Averaged Modeling of Non-ideal Boost Converter Operating ... - ijcee

International Journal of Computer and Electrical Engineering, Vol.3, No.1, February, 2011
1793-8163
di dTv ( D+ dˆ
) ( V+
vˆ
) T
d d 2L 2L
d vap
d d vap
D dˆ
Vap
vˆ
1
+ 1−
1
−
2
=
1
+
1
+
⎡ 2L( IL
+ iˆ
L
) ⎤
+ ⎢1
−
( Vzp
+ vˆ
zp)
( D d )( Vi
vˆ
⎥
⎣ 1
+
1
+
i
) Ts
⎦
2 2
1 L
1
s i 1 1 i i s
= =
1 2
+ (3)
( ) ( )( )
ap
(4)
C. AC Small-Signal Model
If we suppose that, the small-signal component is much
less than the direct component, make the direct component is
zero, and neglect the secondary small-signal product term.
Then the (3) and (4) can be predigested to the (5) and (6). So
we can get the small-signal linear equivalent model of
non-ideal Boost converter in discontinuous conduction
model, shown in fig.2.
( )
2
2
d i d T v D + dˆ
( V + vˆ
)
1
1 L
d + d
1
2
2
1
= a vˆ
+ a dˆ
i
=
2L
1
s
i
=
dv 1 ap + (1 −d1−d2)
vzp
= −
⎡
⎢1
−
⎢⎣
( D + dˆ
)( V + vˆ
)
1
( I + iˆ
)
L L
( D + dˆ
)( V + vˆ
)
1
2L
[( V − V − V − I R ) + ( vˆ
− vˆ
− R iˆ
)]
i
1
1
E
1
2 L
1
o
o
i
L
o
i
E
+
T
∧ ∧ ∧ ∧
= b vi
+ b2
d1+
b3
iL
+ b4
vo
s
⎤
⎥
⎥⎦
1
(6)
Fig.2. small-signal model of Boost converter in DCM
Form fig.2, we can calculate the input-to-output transfer
function
Gvi( s)
and the control-to-output transfer function
Gvd( s)
are
vˆ o( s)
Gvi( s)
=
vˆ i( s )
dˆ1( s ) = 0
i
i
o
i
T
s
E
L
(5)
1 −b1− a1( sL+ RE
+ b3)
E 3 C
=
( sL + R + b )( sR C + sRC + 1)
1+ b4
+
RsRC ( C + 1)
tt 1 4 (1 + swz1)(1 + swz2)
= ⋅
tt 2 3+
tt 0 4 s s 2
1 + + ( )
Qw0 w0
(7)
G
vd
vˆ () s
() s =
o
dˆ 1 ()ˆ s v i () s = 0
b2+ a2( sL+ RE
+ b3)
=−
( sL + RE+ b3)( sRCC + sRC + 1)
1+ b4
+
RsRC ( C + 1)
tt 4 5 (1 + swz2)(1 −swz3)
= ⋅
tt 2 3+
tt 0 4 s s 2
1 + + ( )
Qw0 w0
(8)
The parameters of the actual Boost converter as following:
input-voltage Vi
= 5V
, output-voltage Vo
= 10V
, Load
current Io
= 0.2A
, R = 50Ω ,
L=
19.2μH
,
RL
= 3.6mΩ ,
C = 1000μF
, RC
= 2mΩ . The model
number of the MOSFET which we used in this paper is
2SK2690. Ron
= 10mΩ , for Schottky diode RF
= 1mΩ ,
switching frequency
fs
= 50kHz
, PWM outputs the Peak
voltage is Vm
= 1V
. Then there parameters are put into the
formulas as following
R
V
E
4 DR 1 on DR 2 F
= ( RL+ + )
3( D1+ D2)
D1+ D2 D1+ D2
,
ap =− Vo
,
2LIL
2= −D1
1 s i
IO
D
IL
= ( D1+
D2)
DTV and D2
R is the total equivalent resistance of the three parasitic
E
resistances which were calculated to branch of inductance.
Then we can get
D1 ≈ 0.35 , RE
≈22.69mΩ , VE
≈ 0.184V
,
IL
≈ 0.43A
.
Put the four parameters into formula(7)and (8), then we
can get formula (9) and (10).
s s
(1 + )(1 − )
G ( ) 2.112 500000 913333
vi s = ×
2
s ⎛ s ⎞
1+ + ⎜ ⎟
58.42 ⎝ 4502 ⎠ (9)
s s
(1 + )(1 − )
G ( ) 23.696 500000 307433
vd s = ×
2
s ⎛ s ⎞
1+ + ⎜ ⎟
58.42 ⎝ 4502 ⎠ (10)
80