Internati
o
nal
Journal of P
o
wer Elect
roni
cs an
d
Drive
S
y
ste
m
(I
JPE
D
S)
V
o
l.
3, N
o
. 1
,
Mar
c
h
20
13
,
pp
. 30
~40
I
S
SN
: 208
8-8
6
9
4
30
Jo
urn
a
l
h
o
me
pa
ge
: h
ttp
://iaesjo
u
r
na
l.com/
o
n
lin
e/ind
e
x.ph
p
/
IJPEDS
A High Regulated Low Ripple DC
Power Supply Based on LC
Filter and IGBT
Ay
an
Mi
tra
,
Abhisek
R
o
y
Department o
f
Electrical Engin
e
ering,
Jad
a
vpur U
n
iversity
Ko
lkata, Ind
i
a
Article Info
A
B
STRAC
T
Article histo
r
y:
Received
J
u
n 26, 2012
Rev
i
sed
D
ec 10
, 20
12
Accepte
d
Ja
n 16, 2013
Filaments operating at
a high temperat
ure reg
i
o
n
would require ripple free
current constant power source
for
their h
eatin
g purposes. In
tr
oduction
of
ripple in the p
o
wer supply
would re
sult in high power loss, aberrant,
unpredictable temperature ris
e
etc. High
cu
rrent r
i
pple
can also ex
tenuate th
e
life of electro
l
y
t
ic cap
acitors used in
the circuit.
In this paper we propose a
scheme in which we would not
only
us
e s
i
m
p
le, low cos
t
, eas
i
l
y
av
ail
a
bl
e
electronic
components but also r
e
duce the
value
of ripple factor
to <0.1%
an
d
load r
e
gulation
to 0.1%
.
Keyword:
IGBT
LC Filter
Op
eration
a
l am
p
l
ifier
Ripple fact
or
Copyright ©
201
3 Institut
e
o
f
Ad
vanced
Engin
eer
ing and S
c
i
e
nce.
All rights re
se
rve
d
.
Co
rresp
ond
i
ng
Autho
r
:
Ay
an
Mi
tra
Depa
rtem
ent of Elect
ri
cal
E
n
gi
nee
r
i
n
g,
Jada
vp
ur
U
n
i
v
ersi
t
y
,
Sab
u
jay
a
n a
b
a
s
an,
Ka
bar
d
a
n
ga,
K
o
l
k
at
a
70
01
0
4
.
Em
a
il: ayan
mit
r
a.j
u
@g
m
a
il.co
m
1.
INTRODUCTION
Ad
va
nces i
n
f
a
st
-swi
t
c
hi
ng
i
n
sul
a
t
e
d-
gat
e
bi
p
o
l
a
r
tr
a
n
s
i
s
t
o
r
(I
G
B
T)
te
ch
no
log
y
and
co
s
t
-
e
f
f
e
c
tiv
e,
hi
g
h
-s
pee
d
c
o
nt
r
o
l
pr
oces
so
r
s
ha
ve si
g
n
i
f
i
c
ant
l
y
m
a
t
u
red
con
s
t
a
nt
cu
rre
nt
v
o
l
t
a
ge
po
w
e
r so
u
r
ce t
ech
nol
ogy
.
There
i
s
al
s
o
c
u
r
r
ent
l
y
a
n
i
n
c
r
easi
n
g
dem
a
nd
fo
r
hi
g
h
cu
rr
ent
,
l
o
w
ri
ppl
e
hi
ghl
y
reg
u
l
a
t
e
d c
o
nt
rol
l
e
d
p
o
we
r
su
pp
lies fo
r t
h
e h
eatin
g
pu
rpo
s
es in
high t
e
m
p
erature fila
m
e
nts. To
meet these re
qui
rem
e
nts has been a
chal
l
e
ng
e i
n
t
h
e de
si
g
n
o
f
f
i
l
a
m
e
nt
powe
r
su
ppl
i
e
s,
consid
er
ing
no
n-
id
eal ch
ar
acteri
s
tics of the
de
vices,
pert
ur
bat
i
o
n
s
i
n
t
h
e p
o
w
er
sou
r
ce, a
nd
v
a
ri
at
i
ons i
n
the lo
ad
.
As the research
in
h
i
gh
-en
e
rg
y ph
ysics
progresses a
n
d as the particle accelerators fi
nd m
o
re new
a
pplications in industrial
and medical areas, better
regu
lated
filamen
t
power supplies are in
con
s
tan
t
d
e
m
a
n
d
.
Thi
s
pa
per
pre
s
ent
s
a ne
w sc
hem
e
t
o
desi
gn a hi
g
h
c
u
r
r
e
n
t
co
nst
a
nt
dc
po
we
r su
ppl
y
.
The sy
st
em
uses elem
ents from
analog ele
c
tronic
s for con
t
ro
llin
g
pu
rposes.
We m
a
in
ly e
m
p
l
o
y
th
e
prop
erty
o
f
IGB
T
th
at
ch
ang
e
i
n
t
h
e
Gate to em
i
tter vo
ltag
e
can
al
ter th
e curren
t
th
ro
ugh
it in
t
h
e lin
ear
reg
i
on
o
f
its op
eration. Th
e
co
n
t
ro
ller circu
it is d
e
sig
n
e
d su
ch
t
h
at b
y
sen
s
ing
th
e ch
an
g
e
i
n
th
e ou
tpu
t
vo
ltag
e
it will g
e
n
e
rate sufficien
t
a
m
ount
of
gat
e
-to-em
itter voltage whic
h wi
ll change t
h
e c
u
rrent through it, thus c
h
angi
ng t
h
e loa
d
voltage.
