 4x?40 IC · (314l +30')= 264.772 sin(3141 + 30°)
"Aa
1
= - cos!:sin (J)I+- ~..:--=-cos6stn
tr 6 6 IC
4V 5rr . ( lf)-~cos~sin5(3141 +30')
vAss=
5
7rcos
6
sm5 w1+6- 51! 6
=-52.954 sin 5(3141 +30' ) o
)
4 240
7Jr .
7
(
3141
+30°) =-37.82sin7(3141+30)
4V 77r ( IC -~cos-stn
•As7 =
7
1fcos
6
sin7WI+6- 7Jr 6
"ABII =~cos.!..!!:sin I {w1 +~)lltr 6 -?407sin11(3141+30o)
=4 x 240 cos.!..!!:sin 11(3141 +30')-- .
lltr 6

714 ,.,_, El«tronics
~ instantanrou lint' to line ,-oltagc in Fourier saics
, .AB = 264.77_ >in(3 14r +.10•1- 52.954 sin 5(3 14r + 30°) - 37.82 sin 7(314r + 30°) ..
... + 24.0 sin llt.,l 4r+ .l0' + .. -
Zt= ~Ltan·1( 11~L)
· (14 IOx iO· -')Ztt = 5' + (.114 10 10-'I:L tan-1 .
5
= 5.90L32.21 °
Zt_,= ,/5:+(5 314 lO x lo-.>1: -1(5 x3 14 x iOx iO--' )- • tnn
5
=16.476L72.JJ•
ZL-=Js: .._, .114 x iOx lO--'I: L -1( 7x3 !4 x !Ox !O-·')tnn
5
= 22.54L 77.J S•
ZLII = J 5:+(1l x.114 x l0 10· -'):L t -l( ll x J!4 x lO x iO· -')an 5 = 34.9LM !.76•
The: mstzmtnnt'CUS line to li n~ cum·nts in Fourier series
;,« = 44,87sin(3 14r- ~.2 1 • -J.l 14 sin(5 x 3 14r +7767•- 1 6779 ·' (7 314
.. · + 0.689 sin(ll .11
41
+
248
_
24
•
1
+ · · sm x r + 132.82•)..
(b m1s line-to-lone C>utput ''OIIagc
' - 2 .
Lorn» >- h =0.SI65 V=0.8165 x 240 = 195.96 y
( ) mS 'nluc line-to-phu c output 'llhu~c is
v,.,..,l..,,,= VL<m" ' = O.Sl65V - 0.8165 x 240
(d) T ~- ~= 113. 14 V
Fundamental output 'Ohngc in l<'mls of line 'Ohu~c
V - 4V rr -
Lloom,,---:rr;cosb =0.7797 V=0.7797 X 240 = 187.128 V
Fundnmcntnl output 'Ohugc in t<nns of I .
V P1nsc voltugc
_ 1. 1 mh l 187.P8
11
""''---r=-:rr--=108.04 V
11.6.2 120o Conduction Mode
In this mod,-, cnch ' " lll'h conduct .
m each c~dc of the output '"lttP·' lor_120• durnlion s,.S,.I S,,S2 ! S, S21 SJ. S• I S•. S, I S,.S~
phnsc lncrtcr COnSI'IS ul l 0 'II /~·~ •~nch leg OJ' lhrce
pO. Il iVC grou h !lites, onl' IS 'l p rt . - - - - /P " ' llc C> nnd othct · ' '1 ol _...,. Sequence ofswitching ofsem ·
group 'wilehe' 1h " u pull ot 11 •f c gat1ng tgnuls of
1
cguttve condvctor switches
o scnllconductor swnchcs ~ . swnc tcs arc given fo .
only !wn S1 llchc, cond ' .. I· 5:- s,..,,S,. S,, arc de I . , t r el:~ry 60o Tho.: sequence of swttchrng
other from the ll(·gutJ v t ct 1111111>' mstunl of tune onp c
1
_lct 111
hg. 11 .27. In lhis swttching schctllC•
•
1
c group swn ·I . T · c rotn Ihe p . . d the
stx mN cs and each mode operot. c 1cs he ont• eolllplctc .
1
n~tttvc group of swttchcs nn 10
Table II 6 The swncht cs only for 60• dunn . eye c of swttdung can be: opcrutcd tn.
ng sequence . h ton 1he drlli · n 11115 s own 111 F
1
g
11
_
27
crcnt opcrnttng modes are gtve
Inverters 715
O s wr s~ s,. s.3
Mode-ll 1C 2TC
3s wrs
3
s,, s2
1lu<k-111
2rr
3s wr s rr s .s2
tvlo<k-I V rr S cors~ s3.s.
3
Mode-V
4TC 5rr
-S M S- s•.sl
3 3
Molk-VI
5TC
-s cor s2rr sl.s..
3
It is clear from Table 11 .6 that during each interval. only two switches conducts and one from the
positive group and olh<.:r one J'mm negative group. Assume thut the load is connected in star and the
phase voltages arc F_
1
_
1
• t·
11
_
1
ami t'n - The linc-1<>-linc voltugcs <Jrc l'w· Vue and ~'c.·t· The equivalent
circuits of inverter J'or six ditl'crcnl modes arc ~iven below.
1. Mode I ( 0 $ co1 $ ~) During mode I. switches S1 and 50 conduct J'or 0 $ cor S ~ and the equiva-
lent circuit is shown in Fig. 11 .28.
. fFI 11.20duling os wr
WitJII Equivalent Circuit 0
g.
The currem i, is given by
v v
i.,=R;R=Ui
R

v . v
VAN = inR = 2 as 'o= 2R
v v
VN8 = i0
R =
2
then V8N= -VN8 = -2, VcN =0
2. Mode II (~$Wt$ 2
;) In mode II, switches S1 and S2 conduct for !!_ s Wt < 2n:
3 -3 and the
equivalent circuit is shown in Fig. 11 .29.
The current i0
is given by
. v v
10
= R + R = 2R
asRIIR=!!_
2
VAN= i R ='!'_ as i0
=~!'_,and V8N = 0
" 2 3 R
V . R V V
Nc='o =2 then VcN =-VNc=-
2
R
I8IID Equivalent circuit af Fig 11 20 d . 1r 21r
· · urmg 3$wt $3 when 51 and 51
are closed
3. Mode 111 ( Z1r $ wr < 1r) .3 - Dunng mode Ill sw· h 21t"
equivalent circu"t . h . . , ltc es s3and s2conduct for - $ (.()( $1t" and the
Th . . I IS s own m FJg. 11.30. 3
e current '" IS given by
io=-v-=_x_
R+R 2R
Then
Inverters 117
v
v
R
@Ilui Equivalent circuit of Fig. 11.20 during !.!!_ swt s rr when 53 ond 52 are closed
3
Similarly, during mode IV ( Tr S Wt S ¥),S3 and S4 are closed and the line to neutral voltages are
v v
VAN= -2,VBN= 2and Vc,v =0
In mode V (
4
; $ Wt $
5
;) , S5 and 54 are closed and the line to neutral voltages are
v v
VAN=-
2
,V8,v =Oand Vc,v=
2
During mode VI ( 5; $ wt S 2Tr), 55
and S6 are closed and the line to neutral voltages are
v v
VAN =0,Vs,v =-2and Vc,v =2
As the lines to neutral voltages are known for a complete cycle. the line to line voltages are computed
by using following expressions:
VAB =VAN - VnN• VBc = VBN- Vc,v and VcA = Vc,v - VAN . .
The output phase voltages V~s· v8
s, Vcs and line voltage V~8, V8c, VCA are shown m F1g. 1. 1.~I. The
instantaneous phase voltages can be expressed m Founer senes smce these voltages are penod1c non-
sinusoidal waveform as given below.
~ ?V nTr ( nTr)
vAN = L =-cos6sin n(J)l +6
n=J.3.5... nTr
2V 11Tr ( ntr )
Vn,v = L ;;cos6 sin n(J)l-2
n= l.35..
2V nTr ( 5nTr)
_ ~ -cos-sin n(J)l + 6
Vc,v - £.. nTr 6
"l'L ~~~~ J 5... . voltage waveform V~B is given by
•ne Fourier series expressiOn of hne
vAn= L ~sin(nwt+
11
;)
n=6t:: l
t=0.1.2.3...

