ADP1173
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 ADP1073 should be
considered. This device has a 72% duty cycle, so more energy is
stored in the inductor on each cycle. This results in greater
output power.
–R′t
L
VIN
R′
IL (t)=
1– e
(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, where the voltage drop across
the switch is small compared to VIN , 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, then the 1.5 A limit may be ex-
ceeded at the maximum value of VIN. In this case, the ADP1173’s
current limit feature can be used to limit switch current. Simply
select a resistor (using Figure 4) 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 con-
stant IPEAK as VIN increases. See the Limiting the Switch Current
section of this data sheet for more information.
VIN
L
IL (t)=
t
(4)
Replacing “t” in the above equation with the ON time of the
ADP1173 (23 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
=
LI2
(5)
PEAK
As previously mentioned, EL must be greater than PL/fOSC so the
ADP1173 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 pos-
sible to minimize losses in the switch, inductor and diode.
Inductor Selection—Step-Down Converter
The step-down mode of operation is shown in Figure 15. Unlike
the step-up mode, the ADP1173’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. If higher output
current is required, the ADP1111 should be considered.
In practice, the inductor value is easily selected using the equa-
tions above. For example, consider a supply that will generate
9 V at 50 mA from a 3 V source. 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 =(9V +0.5V –3V )×(50 mA)= 325 mW
On each switching cycle, the inductor must supply:
PL 325 mW
2IOUT
DC VIN –VSW +VD
VOUT +VD
IPEAK
=
(6)
=
=13.5µJ
where DC = duty cycle (0.55 for the ADP1173)
SW = voltage drop across the switch
VD = diode drop (0.5 V for a 1N5818)
OUT = output current
fOSC 24 kHz
V
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:
I
VOUT = the output voltage
VIN = the minimum input voltage
VIN
3V
L =
t =
23 µs =138 µH
As previously mentioned, the switch voltage is higher in step-
down mode than step-up mode. VSW is a function of switch
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:
current and is therefore a function of VIN, L, time and VOUT
.
For most applications, a VSW value of 1.5 V is recommended.
The inductor value can now be calculated:
VIN(MIN) –VSW –VOUT
IPEAK
where tON = switch ON time (23 µs)
–1.0Ω × 23µs
3V
100 µH
IPEAK
=
1– e
=616 mA
1. 0 Ω
L =
× tON
(7)
Once the peak current is known, the inductor energy can be
calculated from Equation 5:
If the input voltage will vary (such as an application that must
operate from a 12 V to 24 V source) an RLIM resistor should be
selected from Figure 5. The RLIM resistor will keep switch cur-
rent constant as the input voltage rises. Note that there are separate
1
2
EL
=
(100 µH)×(616 mA)2 =19 µJ
The inductor energy of 19 µJ is greater than the PL/fOSC re-
quirement of 13.5 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.
RLIM values for step-up and step-down modes of operation.
REV. 0
–6–
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