AIC1638 데이터 시트보기 (PDF) - Analog Intergrations
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AIC1638 Datasheet PDF : 15 Pages
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At the boundary between continuous and
discontinuous mode, output current (IOB) is
determined by
IOB
=
VIN
VOUT +
VD
1
2
VIN
L
TON(1− x)
where VD is the diode drop,
x = (RON+Rs)Ton/L.
RON= Switch turn on resistance, Rs= Inductor DC
resistance
TON = Switch ON time
In the discontinuous mode, the switching frequency
(fsw) is
fsw
=
2(L)(VOUT +
VIN2
VD − VIN)(IOUT)
× TON2
(1+
x
In the continuous mode, the switching frequency is
fsw
=
1
TON
(VOUT + VD − VIN)
(VOUT + VD − VSW)
[1+
x
2
( VIN − VSW )]
VOUT + VD − VSW
≅
1
TON
VOUT + VD − VIN
VOUT + VD − VSW
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
AIC1638
3-PIN ONE-CELL STEP-UP DC/DC CONVERTER
SPEC NO: DS-1638-00 03/01/00
IPK
=
Vin
Ron + Rs
1−
exp(−
Ron +
L
Rs
Ton)
≅
VIN
L
(TON)1
−
x
2
≅
VIN
L
TON
(Simple losses equation),
where x=(RON+RS)TON/L
EL
=
1
2
L
× Ipk2
Power required from the inductor per cycle must be
equal or greater than
PL/fSW
=
(VOUT
+
VD
−
VIN)(IOUT)(
1
fSW
)
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
IPK
=
VOUT + VD − VSW
VIN − VSW
−
x
2
IOUT
+
V IN
− VSW
2L
TON1−
x
2
Valley current (Iv) is
Iv
=
VOUT + VD − VSW
VIN − VSW
−
x
2
IOUT
−
VIN
− VSW
2L
×
TON1−
x
2
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
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