Th
is pro
cess co
n
tinu
e
s
u
n
til th
e lo
ad
vo
ltag
e
h
a
s reach
e
d
to
th
e d
e
sired
v
a
lu
e. So
m
e
flu
c
tu
ation
s
will b
e
p
r
esen
t lo
ad
vo
ltag
e
bu
t th
ese flu
c
tu
ation
s
are no
t co
m
p
ar
ab
le to
th
e l
o
ad
vo
ltag
e
, i.e. rip
p
l
e
facto
r
i
s
v
e
ry
sm
a
ll. Also
fo
r an
y su
dd
en
chan
g
e
in
lo
ad
t
h
e ou
tpu
t
vo
ltage do
es
n
o
t
flu
c
tu
ate to
o
m
u
ch fro
m
th
e set valu
e,
i.e. regu
latio
n
i
s
also
sm
all.
Evaluation Warning : The document was created with Spire.PDF for Python.
31
I
S
SN
: 2
088
-86
94
I
J
PED
S
Vo
l. 3,
No
.
1,
Mar
c
h
2
013
:
30
–
4
0
2.
R
E
SEARC
H M
ETHOD
circuit topology and principle
Fi
gu
re
1.
C
i
rcu
i
t
t
opol
ogy
an
d
p
r
i
n
ci
pl
e
The ci
rc
ui
t
t
o
pol
ogy
o
f
t
h
e
pr
o
pose
d
fi
l
a
m
e
nt
po
we
r s
u
p
p
l
y
sc
hem
e
i
s
sh
ow
n a
b
ov
e. T
h
e ci
rc
ui
t
con
s
i
s
t
s
o
f
a
d
i
ode c
o
nve
rt
er
,
L-t
y
pe
LC
fi
l
t
er,
bl
eede
r
re
s
i
st
ance, I
G
B
T
,
p
o
t
e
nt
i
a
l
di
vi
der
,
a
n
o
n
-
i
n
v
e
rt
i
ng
bu
ffe
r,
in
verter
, a
diffe
re
ntial am
plifier & R
E
F0
1.
Fo
r our
ex
p
e
r
i
men
t
atio
n
purp
o
s
e a
1
80V
A, 23
0V
/18V
tr
an
sfo
r
m
e
r
is used
to pr
odu
ce
an
o
u
t
p
u
t
o
f
10
V,
10
A.
An
excess o
f
8
V
i
s
t
a
ken i
n
t
o
d
e
si
gn co
nsi
d
erat
i
on t
o
com
p
en
sat
e
for t
h
e f
o
r
w
ar
d v
o
l
t
a
ge d
r
o
p
i
n
rectifier & IGBT. Fu
ll wav
e
rectifier Bridg
e
is u
s
ed
to
rectify th
e AC o
u
t
p
u
t
o
f
th
e t
r
an
sfo
r
m
e
r to
a DC
v
o
ltag
e
. A L t
y
p
e
LC filter is u
s
ed
to
reduce h
a
rm
o
n
i
cs in
th
e
o
u
t
p
u
t
wav
e
fo
rm
o
f
th
e rectifier.
A b
l
eed
er
resistan
ce is al
so
p
r
ov
id
ed
to sup
p
l
y a
p
a
th
for th
e cap
acit
o
r
of th
e filter
to
d
i
sch
a
rg
e
wh
en power
supp
ly is
t
u
r
n
ed
of
f.
G
4
PH
50
U
D
I
G
B
T
i
s
use
d
f
o
r
t
h
e p
u
r
p
ose
of
keepi
n
g
t
h
e
out
put
vol
t
a
ge
con
s
t
a
nt
. T
w
o 1
0
k
resistors, c
o
nnected in se
ries, are c
o
nnected in pa
ralle
l with
th
e l
o
ad &
vo
ltag
e
feedb
a
ck
is tak
e
n
from
th
e
mid
d
l
e po
in
t.
In t
h
e v
o
l
t
a
ge
cont
rol
l
e
r ci
rc
ui
t
,
t
h
e bu
ffe
r i
s
used t
o
av
oi
d l
o
adi
ng e
ffec
t
s. Si
nce we ar
e t
a
ki
ng t
h
e
feedb
a
ck
v
o
ltag
e
with
resp
ect to
th
e h
i
g
h
po
ten
tial p
o
i
n
t
o
f
th
e lo
ad
, the feed
b
a
ck
vo
l
t
ag
e will b
e
n
e
g
a
tiv
e
w.r.t th
e
g
r
ou
nd
of th
e lo
ad
circu
it.
So
we req
u
i
re th
is inverter to
h
a
v
e
t
h
e ou
t
p
u
t
v
o
l
t
a
g
e
p
o
s
itiv
e
w.r.t th
e
v
o
ltag
e
o
f
the e
m
it
ter o
f
th
e IGBT, so
th
at th
e IGBT can
be fired
Th
e inverted
ou
tpu
t
of th
e p
r
ev
iou
s
stag
e is
Evaluation Warning : The document was created with Spire.PDF for Python.
