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LT3430_02

High Voltage, 3A, 200kHz Step-Down Switching Regulator

Linear Technology 

LT3430

APPLICATIO S I FOR ATIO

tors and lighter loads, so don’t omit this step. Powdered

iron cores are forgiving because they saturate softly,

whereas ferrite cores saturate abruptly. Other core mate-

rials fall somewhere in between. The following formula

assumes continuous mode of operation, but errs only

slightly on the high side for discontinuous mode, so it can

be used for all conditions.

IPEAK

= IOUT

+

ILP-P

2

= IOUT

+

(VOUT )(VIN – VOUT )

(2)(VIN)(f)(L)

EMI

Decide if the design can tolerate an “open” core geometry

like a rod or barrel, which have high magnetic field

radiation, or whether it needs a closed core like a toroid to

prevent EMI problems. This is a tough decision because

the rods or barrels are temptingly cheap and small and

there are no helpful guidelines to calculate when the

magnetic field radiation will be a problem.

Additional Considerations

After making an initial choice, consider additional factors

such as core losses and second sourcing, etc. Use the

experts in Linear Technology’s Applications department if

you feel uncertain about the final choice. They have

experience with a wide range of inductor types and can tell

you about the latest developments in low profile, surface

mounting, etc.

Maximum Output Load Current

Maximum load current for a buck converter is limited by

the maximum switch current rating (IP). The current rating

for the LT3430 is 3A. Unlike most current mode convert-

ers, the LT3430 maximum switch current limit does not

fall off at high duty cycles. Most current mode converters

suffer a drop off of peak switch current for duty cycles

above 50%. This is due to the effects of slope compensa-

tion required to prevent subharmonic oscillations in cur-

rent mode converters. (For detailed analysis, see Applica-

tion Note 19.)

The LT3430 is able to maintain peak switch current limit

over the full duty cycle range by using patented circuitry to

cancel the effects of slope compensation on peak switch

current without affecting the frequency compensation it

provides.

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 (ILP-P). The follow-

ing formula assumes continuous mode operation, imply-

ing that the term on the right is less than one-half of IP.

IOUT(MAX) =

Continuous Mode

IP

–

ILP-P

2

= IP

−

(VOUT

+

VF )(VIN − VOUT

2(L)(f)(VIN)

–

VF )

For VOUT = 5V, VIN = 12V, VF(D1) = 0.52V, f = 200kHz

and L = 15µH:

( )( ) IOUT(MAX)

=

3

−

(5 + 0.52)(12 − 5 – 0.52)

2 15 •10−6 200•103 (12)

= 3 − 0.5 = 2.5A

Note that there is less load current available at the higher

input voltage because inductor ripple current increases. At

VIN = 24V, duty cycle is 23% and for the same set of

conditions:

( )( ) IOUT(MAX)

=

3

−

(5 + 0.52)(24 − 5 – 0.52)

2 15 •10−6 200•103 (24)

To

cal-

= 3 − 0.71= 2.29A

culate actual peak switch current with a given set of

conditions, use:

ISW(PEAK)

=

IOUT

+

ILP-P

2

=

IOUT

+

(VOUT

+

( VF ) VIN − VOUT

2(L)(f)(VIN)

–

VF )

sn3430 3430is

11

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