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ADP1073AN-33 Datasheet(PDF) 7 Page - Analog Devices

No. de pieza ADP1073AN-33
Descripción Electrónicos  Micropower DC.DC Converter Adjustable and Fixed 3.3 V, 5 V, 12 V
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ADP1073AN-33 Datasheet(HTML) 7 Page - Analog Devices

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ADP1073
–7–
REV. 0
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 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.
Inductor Selection—Step-Down Converter
The step-down mode of operation is shown in Figure 16. Unlike
the step-up mode, the ADP1073’s power switch does not satu-
rate when operating in the step-down mode. Switch current
should therefore be limited to 600 mA for best performance 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.
The first step in selecting the step-down inductor is to calculate
the peak switch current as follows:
IPEAK =
2
× I
OUT
DC
VOUT +V D
V IN –VSW +V D


(6)
where DC = duty cycle (0.72 for the ADP1073)
VSW = voltage drop across the switch
VD = diode drop (0.5 V for a 1N5818)
IOUT = output current
VOUT = the output voltage
VIN = the minimum input voltage
As previously mentioned, the switch voltage is higher in step-
down mode than in step-up mode. VSW is a function of switch
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:
L
=
V IN(MIN) –VSW –VOUT
IPEAK
× t
ON
(7)
where tON = switch ON time (38
µs)
If the input voltage will vary (such as an application which must
operate from a battery), an RLIM resistor should be selected
from Figure 4. The RLIM resistor will keep switch current con-
stant as the input voltage rises. Note that there are separate
RLIM values for step-up and step-down modes of operation.
For example, assume that +3.3 V at 150 mA is required from a
9 V battery with a 6 V end-of-life voltage. Deriving the peak
current from Equation 6 yields:
IPEAK =
2
×150 mA
0.72
3.3
+ 0.5
6– 1.5
+ 0.5


= 317 mA
The peak current can than be inserted into Equation 7 to calcu-
late the inductor value:
L
=
6–1.5 – 3.3
317 mA
×38 µs =144 µH
Since 144
µH is not a standard value, the next lower standard
value of 100
µH would be specified.
To avoid exceeding the maximum switch current when the
input voltage is at +9 V, an RLIM resistor should be specified.
Inductor Selection—Positive-to-Negative Converter
The configuration for a positive-to-negative converter using the
ADP1073 is shown in Figure 17. As with the step-up converter,
all of the output power for the inverting circuit must be supplied
by the inductor. The required inductor power is derived from
the formula:
PL = |VOUT|+V D
()× I
OUT
()
(8)
The ADP1073 power switch does not saturate in positive-to-
negative mode. The voltage drop across the switch can be
modeled as a 0.75 V base-emitter diode in series with a 0.65
resistor. When the switch turns on, inductor current will rise at
a rate determined by:
IL (t) =
V L
R'
1– e
–R't
L


(9)
where
R' = 0.65
Ω + R
L(DC)
VL = VIN – 0.75 V
For example, assume that a –5 V output at 75 mA is to be gen-
erated from a +4.5 V to +5.5 V source. The power in the induc-
tor is calculated from Equation 8:
PL = |−5V|+ 0.5V
()×(75mA)=413mW
During each switching cycle, the inductor must supply the fol-
lowing energy:
PL
f OSC
=
413 mW
19 kHz
= 21.7 µJ
Using a standard inductor value of 330
µH, with 1 Ω dc resis-
tance, will produce a peak switch current of:
IPEAK =
4.5V –0.75V
0.65
Ω+1Ω
1– e
–1.65
Ω× 38 µs
330
µH
 = 393 mA
Once the peak current is known, the inductor energy can be
calculated from Equation 9:
EL =
1
2
(330
µH) × (393 mA)2 = 25.5 µJ
The inductor energy of 25.5
µJ is greater than the P
L/fOSC
requirement of 21.7
µJ, so the 330 µH inductor will work in this
application.
The input voltage varies between only 4.5 V and 5.5 V in this
example. Therefore, the peak current will not change enough to
require an RLIM resistor and the ILIM pin can be connected di-
rectly to VIN. Care should be taken, of course, to ensure that the
peak current does not exceed 800 mA.


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