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voltage drop due to the output capacitor discharge is given by the following equation:
Where D
MAX
is the maximum duty cycle value. The lower is the ESR, the lower is the output drop during load
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
IL between 20% and 30% of the maximum output current. The inductance value can be cal-
culated with this relationship:
V
V
–
fs
Where f
S
is the switching frequency, V
IN
is the input voltage and V
OUT
is the output voltage.
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 loadthe 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 10. Inductor ripple current vs V
out
V
OUT
I
2
L
2 C
OUT
V
INMIN
D
MAX
V
OUT
–
)
--------------------------------------------------------------------------------------------
=
L
I
L
------------------------------
V
V
IN
---------------
=
t
application
V
OUT
V
IN
–
------------------------------
=
t
removal
V
OUT
I
---------------
=
0
1
2
3
4
5
6
7
8
9
0.5
1.5
2.5
3.5
Output Voltage [V]
I
L=3
μ
H,
Vin=12V
L=2
μ
H,
Vin=12V
L=1.5
μ
H, Vin=12V
L=2
μ
H,
Vin=5V
L=1.5
μ
H,
Vin=5V
L=3
μ
H, Vin=5V