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参数资料
| 型号： | LT1507IS8#TR |
| 厂商： | LINEAR TECHNOLOGY CORP |
| 元件分类： | 稳压器 |
| 英文描述： | 3 A SWITCHING REGULATOR, 570 kHz SWITCHING FREQ-MAX, PDSO8 |
| 封装： | 0.150 INCH, PLASTIC, SO-8 |
| 文件页数： | 4/20页 |
| 文件大小： | 291K |
| 代理商： | LT1507IS8#TR |
12
LT1507
APPLICATIONS INFORMATION
WU
U
Example: with VOUT = 3.3V, VIN = 5V;
VOUT/VIN = 3.3/5 = 0.67
IP = 1.75 – (0.5)(0.66) = 1.42A
Maximum load current would be equal to maximum
switch current
for an infinitely large inductor, but with
finite inductor size, maximum load current is reduced by
one half peak-to-peak inductor current. The following
formula assumes continuous mode operation; the term on
the right must be less than one half of IP.
Continuous mode:
II
VV
V
Lf V
OUT MAX
P
OUT
IN
OUT
IN
()
–
()(
–
)
()( )(
)
=
2
For the conditions above, with L = 5
H and f = 500kHz;
I
A
OUT MAX
()
.–
( . )( – . )
.– .
.
=
()
()
==
142
33 5 3 3
2 5 10
500 10
5
142 0 22
12
63
At VIN = 8V, VOUT/VIN = 0.41, so IP is equal to 1.5A and
IOUT(MAX) is equal to;
15
33 8 3 3
2 5 10
500 10
8
15 039
1 11
63
.–
( . )( – . )
.– .
.
()
()
==
A
Note that there is less load current available at the higher
input voltage because inductor ripple current increases.
This is not always the case. Certain combinations of
inductor value and input voltage range may yield lower
available load current at the lowest input voltage due to
reduced peak switch current at high duty cycles. If load
current is close to the maximum available, please check
maximum available current at both input voltage
extremes. To calculate actual peak switch current with a
given set of conditions, use:
II
VV
V
Lf V
SWITCH PEAK
OUT
IN
OUT
IN
()
(–
)
()( )(
)
=+
2
For lighter loads where discontinuous mode operation can
be used, maximum load current is equal to:
Discontinuous mode:
I
If L V
VV
V
OUT MAX
PIN
OUT
IN
OUT
()
() ( )( )(
)
()(
–
)
=
2
Example: with L = 2
H, VOUT = 5V and VIN(MAX) = 15V;
I
mA
OUT MAX
()
(. )
()(
– )
=
()
()
=
1 5
500 10
2 10
15
2 5 15 5
338
23
6
The main reason for using such a tiny inductor is that it is
physically very small, but keep in mind that peak-to-peak
inductor current will be very high. This will increase output
ripple voltage. If the output capacitor has to be made larger
to reduce ripple voltage, the overall circuit could actually
be larger.
CATCH DIODE
The suggested catch diode (D1) is a 1N5818 Schottky or
its Motorola equivalent, MBR130. It is rated at 1A average
forward current and 30V reverse voltage. Typical forward
voltage is 0.42V at 1A. The diode conducts current only
during switch OFF time. Peak reverse voltage is equal to
regulator input voltage. Average forward current in normal
operation can be calculated from:
I
IV
V
D AVG
OUT
IN
OUT
IN
()
(–
)
=
This formula will not yield values higher than 1A with
maximum load current of 1.25A unless the ratio of input to
output voltage exceeds 5:1. The only reason to consider a
larger diode is the worst-case condition of a high input
voltage and
overloaded (not shorted) output. Under short-
circuit conditions, foldback current limit will reduce diode
current to less than 1A, but if the output is overloaded and
does not fall to less than 1/3 of nominal output voltage,
foldback will not take effect. With the overloaded condi-
tion, output current will increase to a typical value of 1.8A,
determined by peak switch current limit of 2A. With VIN =
10V, VOUT = 2V (3.3V overloaded) and IOUT = 1.8A:
IA
D AVG
()
.(
– )
.
==
1 8 10 2
10
144
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