LT3508
APPLICATIONS INFORMATION
The optimum inductor for a given application may differ
fromtheoneindicatedbythissimpledesignguide.Alarger
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 efficiency.
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
maximumloadcurrent. Fordetailsofdiscontinuousmode
operation, see Application Note 44. Finally, for duty cycles
to 1.55A at DC = 90%. The maximum output current is a
function of the chosen inductor value:
ΔIL
2
ΔIL
2
IOUT(MAX) = ILIM
–
= 2A • 1– 0.25 •DC –
(
)
Choosing an inductor value so that the ripple current is
smallwillallowamaximumoutputcurrentneartheswitch
current limit.
One approach to choosing the inductor is to start with the
simplerulegivenabove,lookattheavailableinductors,and
choose one to meet cost or space goals. Then use these
equations to check that the LT3508 will be able to deliver
therequiredoutputcurrent.Noteagainthattheseequations
assumethattheinductorcurrentiscontinuous.Discontinu-
greaterthan50%(V /V >0.5), aminimuminductance
OUT IN
is required to avoid sub-harmonic oscillations:
ous operation occurs when I
is less than ∆I /2.
OUT
L
0.8µH
LMIN = V
+ V •
Input Capacitor Selection
(
)
F
OUT
f
Bypass the V pins of the LT3508 circuit with a ceramic
IN
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.
capacitor of X7R or X5R type. For switching frequen-
cies above 500kHz, use a 4.7ꢀF capacitor or greater. For
switchingfrequenciesbelow500kHz, usea10ꢀForhigher
capacitor. If the V pins are tied together only a single
IN
capacitor is necessary. If the V pins are separated, each
IN
pin will need its own bypass. The following paragraphs
describetheinputcapacitorconsiderationsinmoredetail.
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.Withtwoswitchersoperatingatthesamefrequency
but with different phases and duty cycles, calculating the
input capacitor RMS current is not simple. However, a
conservativevalueistheRMSinputcurrentforthechannel
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:
1–DC V
+ V
F
(
)(
)
OUT
ΔIL =
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:
that is delivering most power (V
times I ):
OUT
OUT
ΔIL
2
ISW(PK) = IL(PK) = IOUT
+
VOUT V – V
(
)
IOUT
OUT
IN
ICIN(RMS) = IOUT
•
<
V
2
IN
To maintain output regulation, this peak current must be
less than the LT3508’s switch current limit I . I is
LIM LIM
and is largest when V = 2V
(50% duty cycle). As
IN
OUT
at least 2A for at low duty cycles and decreases linearly
the second, lower power channel draws input current,
3508fd
11