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LTM4600
APPLICATIONS INFORMATION
(DC) DUTY CYCLE = ON
DC = ON = OUT
t OFF t ON
t
t s
t s V IN
FREQ =
DC
t ON
t V
t ON = 0.41 ? 1μs ? 410ns
t OFF = 1μs – 410ns ? 590ns
t ON and t OFF are above the minimums with adequate guard
4600 F21
PERIOD t s
Figure 17. LTM4600 Switching Period
The LTM4600 has a minimum (t ON ) on time of 100 nanosec-
onds and a minimum (t OFF ) off time of 400 nanoseconds.
The 2.4V clamp on the ramp threshold as a function of
V OUT will cause the switching frequency to increase by the
ratio of V OUT /2.4V for 3.3V and 5V outputs. This is due to
the fact the on time will not increase as V OUT increases
past 2.4V. Therefore, if the nominal switching frequency
is 850kHz, then the switching frequency will increase
to ~1.2MHz for 3.3V, and ~1.7MHz for 5V outputs due
to Frequency = (DC/t ON ) When the switching frequency
increases to 1.2MHz, then the time period t S is reduced
to ~833 nanoseconds and at 1.7MHz the switching period
reduces to ~588 nanoseconds. When higher duty cycle
conversions like 5V to 3.3V and 12V to 5V need to be
accommodated, then the switching frequency can be
lowered to alleviate the violation of the 400ns minimum
off time. Since the total switching period is t S = t ON + t OFF ,
t OFF will be below the 400ns minimum off time. A resistor
from the f ADJ pin to ground can shunt current away from
the on time generator, thus allowing for a longer on time
and a lower switching frequency. 12V to 5V and 5V to
3.3V derivations are explained in the data sheet to lower
switching frequency and accommodate these step-down
conversions.
Equations for setting frequency for 12V to 5V:
I ON = (V IN – 0.7V)/110k; I ON = 103μA
frequency = (I ON /[2.4V ? 10pF]) ? DC = 1.79MHz;
DC = duty cycle, duty cycle is (V OUT /V IN )
t S = t ON + t OFF , t ON = on-time, t OFF = off-time of the
switching period; t S = 1/frequency
t OFF must be greater than 400ns, or t S – t ON > 400ns.
t ON = DC ? t S
1MHz frequency or 1μs period is chosen for 12V to 5V.
band.
Using the frequency = (I ON /[2.4V ? 10pF]) ? DC, solve for
I ON = (1MHz ? 2.4V ? 10pF) ? (1/0.41) ? 58μA. I ON current
calculated from 12V input was 103μA, so a resistor from
f ADJ to ground = (0.7V/15k) = 46μA. 103μA – 46μA =
57μA, sets the adequate I ON current for proper frequency
range for the higher duty cycle conversion of 12V to
5V. Input voltage range is limited to 9V to 16V. Higher
input voltages can be used without the 15k on f ADJ . The
inductor ripple current gets too high above 16V, and the
400ns minimum off-time is limited below 9V.
Equations for setting frequency for 5V to 3.3V:
I ON = (V IN – 0.7V)/110k; I ON = 39μA
frequency = (I ON /[2.4V ? 10pF]) ? DC = 1.07MHz;
DC = duty cycle, duty cycle is (V OUT /V IN )
t S = t ON + t OFF , t ON = DC ? t S , t OFF = off-time of the
switching period; t S = 1/frequency
t OFF must be greater than 400ns, or t S – t ON > 400ns.
The ~450kHz frequency or 2.22μs period is chosen for
5V to 3.3V. Frequency range is about 450kHz to 650kHz
from 4.5V to 7V input.
t ON = 0.66 ? 2.22μs ? 1.46 μ s
t OFF = 2.22 μ s – 1.46 μ s ? 760ns
t ON and t OFF are above the minimums with adequate guard
band.
Using the frequency = (I ON /[2.4V ? 10pF]) ? DC, solve for
I ON = (450kHz ? 2.4V ? 10pF) ? (1/0.66) ? 16μA. I ON current
calculated from 5V input was 39μA, so a resistor from f ADJ
to ground = (0.7V/30.1k) = 23μA. 39μA – 23μA = 16μA,
sets the adequate I ON current for proper frequency range
for the higher duty cycle conversion of 5V to 3.3V. Input
voltage range is limited to 4.5V to 7V. Higher input voltages
can be used without the 30.1k on f ADJ . The inductor ripple
current gets too high above 7V, and the 400ns minimum
off-time is limited below 4.5V.
4600fc
17
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