IJPE
D
fed t
o
am
p
b
p
art
o
p
urp
o
whic
h
b
y u
s
i
th
e n
o
p
aral
l
out
p
u
ot
he
r
w
is co
n
Filte
r
filter
com
b
The
d
flo
w
f
and i
n
val
u
e
,
If t
h
e
D
S
o
the in
v
e
rtin
g
b
y
a po
t. By
c
h
If we fi
x
t
o
f t
h
e circ
uit
w
o
se t
h
e pot r
e
q
h
will g
e
nerat
e
i
ng a stable 1
0
o
n
-
i
n
vert
i
ng i
n
l
el com
b
inati
o
u
t, thu
s
stab
ili
z
Hig
h
er va
w
ise it will l
o
n
necte
d
ac
ros
s
r
Desi
g
n
For an in
d
on t
h
e
ot
her
h
ined as
in an
L
If a
n
LC
fi
For a sm
a
d
i
ode c
u
r
r
e
nt
f
or
long
er ti
m
As L exc
e
n
put
c
u
r
r
e
nt
I
,
th
e i
n
p
u
t v
o
l
indu
ctor
is
p
e
A hi
gh
re
g
g
term
in
al o
f
t
h
h
anging t
h
e o
t
h
e
po
t v
o
lta
g
w
ill try to
kee
p
q
ui
re a fi
xe
d
v
e
10
V D
C
w
h
0
V source R
E
n
pu
t ch
an
ges
t
o
n of
r
e
sisto
r
z
es the
volt
a
g
e
lue of
resisto
r
o
ad the prima
r
s
th
e inv
e
rtin
g
d
u
c
t
o
r filter,
t
h
an
d t
h
e
ri
ppl
e
L
-ty
p
e LC fil
t
fi
lter i
s
conn
e
c
ll v
a
l
u
e of L,
will th
en
flo
w
m
es with
d
i
mi
n
e
e
d
s a
critical
I
are
fu
ll-wa
v
l
tag
e
v to th
e
f
2
e
r
f
ect
havi
ng
n
I
g
ul
at
ed l
o
w
r
ip
h
e op am
p a
n
d
u
t
pu
t of
t
h
e
p
g
e at certain l
e
p
th
is
vo
ltage
v
o
ltag
e
so
urc
e
h
en
w
e
su
pp
l
y
E
F01 chi
p
.
U
s
t
he out
put
v
a
r
r
a
nd
a capa
c
e
ac
ros
s
the
l
o
r
s
ar
e
u
s
ed
to
r
y load circui
t
g
an
d
o
u
t
p
ut
t
e
t
he ri
p
p
le fac
t
e
fact
or
dec
r
e
t
er, the
rippl
e
Fig
u
c
t
e
d
t
o
an out
p
Fi
gu
re 3.
Th
e
the ca
pacitor
w
in
sh
ort pu
n
i
s
hed am
pl
i
t
u
i
nduct
a
nce L
C
v
e rectified s
i
f
ilter can b
e
a
p
.
n
o dc resi
st
a
n
SSN
:
208
8-8
6
ip
pl
e DC
p
o
w
d
refe
rence v
o
ot
we ca
n c
h
a
e
v
e
l th
e lo
ad
co
nst
a
nt
as
f
a
e
. Here
we a
r
e
y
+1
2V to
it.
s
i
ng a
10
K re
s
r
ies accordin
g
ito
r is u
s
ed.
o
ad.
a
voi
d f
o
l
l
o
w
i
t
and u
n
e
xpe
c
e
rm
in
al o
f
th
e
t
or i
n
creases
a
e
ases with t
h
e
f
actor
sho
u
l
d
u
r
e
2
.
L-
type
L
p
u
t
of
a fu
ll-
w
e
out
put
o
f
t
h
e
C will b
e
ch
a
lses. As L in
c
u
des.
C
, on
e dio
d
e
o
i
ne-
w
a
v
es. H
p
pr
o
x
i
m
at
ed
b
n
ce,
t
h
e dc out
p
6
94
w
er supp
ly ba
s
e
o
ltag
e
is fed
t
o
a
n
g
e
th
e
vo
lta
g
vo
ltag
e
will
r
a
r as
po
ssi
bl
e
e
usi
n
g R
E
F
0
1
Refere
nce in
p
s
isto
r in
serie
s
g
ly. Between
t
Th
is help
s i
n
i
ng o
f
hi
ghe
r
c
c
ted
resu
lt wi
l
op
am
p t
o
re
d
a
s th
e load
r
e
increase of l
o
n
o
t
de
pe
nd
o
n
L
C filte
r
w
av
e rectifier,
e
L-
typ
e
L
C
fi
a
rge
d
to t
h
e p
c
reases, the
c
o
r the ot
h
e
r a
l
ence
with an
b
y
p
u
t
v
o
ltage
is
e
d
on
LC filt
e
o
t
h
e no
n-i
n
v
e
g
e
across
of t
h
r
e
m
ain at
a c
e
with
ch
ang
e
i
1
c
h
ip as a c
o
n
p
ut
i
s
fe
d i
n
t
o
s
with
a p
o
t
t
h
t
he out
p
u
t
o
f
t
n
redu
ction
o
f
c
ur
rent
t
h
ro
u
g
l
l
be obt
ai
ne
d
.
d
uce the
oscil
l
e
sistance i
n
cr
e
o
ad resistanc
e
n
load resista
n
t
h
e out
put
o
f
fi
lte
r
eak
vo
ltage
V
c
u
rre
nt p
u
lses
l
way
s
co
nd
uc
t
inductance
g
.
e
r &
IGB
T
(
Ay
e
rt
in
g
term
in
a
l
h
e l
o
ad.
e
rt
ai
n l
e
vel
a
n
i
n
load c
u
rre
n
n
stan
t vo
ltag
e
o
th
e vo
ltag
e
h
e i
n
put
i
s
va
r
t
he
pot
an
d t
h
e
f
flu
c
tu
ation
o
g
h t
h
e cont
rol
l
.