718 Power El«tronics
.
1
rly line voltages V8c and vSiJ!ll a , _ CA are represented by
Vnc= L ~sinn( 1r 21r) ·n=6ktl n1r WI+--- = L 3V · ( TC )
<=0.1.2.3... 3 3 •=6Hl -;;;smn Wt- 3
3V ( t=O.I.2.3...
vcA = L -sinn (!)(-If 2~r) • 3V
n=6k±l lllr 3-3 = L, -sin n(Wt - TC)
k=O.I,2.3 . n=6Hl fiTC
The rms value of phase voltage is expressed by k:O,I.l.L
[
)lf(y)2 ]Ill [ l Ill
Vph~«(nns)= ; !2 dwt = ~(f) (3f-o)]
.fi. v v
= ,[32=16=0.4082 v
The rms value of line voltages is
VL(nns) = J3vph(nns) = J3 Xi= ~ =0.707 V
The rms value of nth component of the line voltage is given by
3V
VLnlnns) =.finn
The rms value of fundamental component of the line voltage is
3V
VLI(mlS) =.fi.tr = 0.6752 V
The power output of inverter is ' <•i"
v;h(nns) - yl v
?,,=3-R--
2
R as Vphnse<nnsl = J6
TI1e rms current flows through switch (transistor) is
I ph<nml _ _ V_
/ swioch(nnsl= J2 - 2J3R
vph(nn<)
as Iphnse!nn<l =--R-
Inverters 719
, , • • .d ,c inverter is f~d from 600 V de supply. If the sem~conduclor switches

(b) rms value of load currem per phase
I = Vph!nmJ= 244.92 =
24
.4
92
A
ph R 10
(c) rms value of current tlows through transistors
v 600
I rms(tr.msistor) = --r;:- = J3 = 17.320 A
. 2-J3R 2 3 X 10
(d) Power delivered to load
Po= 31~h<rms) R = 3 X 24.492
2
X 10 Watt= 17.9957 kW
(e) Average source current is
I =P0 =17995.7 =29
_
992
A
av V 600
Example 11.16 A three-phase bridge inverter operates is operated by de supply and it operates in 12
conduction mode, determine (a) input de voltage for fundamental line voltage of 420 V, (b) rms value line-to-li
output voltage, (c) rms value line-to-phase output voltage and (d) transistor voltage rating.
Solution
Given: Conduction of transistor is 120°, VLI(rms> = 420 Y . H
(a) The rms value of fundamental component of the line voltage is
3V
VLI(rmsJ =J2n =0.6752 V =420
Therefore, de supply is V =
420
V = 622 V m
0.6752
,_
(b) rms value line-to-line output voltage is .r11lU •
VL;rms) =J3Vph(mls) =J3 X ~ = ; =0.707 y =0.707 X 62~ ~~~39(754 V
VO VL , , .
(c) rms value line-to-phase output voltage -J _ <· ron ~~'-/ .
,. --- - - --
J2 v v (' ~ f
Vphase{rmsl = J3 2 =J6 =0.4082 V =0.4082 X 622 =253.90 V
(d) Transistor voltage rating VcEO ~ I.5V ~ 1.5 x 622 =933 V
11.7 PULSE WIDTH MODULATED INVERTERS
Usually the output voltage of single-phase inverter is
a square wave. T!:e majo~~aY_b~s of square wave 5' 0.4
inverters are: ~ - ------:g 0.3
1. Output voltage of inverter is constant and it is -~
equal to ~~pply voltage V. 1~is output volt~~ 0.2
can not be controlled. :J:
2. Output voltage consists of third harmonic and 0.1
other harmonics as shown in Ftg. I1.32. The rms
'Value of nth harmonic component is
V =
4
~ forn=l,3,5, 7...
n mr 2
3 5 7 9 II 13 15
Order of harmonics
Harmonic spectrum ofsquare wave
inverter

~-------------------------------------------------------------~'~~~~~~· ~s_7Zl
'h~n n == I. the nns value of fundamental . 4V v.
..n..> harm nic factor is equal to component IS ~ = r;: = 0.9 V . Then the ratio -
1
= 0.9 .
,.~ nv2 V
I
HFn=- = - for n = 3 5 7 9
I Tl , • ' ...
Arter substituting the values of n. we get
1 I
HF3 = V=- = 0.333.
I 3
He- V5 1
r s,=-=-=0 J
v. ·-·I 5
V7 1
HF7 =-=-=0.143.
v; 7
and
9 I
HF9 = -V. =- = 0.111,
I 9
V1- I
H -~I- }
F.ll----=0.091
v; 11 '
HF.I5 = _) =-- = 0.0666
v. 15 .I
V. I
HF13
= __.!1. =- = 0.0769
v; 13
To control the output voltage. inverter must be fed from an ac-to-de converter or de-to-de converter.
However to control output ·oltage as '"ell as to reduce harmonics in the inverter output voltage, the
pulse width modulation (PWM) technique should be used in inverter.. In PWM control, the output
pulse duration is modulated or ·aried to control the output voltage. There are different methods of
mOdulations. but the most commonly used modulation techmques m mverter are
I. Single-pulse width modulation (SPWM)
2. Multi-pulse idth modulation (MPWM)
3. Sinusoidal Pulse Width Modulation (SinPWM)
4. Modified Sinusoidal Pulse Width Modulation
In this section, the above modulation techniques are discussed elaborately.
11.7.1 Single-Pulse Width Modulation (SPWM)
In a single-pulse ~id~h modulation inverter. there
is only one pulse on each half cycle of ou~ut
voltage. The width of the pulse should be vane~
to-control the inverter output voltage. Figure IJ.J_,
h . . d th output voltaoe V
s ows the gatmg s1gnals an e . ~
Waveform of a single-phase full-bridge mverter.
· h (BJTs MOSFETs,
The 2:ating signals ofs~lt~ _es_ ' .
1GB «::..:.:.:- TO tc ) are determme.dT, Thyristors and G s, e · - --- -- -.
b · al v and a earner
+
Ycomparing a reference s1gn ,. .
Signal v .. The maximum amplitude of.reference QWIIfl Single-phase full-bridge invert
and carrier signals are A,. and Ac respectlve~Y·..,.The ti d I s:: er
c: d ·er signal 1s same as un amenta •requency of output 1•requenc~ of reference an earn - - ·d h f 1 --; vo tage The
OUtpUt ~led by controlling t~ ~~ t obpu.§;dwllic.h depends uoon the amoft d
vo tage cane . d dulation mdex can be determin d fi h 1
u e
lllodulation index (M). The amplltu e 010
e rom t e expression -
- - Ar
M=-
Ac

