ADP1111
INDUCTOR SELECTION–STEP-UP CONVERTER
In a step-up or boost converter (Figure 18), the inductor must
store enough power to make up the difference between the input
voltage and the output voltage. The power that must be stored
is calculated from the equation:
Substituting a standard inductor value of 68 µH with 0.2 Ω dc
resistance will produce a peak switch current of:
−1.0 Ω• 7µs
68 µH
6V
1.0 Ω
IPEAK
=
1 − e
= 587 mA
PL = V
+VD −VIN(MIN) • I
(Equation 1)
(
)
(
)
OUT
OUT
Once the peak current is known, the inductor energy can be
calculated from Equation 5:
where VD is the diode forward voltage (0.5 V for a 1N5818
Schottky). Because energy is only stored in the inductor while
the ADP1111 switch is ON, the energy stored in the inductor
on each switching cycle must be equal to or greater than:
1
EL
=
68 µH • 587 mA 2 =11.7µJ
) (
(
)
2
PL
(Equation 2)
Since the inductor energy of 11.7 µJ is greater than the PL/fOSC
requirement of 3.6 µJ, the 68 µH inductor will work in this
application. By substituting other inductor values into the same
equations, the optimum inductor value can be selected.
fOSC
in order for the ADP1111 to regulate the output voltage.
When the internal power switch turns ON, current flow in the
inductor increases at the rate of:
When selecting an inductor, the peak current must not exceed
the maximum switch current of 1.5 A. If the equations shown
above result in peak currents > 1.5 A, the ADP1110 should be
considered. Since this device has a 70% duty cycle, more energy
is stored in the inductor on each cycle. This results is greater
output power.
−R't
L
VIN
R'
IL t =
( )
1−e
(Equation 3)
where L is in Henrys and R' is the sum of the switch equivalent
resistance (typically 0.8 Ω at +25°C) and the dc resistance of
the inductor. In most applications, the voltage drop across the
switch is small compared to VIN so a simpler equation can be
used:
The 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 ADP1111’s current
limit feature can be used to limit switch current. Simply select a
resistor (using Figure 6) that will limit the maximum switch
current to the IPEAK value calculated for the minimum value of
VIN. This 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.
V
L
IL t = IN t
(Equation 4)
( )
Replacing ‘t’ in the above equation with the ON time of the
ADP1111 (7 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:
Note that the switch current limit feature does not protect the
circuit if the output is shorted to ground. In this case, current is
only limited by the dc resistance of the inductor and the forward
voltage of the diode.
1
2
EL
=
L • I2 PEAK
(Equation 5)
As previously mentioned, EL must be greater than PL/fOSC so
that the ADP1111 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 saturation 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-DOWN CONVERTER
The step-down mode of operation is shown in Figure 19.
Unlike the step-up mode, the ADP1111’s power switch does not
saturate when operating in the step-down mode; therefore,
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 21.
In practice, the inductor value is easily selected using the
equations above. For example, consider a supply that will
generate 12 V at 40 mA from a 9 V battery, assuming a 6 V
end-of-life voltage. The inductor power required is, from
Equation 1:
The first step in selecting the step-down inductor is to calculate
the peak switch current as follows:
PL = 12V +0.5V − 6V • 40 mA = 260 mW
(
) (
)
2 IOUT
DC
VOUT + VD
IN −VSW +VD
IPEAK
=
(Equation 6)
On each switching cycle, the inductor must supply:
V
PL 260 mW
fOSC 72 kHz
=
= 3.6 µJ
where DC = duty cycle (0.5 for the ADP1111)
VSW = voltage drop across the switch
VD = diode drop (0.5 V for a 1N5818)
IOUT = output current
Since the required inductor power is fairly low in this example,
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:
VOUT = the output voltage
VIN
6V
VIN = the minimum input voltage
L =
t =
7µs = 84 µH
IL(MAX) 500 mA
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