LT1507
APPLICATIONS INFORMATION
Example: with V
OUT
= 3.3V, V
IN
= 5V;
V
OUT
/V
IN
= 3.3/5 = 0.67
I
P
= 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 I
P
.
Continuous mode:
(V
OUT
)(V
IN
– V
OUT
)
2(L)( f)(V
IN
)
For the conditions above, with L = 5µH and f = 500kHz;
I
OUT (MAX)
=
I
P
–
I
OUT (MAX)
=
1.42 –
(3.3)(5 – 3.3)
2
5 10
−
6
500 10
3
=
1.42 – 0.22
=
1.2A
( )
( )
5
At V
IN
= 8V, V
OUT
/ V
IN
= 0.41, so I
P
is equal to 1.5A and
I
OUT(MAX)
is equal to;
1.5 –
(3.3)(8 – 3.3)
2
5 10
−
6
500 10
3
=
1.5 – 0.39
=
1.11A
( )
( )
8
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:
V (V – V )
I
SWITCH(PEAK)
=
I
OUT
+
OUT IN OUT
2(L)(f)(V
IN
)
For lighter loads where discontinuous mode operation can
be used, maximum load current is equal to:
12
U
W
U
U
Discontinuous mode:
(I
P
)
2
(f)(L)(V
IN
)
I
OUT (MAX)
=
2(V
OUT
)(V
IN
– V
OUT
)
Example: with L = 2µH, V
OUT
= 5V and V
IN(MAX)
= 15V;
(1.5)
2
500 10
3
2 10
−
6
I
OUT (MAX)
=
2(5)(15 – 5)
=
338 mA
( ) ( )
15
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 (V – V )
I
D(AVG)
=
OUT IN OUT
V
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 V
IN
=
10V, V
OUT
= 2V (3.3V overloaded) and I
OUT
= 1.8A:
I
D(AVG)
=
1.8 (10 – 2)
=
1.44A
10
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