Voltage Reference
Gating
signal
ofsl & s2
Gating
signal
ofs3& s4
Output
voltage
l g3
l g4
Va
I
7r
2
~d---+I I
I
I
0 I
'~--~------+---~--+-------~--~~~--------t---:~Time1-: 1 ~ 2Jr {l)(
I
dJrl
2>-- 2
g ~ q
2 1
2
0
~--+-------~---+--~--------+---~~~--------~--!-~TimeI& 2Jr (l)/
v
0
aJ
~--j_------~--~~--~-------T----~--~------~L---~~Time
n 2n rot
- V ------- - - - --------- - --- -- -- - - L----...1
Conduction
ofdevices
(a) Carrier signal and reference signal, (b) Gating signals of 51 and 52, (c) Gating signals of 53 and S4
(d) Output voltage of single phase full-bridge inverter with one pulse per half cycle
The angle and time of intersections for reference and carrier signals are
a T
a1
=wt1 or, t1 =--1
=(1- M)- and
(1) 4
a T
a2 =wt2 or, t2 =- 2
=(1 + M)-
where, T is the time period of signals
The pulse width in angle is d = a2 - a 1
w 4
T
The pulse width in time is t2 - t1= M 2
The rms value of output voltage is
[
(lr+d) ]1/21- ,- lVo(nns) = - J V
2
dt =V -
n (lr - d ) n
-r
When the amplitud~of reference signal Ar is v~f!:om 0 to A, , the pulse width d can be changed
__~o!!l_9°.to 18.0° or .~:1££Qrdi~gl t~e_ ~~ va1:1-~ ?__?u p t Y?lt~~_V,l([llls~ c~~e ~arieo~ffii_ro_m_.....--:-::
'----The output voltage waveform ts deptcte m Ftg. 11.Kine snape of output voltage 1s ca e quasi-
square wave which can be expressed in Fourier series.

~----~--------~
Inverters 723
DF
10.0
' '
' OF :
:---------~---------~---------~---------
1 ' '
I t
1.0
0.8
OF 6.0
1
I I
I I
----- r --- ------~-- -- - --- -
1 I
I
4.0
I I
- ~ --- -- - - --~ --- - --- - -
' I
I
_.o
0.0 0.2 0.4 0.6 0.8 1.0
Modulation index (M)
IMiifl Harmonic profile of single pulse width modulation inverter
Due to half-wa c ~ mmetry of output voltage, the even harmonics are ab ent and only odd b
components are present. The alues of h, are gi en by "
? (tr +- rf ,,_
b = .:_ J V sin 11WI · dwr
II
rr ,n- r1 ,,
4V . nrr . ml== - s
1
n- sm - where. II = I,3. 5...
111C 2 2
4V · nd I 59b =-sm- wherc.11 == .. ...
11
nrr 2or
and
4V . nd 3 7 11
n
where. n == . . ...
b :: - - Sl -
11 n1C 2
The output voltage can be expres ed a
"" 4V n1C . nd . h_ " - sin-sin-smnwt w ere, n=l, 3, 5 ...
v0
- .t.- n1C 2 2
11 == J.J.5....
4v [ d 1 . 3d . 3 1 . 5d 1
or ·
11
-sin wt- -sm-sm wr +-sm-sin Swt . 7d ]
vo=-;- Sl 2 3 2 5 2 -7sm2sin7(l)f+ ...
nd
To eliminate nth harmonic. sin 2 =
0

or
nd
- =kn where, k is an integer
2
Th c h 'd h f I . 2kn 360k . d -r , . . t th. d here1ore, t e w1 t o pu se IS d =-- =-- 111 egree. 1o e 1mma e 1r armonic, the
n n
. d I 'd h . 360k 360 . k I d 3 ThreqUire pu se WI t IS equal to d =--=-- =120° assummg = an n = · e variations
n 3
of fundamental component and other harmonic components and the rms value of output voltage is
depicted in Fig. 11.35.
The disadvantages of single pulse modulation are as follows:
1. Harmonic content is high.
2. The maximum rms value offundamental component is only about 90.09% ofdc input voltageY.
3. Third harmonic dominate.
Example 11.17 A single-phase PWM inverter is fed from a 220 V de supply and it is connected to a RL
load with R = I0 .Q and L = I0 mH. Determine the total harmonic distortion in the load current. Assume width
of each pulse is 7r and the output frequency is 50 Hz.
2
Solution
Given: V= 220 V, R = 10 .Q L = 10 mH d= n =90°
' ' 2
The rms value of nth harmonic component of the output voltage is
V
4V . nd
n = J2nn sm2
The impedance offered to the nth harmonic current is
Zn =)R2
+(nmL)
2
Therefore,
Similarly,
_ 4V . d _ 4 X 220 . 90 =139 9999
V
Vj - r;; sm - r;; sin .
"1!27r 2 "1!27r 2
and
Z1
=)R2 + (mL)2
= )102
+ (27r X 50 X 10 X 10-3
)
2
= 10.482 .Q
4V . 3d 4 X 220 . 3 X 90
46 666V3 =~sm2 =
3
J2n sm-
2
-= . 6 V and
Z3
=)R2 + (3mL)2
= )102
+ (27r X 3 X 50 X 10 X 10-3
)
2
= 13.744 .Q
4V 5d 4 X 220 . 5 X 90
V5
= -r,::-sin- = r;; sin -
2
- = 27.9999 V and
5"1!27r 2 5"1!2n
Z5
= )R2
+ (5mL)2
= )102
+ (27r X 5 X 50 X 10 X 10- 3
)
2
=18.626 .Q
4V 7d 4 X 220 . 7 X 90
V7 = -r,::-sin- = r;; sm-
2
- = 19.9999 V and
7..;2n 2 7"1!2n
Z7 =)R2
+ (7mL)2
= )102
+ (27r X 7 X 50 X 10 X 10-3
)
2
= 24.165 Q
4V . 9d 4 X 220 . 9 X 90
V9 =~sm -= r;; sm--= 15.5555 V and
9"1!27r 2 9"1!27r 2
Z9= )R2
+ (9mL)
2
= J10
2
+ (27r X 9 X 50 X 10 X 10-3) 2 = 30.00 Q

