LT1506CS8 datasheet(9/24 Pages) LINER

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LT1506

APPLICATIONS INFORMATION

The internal circuitry which forces reduced switching

frequency also causes current to flow out of the feedback

pin when output voltage is low. The equivalent circuitry is

shown in Figure 2. Q1 is completely off during normal

operation. If the FB pin falls below 1V, Q1 begins to

conduct current and reduces frequency at the rate of

approximately 5kHz/

µA. To ensure adequate frequency

foldback (under worst-case short-circuit conditions), the

external divider Thevinin resistance must be low enough

to pull 150

µA out of the FB pin with 0.6V on the pin (RDIV

≤ 4k). The net result is that reductions in frequency and

current limit are affected by output voltage divider imped-

ance. Although divider impedance is not critical, caution

should be used if resistors are increased beyond the

suggested values and short-circuit conditions will occur

with high input voltage. High frequency pickup will

increase and the protection accorded by frequency and

current foldback will decrease.

MAXIMUM OUTPUT LOAD CURRENT

Maximum load current for a buck converter is limited by

the maximum switch current rating (IP) of the LT1506.

This current rating is 4.5A up to 50% duty cycle (DC),

decreasing to 3.7A at 80% duty cycle. This is shown

graphically in Typical Performance Characteristics and as

shown in the formula below:

IP = 4.5A for DC ≤ 50%

IP = 3.21 + 5.95(DC) – 6.75(DC)

2 for 50% < DC < 90%

DC = Duty cycle = VOUT/VIN

Example: with VOUT = 5V, VIN = 8V; DC = 5/8 = 0.625, and;

ISW(MAX) = 3.21 + 5.95(0.625) – 6.75(0.625)

2 = 4.3A

Current rating decreases with duty cycle because the

LT1506 has internal slope compensation to prevent cur-

rent mode subharmonic switching. For more details, read

Application Note 19. The LT1506 is a little unusual in this

regard because it has nonlinear slope compensation which

gives better compensation with less reduction in current

limit.

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, implying

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

IOUT(MAX) =

Continuous Mode

For the conditions above and L = 3.3

µH,

OUT MAX

( )

−

=−

() −

()















 ()

=−

58 5

2 3 3 10

500 10

057

3 73

.•

•

At VIN = 15V, duty cycle is 33%, so IP is just equal to a fixed

4.5A, and IOUT(MAX) is equal to:

515 5

2 3 3 10

500 10

45 101 3 49

.•

•

−

() −

()















 ()

=−

−

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 ex-

tremes. To calculate actual peak switch current with a

given set of conditions, use:

Lf V

SW PEAK

OUT

( ) =+

−

()

()( )( )

IP − ()

−

()

()( )( )

Lf V

OUT

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