Poweer Electroniccs
Prof. B.
B G. Fernan
ndes
Department Off Electrical E
Engineeringg
India
an Institute of Technoloogy, Bombaay
Lecturee No – 12
Let me recapitulate
r
whatever
w
I did
d in my laast class. Wee discussed, full wave reectification uusing
centre tap
p transformeer. We found
d that and th
hat 1 diode cconducts at a time. 1 dioode conductts at a
time and
d voltage ratting of the diode
d
is twiice the peakk value of tthe secondarry voltage oof the
transform
mer. See, wh
hy am I say
ying, 1 diodee conducts aat a time? Inn other conffigurations, there
may be more than 1 device co
onducting att a time. Seee, these aree the issues that have tto be
discussed
d or these are
a to be tak
ken into acccount while calculating the efficienncy. See, 1 ddiode
conducts at a time. So,
S the voltaage across th
he diode is oof the order of 1.5 voltss or so depennding
upon thee power ratin
ng. So, if th
he input volltage itself iis low, moree than 1, if the 2 diodees are
conductin
ng, 3 volts is
i the drop across
a
the diiodes, so thee input voltaage is high thhat 1.5 voltss or 3
volts may
y be insignifficant value.
So, it is all
a depends on
o the inputt value inputt value of thee voltage annd the numbeer of devices that
are com
ming in the main path of the po
ower flow. The secondd 1 is a ccommon catthode
configuraation. We fo
ound that dio
ode whose anode
a
potenttial is maxim
mum, will coonduct. The third
point in common
c
ano
ode configurration, diode whose cathoode potentiaal is least willl conduct.
(Refer Sllide Time: 2:58)
d bridge, sin
ngle phased bridge, we found that 2 diodes coonduct at a time.
Then in uncontrolled
Voltage rating of th
he diode is peak
p
of the input voltagge. See, jusst the oppossite. In full wave
c
tap traansformer; 1 diode condducts at a tim
me, voltage rrating is twicce the
rectificattion using a centre
1 peak of the
t secondarry voltage, whereas
w
heree, 1 2 diodess are connectting at a tim
me. Voltage rrating
of the dio
ode is peak of
o the input voltage.
v
(Refer Sllide Time: 3:29)
We also discussed 2 configurattions in sing
gle phase seemi converteer. In case 1 shown heere, 2
thryristorrs are in the same leg an
nd 2 diodes are
a in the saame leg. Alppha to pi, T1 and D2 condducts.
From pi to pi plus allpha, D2 and
d D3 conduccts. When thhese 2 diodess conduct, vvoltage appliied to
the load is 0. Sourcee does not supply
s
poweer to the loaad. When freeewheeling diode, D2 D3 are
conductin
ng, source does
d
not supp
ply power to
o the load. S
Source suppllies power oonly when T1 and
D2 are co
onducting, in
n the positivee half.
wer, even in the
t negative half when T 4 and D3 coonduct from ppi plus alphaa to 2
A sourcee supply pow
pi and frrom 2 pi to 2 pi plus alpha,
a
again D2 and D3 conduct. Looad current is unidirectiional,
whereas, source current flows on
nly from alph
ha to pi, in thhe positive hhalf and from
m pi plus alppha to
2 pi, it is a almost a square
s
wave.
me a currentt, if I assumee the load cu
urrent to be a constant annd ripple freee, source cuurrent
If I assum
is going to be a squ
uare wave. So,
S what con
nclusion didd we draw inn the last cllass? Conduuction
period fo
or thyristors is pi minus alpha radian
ns. So, as alppha increasees, we foundd that the durration
for which
h thyristor conducts
c
alsso reduces, whereas,
w
thee conductionn period forr diode is pii plus
alpha. Ass alpha increeases, condu
uction the du
uration for w
which the diiode conduccts also increeases.
So in oth
her words, average
a
curreent rating off the thyristoors is less coompared to tthat of diodee, for
any valuee of alpha otther than 0. The
T average current becoomes same oonly when alpha is 0.
