LT3508
11
3508fd
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 discontinuous mode
operation, see 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
MIN
= V
OUT
+ V
F
(
)
•
0.8µH
f
where f is in MHz. The current in the inductor is a triangle
wave with an average value equal to the load current. The
peak switch current is equal to the output current plus
half the peak-to-peak inductor ripple current. The LT3508
limits its switch current in order to protect itself and the
system from overload faults. Therefore, the maximum
output current that the LT3508 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:
∆I
L
=
1–DC
(
)
V
OUT
+ V
F
(
)
L•f
where f is the switching frequency of the LT3508 and L
is the value of the inductor. The peak inductor and switch
current is:
I
SW(PK)
= I
L(PK)
= I
OUT
+
∆I
L
2
To maintain output regulation, this peak current must be
less than the LT3508’s switch current limit I
LIM
. I
LIM
is
at least 2A for at low duty cycles and decreases linearly
to 1.55A at DC = 90%. The maximum output current is a
function of the chosen inductor value:
I
OUT(MAX)
= I
LIM
–
∆I
L
2
= 2A • 1– 0.25 • DC
(
)
–
∆I
L
2
Choosing an inductor value so that the ripple current is
small will allow a maximum output current near the switch
current limit.
One approach to choosing the inductor is to start with the
simple rule given above, look at the available inductors, and
choose one to meet cost or space goals. Then use these
equations to check that the LT3508 will be able to deliver
the required output current. Note again that these equations
assume that the inductor current is continuous. Discontinu-
ous operation occurs when I
OUT
is less than ∆I
L
/2.
Input Capacitor Selection
Bypass the V
IN
pins of the LT3508 circuit with a ceramic
capacitor of X7R or X5R type. For switching frequen-
cies above 500kHz, use a 4.7µF capacitor or greater. For
switching frequencies below 500kHz, use a 10µF or higher
capacitor. If the V
IN
pins are tied together only a single
capacitor is necessary. If the V
IN
pins are separated, each
pin will need its own bypass. The following paragraphs
describe the input capacitor considerations in more detail.
Step-down regulators draw current from the input supply
in pulses with very fast rise and fall times. The input ca-
pacitor is required to reduce the resulting voltage ripple at
the LT3508 input and to force this switching current into a
tight local loop, minimizing EMI. The input capacitor must
have low impedance at the switching frequency to do this
effectively, and it must have an adequate ripple current
rating. With two switchers operating at the same frequency
but with different phases and duty cycles, calculating the
input capacitor RMS current is not simple. However, a
conservative value is the RMS input current for the channel
that is delivering most power (V
OUT
times I
OUT
):
I
C
IN(RMS)
= I
OUT
•
V
OUT
V
IN
–V
OUT
(
)
V
IN
<
I
OUT
2
and is largest when V
IN
= 2V
OUT
(50% duty cycle). As
the second, lower power channel draws input current,
APPLICATIONS INFORMATION
