Datasheet

LM27313
www.ti.com SNVS487D –DECEMBER 2006–REVISED APRIL 2013
INDUCTANCE VALUE
The first question we are usually asked is: “How small can I make the inductor?” (because they are the largest
sized component and usually the most costly). The answer is not simple and involves trade-offs in performance.
More inductance means less inductor ripple current and less output voltage ripple (for a given size of output
capacitor). More inductance also means more load power can be delivered because the energy stored during
each switching cycle is:
E = L/2 x (lp)
2
where
• “lp” is the peak inductor current. (5)
An important point to observe is that the LM27313 will limit its switch current based on peak current. This means
that since lp(max) is fixed, increasing L will increase the maximum amount of power available to the load.
Conversely, using too little inductance may limit the amount of load current which can be drawn from the output.
Best performance is usually obtained when the converter is operated in “continuous” mode at the load current
range of interest, typically giving better load regulation and less output ripple. Continuous operation is defined as
not allowing the inductor current to drop to zero during the cycle. It should be noted that all boost converters shift
over to discontinuous operation as the output load is reduced far enough, but a larger inductor stays “continuous”
over a wider load current range.
To better understand these tradeoffs, a typical application circuit (5V to 12V boost with a 10 µH inductor) will be
analyzed.
Since the LM27313 typical switching frequency is 1.6 MHz, the typical period is equal to 1/f
SW(TYP)
, or
approximately 0.625 µs.
We will assume: V
IN
= 5V, V
OUT
= 12V, V
DIODE
= 0.5V, V
SW
= 0.5V. The duty cycle is:
Duty Cycle = ((12V + 0.5V - 5V) / (12V + 0.5V - 0.5V)) = 62.5% (6)
The typical ON time of the switch is:
(62.5% x 0.625 µs) = 0.390 µs (7)
It should be noted that when the switch is ON, the voltage across the inductor is approximately 4.5V.
Using the equation:
V = L (di/dt) (8)
We can then calculate the di/dt rate of the inductor which is found to be 0.45 A/µs during the ON time. Using
these facts, we can then show what the inductor current will look like during operation:
Figure 14. 10 µH Inductor Current, 5V–12V Boost
During the 0.390 µs ON time, the inductor current ramps up 0.176A and ramps down an equal amount during the
OFF time. This is defined as the inductor “ripple current”. It can also be seen that if the load current drops to
about 33 mA, the inductor current will begin touching the zero axis which means it will be in discontinuous mode.
A similar analysis can be performed on any boost converter, to make sure the ripple current is reasonable and
continuous operation will be maintained at the typical load current values.
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