ADP1108
On each switching cycle, the inductor must supply:
Calculating the Inductor Value
Selecting the proper inductor value is a simple three-step
process:
PL 315 mW
=
=16.6 µ J
1. Define the operating parameters: minimum input voltage,
maximum input voltage, output voltage and output current.
fOSC 19 kHz
T he required inductor power is fairly low in this example, so the
peak current can also be low. Assuming a peak current of
500 mA as a starting point, Equation 4 can be rearranged to
recommend an inductor value:
2. Select the appropriate conversion topology (step-up, step-
down or inverting).
3. Calculate the inductor value, using the equations in the fol-
lowing sections.
VIN
IL(MAX )
2 V
500 mA
Inductor Selection—Step-Up Conver ter
L =
t =
36 µs = 144 µH
In a step-up or boost converter (Figure 15), the inductor must
store enough power to make up the difference between the input
voltage and the output voltage. T he inductor power is calculated
from the equation:
Substituting a standard inductor value of 100 µH with 0.2 Ω dc
resistance, will produce a peak switch current of:
–1.0 Ω×36 µs
100 µH
PL = VOUT +VD −VIN
× I
OUT
(Equation 1)
2 V
1. 0 Ω
(
)
(
)
MIN
(
)
IPEAK
=
1 – e
= 605 mA
where VD is the diode forward voltage (≈ 0.5 V for a 1N5818
Schottky). Energy is only stored in the inductor while the
ADP1108 switch is ON, so the energy stored in the inductor on
each switching cycle must be equal to or greater than:
PL
Once the peak current is known, the inductor energy can be
calculated from Equation 5:
1
2
2
EL
=
100 µH × 605 mA =18.3 µ J
(Equation 2)
(
)
fOSC
in order for the ADP1108 to regulate the output voltage.
T he inductor energy of 18.3 µJ is greater than the PL/fOSC
requirement of 16.6 µJ, so the 100 µH inductor will work in this
application. By substituting other inductor values into the same
equations, the optimum inductor value can be selected. When
selecting an inductor, the peak current must not exceed the
maximum switch current of 1.5 A. If the calculated peak current
is greater than 1.5 A, either the ADP3000 should be considered
or an external power transistor can be used.
When the internal power switch turns ON, current flow in the
inductor increases at the rate of:
–R't
L
VIN
R'
IL (t) =
1− e
(Equation 3)
where L is in henrys and RЈ is the sum of the switch equivalent
resistance (typically 0.8 Ω at +25°C) and the dc resistance of
the inductor. If the voltage drop across the switch is small
compared to VIN, a simpler equation can be used:
T he peak current must be evaluated for both minimum and
maximum values of input voltage. If the switch current is high
when VIN is at its minimum, the 1.5 A limit may be exceeded at the
maximum value of VIN. In this case, the current limit feature of
the ADP1108 can be used to limit switch current. Simply select
a resistor (using Figure 3) that will limit the maximum switch
current to the IPEAK value calculated for the minimum value of
VIN
L
IL (t) =
t
(Equation 4)
Replacing t in the above equation with the ON time of the
ADP1108 (36 µs, typical) will define the peak current for a
given inductor value and input voltage. At this point, the
inductor energy can be calculated as follows:
V
IN . T his will improve efficiency by producing a constant
IPEAK as VIN increases. See the Limiting the Switch Current
section of this data sheet for more information.
Note that the switch current limit feature does not protect the
circuit if the output is shorted to ground. In this case, current is
limited only by the dc resistance of the inductor and the forward
voltage of the diode.
1
2
EL
=
L × I 2
(Equation 5)
PEAK
As previously mentioned, EL must be greater than PL/fOSC so the
ADP1108 can deliver the necessary power to the load. For best
efficiency, peak current should be limited to 1 A or less. Higher
switch currents will reduce efficiency because of increased satura-
tion voltage in the switch. High peak current also increases output
ripple. As a general rule, keep peak current as low as possible to
minimize losses in the switch, inductor and diode.
Inductor Selection—Step-D own Conver ter
T he step-down mode of operation is shown in Figure 16. Unlike
the step-up mode, the ADP1108’s power switch does not
saturate when operating in the step-down mode. T herefore,
switch current should be limited to 650 mA in this mode. If the
input voltage will vary over a wide range, the ILIM pin can be
used to limit the maximum switch current. Higher switch current
is possible by adding an external switching transistor, as shown
in Figure 18.
In practice, the inductor value is easily selected using the equations
above. For example, consider a supply that will generate 12 V
at 30 mA from a 3 V battery, assuming a 2 V end-of-life voltage.
T he inductor power required is from Equation 1:
T he first step in selecting the step-down inductor is to calculate
the peak switch current as follows:
PL = 12 V + 0.5V – 2 V × 30 mA = 315 mW
(
)
(
)
2 IOUT
DC
VOUT +VD
VIN –VSW +VD
IPEAK
=
(Equation 6)
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