LT1108
O U
W
U
PPLICATI
A
S I FOR ATIO
where DC = duty cycle (0.60)
VSW = switch drop in step-down mode
VD = diode drop (0.5V for a 1N5818)
IOUT = output current
Energy required from the inductor is
P
315mW
19kHz
L
=
= 16.6µJ
(07)
f
OSC
VOUT = output voltage
VIN = minimum input voltage
Picking an inductor value of 100µH with 0.2Ω DCR results
in a peak switch current of
VSW is actually a function of switch current which is in turn
a function of VIN, L, time, and VOUT. To simplify, 1.5V can
be used for VSW as a very conservative value.
–1.0Ω × 36µs
2V
1.0Ω
I
=
1– e
= 605mA
(08)
100µH
PEAK
Once IPEAK is known, inductor value can be derived from
Substituting IPEAK into Equation 04 results in
V
− V
− V
OUT
IN MIN
SW
1
2
2
L =
× t
(11)
ON
E = 100µH 6.605A = 18.3µJ
) (
(09)
(
)
L
I
PEAK
Since 18.3µJ > 16.6µJ, the 100µH inductor will work. This
trial-and-error approach can be used to select the optimum
inductor. Keep in mind the switch current maximum rating
of 1.5A. If the calculated peak current exceeds this, an
external power transistor can be used.
where tON = switch-ON time (36µs).
Next, the current limit resistor RLIM is selected to give IPEAK
from the RLIM Step-Down Mode curve. The addition of this
resistor keeps maximum switch current constant as the
input voltage is increased.
A resistor can be added in series with the ILIM pin to invoke
switch current limit. The resistor should be picked so the
calculated IPEAK at minimum VIN is equal to the Maximum
Switch Current (from Typical Performance Characteristic
curves). Then, as VIN increases, switch current is held
constant, resulting in increasing efficiency.
As an example, suppose 5V at 300mA is to be generated
from a 12V to 24V input. Recalling Equation (10),
2 300mA
(
)
5 + 0.5
I
=
= 500mA (12)
PEAK
0.60
12 – 1.5 + 0.5
Step-Down Converter
Next, inductor value is calculated using Equation (11)
The step-down case (Figure 2) differs from the step-up in
thattheinductorcurrentflowsthroughtheloadduringboth
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 3). The ILIM pin is the key to successful
operation over varying inputs.
12 – 1.5 – 5
L =
36µs = 396µH
(13)
500mA
Use the next lowest standard value (330µH).
Then pick RLIM from the curve. For IPEAK = 500mA,
RLIM = 220Ω.
After establishing output voltage, output current and input
voltage range, peak switch current can be calculated by the
formula:
Positive-to-Negative Converter
Figure 4 shows hookup for positive-to-negative conver-
sion. All of the output power must come from the inductor.
In this case,
2I
V
+ V
OUT D
OUT
I
=
(10)
PEAK
DC
V
– V
+ V
IN
SW D
PL = ( VOUT + VD)(IOUT)
(14)
7
Linear [ Linear ]
