7
L T 1111
1111fd
Inductor Selection — General
A DC/DC converter operates by storing energy as mag-
netic flux in an inductor core, and then switching this
energy into the load. Since it is flux, not charge, that is
stored, the output voltage can be higher, lower, or oppo-
site in polarity to the input voltage by choosing an
appropriate switching topology. To operate as an efficient
energy transfer element, the inductor must fulfill three
requirements. First, the inductance must be low enough
for the inductor to store adequate energy under the worst
case condition of minimum input voltage and switch-on
time. The inductance must also be high enough so maxi-
mum current ratings of the LT1111 and inductor are not
exceeded at the other worst case condition of maximum
input voltage and ON time. Additionally, the inductor core
must be able to store the required flux; i.e., it must not
saturate. At power levels generally encountered with
LT1111 based designs, small surface mount ferrite core
units with saturation current ratings in the 300mA to 1A
range and DCR less than 0.4Ω (depending on application)
are adequate. Lastly, the inductor must have sufficiently
low DC resistance so excessive power is not lost as heat
in the windings. An additional consideration is Electro-
Magnetic Interference (EMI). Toroid and pot core type
inductors are recommended in applications where EMI
must be kept to a minimum; for example, where there are
sensitive analog circuitry or transducers nearby. Rod core
types are a less expensive choice where EMI is not a
problem. Minimum and maximum input voltage, output
voltage and output current must be established before an
inductor can be selected.
Inductor Selection — Step-Up Converter
In a step-up, or boost converter (Figure 4), power gener-
ated by the inductor makes up the difference between
input and output. Power required from the inductor is
determined by:
PV VV I
L OUT D IN
MIN
OUT
=+
()()
–()1
where V
D
is the diode drop (0.5V for a 1N5818 Schottky).
Energy required by the inductor per cycle must be equal or
greater than:
U
S
A
O
PP
L
IC
AT
I
WU
U
I FOR ATIO
P
f
L OSC
/()2
in order for the converter to regulate the output.
When the switch is closed, current in the inductor builds
according to:
It
V
R
e
L
IN
Rt
L
() – ( )
–
=
′
′
13
where R′ is the sum of the switch equivalent resistance
(0.8Ω typical at 25°C) and the inductor DC resistance.
When the drop across the switch is small compared to V
IN
,
the simple lossless equation:
It
V
L
t
L
IN
()
= ()4
can be used. These equations assume that at t = 0,
inductor current is zero. This situation is called “discon-
tinuous mode operation” in switching regulator parlance.
Setting “t” to the switch-on time from the LT1111 speci-
fication table (typically 7µs) will yield I
PEAK
for a specific
“L” and V
IN
. Once I
PEAK
is known, energy in the inductor
at the end of the switch-on time can be calculated as:
ELI
L
PEAK
=
1
2
5
2
()
E
L
must be greater than P
L
/f
OSC
for the converter to deliver
the required power. For best efficiency I
PEAK
should be
kept to 1A or less. Higher switch currents will cause
excessive drop across the switch resulting in reduced
efficiency. In general, switch current should be held to as
low a value as possible in order to keep switch, diode and
inductor losses at a minimum.
As an example, suppose 12V at 60mA is to be generated
from a 4.5V to 8V input. Recalling equation (1),
P V V V mA mW
L
=+
()()
=12 0 5 4 5 60 480 6.–. ()
Energy required from the inductor is
P
f
mW
kHz
J
L
OSC
==
480
72
67 7.()µ