2 (Refer Sllide Time: 6:12)
In the casse 2, second
d case, we haave 2 thyristo
ors forming a common ccathode conffiguration. So, we
have alm
most a symm
metrical conffiguration. Thyristor
T
andd diode in 1 leg, even inn another leeg we
have ano
other thyristo
or and a diod
de. Again in
n the positivee half, at alppha T1 and D2 are at alphha T1
is triggerred, from alp
pha to pi, T1 and D2 conduct. Sourcce supplies power to the load, at pii plus
alpha D3 and D4 are triggered. So,
S what hap
ppens from ppi to pi pluss alpha? Wee have a com
mmon
anode co
onfiguration here and 2 uncontrolleed devices, rremember, w
whereas we have a com
mmon
cathode configuration
c
n, but then both
b
are con
ntrolled devicces. So till yyou triggeredd them, theyy as if
or they arre in forward
d blocking mode,
m
they block
b
the volltage.
Since wee have 2 unccontrolled deevices formiing a commoon anode coonfiguration, the diode w
whose
cathode potential
p
is least
l
will staart conductin
ng. So, D4 sttarts conducting at omegga t is equal to pi
plus. So therefore,
t
fro
om pi to pi plus
p alpha, current
c
free w
wheels throuugh T1, load and D4. So, again
from 2 pi
p to 2 pi plu
us alpha, cu
urrent free wheels
w
througgh T3 and D2. So, wavee forms of ooutput
voltage, the source current
c
are the same. Only
O
thing iss the devicee, all the deevices conduuct or
duration for which th
he device con
nduct is the same. Averaage value off the output vvoltage, we ffound
that is allways positiv
ve. It is giv
ven by Vm pi
p into 1 pluus cos alpha,, where alphha is a trigggering
angle. It can take a value anywheere from 0 to
o pi radians.
When it is
i 0, it is equ
uivalent to uncontrolled
u
bridge, wheen it is pi is eequal to alphha, output vooltage
is 0. sorrry when th
he trigger angle
a
alpha is equal too pi, outputt voltage is 0. What is the
displacem
ment factor? Or the displacement an
ngle is the anngle betweenn the fundam
mental compoonent
of the in
nput voltage and the fun
ndamental co
omponent off the sourcee current. W
We found thaat this
angle forr half controllled bridge or
o a semi con
ntrolled briddge is alpha bby 2. So, dissplacement ffactor
is cos minus
m
alphaa by 2. Thiis minus siign indicatees that pow
wer factor iis lagging, sorry
displacem
ment factor is
i lagging.
So, how do we turn off either T1 or T3? So, T1 is turnedd off by turnning on T3. S
So, when wee turn
on T3, a negative liine voltage, input voltaage appears across T1 aand in turn,, it turns offf the
3 conductin
ng thyristor.. Hence the name,
n
line commutated
c
convertor bbecause the iinput line vooltage
itself is used
u
to turn off
o the condu
ucting thyrisstor. Hence tthe name, linne commutatted convertoor.
So, how many pulsees are there in 1 cycle, from 0 to 2 pi? There is 1 pulse ffrom alpha tto pi,
p
from pi
p plus alph
ha to 2 pi. So,
S there aree 2 pulses inn 1 cycle. S
So thereforee, this
another pulse
convertorr is also kno
own as 2 pu
ulse converto
or. I have too show som
mething for yyou. So, heree is a
single ph
hase uncontro
olled bridge.
(Refer Sllide Time: 11:08)
There aree 4 terminalss. These 2 arre inputs, positive DC buus and negattive DC bus.. Current ratiing is
25 amperres, 600 volts. When I am
a saying 25 amperes, iit can carry 25 amperes of current aand it
can block 600 volts. So input voltage,
v
theo
oretically, peak value should not eexceed 600 vvolts.
These aree all ideal vaalues, theoreetical values.
4 (Refer Sllide Time: 11:56)
Here is a high curren
nt, high volttage thyristo
or. See the ssize, it can ccarry 400 am
mperes and iit can
block 2 KV.
