LT1374-5 Ver la hoja de datos (PDF) - Linear Technology

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LT1374-5

4.5A, 500kHz Step-Down Switching Regulator

Linear Technology 

LT1374

APPLICATIONS INFORMATION

Operating duty cycle:

DC =

VOUT + VF

VIN − 0.3 + VOUT + VF

(This formula uses an average value for switch loss, so it

may be several percent in error.)

With the conditions above:

DC =

5 + 0.5

= 51%

5.5 − 0.3 + 5 + 0.5

This duty cycle is close enough to 50% that IP can be

assumed to be 4.5A.

OUTPUT DIVIDER

If the adjustable part is used, the resistor connected to

VOUT (R2) should be set to approximately 5k. R1 is

calculated from:

( ) R1= R2 VOUT – 2.42

2.42

INDUCTOR VALUE

Unlike buck converters, positive-to-negative converters

cannot use large inductor values to reduce output ripple

voltage. At 500kHz, values larger than 25µH make almost

no change in output ripple. The graph in Figure 16 shows

peak-to-peak output ripple voltage for a 5V to – 5V con-

verter versus inductor value. The criteria for choosing the

inductor is therefore typically based on ensuring that peak

switch current rating is not exceeded. This gives the

lowest value of inductance that can be used, but in some

cases (lower output load currents) it may give a value that

creates unnecessarily high output ripple voltage. A com-

promise value is often chosen that reduces output ripple.

As you can see from the graph, large inductors will not

250

5V TO – 5V CONVERTER

OUTPUT CAPACITOR’S

200

ESR = 0.05Ω

150

ILOAD = 1A

100

50

ILOAD = 0.25A

0

0

5

10

15

20

INDUCTOR SIZE (µH)

1374 F16

Figure 16. Ripple Voltage on Positive-to-Negative Converter

give arbitrarily low ripple, but small inductors can give

high ripple.

The difficulty in calculating the minimum inductor size

needed is that you must first know whether the switcher

will be in continuous or discontinuous mode at the critical

point where switch current is 4.5A. The first step is to use

the following formula to calculate the load current where

the switcher must use continuous mode. If your load

current is less than this, use the discontinuous mode

formula to calculate minimum inductor needed. If load

current is higher, use the continuous mode formula.

Output current where continuous mode is needed:

( ( )() ( ) ) ICONT =

22

VIN IP

4 VIN + VOUT VIN + VOUT + VF

Minimum inductor discontinuous mode:

( )( ) LMIN = 2 VOUT

IOUT

2

( )( )f IP

24

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