Substituting these equations in Eqs. ( 4.19) and ( 4.20), we obtain
Hence, the current gain is given by
I
=
DVin
o
R
D2Vin
Jin=~
(4.21)
(4.22)
(4.23)
This relation can be obtained by equating the average input and output power, to yield
From Eqs. (4.15) and (4.23), it
is
clear that the current and voltage relations for
the converter are equivalent to a de transformer model with a ratio
of
D,
as shown in
Fig. 4.13. The sinusoidal curve and straight line drawn across the transformer wind
ings indicate that the transformer is capable
of
transferring ac and de, respectively.
Critical Inductor Value
It
is
clear that for
I
Lmin
;;t:.
0 , the converter will operate in the
continuous conduction
mode (cern).
To
find the minimum inductor value that is needed to keep the converter
in the cern, we set
JLmin
to zero and solve for
L:
I
:D
+
_
(1
(1D)TJ
_
]Lmin

D
vin
R
2L

0
/()
(
1D)
Lcrit
=
2
TR
(4.24)
+
Figure 4.13
Equivalent circuit representation for the
buck converter, referred
to
as a dede transformer.
142
Chapter 4
Nonisolated SwitchMode dede Converters
where
Lcrit
is
the critical inductance minimum value for a given
D,
T,
and
R
converter enters the discontinuous conduction mode (dcm)
of
operation.
Output Voltage Ripple
Since we have assumed that the output voltage has no ripple, the entire
ac
current from the inductor passes through the parallel capacitor, and only de
rent is delivered to the load resistor. In practice, the value
of
the output
is an important design parameter since it influences the overall size
of
the
de converter and how much
of
the switching frequency ripple is being
Having said that, it is design practice to choose a larger output capacitor
in
to limit the ac ripple across
V
0
•
Theoretically speaking,
if
C
~
oo,
the
acts like a short circuit to the ac ripple, resulting in zero output voltage
we assume
Cis
finite, then there exists a voltage ripple superimposed on the
age output voltage. In order to derive an expression for the capacitor ripple
age, we first obtain an expression for the capacitor current, which is given
by
following relation:
As a result, the initial capacitor current at
t
=
0
is
given by
and at
t
=
DT,
Jc(DT)
=
JL(DT)
1
0
=
+
(/Lmax;
/Lmin)
The resultant capacitor current and voltage are shown in Fig. 4.14.
DT
T
DT
l+DT
2
T
Figure 4.14
Capacitor current and voltage waveforms.
4.2
Continuous Conduction Mode
143
The
instantaneous
capacitor
current
can
be
expressed
in terms
of
/).I
from
Eqs.
(
4.21) and ( 4.22), as shown in the following equations:
i
{t)
=
ILmax
ILmin!
_
ILmax
ILmin
=
/).I!
_/).I
e
DT
2
DT
2
0
$
t
<
DT
(4.25a)
I
I
.
IL
IL
·
i
(!}
= _
Lmax
Lmm{tDT)
+
max
mm
e
(I
D)T
2
=
/).I
(tDT)+/).I
(I
D)T
2
0$
t
<
DT
(4.25b)
where
/).I=
(
Vin(I
D)TD)/
L.
From the capacitor voltagecurrent relation,
ie
=
C(dv/
dt),
the capacitor volt
age,
vel(!),
can be expressed by the following integral
fort
2::0:
Vel
(t)
=
.!.Jt
ie
dt
+
Ve(O)
c
0
where
Ve(O)
is the initial capacitor voltage at
t
=
0.
Substituting for
ie(t)
from
Eq.
(4.25a),
we obtain the following equation:
Vel(!)=
bf:
(~~~~I)
dt+
Ve(O)
Evaluating this integral yields
1
/).I
t
2
/).I
v
l(t)
=

t+
v
(0)
e
CDT2
2C
e
0$
t
<
DT
Similarly, for
t
2::
D
T,
the capacitor voltage is given by
vc
2
(t)
=
.!.Jt
ie
dt
+
Ve(DT)
C
DT
(4.26)
where
Ve(DT)
is the initial capacitor voltage when the switch is turned
off
at
t
=
DT.
From Eq. (4.27)
vc
2
(t)
is given by
v
(t)
=
/).I
(tDT)
2
+
/).I(tDT)+
V
(DT)
DT$t
<
T
e
2
C(l
D)T
2
2C
e
(4.27)
Since the capacitor voltage is in the steady state, we have
v
e
1
(t
=
D
T)
=
ve
2
(t
=
DT)
and
vel
(0)
=
ve
2
(T),