K Cathod
de anode, it is
i generally mounted onn a heat sinkk and this w
was a gate siignal,
wire for wire to supp
ply the gate current. Seee the size off the cross ssection area of the condductor
y the gate cu
urrent, very thin.
t
So, veryy small curreent that is reequired to tuurn on
that is ussed to supply
an SCR. See the cro
oss section area
a
of the conductor
c
thhat is used tto that is ussed in the foor the
cathode circuit.
c
The main
n power flow
ws in this con
nductor wheereas, a smalll current floowing througgh the gate. IIt can
carry theeoretically, as
a per the ratting, 400 am
mperes and itt can block 2 KV. So, yoou can use inn any
circuit, wherein
w
that voltage doees not exceeed 2 KV at any given iinstant. Thesse are theoretical
values bu
ut this the values
v
that sh
hould be tak
ken into accoount while ddesigning, w
we will discuuss at
the appro
opriate time, all this issues, we will discuss.
d
What is another
a
issuee we discusssed? Let me discuss RLE
E load and ccurrent is conntinuous. I ssaid if
the curreent is contin
nuous, wheth
her the load is RLE or not, alpha m
minimum caan be 0. In other
words, diiode starts co
onducting, diode
d
that is coming in thhe forward ppath or in thiis bridge.
5 (Refer Sllide Time: 13
3:54)
If I have an uncontro
olled bridge, diode D1 an
nd D2 will sttart conductiing in positivve 0 crossingg and
if I have,, since I havee thyristor here, I can triigger this thyyristor at alppha is equal tto 0, indepenndent
of the loaad. If curren
nt is continuo
ous remember, thyristor can be triggger at alpha is equal to 00. But
then, wh
hat happens to the curreent? Whether current is increasing oor decreasinng or how thhat is
possible?? Take for example, th
he equivalen
nt circuit w
when T1 andd D2 are coonducting, soo the
equivalen
nt circuit is given heree. Source, we
w have a ssource, T1 iis a short circuit when it is
conductin
ng, R, L and
d E, D2, back
k to source.
Current is
i continuous, so this is the
t path, KV
VL has to hoold good. Soo, KVL givess Vi is equall to R
into i plu
us L di by dtt plus E. So therefore, dii by dt is Vi minus E miinus Ri dividded by L. Soo, if I
neglect R,
R it is given
n by V minuss E divided by
b L. So, di by dt is possitive when tthe instantanneous
value of the input vo
oltage is high
her than E. So
S remembeer, I might have said yess, if the current is
us, thyristorr can be trigg
gered at the positive 0 crrossing. Thaat does not m
mean that di by dt
continuou
is positiv
ve. So, firstt thing that comes to my mind iss when the positive off D1 and D2 are
conductin
ng, when I say
s that, I caan assume th
hat current m
may be increaasing or in oother words, di by
dt is posiitive. It is no
ot necessarily
y true.
or that matteer, even if the load is RL
L or RLE, di by dt need nnot be positiive or
If the loaad is, even fo
it is not positive,
p
thee moment yo
ou trigger thee thyristors. It all depennds on the vvalue of L, itt also
depends on E. So, in this case wee found that till omega t is equal to sine inversee E divided bby Vi,
E is greaater than Vi. So, cathode potential neeed not be, s o, di by dt is always, di by dt is neggative
till this reegion.
6 See the wave
w
form, voltage,
v
the current.
c
I miight have tri ggered, the current is floowing, di byy dt is
decreasin
ng. So, at th
his instant it starts increaasing, of couurse, providded R is 0. If R is finite, it is
beyond this
t
region, beyond this point, curreent starts inccreasing. Cuurrent increaases again aat this
point, ag
gain, Vi minu
us E divided
d L, they aree equal. So, di by dt is 0 that meanns this peak starts
decreasin
ng. So curren
nt is, how th
he current is flowing eveen though maagnitude of Vi is less thaan E?
It is becaause of L di by
b dt and L di
d by dt is neegative, di bby dt is negattive, current is decreasinng.
(Refer Sllide Time: 18:04)
(Refer Slide
S
Time: 18:10)
1
blem starts, what
w
happen
ns when the current disccontinuous? If the curreent is continnuous,
The prob
you do not
n need to know
k
the typ
pe of load. You
Y can triggger the thyriistor wherevver you wannt. So,
7 di by dt is positive or
o negative, depends on
n instantaneoous value off E and the load param
meters.
And, wheen the devicces are condu
ucting, inputt voltage is ssame as the output voltaage or V0 is same
as Vi and
d if the freew
wheeling diodes are cond
ducting, outpput voltage iis 0. We do nnot need to kknow
the type of
o load at alll, if the curreent is continu
uous.