A
capacitor
l
ati
o
n of
t
h
e
o
e
ase
s
. For th
e
e
. If these two
n
ce
(
R
L
)
.
th
e filter as f
o
V
m
to
cu
t o
f
f t
h
ar
e sm
oot
he
n
t
s. Thus inpu
t
g
reater tha
n
t
h
32
y
an
Mitra
)
l
of t
h
e op
n
d t
h
e re
st
n
t. Fo
r
th
is
e
ge
ne
rat
o
r
contr
o
ller
r
i
e
d a
nd
a
s
e
g
r
ou
nd
a
o
f
th
e
p
o
t
l
er circ
uit,
of 0.
1
m
F
o
ut
pu
t
.
e
ca
pacitor
filters are
o
llo
ws:
h
e diodes.
n
ed out
t
o
t
v
o
ltage v
h
e critical
Evaluation Warning : The document was created with Spire.PDF for Python.
33
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S
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: 2
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94
I
J
PED
S
Vo
l. 3,
No
.
1,
Mar
c
h
2
013
:
30
–
4
0
A.
R
e
gul
at
i
o
n:
In
p
r
actise, the in
du
ctor is im
p
e
rfect. If R
is th
e to
tal resistan
ce of on
e-h
a
lf
o
f
th
e tran
sform
e
r
secon
d
a
ry, th
e
d
i
od
e an
d th
e in
du
ctor i
n
series, th
e d
c
ou
tpu
t
vo
ltag
e
across th
e l
o
ad will b
e
The
perce
n
t
a
ge
re
gul
at
i
o
n i
s
.
100
.
100
B.
Ri
ppl
e f
a
ct
or
The ri
ppl
e c
u
r
r
ent
fl
owi
ng t
h
r
o
ug
h L
doe
s
not
gi
ve a
p
pr
eci
abl
e
ri
p
p
l
e
vol
t
a
ge
acr
oss
R
L
pr
ovi
de
d
the reacta
n
ce
X
C
of C at t
h
e
ripple freque
nc
y is m
u
ch less
than R
L
.
So
, fo
r an LC
Filter,
X
C
<< R
L
and
X
L
>> X
C
at the angular freque
ncy 2
ω
. Here
,
X
L
=2
ω
L
is the reacta
n
c
e
of L at
2
ω
.
T
h
ere
f
o
r
e, t
h
e a
c
com
pone
nt
o
f
th
e cu
rren
t th
rou
g
h
t
h
e L is m
a
in
ly d
e
termin
ed
b
y
X
L
, s
o
that R.M.S.
ri
pple curre
nt is
I’
rm
s
=
√
whe
r
e V
dc
=
2V
m
/
π
Th
e
ou
tpu
t
r
i
pp
le vo
ltag
e
pr
od
u
c
ed
b
y
I
’
r
m
s
f
l
ow
ing
t
h
ro
ug
h th
e C
,
is
V’
rm
s = I’
rm
sX
C
= (
√
2 X
c
)/(
2
X
L
) V
dc
Th
e
ripp
le fact
o
r
is :
√
√
Thus
γ
i
s
i
n
de
p
e
nde
nt
of
R
L
, a
s
expected.
C. Critica
l
Indu
ctan
ce
If L is larg
er th
an
its critical
v
a
lu
e LC, as assu
m
e
d in our analysis, the curr
en
t I
do
es
n
o
t
dr
op
to
zero
.
The cu
r
r
ent
I has a dc com
pone
nt
I
dc
=V
dc
/R
L
and a si
n
u
soi
d
al
ac com
ponent
o
f
pea
k
val
u
e
(4
V
m
)/(3
π
X
L
) . He
nce,
I
dc
m
u
st
excee
d t
h
e
n
e
gat
i
v
e
pea
k
v
a
l
u
e o
f
t
h
e ac c
o
m
pone
nt
, i
.
e
.
⩾
or
,
Or,
L=L
C
&
as w
e
k
now
X
L=
2
ω
L
,
at
l
i
n
e f
r
e
q
ue
ncy
of
5
0
H
z
For
o
u
r
e
xpe
ri
m
e
nt
at
i
on,
V
dc
=10V a
n
d
I
dc
=10A
W
e
k
now
, V
dc
=
.
So,
, V
dc
=15
.
70
8V
So,
V
R.M.S.
=
/√2
=1
1.
1 V
Loa
d
re
sistanc
e
R
L
= V
dc
/ I
dc
=1
Ω
It is disc
usse
d t
h
at
. Fro
m
th
is L
C
(
c
r
itical in
du
ctan
ce)
co
m
e
s ou
t to b
e
1.06m
H
.
W
e
choo
se L=2
m
H
.
An estim
a
tio
n
o
f
th
e
v
a
l
u
e
o
f
cap
acitor can
b
e
m
a
d
e
fro
m
th
is cond
itio
n
X
C
<< R
L
.
From
th
is eq
u
a
tion
v
a
lu
e
of C
mi
n
= 1.
6
m
F. So we
sh
oul
d c
h
o
o
se
c
a
paci
t
o
r
o
f
val
u
e
great
er
t
h
a
n
C
mi
n
.From
t
h
e e
x
p
r
essi
on
of
ri
p
p
l
e
factor, we know,
√
Ch
oo
sing
r
i
pp
l
e
f
actor
, w
e
can
g
e
t
the val
u
e of
capacitor re
qui
red.
