7
LT1107
1107fa
Inductor Selection –– Step-Up Converter
In a step-up, or boost converter (Figure 1), power gener-
ated by the inductor makes up the difference between
input and output. Power required from the inductor is
determined by:
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:
in order for the converter to regulate the output.
When the switch is closed, current in the inductor builds
according to:
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:
As an example, suppose 12V at 60mA is to be generated
from a 3V to 6V input. Recalling equation (1),
Energy required from the inductor is:
Picking an inductor value of 33µH with 0.2Ω DCR results
in a peak switch current of:
Substituting I
PEAK
into Equation 4 results in:
Since 11.9µJ > 9.05µJ, the 33µH inductor will work. This
trial-and-error approach can be used to select the opti-
mum inductor.
A resistor can be added in series with the I
LIM
pin to invoke
switch current limit. The resistor should be picked so the
calculated I
PEAK
at minimum V
IN
is equal to the Maximum
Switch Current (from Typical Performance Characteristic
curves). Then, as V
IN
increases, peak switch current is
held constant, resulting in increasing efficiency.
Inductor Selection –– Step-Down Converter
The step-down case (Figure 2) differs from the step-up in
that the inductor current flows through the load during
both the charge and discharge periods of the inductor.
Current through the switch should be limited to ~650mA
in this mode. Higher current can be obtained by using an
external switch (see LT1111 and LT1110 data sheets). The
I
LIM
pin is the key to successful operation over varying
inputs.
After establishing output voltage, output current and input
voltage range, peak switch current can be calculated by the
formula:
U
S
A
O
PP
L
IC
AT
I
WU
U
I FOR ATIO
P
f
L OSC
/()2
It
V
R
e
L
IN
Rt
L
() ( )=
′
−
−
′
13
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 LT1107 speci-
fication table (typically 11µs) 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:
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.
P
f
mW
kHz
J
L
OSC
==
570
63
905 7.()µ
EHA J
L
=
()()
=
1
2
33 085 1191 9
2
µµ.. ()
ELI
L
PEAK
=
1
2
5
2
()
PVVVmA mW
L
=+−
()()
=12 0 5 3 60 570 6.()
I
V
emA
PEAK
s
H
=−
=
−
3
1
1 850 8
111
33
Ω
Ω•
()
µ
µ
PV VV I
L OUT D
IN
MIN
OUT
=+−
()
()
()1
I
I
DC
VV
VV V
PEAK
OUT OUT D
IN SW D
=
+
−+
2
10()
