LT3695 Series
14
3695fa
Inductor Selection and Maximum Output Current
A good fi rst choice for the inductor value is:
L
V
V
f
OUT
D
SW
=
+
(
) •
.
1 8
where f
SW
is the switching frequency in MHz, V
OUT
is the
output voltage, V
D
is the catch diode drop (~0.5V) and L
is the inductor value in μH.
The inductor’s RMS current rating must be greater than the
maximum load current and its saturation current should be
about 30% higher. To keep the effi ciency high, the series
resistance (DCR) should be less than 0.1Ω, and the core
material should be intended for high frequency applications.
Table 2 lists several vendors and suitable types.
For robust operation in fault conditions (start-up or short-
circuit) and high input voltage (>30V), the saturation
current should be chosen high enough to ensure that the
inductor peak current does not exceed 3.5A. For example,
an application running from an input voltage of 36V
using a 10μH inductor with a saturation current of 2.5A
will tolerate the mentioned fault conditions.
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 effi ciency.
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% (V
OUT
/V
IN
> 0.5), a minimum inductance
is required to avoid sub-harmonic oscillations:
L
V
V
f
MIN
OUT
D
SW
=
+
(
) •
.
1 2
The current in the inductor is a triangle wave with an av-
erage value equal to the load current. The peak inductor
and switch current is:
I
I
I
I
SW PEAK
L PEAK
OUT MAX
L
(
)
(
)
(
)
=
=
+
Δ
2
where I
L(PEAK)
is the peak inductor current, I
OUT(MAX)
is
the maximum output load current and
Δ
I
L
is the inductor
ripple current. The LT3695 regulators limit their switch
current in order to protect themselves and the system
from overload faults. Therefore, the maximum output
current that the LT3695 regulators 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:
Δ
I
DC
V
V
L f
L
OUT
D
SW
=
−
+
(
) •(
)
•
1
where f
SW
is the switching frequency of the LT3695
regulators, DC is the duty cycle and L is the value of the
inductor.
To maintain output regulation, the inductor peak current
must be less than the LT3695 regulators’ switch current
limit, I
LIM
. If the SYNC pin is grounded, I
LIM
is at least
1.45A at low duty cycles and decreases to 1.1A at DC =
90%. If the SYNC pin is tied to 0.8V or more or if it is
tied to a clock source for synchronization, I
LIM
is at least
1.18A at low duty cycles and decreases to 0.85A at DC =
90%. The maximum output current is also a function of
the chosen inductor value and can be approximated by
the following expressions depending on the SYNC pin
confi guration:
For the SYNC pin grounded:
I
I
I
A
DC
I
OUT MAX
LIM
L
L
(
)
.
•(
.
•
)
=
−
=
−
−
Δ
Δ
2
1 45
1 0 24
2
For the SYNC pin tied to 0.8V or more, or tied to a clock
source for synchronization:
I
I
I
A
DC
I
OUT MAX
LIM
L
L
(
)
.
•(
.
•
)
=
−
=
−
−
Δ
Δ
2
1 18
1 0 29
2
APPLICATIONS INFORMATION