--------------------------------~~~~
'~ = _l_ =~- ~~
- z~
= - """.J -
:. I .
1-= -=--- -~ = 0
z_
ba:rm
/~
11.7.2 Multi-Pulse Width Modulation (MPWM)
w-here m
1
=fc is the frequenc)· modulation rario.
fr
Hence, the frequency modulazio · ex m.-cornrots
also called uniform pul5e-r;~idrh modu a 'o CL1' . f).
·oltage. This _-pe of modulation is
Figures 11 .36 and l I3 show the ~ - · and
full-bridge inverter ~itb sin~ e .
The rrns output ·ohage is
·oh:age ,-a,·d onn ofa single phase
lse "'-i dth modulation respectively.
pd
where._ p is number of pulses per half . dura io of ea h Ise. V"'
. With the variation ofmodulation !!.Jdex · I fro 0 10 L e pulse ,; dlh ·aries from 0 to !:_
~ t~~output ..-oh:age ·aries from 0 to -_ p
180°
or-
p
10 half-wave svmmeuy o o ·ohage. the e ·en harmo
components are presen~ The ,-al of ~ are giYm by
are absent and onlv odd b
• n
or
2Jt
b =-J'o sin nlJJl . d(J)I
lf Tr
0
tr ..-! )
2 .
b. = p x - J
n lr-~
. d 2p I 1 - .l
sm niiJ · (fJl =-- cos n(J)rI -
me r-~

(a)
(b)
(c)
(d)
oltage
Ac
Carrier
signal
Reference
signal v,.
Ar 1----~..-----~--~~--T----i
I
I
~d~
I
I I
l gl
:..-d-+: 7CI I
Gating I I
l g2
I I
signal I
ofsl & s2
0
0 al (12 (13 (14 7C
Gating
l g3
signal l g4
ofs 3 & s4
0 I I
I I
:+- d ---+-1
I
: d
Output Vo :CXrn+2I
voltage I
v
lX<j 17C0 a1 (12 (13
I
r I
I
-V ------~------------------------- -:----
lXm
I
7C+ Clnl
Conduction
OJ []]ofdevices 22
I
27C
in
Time
(J)f
Time
27C (J)f
Time
27C 27C (J)/
[I]4
rn4
Q&lllj (a) Carrier signal and reference signal, (b) Gating signals of 51cmd 52
, (c) Gating signals of 5
3
and 541
(d) Output voltage ofsingle phase full-bridge inverter with two pulse per half cycle
d
where, p is number of pulse per half cycle and r =am+ 2
4pV . . nd
b = --sm ny sm- where, n = I, 3, 5
" nn 2
Then
8V . . nd h
lncaseof p=2,b,=-smnysm- · w ere,n= 1, 3,5 ....·
' nn 2
Then output voltage can be expressed by Fourier series as given below.
~ 8V . . nd .
V0 = "-- -sm ny sm-sm nwt where, n = l, 3, 5...
n=l.3.5... nn 2

.----- Inverters 111
Voltage
Ac
Ar
(a)
I I
Gating f gl
~d~:
I
signal lg2
ofS, & Sz
I I
I I
~d~I
(b)
Gating fg3
signal lg4
of53 & S4
Time(c)
r+d-+i
I I
TC : 2TC (JJ/
d ' I I
Output vo
voltage
v
(d)
- V
Conduction
ofdevices
I
o;,,
~
~
o;,, ~d~:
2 ~ I
I I
I I
TC + a,n
rs:l
~
fS;I
~
~
~
{S;I
~
js;j
~
I
I
Time
: 2TC
(JJ/
QMilp (a) Carrier signal and reference signal, (b) Gating signals of 51and 52, (c) Gating signals of 53 and s
(d) Output voltage of single phase full-bridge inverter with two pulse per halfcycle 4'
or
8V [ d . I .
3
. 3d . 3
I .
5
. 5d .
v =- sinysin - smwt+ - sm y sm-sm wt+ - sm y sm -sm5wr
(} Jr 2 3 2 5 2
when p = 2
I . 7 . 7d ]
+ 7sm Y sm2sin 7wr + ...
The amplitude of the nth harmonic voltage is equal to
8V . .· nd 2 d I 3 5V = -Si ll ny sm - when p = an n = , , ...
" nTC 2
It is cl t' b uation that the amplitude of the nth harmonic voltage depends
. ear rom a ove eq _ on rand d To
ehminate nth harmonic, sin y = 0 or, ny - TC •
• 11d nd =k"" h k · · t
and sm - = 0 or. ? '" w ere, 1s 111 eger
2 -
. . 2kTC 360k
Ther lC' 7r 180o d the width of pulse I S d = - - = - In deg·ree.
elore, r = - = - an 11 11
11 11 .
the variations of fundamental component and other harmomc components and the rm
Voltage is depicted in Fig. 11 .38. · s value of output

!
i
I
II
ns
I '
!L
I'
..''. 'I
'·-'
0.~
I
I
·II
'I
I
,.
'I
... ......I
''
-'
'I
I
'
0.4 0.6 O.K
-----'..,. 1ltoduluti~.ln index (i1)
IJF
in ''"
QWf@ HomlOfliC profil of multi puis modulation with P = 5
The angle and time of intersecti~)ns for rt'f~t'l!nC~ und <.:urrier signuls ar~
lfV - ror 'lr 1 =a,. =,,,, --M).!_ form =l. 3. 5.... 2p and..,., - . Ill ' • Ill (l) 4p
am l M) T t' "> 4 6 .,a - ,.,,, or. r =- =1m - + - or "' = .... . . ... -11Ill - l.U Ill Ill
(l) 4p
where. Tis the time pt!riod ~lf signuls.
The pulse width in angle is .d = am+t- a,
r
The pulse width in time is tm+t- 1111 =M-
2p
Example 11.18 The output voltage of multi-pulse modulation invertt•r is shown in Fig. 1 ..W. Determine
(a) the nns alu~ of output oltage. (b) the nns value of fundamental ~omponent of output voltage. ld the llltal
harmonic distortion.
Output 1()
voltage
200 V
-lOOV
--.---- .---
Tr
4
~ Tr ~ Tr
!! 2 !!
4 4
--------------------- ......__
Tr
4
3Tr
2
..__
2Tr
Time
(t)(
QM!Iijl Output voltage of multi-pulse modulation inverter
Solution
(a) The nns alue ofoutput voltage is equal to
[
2()()2 XTr Jlr2
v.= ~ 4
=2ooH=I41.42v

