L6997S

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6

STEP BY STEP DESIGN

6.1 Input capacitor.

A pulsed current (with zero average value) flows through the input capacitor of a buck converter. The AC com-

ponent of this current is quite high and dissipates a considerable amount of power on the ESR of the capacitor:

(17)

The RMS current, which the capacitor must provide, is given by:

(18)

Where

δ is the duty cycle of the application

Neglecting the last term, the equation reduces to:

(19)

which maximum value corresponds to to

Therefore, in worst case, the input capacitors should be selected with a RMS ripple current rating as high as

half the respective maximum output current.

Electrolytic capacitors are the most used because theyare the cheapest ones and are available with a wide

range of RMS current ratings. The only drawback is that, for a givenripple current rating, they are physically larg-

er than other capacitors. Very good tantalum capacitors are coming available, with very low ESR and small size.

The only problem is that they occasionally can burn if subjected to very high current during the charge. So, it is

better avoid this type of capacitors for the input filter of the device. In fact, they can be subjected to high surge

current when connected to the power supply. If available for the requested capacitance value and voltage rating,

the ceramic capacitors have usually a higher RMS current rating for a given physical dimension (due to the very

low ESR). The drawback is the quite high cost. Possible solutions:

With our parameter from the equation 3 it is found:

6.2 Inductor

To define the inductor, it is necessary to determine firstly the inductance value. Its minimum value is given by:

(20)

where RF =

10

µF

C34Y5U1E106ZTE12 TOKIN

22

µF

JMK325BJ226MM

TAIYO-YUDEN

47

µF

ECJ4XF0J476Z

PANASONIC

33

µF

C3225X5R0J476M

TDK

P

CIN

ESR

CIN

Iout

2

Vin

Vin

Vout

–

()

⋅

Vin

2

------------------------------------------------

⋅⋅

=

Icin

rms

Iout

–

()

δ

12

------

I

L

∆

()

2

+

=

Icin

rms

Iout

δ 1 δ

–

()

=

Lmin

V

o

Vin

max

V

o

–

()

⋅

F

SW

I

out

RF Vin

max

⋅

⋅⋅

---------------------------------------------------------------

≥