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
maximum load current. For details of maximum output
current and discontinuous mode operation, see Linear
Technology’s 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-
ous operation occurs when I
is less than ΔI /2.
OUT
L
greaterthan50%(V /V >0.5), aminimuminductance
OUT IN
is required to avoid sub-harmonic oscillations:
Input Capacitor Selection
800kHz
Bypass the V pins of the LT3508 circuit with a ceramic
LMIN = V
+ V •
(
)
IN
OUT
F
f
capacitor of X7R or X5R type. For switching frequencies
above500kHz,usea4.7ꢀFcapacitororgreater.Forswitch-
ingfrequenciesbelow500kHz,usea10ꢀForhighercapaci-
Thecurrentintheinductorisatrianglewavewithanaverage
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 cur-
rent 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.
tor. If the V pins are tied together only a single capacitor
IN
is necessary. If the V pins are separated, each pin will
IN
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.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
2
IN
OUT
CIN(RMS) = IOUT
•
<
V
IN
To maintain output regulation, this peak current must be
less than the LT3508’s switch current limit I . I is
at least 2A for at low duty cycles and decreases linearly
LIM LIM
and is largest when V = 2V
(50% duty cycle). As
IN
OUT
the second, lower power channel draws input current,
3508fb
11
Linear [ Linear ]