The prob
blem starts when
w
the load
d is discontin
nuous. Thinngs are not veery straight forward, listten to
me carefu
fully. 1 assum
mption that I am making
g is assume that i becom
mes 0 after ppi. I am assuuming
it, see, neeed not be, current
c
may become
b
0 an
nywhere. Seee it attends a peak valuee here and it starts
decreasin
ng. If I havee a high valu
ue of E, it cu
urrent may bbecome 0 m
much before pi. It is all, what
you need
d to find out is we need to know the value
v
of L, E and R.
(Refer Sllide Time: 19
9:54)
uming that current
c
becom
mes 0 after pi.
p Positive L di by dt sshould be eqqual to negattive L
I am assu
di by dt and
a equate th
he areas or solve
s
the diffferential equuation and yoou can find oout where exxactly
the or you can find out the point at which thee current beccomes 0.
8 (Refer Sllide Time: 19
9:58)
So, at alp
pha which is higher than
n sin inverse E divided byy Vi Vm, I am
m triggeringg T1.
(Refer Sllide Time: 20
0:17)
The mom
ment I am triggering
t
T1, they startt conductingg, T1 and D2 starts connducting. Soource
supplies power to thee load and sttarts flowing
g. What happpens at omegga t is equal to pi plus? IIn the
f
a coommon cathoode configurration. So, om
mega
upper half, we have a diode and a thyristor forming
t is equaal to pi pluss, D3 gets or
o anode po
otential of B is higher than the pootential of A
A. So
immediattely, D3 starrts conducting, T1 turnss off. In thee lower halff, T4 is not triggered. Itt gets
triggered
d only at pi plus
p alpha. Till then, this is in the bloocking modee.
9 So, curreent start flow
wing through
h the load, D2, D3. In otther words, lload is freew
wheeling thrrough
D2, D3. Source
S
curren
nt is 0, sourcce current so
orry source current is 0,, source currrent is 0 andd load
voltage is
i also 0. Iff the current is continuo
ous, D2 and D3 would haave continueed to conduct till
you trigg
ger T4. The moment T4 starts, the moment
m
youu trigger T4, T4, D3 starts conductingg and
source su
upplies pow
wer. Unfortu
unately, in this
t
case, cuurrent has bbecome 0, m
much beforee you
trigger T4.
So, what happens now? What is the load, wh
hat is the volltage appeariing across thhe load? Listten to
me, if the load was, if the load is
i passive, there is no ccurrent, so thhere is no vooltage. Since, we
have con
nsidered load
d to be RLE, voltage app
pearing acrosss the load iss E. So, the iinstance from
m the
current becoming
b
thee 0, till you trigger
t
the otther thyristor, that is T4, voltage applied to the looad is
E. See heere, somewh
here at this in
nstant curren
nt has becom
me 0, next thyyristor triggeering is onlyy at pi
plus alph
ha. So, from
m this instantt to pi plus alpha, voltagge appearingg across thee load is E. At pi
plus alph
ha, you trigg
ger T4, T4 D3 starts condu
ucting. Voltaage applied to the load iis again the input
voltage, is
i a sinusoid
dal.
oad current, unidirection
nal, source current,
c
sorryy, source currrent is 0 herre, it shouldd have
See the lo
been, this is the sourrce current wave
w
form. So, you shoould be extreemely carefu
ful while draawing
the load voltage
v
wav
ve form when
n the thyristo
or is, when tthe current is discontinuuous. You neeed to
know thee type of loaad. If the cu
urrent is con
ntinuous, youu do not neeed to know the type of load,
because the load volltage is eith
her the inputt voltage or 0. It depennds on whethher thyristorrs are
ng or the freeewheeling mode.
m
conductin
(Refer Sllide Time: 25
5:00)
Now we discussed case
c
1 and case 2, 2 diifferent conffigurations. 2 thyristorss in 1 leg annd in
another second
s
case second case is 1 thyristo
or and 1 diodde in 1 leg, oother then gaama being noot the
same. What is the difference betw
ween these 2?
2 What is tthe differencce between tthese 2 bridgges or
when do you prefer?? or what I saaid, both aree line comm
mutated convertors, line vvoltage is ussed to
turn off the
t conductin
ng thyristor.
