LT3508 查看數據表（PDF） - Linear Technology

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LT3508

Dual Monolithic 1.4A Step-Down Switching Regulator

Linear Technology 

LT3508

APPLICATIONS INFORMATION

The optimum inductor for a given application may differ

from the one indicated by this simple design guide. A larger

value inductor provides a higher maximum load current,

and reduces the output voltage ripple. If your load is lower

than the maximum load current, then you can relax the

value of the inductor and operate with higher ripple cur-

rent. This allows you to use a physically smaller inductor,

or one with a lower DCR resulting in higher efficiency.

Be aware that if the inductance differs from the simple

rule above, then the maximum load current will depend

on input voltage. In addition, low inductance may result

in discontinuous mode operation, which further reduces

maximum load current. For details of maximum output

current and discontinuous mode operation, see Linear

Technology’s Application Note 44. Finally, for duty cycles

greater than 50% (VOUT/VIN > 0.5), a minimum inductance

is required to avoid sub-harmonic oscillations:

( ) LMIN =

VOUT + VF

• 800kHz

f

The current in the inductor is a triangle wave with an average

value equal to the load current. The peak switch current

is equal to the output current plus half the peak-to-peak

inductor ripple current. The LT3508 limits its switch cur-

rent in order to protect itself and the system from overload

faults. Therefore, the maximum output current that the

LT3508 will deliver depends on the switch current limit,

the inductor value, and the input and output voltages.

When the switch is off, the potential across the inductor

is the output voltage plus the catch diode drop. This gives

the peak-to-peak ripple current in the inductor:

ΔIL

=

(1–

DC)(VOUT

L•f

+

VF)

where f is the switching frequency of the LT3508 and L

is the value of the inductor. The peak inductor and switch

current is:

ISW(PK)

=

IL(PK)

=

IOUT

+

ΔIL

2

To maintain output regulation, this peak current must be

less than the LT3508’s switch current limit ILIM. ILIM is

at least 2A for at low duty cycles and decreases linearly

to 1.55A at DC = 90%. The maximum output current is a

function of the chosen inductor value:

IOUT(MAX)

= ILIM

–

ΔIL

2

=

2A

• (1– 0.25 • DC) –

ΔIL

2

Choosing an inductor value so that the ripple current is

small will allow a maximum output current near the switch

current limit.

One approach to choosing the inductor is to start with the

simple rule given above, look at the available inductors, and

choose one to meet cost or space goals. Then use these

equations to check that the LT3508 will be able to deliver

the required output current. Note again that these equations

assume that the inductor current is continuous. Discontinu-

ous operation occurs when IOUT is less than ΔIL/2.

Input Capacitor Selection

Bypass the VIN pins of the LT3508 circuit with a ceramic

capacitor of X7R or X5R type. For switching frequencies

above 500kHz, use a 4.7μF capacitor or greater. For switch-

ing frequencies below 500kHz, use a 10μF or higher capaci-

tor. If the VIN pins are tied together only a single capacitor

is necessary. If the VIN pins are separated, each pin will

need its own bypass. The following paragraphs describe

the input capacitor considerations in more detail.

Step-down regulators draw current from the input supply

in pulses with very fast rise and fall times. The input ca-

pacitor is required to reduce the resulting voltage ripple at

the LT3508 input and to force this switching current into a

tight local loop, minimizing EMI. The input capacitor must

have low impedance at the switching frequency to do this

effectively, and it must have an adequate ripple current

rating. With two switchers operating at the same frequency

but with different phases and duty cycles, calculating the

input capacitor RMS current is not simple. However, a

conservative value is the RMS input current for the channel

that is delivering most power (VOUT times IOUT):

( ) CIN(RMS) = IOUT •

VOUT VIN – VOUT < IOUT

VIN

2

and is largest when VIN = 2VOUT (50% duty cycle). As

the second, lower power channel draws input current,

3508fb

11

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