LT1507CS8 datasheet(12/20 Pages) LINER

12 / 20 page

LT1507

APPLICATIONS INFORMATION

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:

Lf V

OUT MAX

OUT

()

–

()(

–

)

()( )(

)

For the conditions above, with L = 5

µH and f = 500kHz;

OUT MAX

()

.–

( . )( – . )

.– .

()









()









−

142

33 5 3 3

2 5 10

500 10

142 0 22

At VIN = 8V, VOUT/VIN = 0.41, so IP is equal to 1.5A and

IOUT(MAX) is equal to;

33 8 3 3

2 5 10

500 10

15 039

1 11

.–

( . )( – . )

.– .

−

()









()









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:

Lf V

SWITCH PEAK

OUT

()

(–

)

()( )(

)

For lighter loads where discontinuous mode operation can

be used, maximum load current is equal to:

Discontinuous mode:

If L V

OUT MAX

OUT

()

() ( )( )(

)

()(

–

)

Example: with L = 2

µH, VOUT = 5V and VIN(MAX) = 15V;

OUT MAX

()

(. )

()(

– )

()







 ()









−

1 5

500 10

2 10

2 5 15 5

338

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:

D AVG

OUT

()

(–

)

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:

D AVG

()

– )

1 8 10 2

144

LT1507CS8 Datasheet(PDF) 12 Page - Linear Technology

LT1507CS8 Datasheet(HTML) 12 Page - Linear Technology

Html Pages

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Datasheet Download

Link URL