10 Rememb
ber, I discusssed this poiint in the while
w
the divvided whilee discussing the devicess that
device sh
hould be com
mpletely turn
ned off befo
ore it is forw
ward biased aagain. I repeeat, device shhould
attain its blocking mo
ode before itt is forward biased againn. What happpens in casee 1? At omegga t is
p plus, D3 start
s
conductting. In the positive
p
half,
f, omega t is equal to pi pplus, in the ssense,
equal to pi
positive half
h end set pi, at omegaa t pi plus, D3 starts condducting, indeependent of the instant w
where
you trigg
ger T1. Wheth
her alpha is 0 or alpha is pi by 2, alp
lpha approacches pi, it dooes not matteer. At
omega t is
i equal to pi plus, D3 tak
kes over from
m T1.
What hap
ppens in casse 2? We haave a comm
mon annual cconfigurationn. In lower hhalf, assumee that
alpha app
proaches pi.
(Refer Sllide Time: 27
7:24)
gger T1. In the positivee half, prior to T1 triggeering,
So, someewhere nearring pi, you want to trig
current was
w flowing through D2 and T3. In the positivee half, sincee alpha is appproaching ppi, so
entire po
ositive half, T3 and D2 were
w
conductting. So, loaad voltage iss 0, current w
was freewheeeling
through T3 and D2, source currrent is 0, alll most the eentire part oof positive hhalf. At om
mega t
hes pi, you want to triigger T1. You
Y
have appplied a gatte signal too T1, somew
where
approach
approxim
mately when approximattely near to pi.
p T1 tries tto turn on. S
So, potential of A is sam
me the
cathode potential.
p
A small posittive voltage appears accross T3, sorrry, a small negative vooltage
appears across
a
T3, beecause potenttial of B is slightly
s
higheer, slightly less than potential of A.
I will reepeat, I mad
de a when you
y trigger th
he T1, potenntial, cathodde potential of T3 is sam
me as
potential of A. Alphaa is approach
hing pi. We find that, yoou know, pottential of A is approachiing 0,
same as the
t potentiall of B is also
o approachin
ng 0. At omeega t is equaal to pi, both are 0. So, as you
approach
h the negativ
ve 0 crossing
g, voltage ap
ppearing acrross, the maggnitude of vvoltage appeearing
across T3 also reduces. A smaall negative voltage is appearing aacross T3. T
The device iis, in
principall, it is reversse biased. Tiime to turn off,
o I said, ddevice is nonn ideal. It haas, it will takke its
own timee to turn offf. Basically, thyristors arre slow devices. As it ttries to turn off at omegga t is
11 equal to pi
p plus, agaiin a positive voltage app
pears across iit, isn’t it? A
At omega t iss equal to pi plus,
what hap
ppens? Poten
ntial of B is higher
h
than potential off A. We mighht have trigggered T3 justt near
the negattive 0 crossin
ng.
0:38)
(Refer Sllide Time: 30
vice is tryinng to turn ooff. At that instant, maay be,
A negatiive voltage appeared accross T3, dev
instantan
neous value of
o the input voltage beccome 0, omeega t is equaal to pi plus and immediiately
after thatt potential off B becomes higher than
n potential off A. A positiive voltage aappears across T3.
Unfortun
nately, it may
y so happen that T3 may not have atttained a posiitive blockinng capabilityy.
So, it waas trying to turn off, som
me current was flowing, current was increasing, immediatelyy it is
started geetting forwarrd biased. So
o therefore, T3 continuess to conductt, T1 cannot tturn on. … IIt was
trying to, T3 was try
ying to turn off because you have trriggered T1, T1 was tryinng to turn oon, T3
was tryin
ng to turn offf. But then, as
a it was turn
ning off, it w
was getting fforward biassed. So, somee sort
of a commutation faiilure we hav
ve here. Thyrristor does nnot turn off, sso it continuues to conducct. T1
does nott turn off. So,
S what did happen from
fr
0 to p i till 0 to aalpha wheree that alphaa was
approach
hing pi? Currrent was flow
wing through
h T3 and D2.