Evaluation Warning : The document was created with Spire.PDF for Python.
I
J
PED
S
I
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:
208
8-8
6
9
4
34
A h
i
g
h
regu
la
t
e
d
lo
w
ripp
le
DC po
wer supp
ly ba
sed
on
LC filter
&
IGBT (Ayan
Mitra
)
Table
1. T
h
e
value of ca
pacitor re
qui
red
Ripple Factor
Capacitor
r
e
quir
e
d
5% 11.
9
m
F
2% 29.
8
m
F
1% 59.
7
m
F
We ch
o
o
se cap
aci
t
o
r o
f
5
4
m
F
& t
h
e ri
p
p
l
e
cause
d by
t
h
i
s
1.
1% w
h
i
c
h ca
n be f
u
rt
her i
m
pr
o
v
ed as
a
v
o
ltag
e
regu
lato
r is em
p
l
o
y
ed
.
So,
we
ha
ve c
h
ose
n
t
h
e
val
u
e
of
t
h
e i
n
d
u
ct
o
r
t
o
be
2m
H an
d
a bl
ee
der
resi
s
t
ance o
f
val
u
e
1k
Ω
.
Ope
r
at
i
o
n
of t
h
e I
G
B
T
The t
u
rni
ng
o
n
of t
h
e
devi
ce i
s
achi
e
ve
d by
i
n
creasi
ng t
h
e g
a
t
e
vol
t
a
ge
V
GE
so
th
at it is g
r
eater th
an
th
e th
resho
l
d
vo
ltag
e
V
th
. Th
i
s
resu
lts in
an
in
v
e
rsion
layer fo
rm
in
g
und
er th
e g
a
te wh
ich
prov
id
es a ch
ann
e
l
lin
k
i
ng
th
e source to
th
e
drift reg
i
o
n
o
f
t
h
e
device.
Electrons are t
h
en i
n
j
ected from
the source int
o
the drift
reg
i
o
n
wh
ile at th
e sam
e
ti
me j
u
n
c
tion
J
3
, which is forward
biased, injects
hol
es i
n
to
t
h
e
n-
dop
ed dr
if
t r
e
g
i
on
.
This injection
causes c
o
nduct
i
vity
m
odulation of the
drift
regi
on
wher
e
bot
h
the
electron
and hole de
nsities
are se
veral
o
r
d
e
rs
of
m
a
gni
t
ude
hi
g
h
er
t
h
a
n
t
h
e
ori
g
i
n
al
n-
d
opi
ng
.
It
i
s
t
h
i
s
c
o
n
d
u
ct
i
v
i
t
y
m
odul
at
i
o
n
whi
c
h
gi
ves t
h
e I
G
B
T
i
t
s
l
o
w o
n
-
st
at
e vol
t
a
ge
be
cause
of t
h
e
re
duce
d
resi
st
an
ce of t
h
e d
r
i
f
t
regi
on
. S
o
m
e
of t
h
e
in
j
ected
ho
les
will reco
m
b
in
e in
th
e drift regio
n
,
wh
ile
o
t
h
e
rs will cross the reg
i
o
n
v
i
a drift an
d
d
i
ffu
s
i
o
n
and
will reach the j
u
nction with
the p-ty
pe regi
on
whe
r
e they will be co
llec
t
ed. The
ope
ra
tion of the IGBT ca
n
t
h
eref
o
r
e be c
onsi
d
ere
d
l
i
k
e
a wi
de-
b
ase
p-
n-
p t
r
a
n
si
st
o
r
w
hose
base
dri
v
e c
u
r
r
ent
i
s
suppl
i
e
d b
y
t
h
e
MOSFET cu
rren
t
throug
h th
e ch
an
n
e
l.
Calcul
ati
o
n
of the
sys
t
em
be
haviour
a.
Trans
f
er F
unc
tion
Let th
e feed
back vo
lta
g
e
t
o
th
e bu
ffer is V
F
(
w
.r.t
t
h
e gr
ou
nd
of
t
h
e c
ont
r
o
l
l
e
r su
ppl
y)
. The
out
put
of
th
e bu
ffer will b
e
th
en
V
1
=V
F
. Nex
t
on
e is an in
v
e
rting
am
p
lifier. So
its
ou
tp
u
t
will b
e
V
2
= -
V
1
.
Th
e n
e
x
t
op
-am
p
is wo
rk
ing as a d
i
fferen
tial a
m
p
lif
ier. In its n
o
n
-inv
erti
n
g
inpu
t is th
e referen
ce is
fed
.
Let t
h
e re
f
e
rence
v
o
ltage
be
V
C
w.r.t. the gr
oun
d of
the vo
ltag
e
con
t
ro
ller
g
r
o
und
.
Co
n
s
i
d
er
i
n
g V
2
be i
t
s
in
v
e
rtin
g term
i
n
al inp
u
t
, th
e ou
tpu
t
will be,
Vo=
(
1
+
) Vc
-(
2
;
On sim
p
lificati
o
n we g
e
t,
Vo=
(
1
+
)Vc+
(
;
Or,
V
o
=
(
1
+
)V
c+(
;
Here
t
h
e
fo
r
f
eedbac
k
p
u
r
p
o
s
e we
t
a
ke
hal
f
of t
h
e l
o
a
d
v
o
l
t
a
ge.
