LT1533CS datasheet(12/20 Pages) LINER

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LT1533

APPLICATIONS INFORMATION

For continuous operation the inductor ripple current must

be less than twice the output current. The worst case for

this is at maximum input (lowest DC) but we will evaluate

at nominal input since the IOUT/4 is somewhat arbitrary.

Note when both inputs are off, inductor current splits

between outputs and the diode common goes to 0V.

Looking at the inductor current during off time, output

ripple current is:

∆IOUT = 2 • IOUT(MIN)

IOUT(MIN) = IOUT(MAX)/4

OUT

NOM

OUT

−•

()

•

∆

The inductance of the transformer primary should be such

that LO, when reflected into the primary, dominates the

input current. In other words, we want the magnetizing

current of the transformer small with respect to the

current going through the transformer to LO. In general,

then, the inductance of the primary should be at least five

times that of LO. This ensures that most of the power will

be passed through the transformer to the load. It also

increases the power capability of the converter and

reduces the peak currents that the switch will see.

LPRI = 5 • LO/N2

If the magnetizing current is below 100mA, then a smaller

LO can be used.

With the value of LO set, the ripple in the inductor is:

∆I

OUT

−•

()

•

However, the peak inductor current is evaluated at maxi-

mum load and maximum input voltage (minimum DC).

LMAX

OUT MAX

()

∆

The magnetizing ripple current can be shown to be:

∆I

MAG

OUT

PRI

••

and the peak current in the switch is:

ISW(PEAK) = N • ILMAX + ∆IMAG

This should be less than the 1A current limit.

In the push-pull converter the maximum switch voltage

will be 2 • (VIN – VSW) plus a small amount (10%) for

leakage spikes. Because voltage is slew-controlled, the

spikes will be less than normal. So, maximum switch

voltage is:

VSW(MAX) = 2 • VIN • 1.1

This should be below the maximum rated switch voltage.

So, given the turns ratio, primary inductance and current,

the transformer can be designed. As an example:

VIN = 5V ±10%, VOUT = 12V, IOUT(MAX) = 150mA,

VSW = 0.5V, VF = 0.5V, f = 50kHz,

() −

()

12 0 5

2 0 44 45 05

355

•.

Round up so N = 3.6.

For continuous operation at IOUT(MIN) = IOUT(MAX)/4,

inductor ripple is:

∆I

OUT ==

150

•

The duty cycle for nominal input is:

kHz

NOM

OUT

IN NOM

O MIN

()

−

()

() −

()

−

()

12 0 5

23 6 5 0 5

38 6

12 1 2 38 6

730

•

•.

•. %

•

()

Off-the-shelf components can be used for this inductor.

Say we found an 800

µH inductor (Coiltronics CTX200-1

for instance).

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