12 (Refer Sllide Time: 32
2:29)
At omega t was apprroaching pi, you triggereed T1 and I ttold you thaat T4 did not turn off. We had
some sorrt of a comm
mutation faillure there an
nd again it sstarts conduccting in the entire half ccycle.
Now, T3 and D4 startts conducting
g at omega t is equal pi pplus, becausse in the low
wer half we hhave a
common anode conffiguration. D4 immediateely starts coonducting. T
This was tryiing to turn ooff, it
did not tu
urn off. It co
ontinues to co
onduct, so we
w have a powering modde now.
There is power flowss through T3,3 load, D4 an
nd source. Soo the entire nnegative enttire pi radianns, D3
and D4 conduct. See the voltage wave form, we have ann uncontrolleed rectificatiion. May be,, as if
I have on
nly a single diode, condu
ucting for th
he entire pi rradians and here, outputt voltage is 00. So,
this wavee form is sim
milar to thatt of half waave rectifier. Half wave rectifier, sim
milar to thatt of a
half wav
ve rectifier. We
W had a so
ource, 1 diod
de coming iin series andd a resistive load or I haave 1
source, 1 diode, freeewheeling diode
d
and loaad. If I hav e that sort oof a configuuration, I geet this
wave form.
I will rep
peat, if I hav
ve a source, 1 diode and the load or ssource, 1 dioode and a freeewheeling ddiode
and the load,
l
I get th
his sort of a wave form. This is nothhing but a haalf wave recctification. H
Hence
the namee, half wave effect. So, in
n case 2 we have half w
waving effectt which is noot there in caase 1.
So remem
mber, whateever, there are
a certain ad
dvantages orr disadvantaages, nothingg like, this iis the
perfect system and this
t
is a bad
d system, no. Out here, in case 1, thhe average ccurrent ratinng for
alpha no
ot equal to 0 is not the same or th
he conductioon period foor the devices not the ssame,
whereas in this case,, conduction
n period, con
nduction durration for alll the devicess is the samee. But
a
is apprroaching pi, we have hallf waving efffect, whereas, this effectt is totally abbsent.
then, as alpha
So, certain plus pointts, certain minus
m
points. That is the w
way the life is all about.. We have ceertain
nts, certain minus
m
pointss. There is nothing
n
like, this is the pperfect systeem and this iis the
plus poin
totally a useless
u
systeem. In naturee, some of nature,
n
in natture, we do nnot have such a system.
13 Let us seee, let us solv
ve some prob
blems. We will
w start from
m simple prroblems thenn we will inccrease
the comp
plexity. May
y be, will if you
y try to so
olve this probblem, you m
may be able to appreciate this
subject more.
m
6:20)
(Refer Sllide Time: 36
A simplle problem here, halff wave recttification, hhalf wave ccontrolled rrectification, the
freewheeeling diode. I have put th
he thyristor, 1 thyristor, 1 diode or L
LE load. Thee problem saays L
is large. The
T momentt when I say
y L is large, first
f
thing thhat should coome to your m
mind is currrent is
continuou
us and someeone says th
hat L is largee, you can ssafely assum
me that curreent is continnuous,
because we
w know thaat the curren
nt cannot chaange instanttaneously. Soo, L is very large. First of all
L is very
y large, so caan safely asssume, the current is conttinuous. Thee moment cuurrent continnuous,
I am tellling you, pro
oblem solving is going to be very simple, beccause, I knoow that when the
thyristor is conductin
ng, output voltage
v
is saame as the innput voltagee. The freew
wheeling dioode is
conductin
ng, output vo
oltage is 0. The
T problem
m starts whenn the currentt is discontinnuous.
So fortun
nately, L is current
c
is con
ntinuously here.
h
So, the problem is, find the value of alpha, what
should bee the trigger angle if the average value of i is 100 amperes? The averagee value of i is 100
amperes, how do I so
olve that problem? I willl plot the inpput voltage w
which is a sinnusoid, this is the
value of E, this 20 vo
olts. Input iss 230 volts, 50 hertz singgle phase suupply. So, thhis peak is arround
230 into root 2, is of the order off, may be 325
5 volts. E is 20 volts, so somewhere here.