N
o
w
si
nce t
h
e
g
r
ou
n
d
of
p
o
w
er
su
pp
ly
o
f
th
e
v
o
ltag
e
co
n
t
ro
l
l
er is co
nn
ected
to th
e em
i
tte
r
o
f
t
h
e
IGBT
so
t
h
e
g
r
ou
nd
is at V
L
po
ten
t
ial.
So
V
F
= V
L
/2- V
L
.
So,
Vo=
(
1
+
)Vc
-
(
;
Or,
V
o
=
(
1+
)V
c-(
;
Now th
e
vo
ltag
e
V
0
is
with
resp
ect to
th
e
emit
ter. So
t
h
is
voltage can be
considere
d
as the
voltage
b
e
tween
th
e Gate an
d Em
it
ter term
in
al o
f
th
e IGB
T
,
resp
on
sib
l
e fo
r t
h
e
firi
n
g
of t
h
e IGBT.
So,
V
GE
=Vo=(1+
)V
c-(
;
Evaluation Warning : The document was created with Spire.PDF for Python.
35
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: 2
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I
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S
Vo
l. 3,
No
.
1,
Mar
c
h
2
013
:
30
–
4
0
Now t
h
is vo
ltag
e
will d
e
termin
e to
wh
at ex
t
e
n
t
th
e
IGBT
will b
e
forward
b
i
ased
i
.
e. it
s resistan
ce.
Th
e
p
a
ram
e
ter tran
s-cond
u
c
tan
ce
g
i
v
e
s u
s
the ratio
o
f
th
e co
llecto
r
curren
t
and
g
a
te to
emitter v
o
ltag
e
.
So
we ca
n als
o
write
I
L
=G
fe
.V
GE
;
[w
here
G
fe
is th
e tran
s con
ductan
ce of t
h
e
IGBT]
Now si
nce t
h
e
load
resistance
is 1
Ω
,so
th
e lo
ad vo
ltag
e
V
L
=I
L
x 1
Ω
Or,
V
L
[1+
G
fe
(Z
f
/2
R
1
)]
=V
c
.G
fe
.(1+
Z
f
/R
1
);
[w
here
Z
F
i
s
t
h
e fee
dbac
k
pat
h
i
m
pedance
]
Or,
=
;
Taking La
plac
e Tra
n
sform
a
tion,
=
;
Or,
=
.
.
;
Or,
..
.
.
.
;
=
∗
is th
e requ
ired
tran
sfer
fu
n
c
tion
o
f
th
e
vo
ltage con
t
ro
ller circu
it.
whe
r
e k=
G
fe
; a=
..
; b
=
.
.
.
;
b.
Estima
tio
n o
f
Output
fo
r g
i
ven
Inp
u
t:
We
have
al
rea
d
y
o
b
t
a
i
n
e
d
,
..
.
.
.
;
Or,
V
L
(s)
..
.
.
.
N
o
w
con
s
id
er
in
g a
s
t
ep
Inp
u
t
V
C
(s
) =
V
C
/s, we g
e
t
V
L
(s
)
..
.
.
.
Pu
ttin
g th
e
v
a
l
u
es R
1
=1
0K
Ω
; R
2
=3.3M
Ω
;
C
=
0.
1µF a
n
d a
p
pl
y
i
ng
Fi
nal
V
a
l
u
e T
h
eo
rem
,
we
get
l
i
m
→
1
2
1.
2
.
2
21
.
2
21.
2
.
2
.
1
or
,
2
So
, th
e ou
t
p
u
t
v
o
ltag
e
i.e. th
e v
o
ltag
e
acro
s
s th
e lo
ad
will
b
e
2
ti
m
e
s th
at
o
f
th
e ref settin
g, i.e. the
rel
a
t
i
ons
hi
p
be
t
w
een i
n
p
u
t
-
o
u
t
put
s
h
oul
d
be
l
i
n
ear.
Evaluation Warning : The document was created with Spire.PDF for Python.
I
J
PED
S
I
S
SN
:
208
8-8
6
9
4
36
A h
i
g
h
regu
la
t
e
d
lo
w
ripp
le
DC po
wer supp
ly ba
sed
on
LC filter
&
IGBT (Ayan
Mitra
)
c.
Estima
tio
n o
f
Rise
Time:
By d
e
fi
n
itio
n
,
rise tim
e is th
e ti
m
e
requ
ired
for t
h
e
respon
se to
rise
fro
m
1
0
% to
9
0
%
of its stead
y
value. So,
we
can estim
ate
the rise
tim
e th
eo
retically b
y
ap
p
l
ying
a step
change in the ref i
n
put a
n
d ca
n
calcu
late th
e ri
se ti
m
e
. Th
eoretically calcu
la
ted
rise tim
e should be
close
enough to
t
h
e
practical val
u
e
;
exact
match is not expecte
d
as the
r
e is
m
i
s
m
atch betwee
n th
e
va
lue we ass
u
m
e
d and
practical value
of the c
i
rcuit
com
pone
nt
s.
N
o
w
con
s
id
er
in
g a
s
t
ep
Inp
u
t
V
C
(s
) =
V
C
/s;
V
L
(s
) =
.V
C
;
V
L
(s
) =
K
[
]V
C
;
V
L
(s
) =
K
.
V
C
[
.
+ (1-
)(
;
V
L
(t
) =K
.V
C
[
+ (1-
)
∗
;
where u(t)=1 for t>=0;
u(t)=
0
else
whe
r
e.
