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부품명  ADP1108AN5  Analog Devices 
상세내역  Micropower DCDC Converter Adjustable and Fixed 3.3 V, 5 V, 12 V 
ADP1108AN5 Datasheet PDF : 12 Pages

ADP1108
Calculating the Inductor Value
Selecting the proper inductor value is a simple threestep
process:
1. Define the operating parameters: minimum input voltage,
maximum input voltage, output voltage and output current.
2. Select the appropriate conversion topology (stepup, step
down or inverting).
3. Calculate the inductor value, using the equations in the fol
lowing sections.
On each switching cycle, the inductor must supply:
PL = 315 mW =16.6 µ J
f OSC 19 kHz
The 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:
Inductor Selection—StepUp Converter
In a stepup or boost converter (Figure 15), the inductor must
store enough power to make up the difference between the input
voltage and the output voltage. The inductor power is calculated
from the equation:
L = V IN t = 2 V 36 µs = 144 µH
IL(MAX ) 500 mA
Substituting a standard inductor value of 100 µH with 0.2 Ω dc
resistance, will produce a peak switch current of:
( ) ( ) PL = VOUT +V D −V IN(MIN ) × IOUT
(Equation 1)
where VD is the diode forward voltage (≈ 0.5 V for a 1N5818
Schottky). Energy is only stored in the inductor while the
I PEAK
=
2V
1.0 Ω
1 –
e
–1.0 Ω×36
100 µH
µs
= 605 mA
Once the peak current is known, the inductor energy can be
ADP1108 switch is ON, so the energy stored in the inductor on calculated from Equation 5:
each switching cycle must be equal to or greater than:
PL
f OSC
(Equation 2)
( ) EL
=
1
2
100
µH
×
605 mA
2
=18.3 µ J
in order for the ADP1108 to regulate the output voltage.
When the internal power switch turns ON, current flow in the
inductor increases at the rate of:
The 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
IL
(t)
=
V IN
R'
1 −
– R 't
eL
(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:
I
L
(t
)
=
V IN
L
t
(Equation 4)
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.
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 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
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:
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.
Note that the switch current limit feature does not protect the
EL
=
1
L
2
×
I2
PEAK
(Equation 5)
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.
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 endoflife voltage.
The inductor power required is from Equation 1:
PL = (12 V + 0.5V – 2V ) × (30 mA) = 315 mW
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.
Inductor Selection—StepDown Converter
The stepdown mode of operation is shown in Figure 16. Unlike
the stepup mode, the ADP1108’s power switch does not
saturate when operating in the stepdown 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 18.
The first step in selecting the stepdown inductor is to calculate
the peak switch current as follows:
I PEAK
= 2 IOUT
DC
V
VOUT + V D
IN – V SW + V
D
(Equation 6)
–6–
REV. 0

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