LT11
11
APPLICATI
S I FOR ATIO
Picking an inductor value of 47µH with 0.2Ω DCR results
in a peak switch current of:
4.5 V
I
PEAK
=
1.0
Ω
–1.0
Ω ×
7
µ
s
1 – e
47
µ
H
=
623mA .
Substituting I
PEAK
into Equation 04 results in:
E
L
=
1
47
µ
H 0.623 A
2
=
9.1
µ
J
2
(
)(
)
Since 9.1µJ > 6.7µJ, the 47µH inductor will work. This
trial-and-error approach can be used to select the opti-
mum inductor. Keep in mind the switch current maximum
rating of 1.5A. If the calculated peak current exceeds this,
consider using the LT1110. The 70% duty cycle of the
LT1110 allows more energy per cycle to be stored in the
inductor, resulting in more output power.
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, switch current is held
constant, resulting in increasing efficiency.
Inductor Selection — Step-Down Converter
The step-down case (Figure 5) 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 Figure 6). 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:
I
PEAK
=
2 I
OUT
DC
V
OUT
+
V
D
V – V
SW
+
V
D
IN
(10)
where DC = duty cycle (0.50)
V
SW
= switch drop in step-down mode
V
D
= diode drop (0.5V for a 1N5818)
8
U
I
OUT
= output current
V
OUT
= output voltage
V
IN
= minimum input voltage
V
SW
is actually a function of switch current which is in turn
a function of V
IN
, L, time, and V
OUT
. To simplify, 1.5V can
be used for V
SW
as a very conservative value.
Once I
PEAK
is known, inductor value can be derived from:
(09)
(08)
W
U
UO
L
=
V
IN MIN
−
V
SW
−
V
OUT
×
t
ON
I
PEAK
(11)
where t
ON
= switch-on time (7µs).
Next, the current limit resistor R
LIM
is selected to give
I
PEAK
from the R
LIM
Step-Down Mode curve. The addition
of this resistor keeps maximum switch current constant as
the input voltage is increased.
As an example, suppose 5V at 300mA is to be generated
from a 12V to 24V input. Recalling Equation (10),
I
PEAK
=
2 300mA
5
+
0.5
12 – 1.5
+
0.5
=
600mA
0.50
(
)
(12)
Next, inductor value is calculated using Equation (11):
L
=
12 – 1.5 – 5
7
µ
s
=
64
µ
H.
600mA
(13)
Use the next lowest standard value (56µH).
Then pick R
LIM
from the curve. For I
PEAK
= 600mA, R
LIM
= 56Ω.
Inductor Selection — Positive-to-Negative Converter
Figure 7 shows hookup for positive-to-negative conver-
sion. All of the output power must come from the inductor.
In this case,
P
L
=
(
V
OUT
+
V
D
)(
I
OUT
)
(14)
In this mode the switch is arranged in common collector
or step-down mode. The switch drop can be modeled as
a 0.75V source in series with a 0.65Ω resistor. When the
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