~-----------------------------------------------------------------~~~nv~e~n~e~~~729
Due to half-wave symmetry only odd h .
~r/2 ' armomcs are present in the Fourier series.
b - 4 f 0 4 J~r/8
n-- Vo(t)sm nwt . dcm- f 200 . d
1! 0 --; sm nw1 · WI where, n =I, 3, 5
I~ tr/8
4 J~r/8
bl = 1! J 200 sin Wt . dwt = 800[-cos WI]67.5:
Jr/8 1! 22.5
Therefore,
800 .
=-;-[-cos 67.5 +cos 22.5] = 137.75
(b) The rms value of fundamental component of output voltage is
VI= b);;= 137.75 =97 41 v
'J2 J2 .
(c) The total harmonic distortion is equal to
T
'uD _ JV} - Yr2 Jr-i4_1_.4_2_2___9_7_.4_1_2
n - = = 1.052
Yr 97.41
Example 11.19 A single-phase transistorized half bridge bipolar PWM inverter is operated from a center
tap 96 V de input voltage. The fundamental output frequency is 50 Hz and the carrier frequency is I kHz and
modulation index is 0.75. Determine (a) carrier ratio mf, (b) number of pulses per cycle, (c) fundamental output
voltage, (d) distortion factor of output voltage waveform and (e) harmonic factor of output voltage waveform.
Solution
Given: ~ =96 V ,/,. = 50 Hz,.fc. = I kHz and m = 0.75
0 0 0 !, 1000
(a) Carner ratio m1 1s mf =; =5o=20
mf 20
(b) Number of pulses per cycle P = 2 = 2 = 10
I V I
(c) F d I t It e v =-m-=-x0.75x96=50.919Vun amenta outpu vo ag o l(rrns) J2 2 J2
(d) Distortion factor of output voltage waveform
DF = Vol (rms> = 50.919 = 0.5304
96
Example 11.20 A
. 1 -phase transistorized full bridge bipolar PWM inverter is operated tlrom
120smg e d 1 . . d .
0
a V
d b . . d 'th a RL load. If the mo u at1on m ex IS .9, determine (a) rms 0
t
c attery and 1t IS connecte WI •
1 u put voltage
(b) . 1 (c) distortion factor of output vo tage waveform, (d) harmonic f '
fundamental output vo tage, . actor of output
Voltage waveform and (e) gain of mverter.
Solution
Given: V= 120 V and m = 0.9
( ) I v - v =120 va rms output vo tage om1s -
1
1
V =-mV= r;;;X0.9xl20=76379V
(b) Fundamental output voltage o J(rms> .fi v2 · ·
~ DF = Vo l(rms) _ 76.379
(c) Distortion factor of output voltage wave1
orm - = 0 6364vo(rms) 120 .

(d) Ham10nic factor of output voltage wavefonn HF =J1
, - I =J 1
, - I =1.212
DF- 0.6364-
(e) Gain of inverter is G =0.707m =0.707 x 0.9 =0.6363
11.7.3 Sinusoidal Pulse Width Modulation
In sinusoidal pulse width modulation, several pulses are used in each half cycle of output v~Itage but
~of th_~_pulses is not same as in the case of multiple pulse modulations (uniform pulse width
modulation). But ili"bot h sinttsoidalj?.ulse ~Tcit'h;~~d~atio11 ariamulttp e pulse modulations, ~he width
ofe~ch pulse varietwith tile~~n-plit.ude ofsiiw wave reference voltage. In ~ase ofsinusoidal pulse width
modulation, the width of the pulse at the centre of the half cycle is maximum and decreases on either
side. The generation of gating signal of switches by comparing a sinusoidal reference signal v,. with
a triangular carrier signal vc is depicted in Fig. 11.40. A comparator is used to compare a sinusoidal
reference signal v,. with respect to a triangular carrier signal vc. When the amplitude of sinusoidal
reference signal is greater than the amplitude of triangular carrier signal,; comparator output is high.
If the amplitude of sinusoidal reference signal is less than the amplitude of triangular carrier signal,
comparator output is low.
Voltage
l gl
Gating Ig2
signal
ofS1 &S2
Gating lg3
signal lg4
ofS3 & S4
Output v0
voltage
v
- V
Carrier signal Reference signal
Sinusoidal pulse width modulation

Inverters 731
If the frequency of carrier signal ·._ r d · · · h '·
L J c an the frequency of sinusoidal reference stgnal IS j,., t ere 1,
fc ·
re m=- earner pulses per half 1 · 1a 2fr eye e. The number of pulses in each half cycle IS e.qua to
(f--1)= (m -I).
-fr
The modulation index is M = ~r and it controls the rms output voltage and the harmonic content
of the output voltage waveform. c
When dm is the width of the mth pulse, the rms value of output voltage is
[
J p d ]1/2
vo = v L. _..!.!l._
111=1 7r
From the harmonic analysis of output voltage of sinusoidal pulse width modulation inverter, the
following features will be observed:
I. In linear modulation (modulation index is less than I), the largest amplitudes in the output
voltage are associated with harmonics oforder
fc ± 1 or, 2m ± I where, m in the number of carrier pulses per half cycle.
fr
2. By increasing the number of pulses per half cycle, the order of dominant harmonic frequency
can be increases so that the higher order harmonic frequency signals can be filtered easily
and size of filter is minimized. If m = 5, the 9th and lith order harmonics will be significant
in the output voltage and these signal can be filtered easily. But with the increase of m, the
switching frequency of semiconductor devices will be increased. Therefore the switching loss
will be more and inverter efficiency will be reduced. During design of inverter, there should be
a compromise state between the filter requirement and inverter efficiency.
In case of over modulation (modulation index is greater than I), the lower order hannonics wil
be present in output signal and pulse widt~.no longer a _sine function: of ~he angular position of pulse.
The output w ltage waveform of sinusoidal pulse Width modulatiOn mverter can be expressed by
Fourier series as given below.
v (t) = b sin nwt for n= 1,3, 5, 7...0 /1
2p 4V . ndm { . (a + 3dm) - sin n(n +a + dm )} .c
where, b = ' -sm- sm n m
4 m
4 tOr n = I, 3, 5, 7 ...II L.,. ..,. 4 •
m = l 11H a T
The angle a111
= (J)/111 and time '~~~ = ;' = 1
x + m 4(p + 1)
wh . ted from the following equations:ere, tr IS compu
. 4t(p + I) = M sin {w(tx + m T )} for m = l, 3 5 7
2I- T 4(p +I) ' ' ... p
and 4t(p+I)=Msin{w(rr+m T )} form=2,4, 6, ...
2p
T · 4(p+l)
The width of mth pulse in angle is dill = am+I- am
a,+1 a,
The pulse width in time is tm + l - t, = w - -;-
i
I
'I
1
i

d h h monic components and the rms value ofTh~ variations of fundamental component an ot er ar
mltp:·t voltage is depicted in Fig. 11 .41. The variation of Jwith respect to M is illustrated in Fig.
11 .42.
VI
DF
in %
v,
v
1.0
v Linear
modulation
Over
modulation
DF
t
1.0 --- -- f - --- -~--- - --~ ----- t - ----0 • •
- - ~ - - ~ - - - - - ~- - - 0.8
0.8
0.6
0.4
0.2
0.6 v,
v
0.4
t
0.2
0.0 '1 .0 2.0 3.0
0.0 0.0 -~)> Modulation index (M)
0.0 0.2 0.4 0.6 0.8 1.0
-----+ Modulation index (M)
QMIIII Harmonics ofsinusoidal pulse-width
modulation with p =5
vQjllfl The variation of ; with respect to M
11.7.4 Modified Sinusoidal Pulse Width Modulation
It is clear from output voltage of Fig. 11.43 that the widths of the pulses can not change significantly
with the variation of modulation index at the middle ofhalfcycle due to the characteristics of the sine
wave reference signal. When the carrier signal is applied during the fast and last n (60°) interval of
3
each half cycle ( 0 $; mt $; ; and
2
; $; mt $; 1C) ,the width of the pulses will be changed significantly.
This type of modulation is known as modified sinusoidal pulse width modulation.
Due to change in output voltage waveform as shown in Fig. 11.43, the fundamental component is
increased and its harmonics characteristics are improved.As the number ofswitching of power devices
are reduced, the switching losses are also reduced.
4 am T
The angle a, =(J)(, and time t"' =w= t_,.+m
12
(p +
1
) form= 1, 3, 5,7 ... p
where, r.,. is computed from the following equations:
1- 12t(p +1) =M sin{w(t:c +m T )} for m =1 3 5 7 p
T 12(p + 1) ' ' ' ···
and J2t(p +1) = M sin{w(rx+m T )} form= 2, 4, 6,... 2p
T 12(p +1)
The width ofmth pulse in angle is d, = am+l- a,
'd h. . . am+l aThe pulse W I t m time IS tm+l - t = --- -l!L.
m (J) (J)

