L6918 L6918A

20/35

A minimum capacitor value is required to sustain the current during the load transient without discharge it. The

voltage drop due to the output capacitor discharge is given by the following equation:

transient and the lower is the output voltage static ripple.

INDUCTOR DESIGN

The inductance value is defined by a compromise between the transient response time, the efficiency, the cost

and the size. The inductor has to be calculated to sustain the output and the input voltage variation to maintain

the ripple current

culated with this relationship:

Increasing the value of the inductance reduces the ripple current but, at the same time, reduces the converter

response time to a load transient. The response time is the time required by the inductor to change its current

from initial to final value. Since the inductor has not finished its charging time, the output current is supplied by

the output capacitors. Minimizing the response time can minimize the output capacitance required.

The response time to a load transient is different for the application or the removal of the load: if during the ap-

plication of the load the inductor is charged by a voltage equal to the difference between the input and the output

voltage, during the removal it is discharged only by the output voltage. The following expressions give approx-

imate response time for DI load transient in case of enough fast compensation network response:

The worst condition depends on the input voltage available and the output voltage selected. Anyway the worst

case is the response time after removal of the load with the minimum output voltage programmed and the max-

imum input voltage available.

Figure 11. Inductor ripple current vs. Vout

∆V

OU T

∆i

OU T

2

L

⋅

2C

OU T

V

IN m in

D

MA X

V

OU T

–

⋅

()

⋅⋅

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

=

L

V

IN

V

OU T

–

f

SW

∆I

L

⋅

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

V

OU T

V

IN

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

⋅

=

t

a ppli ca tion

L

∆I

⋅

V

IN

V

OU T

–

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

=

t

removal

L

∆I

⋅

V

OUT

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

=

Figure 12 – Inductor ripple current vs. Vout

0

1

2

3

4

5

6

7

8

9

0.5

1 .5

2.5

3 .5

Output V oltage [V ]

L=3

µH,

Vin=12V

L=2

µH,

Vin=12V

L=1.5

µH, Vin=12V

L=2

µH,

Vin=5V

L=1.5

µH,

Vin=5V

L=3

µH, Vin=5V