Datasheet

LMR62014
SNVS735B –OCTOBER 2011–REVISED APRIL 2013 www.ti.com
To better understand these trade-offs, a typical application circuit (5V to 12V boost with a 10 µH inductor) will be
analyzed. We will assume:
V
IN
= 5V, V
OUT
= 12V, V
DIODE
= 0.5V, V
SW
= 0.5V (5)
Since the frequency is 1.6 MHz (nominal), the period is approximately 0.625 µs. The duty cycle will be 62.5%,
which means the ON time of the switch is 0.390 µs. 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) (6)
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 20. 10 µH Inductor Current,
5V–12V Boost (LMR62014X)
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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