--------------------------------------------------------------------~~~n~ve~rt~er~s~13
3
Voltage
I I
I
Tel
I
I
I
I
I
I
I I
I
I
I I
I I
I I ~ ' Time
r:~-:-t---T~:~:----------~:~~--~~:~---TC~~ ----~~~L-~~--------~_l_j~--~~--~-2~1r----~~ GOt
I I I I
I I I
I I I
I I 1 I
I I 1 I
~-+-dm~ ~
I I
I I I
I I I
I I I
I I I
I I I
I I I
I I I
1
I I I
I I
I I I
I I
I I I
I I
I I I
I I
I I I
I I
I I I
I I
I I I I I
Q¥11@1 Modified sinusoidal pulse width modulation inverter
The variations of fundamental component and other harmonic components and the rms value of
output voltage is depicted in Fig. l .44.
11.8 RESONANT CONVERTERS
When pulse width modulation (PWM) techniques
are used for de-to-de and dc-to-ac converters.
the semiconductor switches intl!rrupts the entire
load current durino turn-ON and turn-OFF
Process. Therefore t~ese semiconductor sw.i!£hes
~subjttted to ~igh swi~ s~res:es_ in the
~h-mode operation and the sw~~~11r:!g _power
oss IS considerably high. The _othe~roblem. of
~es-S that this swi,!S.h!Pg operatlo~
~rates electromagnetic interfe~ due
to large ~ and dr . To maintain low switching
~de dt - ----
·-:~ltchmg devices should be ~sed_i!1
swltch · -~ ·- - - -
~~
DF
in %
0.0
8.0
DF 6.0
14.0
0.0
0.0
!",.
v
l.O
0.2
0.2 0 4
. 0.6
-......Modulation ind 0.8
•u.,a~•f•t•·t:!•.• H ex (M)
- - armonics ,~
o.o
l.O
w ·d oJ modif' d
' th modulation ~~thsinusoidal pu/se-
1 P ==s

If the witching frequency is increased in order to reduce the converter size and weight, the
den ity should be increased. The switching stres , switching losses and electromagnetic interferen
(E II) are also increased linearly with switching ti·equency. If a semiconductor switch operates wi
snubl,c1 circuit in a de to de conve11er, trajectories of voltage across device (vd), current through
(id) and power los (p") are depicted in Fig. 11.45. It is clear from Fig. 11.45 that the switching I
are significant through the snubber circuit is present across the device.
Current
ld
Safe operating area
- - - - --- ....... - - - - - - - - - - - - - - - - - - - - - - • I
I
I
:
''
I
':
':
'
'
i
V
1
Voltage
~nax vel
Cu7,ent t·-~~-+/
L - .., Time
I
Voltage tVe~
""---/.::::::::...______--i..,~Time
tPower
pd t.....~-=o::::::------====--~L ~ ..,Time
I
UJWfl~j Trajectories of voltage across device (vd), current through device (id) and power lossyd) with
snubber circuit /
Consequently, to implement high switching frequency in converter circuits, all shortcomings can
be minimized if each semiconductor switch of converter circuit changes its state either from OFF to
ON or ON to OFF wl~olta e across device becomes zero or current through device is zero. This
D'Pe of switching is called zero voltage switching_(ZVS) or zero-cur~e~t switching (Z S). Figure 11.46
shows trajectories of voltage across device (vd), current though device (id) and power loss (pd) with
ZYS or ZCS. By using the concept of ZYS or ZCS, the switching losses, stress on semiconductor
_ devices, gen~ration of EMI will be reduced si&!!i.fu&.nll}L The converter topologies using-the ~S or
ZCS strategies are called resonant convert~rs. Most of the resonant converters operate satisfactory
---using LC resonance circuit.
Current
ld
Current
Safe operating area I, l
~----~~----------~
'_ )o Time
t
f Time)o
t
- -------------- ...- _......-------,
j
'i
I Vo~~get
' --~~--------_.
,
! po;der t
~____......:::::==----..-!....i~Voltage I
V Vmax Vc~ --....:~:::._:~-----........ Time
I
l-
•
wlllj Trajectories of voltage across device (vd), current though device (id) and power Joss (pcJ with ZVS
orZCS

Inverters 735
In a typical resonant converter, the basic operation of switches is to generate ~quare wave ac 1, veforms from de supply and the resonating components (inductor L and capacitor C) are used to
'
8
t d h · · 't must ber.jrer the unwan e arrnomc components from the square wave. The resonating LC cJrcuJ
Jl d h · h· {.', . . · f esonantproper tune ~t t e SWitc mg 1 requency ofconverter. There are different circuit topologtes 0 r
converters. Figure 11.47(a), (b) and (c) show the block diagrams of resonant converters.
+
+v
AC
supply
(Single__......
phase
or three
phase)
+
f Semiconductor Resonant t LOADswitches circuits
J'
v
(a)
Semiconductor Resonant
DC to DC JDC toAC
ro
LOAD
switches circuits
Converters
(b)
Semiconductor Resonant DC to DC tRectifier DCtoAC
ro LOADcircuits switches circuits
Converters
(c)
QWflfl Block diagram ofresonant converters
The LC circuit does not act as a simple low-pass filter, but it operates in ideally undamped or
highly underdamped oscillation of current and/or voltage. For ideally underdamped condition, R = 0
..!_nd the output wayefor'i'n is sinusoidal. For h~ghly un~erda~ped condition,_!! _::_0 and the current
~ough R and the voltage acr~__R are apprOXImately .smusOJd~y_sual~y, ~c. circuits'are tw-o- types
such senes resonant and paniTiel resonant. When a senes Lcresonant c1rcu1t 1s used in converter it
is called~s-resonant .cr:nyerteJ?, If a. ~arallel LC res? nant circuit is used in conver1er, it is called
parallel-resonant converters. In th1s sect1on, both the senes-resonant converters and parallel -re~ . _ ___ - ____,. sonant
.unverters are discussed elaborately.
11.8.1 Undamped Series Resonant Circuits
~undamped series resonant circuit is shown in Fig. I I.48 wher.e V is the
;n~~t voltage at time t0
• The initial current of inductor is IL(O) and the
nn1al voltage of capacitor is Vc(O).
~hen a current iL flow through the circuit, the KVL equation of the
Circuit is
d ' dv
v =L __}:_£_ + Vc and iL =cr_£
r dt dt
~ L,.
Fig. 11.48
An undamped
series resonant
circuit