L is larg
ge, E is very
y small, R is also very
y small, so there are veery good chhances of cuurrent
becoming
g continuou
us, because as
a this value of E com
mes down, thhe duration available foor the
current to
o increase, di
d by dt also increases. Iss only whenn instantaneoous value of the input vooltage
is higherr than E, cu
urrent starts increasing. Let us assuume that thyyristor is triiggered at aalpha.
Current is
i continuou
us, do not wo
orry, so the moment yoou trigger at alpha, inputt voltage apppears
across th
he load. At alpha you have
h
triggerred, input vooltage is sam
me as the ooutput voltagge or
output vo
oltage is sam
me as input voltage.
v
14 At omega t is equal to pi plus, since the load is inductive, current continuous to flow. But then,
the moment instantaneous value of the input voltage is becoming negative, the freewheeling
diode starts conducting, because you cannot have a positive voltage appearing across the diode,
because beyond pi, this potential is higher than this potential.
If T conducts, this potential is same as this potential. So, a positive voltage appears across diode,
not possible, so diode starts conducting, turns of the thyristor T. So, in the entire negative half,
since the current is continuous, current D conducts. See, I have if there was no diode here,
thyristor would have continued to conduct. Beyond pi, a negative voltage appeared across the pi,
decay would have been faster. So, current would have become 0 somewhere in between pi to 2
pi. So, I have achieved, what I did here? I have connected a freewheeling diode, voltage
appearing across the load beyond pi is 0. So, the forcing function is 0, current decays slowly.
Second is, I have used a very large inductor, so current continuous to conduct and that too very
slowly. What happens in the positive half from 0 to pi?
If I had a diode replacing the thyristor, it would have started conducting at omega t is equal to 0
plus. In the second cycle or at steady state in the positive 0 crossing, main diode would have
started conducting. Since the current is continuous, you can trigger the thyristor at omega t is
equal to 0 plus and when as soon as I trigger the thyristor, in the positive half, now a negative
voltage appears across D and diode turns off. So, till you trigger the thyristor, source voltage
does not appear across diode, because we have an open circuit here. We have an open circuit
across the thyristor, because you have not triggered the thyristor, we have an open circuit here.
So, at steady state current is continuous. From pi to 2 pi freewheeling diode was conducting, it
continues to conduct till you trigger the thyristor even in the positive half. You can trigger at
omega t is equal to pi, omega t is equal to 0 plus, you can trigger it. So, till you trigger the
thyristor, diode, freewheeling diode continuous to conduct even in the positive half. Remember
this, because we have an open circuit there, T does not conduct, it is not triggered. So, source
voltage does not appear across the diode, because we have an open circuit, remember. Do not
make a mistake while drawing the wave form. If your wave forms are correct then you need to
find out just the average value of it, you have to integrate it. Anyone can do, even an eleventh
standard or twelfth standard student can do.
The first thing is you need to understand, the how the circuit works? Draw the wave forms,
having drawn the wave forms correctly, you are through, you are done. Let me tell you 1 thing,
answer, you may, you can afford to make a mistake while integrating. As a teacher, I may not
deduct any marks if you make a mistake while integrating. But then, since the, because it is not a
mathematics, course on mathematics, it is a course on power electronics. Fine, you may, you can
make a mistake. You can afford to make a mistake. I may try to overlook that. But the, since, it is
the course on power electronics, I would expect you to draw the correct wave form.
If the wave form is correct, I know that student knows the subject. He must have made a silly
mistake in a hurry, he might have made a mistake while integrating, is pardonable. So, try to the
first point is, draw the wave form, try to understand the functionalities, how the circuit works.
The second is, draw the correct wave forms, you are done, problem.
15 So, diode continuous to conduct, so at steady state, N cycles are over, from 0 to alpha, D1 was
conducting, from alpha to pi, thyristor starts conducting and N pi to N pi plus 1 alpha, D1
continues to conduct. When the source supplies power? Source supplies power only when
thyristor mean thyristor is conducted. So, thyristor conducts from alpha to pi. So, there is only 1
pulse in 1 cycle. Even output voltage also, there is only 1 pulse in 1 cycle. So half wave rectifier,
they also known as single pulse convertor or 1 pulse convertor.
So before solving, we try to see the source current wave form. The moment diode starts
conducting, source current instantaneously become 0. When the thyristor is conducting, the load
current is same as source current, it is. So, at the moment thyristor is triggered, diode turns off.
