L6995

16/25

3

STEP BY STEP DESIGN

In this design it is considered a low profile demoboard, so a great attention is given to the components height.

3.1 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 capacitor:

Eq 17

Eq 18

Neglecting the last term, the equation reduces to:

Eq 19

which maximum value corresponds to

δ = 1/2.

most used because are the cheapest ones and are available with a wide range of RMS current ratings. The only

drawback is that, considering a requested ripple current rating, they are physically larger 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 out 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 value and voltage rating, the ceramic capacitors have usually

a higher RMS current rating for a given physical size (due to the very low ESR). From the equation 18 it is found:

Considering 10

3.2 Inductor

In order to determine the inductor value is necessary considering the maximum output current to decide the in-

ductor current saturation. Once the inductor current saturation it is found automatically is found the inductor val-

ue. In our design it is considered a very low profile inductor.

L = 0.6

µH

The saturation current for this choke is around 25A

3.3 Output capacitor

The output capacitor is chosen by the output voltage static and dynamic accuracy. The static accuracy is related

to the output voltage ripple value, while the dynamic accuracy is related to the output current load step.

If the static precision is around ± 2% for the 1.25V output, the output accuracy is ±25mV.

To determine the ESR value from the output precision is necessary before calculate the ripple current:

P

CIN

ESR

CIN

Iou t

2

Vi n

Vi n

Vo ut

–

()

⋅

Vin

2

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

⋅⋅

=

Icin

rm s

Iout

–

()

δ

12

------

I

L

∆

()

2

+

=

Icin

rm s

Io ut

δ 1 δ

–

()

=