The solution of above equations with time is
. V- Vc(O) .
tl.(r)=IL(O)cosw0 (t-t0 )+ Z smw0 (t-t0 )
0
for l >tv
. I
where. w0 =2nj0 = ~ = resonance frequency
..y L,.C,.
and characteristic impedance Z0
= (I" in n
vc:The voltage across capacitor is
vc(t) =V -IV- Vc(O)}cos w0(t- t0 ) + Z0!L(O)sin w0(t- t0 ) for r > 10
The plot of i1.U) and vc(t) are shown in Fig. 11.49.
11.8.2 Highly Underdamped Series
Resonant Circuits
Figure 11.50 shows a RLC series resonant circuit where R
is load resistance and L,. and C,. are resonating components
(inductance and capacitance).
The KVL equation of the circuit is
V R. L di 1 f·d= 1+ ,.-+- l t
dt c,
Assume Vc(O) = 0 and /L(O) = 0
The solution of the above equation is
or
"( V -"w, .t t) =--e ~ 0 sm w,t
w0 L,
i(t) = _V_e-a' sin w,t
w0 L,
where, W0 = undamped resonance frequency = f"T7
..yL,C,
and
The voltage across capacitor is
QWIIQI iJt) and vcft) waveforms of
a undamped series resonant
circuit
i L,._.. c,.
QWIJiel RLC series resonant
circuit
vc(t) =V- (V,- V)e-w (;,sinro,t +cosw,tl
or vc(l) =V- (V0 - VV"' [{I+(;,rrcos(ro,t+II)]
where, 8 =tan_,( ;J.V, =output voltage

The quality factor Q is equal to
W0 L, 1 Z
Q=--= :-2....
R W0 CrR R where, Zo =characteristic impedance = w L =___!__o r (I) C
o r
The impedance of RLC circuit is
1
Z(s) = R +sL +-
sC
The admittance of RLC circuit is
Y(s)=--= R+sL+-
1
-1 [ ]-1
Z(s) sC
1 2as
R i + 2as +m~
where, roots are s1 = - a+ jmr and s2 =-a-jmr and mr =Jmb- a 2
= m0 J1-~2
At resonance frequency, s = jm0 and Y(s) = ~ Q
4
Q
3
Q
2
Q
1
The impedance of the circuit is a function
of frequency with Q. When R is constant, at
resonance frequency m = ml), the impedance
becomes pure resistance R. At frequenc
w, > W0
and mr < W0
, Z increases with m a
shown in Fig. 11.51.
The voltage gain of RLC series circuit is
equal to
/trwnt!rs 7J7
G( )
_ V0
(s) _ R ~------------~------------------~(()
s - - 1
V(s) R+sL+ -
sC
liJ¥1111 Impedance Z varies with m
or G(jm) = Vo(~m) = R 1 = [ m ]
V(JW) R + jmL- j - 1+JQ _!!!__- _Q_
mC m0 m
or IG(Jm)l = [
2
(
1
)]
112
where,~=~ IG(iro)l
l+Q u--;;
1.0
The variation of gain of an RLC series circuit is
depicted in Fig. I I.52. .
Figure 11.53 shows a RLC senes resonant
circuit with ac input voltage. Wh~n a square
~ave voltage signal is applied to F1g. II.53 as
Input signal, the output voltage and current wave
forms are depicted in Fig. II.54.
0.8
0.6
0.4
0.2
0.1
0.8 1.0 1.2
•.u::•::•1::9.:!il; Gain var· r·---- • ta ton of RLC series circuit

oltagc
I '
I
~
!d O) Vc(O)
R
QWIJjl RLC series resonant circuit with square wove input voltage signal
Voltage
v
v ~-----.
(a) ~----~~----~-------+-------+----------~ Time
' I
(b)
- V
4V
TCR
T
2
I
I
T
4
V ----------L·==-------1 -V
TCR T T
2
3T
2
2T
3T 2T
2
Time
I
QWIJjl (a) Square wave input voltage and (b) Sinusoidal current signal
11.8.3 Parallel Resonant Circuit
An undamped parallel resonant circuit is depicted in Fig.
11.55. This circuit is supplied by a de current/. iL(O) is the
initial current and Vc{O) is the capacitor voltage.
The KCL equation of the circuit is
I =iL +C, dVc and
dt
L diL = v
dt c
The solution of the above equations are
I
h (O) ~
Vc(O)
An undampedpara/Jel resonant
circuit
. v. (0)
lL (t) =I+ [IL(0)- /]cos Wo(t- to)+ T-sin mo(t- 1o ) for 1> 10
0

Inverters 739
dthe voltage across capacitor is
an
Vc (f)= Z,JI- IL (O)]sin w0 (t- r0
) + Vc(O)cos W
0
(t -r
0
)
I Ifhere resonant frequency Wo = JL:c: and the characteristic impedance Z = _r
w ' LC o C
r r r
11.8.4 Highly Underdamped Parallel Resonant Circuits
The RLC parallel resonant circuit is shown in Fig.
11.56 which is supplied by current I.
The quality factor Q =W 0
RC,. =_!!__ =!5_
. woL,. Zo
The amplitude of impedance Z is a function
of frequency with Q as a parameter with R is
constant:
fr(O) ~ +
I
Vc(O)
R
R · - s-
Qmu
QWIIij Parallel resonant circuit
Z(s) =__?,.....:..=----"--
s- s
I+·- + --
m2 Qm0 0
The gain of RLC parallel circuit is
R/jmC,.
. R + 1/jmC,.
G(jm) = RljmC.
. L + I
JW ,. R + I IjwC,.
I
IG(jm)l = -r======:=-
(1- u')' + (~)'
or
=---------=--------
1- w2L C + _}m_L_r ( m )2 mlmo
r r R 1- Wo + j Q
IG(jw)lmax = Q at u =I or W = Wa
r
r
Th · · f · d ce z with respect to m and the current and output voltage waveforms of ae vanatiOn o tmpe_an . . .
1 57parallel resonant circUit are deptcted tn Ftg. I · ·
z
~)
OJ = OJ,
(a)
i
I
- /
T T
2
3T
-
2
2T
Time
t
n ~-~--~~--~----~--~~Time
4/R : : t
T
... -----_,
3T 2T
2
(b)
QWIJjl (a) Impedance Z varies with OJ and (b) Current and output voltage waveforms