R
earra
ngi
ng
t
h
e ex
pres
si
o
n
w
e
get
V
L
-
.
.
.
.
;
t =
ln
.
.
.
;
In our case R
1
=10K
Ω
; R
2
=3.3M
Ω
;
C
=
0.
1µ
F;
No
w
by
t
a
ki
ng
G
fe
=
30;
W
e
g
e
t K
=
30
;
a=8
3
6
.
36
; b
=
12
503
;
Now for
i
n
pu
t v
o
ltag
e
V
C
=4
.9
62V
,
Loa
d
Vol
t
a
ge
of
V
L
=9
.9
7V
;
We g
e
t
t=(
l
n
.
.
.
.
.
;
t
=
7.
46
x
1
0
-4
se
c;
t
=
746
.0
7
µ
s
.
Sin
ce
b
y
d
e
fin
i
tio
n
,
rise tim
e i
s
th
e i
n
terv
al
fo
r th
e
res
p
onse
to
rise from
10% t
o
90%
of i
t
s steady
value
so we
ca
n estim
ate the ri
se ti
m
e
to
b
e
59
6.856
µs.
E
X
PERI
MEN
T
AL RES
U
L
T
S
Variation
o
f
Ou
tpu
t
w.r.t
In
pu
t:
Tabl
e 2. Vari
at
i
on o
f
Out
put
w.r
.
t
I
n
put
Refere
nce in
put
through pot
(v
o
l
ts)
Load
Volt
age
(v
o
l
ts)
Theoretically obtained
Load Volt
age (
volt
s
)
0.
0 0
0.
0
0.
502
0.
980
1.
04
1.
003
2.
000
2.
006
1.
500
3.
014
3.
0
2.
002
4.
037
4.
004
2.
501
5.
055
5.
002
3.
007
6.
085
6.
014
3.
501
7.
09
7.
002
4.
005
8.
12
8.
01
4.
503
9.
14
9.
006
4.
927
10.
0
9.
854
Evaluation Warning : The document was created with Spire.PDF for Python.
37
I
S
SN
: 2
088
-86
94
I
J
PED
S
Vo
l. 3,
No
.
1,
Mar
c
h
2
013
:
30
–
4
0
Fi
gu
re 4.
The
vari
at
i
o
n of o
u
t
put
vol
t
a
g
e
ac
ross
t
h
e
l
o
a
d
Th
e
v
a
riation
of
o
u
t
p
u
t
vo
ltage acro
ss t
h
e lo
ad
is
qu
ite lin
ear wit
h
th
e v
a
riatio
n
o
f
th
e i
n
put
Rise time:
B
y
usi
ng a s
w
i
t
ch at
t
h
e ref i
n
p
u
t
we s
u
dde
nl
y
swi
t
c
h o
n
t
h
e re
f i
n
put
t
o
i
t
s
m
a
xim
u
m
set
t
i
ng. T
h
i
s
o
p
e
ration
can
b
e
treated
as step
ch
ang
e
in
in
pu
t an
d
th
e
o
u
t
p
u
t
ob
tain
ed
will b
e
th
e step
resp
on
se
o
f
th
e
syste
m
. The pl
ot give
s us the
step res
p
onse
of the syst
em
. C
H
1(yello
w) is th
e ou
tpu
t
& C
H
3(
pur
p
l
e)
is th
e ref
in
pu
t
Fi
gu
re 5.
R
i
se t
i
m
e
There
i
s
a si
g
n
i
fi
cant
am
ount
of
del
a
y
bet
w
e
e
n t
h
e
i
n
put
an
d
out
put
.
T
h
i
s
i
s
t
h
e l
o
o
p
del
a
y
i
s
cause
d
due
to t
h
e all form
of ca
pacit
a
nces
(actual
& stray) in
t
h
e
circuit.
Proper wiri
ng
and
usi
n
g m
o
re linear circuit
com
pone
nt
ca
n
re
duce
t
h
i
s
del
a
y
.
Here th
e C
h
ann
e
l 1
is th
e conn
ected
acro
ss t
h
e lo
ad
. Th
e o
s
cillo
sco
p
e
reported
a rise ti
m
e
o
f
608
µs.
our calc
u
lated
value
was
596.856 µs
, cl
ose e
n
ough to t
h
e
practical value.
Evaluation Warning : The document was created with Spire.PDF for Python.
I
J
PED
S
I
S
SN
:
208
8-8
6
9
4
38
A h
i
g
h
regu
la
t
e
d
lo
w
ripp
le
DC po
wer supp
ly ba
sed
on
LC filter
&
IGBT (Ayan
Mitra
)
Lo
ad
Regu
latio
n
:
Fo
r
a referen
c
e in
pu
t of 4.94V th
e
o
u
t
p
u
t
vo
ltag
e
acro
ss l
o
ad
is
10
.0
16V in
fu
ll lo
ad
con
d
ition
.
Now
u
s
ing
a switch
we
d
i
sconn
ect
th
e lo
ad
fro
m
th
e circu
it an
d th
e
v
o
ltag
e
acr
o
ss th
e f
e
ed
back r
e
sistor
s is 10
.02V
i.e. no l
o
ad vol
t
age at
th
at ref settin
g
is 1
0
.05V.
