
AIC1638
3-PIN ONE-CELL STEP-UP DC/DC CONVERTER
SPEC NO: DS-1638-00 03/01/00
ANALOG INTEGRATIONS CORPORATION
www.analog.com.tw
4F, 9 Industry E. 9th Road, Science-based Industrial Park , Hsinchu, Taiwan, R.O.C. TEL: (8863)577-2500 FAX:(8863)577-2510
13
At
discontinuous mode, output current (IOB) is
determined by
the
boundary
between
continuous
and
(
)
x
1
T
L
V
2
1
V
V
V
I
ON
IN
D
OUT
IN
+
OB
=
where V
D
is the diode drop,
x = (R
ON
+Rs)Ton/L.
R
ON
= Switch turn on resistance, Rs= Inductor DC
resistance
T
ON
= Switch ON time
In the discontinuous mode, the switching frequency
(fsw) is
x
(1
2
T
2
V
)
)(I
V
V
2(L)(V
fsw
ON
IN
OUT
IN
D
OUT
+
×
+
=
In the continuous mode, the switching frequency is
(
V
V
(V
T
D
OUT
ON
+
)
)]
V
V
V
V
V
(
2
x
[1
)
V
V
V
1
fsw
SW
D
OUT
SW
IN
+
SW
IN
D
OUT
+
+
=
+
+
SW
D
OUT
IN
D
OUT
ON
V
V
V
V
V
V
T
1
where Vsw = switch drop and proportion to output
current.
INDUCTOR SELECTION
To operate as an efficient energy transfer element,
the inductor must fulfill three requirements. First, the
inductance must be low enough for the inductor to
store adequate energy under the worst case
condition of minimum input voltage and switch ON
time. Second, the inductance must also be high
enough so maximum current rating of AIC1638 and
inductor are not exceed at the other worst case
condition of maximum input voltage and ON time.
Lastly, the inductor must have sufficiently low DC
resistance so excessive power is not lost as heat in
the windings. But unfortunately this is inversely
related to physical size.
Minimum and Maximum input voltage, output voltage
and output current must be established before and
inductor can be selected.
In discontinuous mode operation, at the end of the
switch ON time, peak current and energy in the
inductor build according to
+
+
=
Ton)
L
Rs
Ron
exp(
1
Rs
Ron
Vin
I
PK
(
)
V
2
x
1
T
L
V
ON
IN
ON
IN
T
L
(Simple losses equation),
where x=(R
ON
+R
S
)T
ON
/L
2
Ipk
L
2
1
EL
×
=
Power required from the inductor per cycle must be
equal or greater than
)
f
1
)(
)(I
VI
V
(V
/f
P
SW
OUT
N
D
OUT
SW
L
+
=
in order for the converter to regulate the output.
When loading is over IOB, PFM controller operates
in continuous mode. Inductor peak current can be
derived from
2L
+
I
SW
+
=
2
x
T
V
V
2
x
V
V
V
V
V
I
ON
SW
IN
OUT
IN
SW
D
OUT
PK
Valley current (Iv) is
×
2L
x
SW
+
=
2
T
V
V
I
2
x
V
V
V
V
V
Iv
ON
SW
IN
OUT
IN
SW
D
OUT