11.10 VOLTAGE CONTROL OF INVERTERS
Eep~nding on t?e nature of load, variable ac voltag~or variable voltage with variable frequency are
required at the mput terminals of load. When the output voltage of inverter is applied to load, the
()'"Utput voltage of inverter can be controlled to get desired output from the system. There are different
methods to control the output voltage of inverter. The most commonly used control methods are:
I. External control of de input voltage of inverter
2. External control of ac output voltage of inverter
3. Internal control of inverter
11.10.1 External Control of de Input Voltage of Inverter
By controlling the de input voltage of inverter, the output voltage inverter can be controlled. The
different schemes of external control of de input voltage of inverter is depicted in Fig. 11.62(a), (b),
(c) and (d). In Fig. 11.62(a), the fixed de voltage is applied to a de-to-de converter or chopper to
.Q.btain variable de volta_g_e. When the variable de voltage is applied to an inverter through filter, the
controllable ac output voltage can be obtained from inv~rter.
&...._ _
Constant
de voltage
Variable
de voltage
~-~
Variable frequency
variable amplitude
..------, ac voltage
1--------l
DC to DC
converter
or chopper
Filter
circuits
Inverter v Load0
QWiifl (a} Input voltage control of using de-to-de converter
In F" 11 62
(b) th ac input voltage is initially converted into a variable ac voltage b .tg. . , e . . d . usmg ac
voltage controller an ~_converts mto e ~smg a~ unc-:-ontro~.<!J~ctifier. In this scheme,
~oltage ana variable frequency ae ou!£_ut I_s obtamed JUSt afte_r three s;_onversion stages.
Consequently, the efficiency of system ts poor and the mput power factor IS poor at low voltages.

y
e
Constant frequenc
constant amplitud
ac voltage - ~
Variable
ac voltage
I I •
Variable
de voltage
AC voltage Uncontrolled Filter
controller rectifier circuits
Inverter
Variable fr~:qu~o:m:y
variable ampIitudc
Iac vo tagc
i
v, Load
~ J
QM!Ifj (b) Input voltage control of Inverter using ac voltage controller and uncontrolled rectifier
In Fig. 11.62(c), the variable de voltage is obtained from a controlled rectifier and t)1_en Jhis variable
de voltage is a lied to inverter to get variable ac output voltage frQ!ll inverter. In th,is sch~. two
conversion stages are require an the efficiency of the system _is better than previous schcrm:. For
low output voltage, the input power factor is low. Since the output voltage contains low frequency
harmonics, the size of filter is bulky and the system response will be sluggish.
Similarly in Fig. 11 .62(d), ac voltage is applied to uncontrolled rectifier to get fixed de voltage which
is applied to chopper to obtain variable de voltage. As the chopper operates at very high ti·equency,
the constant de voltage is converted into a variable de ~phage '!_t high- frequency~ Due to very high
frequency, the srze of filter is reduced significantly. The fundamental power factor remains unity for
·all operating conditions but the system loss increases due to an extra converter.
Constant frequency
constant amplitude
ac voltage
_..,
Variable
de voltage
-
Controlled Filter
rectifier circuits
r -
Inverter
Variable frequency
variable amplitude
ac voltage
t
___l_
v" Load
~ ~
QW!Ifl (c) Input voltage control ofinverter using controlled rectifier
Constant frequency
constant amplitude
acvoltage ~
Constant
de voltage
Uncontrolled
Rectifier
Variable
de voltage
DC to DC
converter
or chopper
Filter
circuits Inverter
Variable frequency
variable amplitude
ac voltage
l
j
v, Load
J I
QW!Ifj (d) Input voltage control of Inverter using uncontrolled rectifier and de to de converter
11.10.2 External Control of ac Output Voltage of Inverter
The external control of ac output voltage of inverter can be possible by the following methods:
I. AC voltage controller
2. Series-connected inverters

f
I ~--------------------------·--------------------------------~m~v~er~~~N~741
1
. 14c volt ae controll r Fi 'ltt·~ 11. ,,1
I
,,,·s tlW l'XIl'l'll 1l (01tt•nl nf ll' uUIJ)ltl , 11~ 1 • . ' O ugc
I
. inn!rl'-'1' U~lllg ll' 'Olt IJ.t. -.:ontt''lll··t· I tl .,, . ' . " . n us
111~th,ld. :11 m· 'llltu~~: ·nntrulll:r is inl:,wpnrnt ·d
.,~...•1w~~n th~o: "'utpul volta ••,,rinv ,1•1•r d 1II ~~ . · 1.: llll llC
1,~1 J t·nn11l lis. ~ h~l th~ tirin~ an~k or th ris-
( 'unstunt
de vollngc
v
Variable frequency
variable amplitude
_.....-..---. ac voltage
AC voltage
controller Vo
wllll External control of ac output voltage of
Inverter using ac voltage controller
t,H'S of •~· ''">Ita~~ 1:Hllwlkr is vari.:d. the nri-
:1bl~ t: 'oltmt~ wtll b.: tppli~d h' lund t~rminuls.
~in~·c th~ hannoni~.:s 1.:H~h..·u~ lll' output vnltugl'
11
f :1~· 'tiiHH.l.l: .'UIIIWikr IS lu~lt. this 1ll.)tlod or
"''Ill~~ ~.:ontwl is m)t wid~l. us~·d nnd it is suitable f()r low power applications only.
2. Series-connected Inverters In series connected inverters method of voltage control, two or
nHH'.' inn:rt~:rs :tr~· i.'lllllll'l'tl'd in series. Figure 11 .64 shows that the two square wave inverters are
~utmc~l:d in Sl'riL·s ''' ~ct 11 v~triublc 1u.: output voltage. The output voltage of inverters I and II are
npplied tn th1..· primnry windings nl' two tmnsformcrs b~t · 1c secon-d~1ry-windings of transformers are
llitlt;;;;lcd iU.JL' .· •s,_'fl1~OUtput V(lJtugc or trnnslill'lnCrs_!:;..t and V,,2 have Sll;;;e;;:;ag~~upintS~ <htkrcm·c </J. l1c 1hu"n·sum or'"'" rundamcntal voltages V,,1 and V,,2 provides the: resultant
t'undum~..·ntul Hlltugto' ·--
==,,
1)1Sllll1
(k: voltngo:
+
,. Inverter
I
•
Vo l
1:1
Series connection of two inverters ond its output voltage V0 = V01 + V02
F' II 1:.1:
1
,. the out(llll voltage waveform of series connected inverters. Since the frequency of
tgure .o_l s lo ~ d. 11.. . 1
Out I I
.d I'
1
·s s·
1
n1
e if the phase 1 crencc IS zero1 t 1e outnut voltage is v = v + vput vo tagcs ·" an c>2 • • • • - - ·• • - . . -- :.c.: .. (I . " ' u2;
Whetl the phase ditli:rcm:c ¢_is~· the ou.tp~'.~ volt.agc.~s... : ,.- :•I-v;12 = 0. A_s the phase difference
··a~tl b -
1
, .,
1
;. finn 1 nn'lc ol two mvetlt:ts, th~o: output voltagre can be ,.01111·olled The
-.ln be vaneu oy ~.:nnngtt g ... . . . " ·
vo tage control using seri..:s cunnect..;d '.nvertcrs ~~ also called.mulllple converter control. Since the
harmonics cont~nt in th~ output voltngc.: IS large. thts method ul voltage control is used for low output
voltage levels. i.e.. 25% to 30% or tht: rut~d voltage.

Everything You Need to Know About Inverters

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