Datasheet
LMR64010
SNVS736B –SEPTEMBER 2011–REVISED APRIL 2013
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MINIMUM INDUCTANCE
In some applications where the maximum load current is relatively small, it may be advantageous to use the
smallest possible inductance value for cost and size savings. The converter will operate in discontinuous mode in
such a case.
The minimum inductance should be selected such that the inductor (switch) current peak on each cycle does not
reach the 1A current limit maximum. To understand how to do this, an example will be presented.
In the example, minimum switching frequency of 1.15 MHz will be used. This means the maximum cycle period
is the reciprocal of the minimum frequency:
T
ON(max)
= 1/1.15M = 0.870 µs (12)
We will assume the input voltage is 5V, V
OUT
= 12V, V
SW
= 0.2V, V
DIODE
= 0.3V. The duty cycle is:
Duty Cycle = 60.3%
Therefore, the maximum switch ON time is 0.524 µs. An inductor should be selected with enough inductance to
prevent the switch current from reaching 1A in the 0.524 µs ON time interval (see below):
Figure 22. Discontinuous Design, 5V–12V Boost
The voltage across the inductor during ON time is 4.8V. Minimum inductance value is found by:
V = L X dl/dt, L = V X (dt/dl) = 4.8 (0.524µ/1) = 2.5 µH (13)
In this case, a 2.7 µH inductor could be used assuming it provided at least that much inductance up to the 1A
current value. This same analysis can be used to find the minimum inductance for any boost application.
When selecting an inductor, make certain that the continuous current rating is high enough to avoid saturation at
peak currents. A suitable core type must be used to minimize core (switching) losses, and wire power losses
must be considered when selecting the current rating.
SHUTDOWN PIN OPERATION
The device is turned off by pulling the shutdown pin low. If this function is not going to be used, the pin should be
tied directly to V
IN
. If the SHDN function will be needed, a pull-up resistor must be used to V
IN
(approximately
50k-100kΩ recommended). The SHDN pin must not be left unterminated.
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