ADP1111AN Analog Devices Inc, ADP1111AN Datasheet - Page 7
![no-image](/images/manufacturer_photos/0/0/57/analog_devices_inc_sml.jpg)
ADP1111AN
Manufacturer Part Number
ADP1111AN
Description
IC REG SWITCHING ADJ VLT 8-DIP
Manufacturer
Analog Devices Inc
Type
Step-Down (Buck), Step-Up (Boost), Invertingr
Datasheet
1.ADP1111ANZ-5.pdf
(15 pages)
Specifications of ADP1111AN
Rohs Status
RoHS non-compliant
Internal Switch(s)
Yes
Synchronous Rectifier
No
Number Of Outputs
1
Voltage - Output
1.2 ~ 30 V
Current - Output
200mA
Frequency - Switching
72kHz
Voltage - Input
2 ~ 30 V
Operating Temperature
0°C ~ 70°C
Mounting Type
Through Hole
Package / Case
8-DIP (0.300", 7.62mm)
Power - Output
500mW
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
Company:
Part Number:
ADP1111ANZ-3.3
Manufacturer:
AD
Quantity:
2 100
Company:
Part Number:
ADP1111ANZ-5
Manufacturer:
JOHANSON
Quantity:
24 000
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:
where V
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:
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:
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 V
used:
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:
As previously mentioned, E
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.
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:
On each switching cycle, the inductor must supply:
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:
REV. 0
P
I
I
E
f
L
L
P
L
L
OSC
( )
( )
L
=
= V
D
t
t
1
2
(
is the diode forward voltage (0.5 V for a 1N5818
=
=
P
L • I
V
V
OUT
L
R'
L
IN
IN
= 12V + 0.5V − 6V
2
L =
(
⎛
⎜
⎝
t
+V
PEAK
1− e
I
D
L(MAX )
V
f
−R't
−V
P
OSC
L
IN
L
⎞
⎟
⎠
IN(MIN )
=
t =
L
260 mW
72 kHz
must be greater than P
500 mA
IN
6V
)
so a simpler equation can be
• I
)
(
• 40 mA
(
OUT
= 3.6 μJ
7 μs = 84 μH
)
)
= 260 mW
(Equation 1)
(Equation 2)
(Equation 3)
(Equation 4)
(Equation 5)
L
/f
OSC
so
–7–
Substituting a standard inductor value of 68 μH with 0.2 Ω dc
resistance will produce a peak switch current of:
Once the peak current is known, the inductor energy can be
calculated from Equation 5:
Since the inductor energy of 11.7 μJ is greater than the P
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.
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.
The peak current must be evaluated for both minimum and
maximum values of input voltage. If the switch current is high
when V
the maximum value of V
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 I
V
as V
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
only limited 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 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 I
used to limit the maximum switch current. Higher switch
current is possible by adding an external switching transistor as
shown in Figure 21.
The first step in selecting the step-down inductor is to calculate
the peak switch current as follows:
where DC = duty cycle (0.5 for the ADP1111)
IN
. This will improve efficiency by producing a constant I
IN
I
PEAK
increases. See the “Limiting the Switch Current” section
IN
V
V
I
V
V
OUT
SW
D
OUT
IN
is at its minimum, the 1.5 A limit may be exceeded at
=
= diode drop (0.5 V for a 1N5818)
= the minimum input voltage
2 I
= voltage drop across the switch
I
= output current
= the output voltage
PEAK
DC
E
PEAK
OUT
L
=
=
⎛
⎜
⎝
1
2
value calculated for the minimum value of
V
1.0 Ω
(
6V
68 μH
IN
V
OUT
IN
−V
⎛
⎜
⎝
. In this case, the ADP1111’s current
1− e
SW
)
+ V
• 587 mA
(
+V
−1.0 Ω • 7 μs
D
68 μH
D
⎞
⎟
⎠
)
2
⎞
⎟ = 587 mA
⎠
=11.7 μJ
ADP1111
LIM
pin can be
(Equation 6)
L
/f
PEAK
OSC