LT1507
U
W U U
APPLICATIONS INFORMATION
Example: with VOUT = 3.3V, VIN = 5V;
VOUT/VIN = 3.3/5 = 0.67
Discontinuous mode:
2
(I ) (f)(L)(V )
P
IN
I
=
OUT(MAX)
2(V )(V – V
)
IP = 1.75 – (0.5)(0.66) = 1.42A
OUT IN
OUT
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
formulaassumescontinuousmodeoperation;thetermon
the right must be less than one half of IP.
Example: with L = 2µH, VOUT = 5V and VIN(MAX) = 15V;
2
3
−6
(1.5) 500 10
2 10
15
(
)
(
)
I
=
OUT(MAX)
2(5)(15 – 5)
= 338m
A
Continuous mode:
The main reason for using such a tiny inductor is that it is
physically very small, but keep in mind that peak-to-peak
inductorcurrentwillbeveryhigh. Thiswillincreaseoutput
ripplevoltage.Iftheoutputcapacitorhastobemadelarger
to reduce ripple voltage, the overall circuit could actually
be larger.
(V )(V – V
)
OUT IN
OUT
I
= I –
P
OUT(MAX)
2(L)(f)(V )
IN
For the conditions above, with L = 5µH and f = 500kHz;
(3.3)(5 – 3.3)
I
= 1.42 –
OUT(MAX)
−6
3
2 5 10
500 10
5
(
)
(
)
CATCH DIODE
= 1.42 – 0.22 = 1.2A
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
regulatorinputvoltage.Averageforwardcurrentinnormal
operation can be calculated from:
At VIN = 8V, VOUT/VIN = 0.41, so IP is equal to 1.5A and
IOUT(MAX) is equal to;
(3.3)(8 – 3.3)
1.5 –
−6
3
2 5 10
500 10
8
(
)
(
)
= 1.5 – 0.39 = 1.11A
I
(V – V
)
OUT IN
OUT
I
=
D(AVG)
V
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:
IN
This formula will not yield values higher than 1A with
maximumloadcurrentof1.25Aunlesstheratioofinputto
output voltage exceeds 5:1. The only reason to consider a
larger diode is the worst-case condition of a high input
voltageand overloaded(notshorted)output. Undershort-
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:
V
(V – V
2(L)(f)(V )
)
OUT IN
OUT
I
= I
+
OUT
SWITCH(PEAK)
IN
Forlighterloadswherediscontinuousmodeoperationcan
be used, maximum load current is equal to:
1.8(10 – 2)
I
=
= 1.44A
D(AVG)
10
12
Linear [ Linear ]