So
, th
e
reg
u
l
atio
n at fu
ll lo
ad
co
m
e
s o
u
t
to be,
(V
NL
-V
FL
)/V
FL
.1
00
%=
(
(
1
0
.02-
10
.01
6
)
/
10
.0
16
)x1
00%=0
.10
9
%
Vo
ltag
e
regu
latio
n
is qu
ite go
od
. Th
ere is no
app
r
eciab
le ch
ang
e
in
t
h
e ou
tpu
t
vo
ltag
e
at
m
a
x
i
m
u
m
ref
setting
from
no load to
full lo
ad c
o
ndition,
which is a
necessary
c
o
ndition for voltage regulator.
Lin
e
Regu
lation
:
Lin
e
regu
latio
n
is th
e cap
a
b
ility to
m
a
in
tain
a con
s
tan
t
o
u
t
p
u
t
v
o
ltag
e
level o
n
t
h
e ou
tput ch
ann
e
l
of
a po
wer
su
p
p
l
y
despi
t
e
cha
n
ges t
o
t
h
e i
n
p
u
t
vol
t
a
ge l
e
vel
.
In
ou
r case, t
h
e n
o
m
i
nal
suppl
y
v
o
l
t
a
ge i
s
23
0
V
AC.
Tabl
e 3.
Li
ne r
e
gul
at
i
o
n
Input voltage to t
h
e
t
r
ansf
orm
er (
V
)
%
o
f
no
m
i
nal
volt
age
Volt
age across Load
(V)
200
-
13%
10.
01
210
-
8
.
7
%
10.
01
230(
nom
inal value)
0
10.
01
250
8.
7%
10.
02
260
13%
10.
02
Th
e lin
e r
e
gu
latio
n
is
w
ith
in (1
0.02-
10
.01
)
/10
.
0
1
x 100
% = 0.1%
Response
Tim
e
Un
de
r ful
l
val
u
e of
out
put
v
o
l
t
a
ge o
f
10
V
we ha
ve chec
k
e
d t
h
e res
p
o
n
s
e
t
i
m
e
. In ou
r case, t
h
e sy
st
em
tu
rn
s
o
u
t
t
o
b
e
little
m
o
re slug
g
i
sh
t
h
an normal electr
o
n
i
cs circu
it.C
H
3 (yello
w) co
rresp
ond
s t
o
th
e ou
tpu
t
wave
f
o
rm
&
C
H
1
(
p
u
r
pl
e
)
c
o
r
r
es
po
n
d
s t
o
t
h
e i
n
p
u
t
re
f
vol
t
a
ge wa
ve
fo
rm
.
Fi
gu
re
6.
R
e
sp
ons
e t
i
m
e at
no l
o
ad
Evaluation Warning : The document was created with Spire.PDF for Python.
39
I
S
SN
: 2
088
-86
94
I
J
PED
S
Vo
l. 3,
No
.
1,
Mar
c
h
2
013
:
30
–
4
0
Fig
u
re
7
.
Respo
n
s
e tim
e at fu
ll lo
ad
Ripple Fact
or
C
H
3
(y
el
l
o
w
)
cor
r
es
po
n
d
s t
o
t
h
e
out
put
wa
ve f
o
rm
& C
H
1
(
p
u
r
pl
e) c
o
r
r
es
po
n
d
s t
o
t
h
e i
n
put
ref
vol
t
a
ge
wa
ve
fo
rm
. The ri
ppl
e
i
s
cal
cul
a
t
e
d at
f
u
l
l
val
u
e
o
f
t
h
e
refe
rence
i
n
put
.
Figure
8. Rippl
e Factor at full
load
R
i
ppl
e =
1
2
.
8m
V
pk-pk
=7.8m
V
R.M.S
Ripple f
actor=
γ
=
=
.∗
.
x100%=0.
0
77%
OBSER
VATI
O
N
It can
b
e
ob
serv
ed
th
at du
ri
n
g
tran
sitio
n
from n
o
lo
ad
to
fu
ll lo
ad
th
ere is a d
i
p
in
th
e ou
tpu
t
v
o
ltage.
This can
be explained as si
nc
e at no load t
h
e
r
e is s
m
a
ll cu
rren
t
th
ro
ugh
th
e IGBT, th
e capacito
r is d
i
sch
a
rg
i
n
g
at slo
w
er
rate. Now if we su
dd
en
ly switch
on
th
e lo
ad
th
en
su
dd
en
ly fu
ll lo
ad
cu
rren
t will start flo
w
i
n
g
th
ro
ugh
th
e circu
it cau
sing
the v
o
ltag
e
acro
s
s th
e cap
acito
r to
d
r
o
p
. Th
is
will d
ecrease th
e vo
ltag
e
acro
ss th
e
load which
will change the
feedbac
k
voltage and the voltage regulator
will take necessary
m
easures to
in
crease th
e load
vo
ltag
e
b
a
ck
to
its p
r
ev
i
o
u
s
v
a
l
u
e.
Th
e rev
e
rse case
h
a
p
p
e
n
s
d
u
ring
th
e tran
sition
fro
m
fu
ll
lo
ad
t
o
n
o
lo
ad.
3.
CO
NCL
USI
O
N
The
0.
1µF
ca
paci
t
o
r
&
3.
3
M
Ω
resistors
are c
o
nnecte
d
to the
no
n-in
v
e
rtin
g termin
al of th
e
d
i
fferen
tial o
p
-am
p
to
m
a
tch
th
e i
m
p
e
d
a
n
c
e in
th
e feedback
p
a
th. Th
e id
en
tical i
m
p
e
d
a
n
ce in
bo
th
in
pu
t
Evaluation Warning : The document was created with Spire.PDF for Python.