Whatever the current that is flowing through the diode, starts flowing through the thyristor. So
instantaneously, it jumps to the load current, increases, follows this load current, reaches a peak
then starts coming at pi at pi, omega t is equal to pi plus, D starts conducting. Immediately T1
turns off. This is because I am assuming, in the beginning of my course I told that let us neglect
the source inductance. If you take the source inductance into account, what happens, we will see
sometime later. Let us not worry now.
So, source supplies power in only in 1 cycle. There is only 1 pulse per cycle. Input is a sinusoid,
appears sinusoid, current is a non sinusoid. But then, it supplies only between alpha to pi. There
is only 1 pulse. So, if I try to take the Fourier series of this wave form, it has an average
component. In other words, A0 plus the sigma AN into cos omega t plus BN into sin omega t. So,
if find that if A0 is finite, it has an average component. Average component is nothing but a DC
component. So, I am drawing a DC component from the source. I will repeat I am drawing a DC
component from the source. What happens when I draw a DC component from the source? After
all, this current has to come from transmission lines, transformers, secondary, it as to flow from a
secondary transformers, what happens?
We will discuss this point in detail. I will solve a very interesting problem sometime later. A
very interesting problem: putting a diode across the transformer or putting a diode in the
secondary of the transformer. Apart from the average component, it has other frequency
components also. Write as Fourier series, A0 plus AN cos omega t sigma plus BN sin omega t.
you will find that it has other frequency component also. Only the fundamental component that
D1 sin omega t is responsible for power transfers, because input is odd function that is some sin
omega t. So, source current which a component of a source current whose frequency is same as
omega and sinusoidal is responsible for power transfer.
The remaining higher frequency components flow in the transformer, secondary transmission
lines. The effect of higher frequency components, this higher frequency components are also
know as harmonics, hence forth I will say the higher frequency components has harmonics.
These are nothing but the higher frequency components which are present in the source current
will definitely affect the source and other circuits. The effect of harmonics on other circuits, we
will discuss in detail sometime later. So we found that now let us solve this problem, source
current flows from alpha to pi, load current is continuous, increases, decreases. How do I solve?
16 (Refer Sllide Time: 51:32)
Said, aveerage value of
o current iss average voltage appliedd to the loadd minus E ddivided by R
R. Are
you with
h me? Averaage value of current depeends on the difference bbetween the average vallue of
the outpu
ut voltage minus E. In otther words, V average iss equal to i aaverage into R plus E. L di by
dt averag
ge is 0.
It is only
y RE and V average willl determine the current fflowing in thhe circuit, w
will determinne the
average value
v
of thee current. So
o, average vaalue of i is aaverage volttage across R divided bby the
value of R.
R Value of R is given by
b 0.01 ohmss. So, averagge voltage accross R becoomes 1 volt, since
the valuee of i is 100 amperes.
a
(Refer Sllide Time: 53
3:05)
17 This is the equation, V average is equal to i average into R plus E. So therefore, V average is 21
volts. Isn’t it? E is 20, i average into R is 1. Therefore, E average is 20. So, what is the average
value of the output voltage? 0 to alpha it is 0, alpha to 2 alpha to pi, it is Vm sin omega t and pi to
2 pi, it is 0. So 0 to pi, alpha to pi, Vm sin omega d omega dt, Vm is 230 into root 2, 1 plus cos
alpha. So, alpha you get it as 53 degrees. Alpha is equal to 53 degrees.
So by solving this problem, I have brought out various issues. 1 is in a half wave rectifier or
single pulse converter, average value of the source current is finite. In other words, we are
drawing a DC component from the source, also there are high frequency component in the source
current. Effect of DC, effect of high frequency component in the source, we will discuss
sometime later.
If there is a controlled device here, freewheeling diode conducts even in the positive half cycle, if
the current is continuous, remember. If the current is continuous, freewheeling diode conducts
even in the positive half till you trigger the thyristor, because till you trigger the thyristor, source
voltage does not appear across the diode. So, if the current is continuous is only when the
thyristor is conducting, input voltage is applied to the load and for the entire remaining entire
duration, voltage applied to the load is 0.
Thank you.
18