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參數(shù)資料
| 型號： | ADP1073AN-5 |
| 廠商： | ANALOG DEVICES INC |
| 元件分類： | 穩(wěn)壓器 |
| 英文描述： | Circular Connector; No. of Contacts:100; Series:LJT06R; Body Material:Aluminum; Connecting Termination:Crimp; Connector Shell Size:23; Circular Contact Gender:Socket; Circular Shell Style:Straight Plug; Insert Arrangement:23-35 |
| 中文描述： | 0.4 A SWITCHING REGULATOR, 23 kHz SWITCHING FREQ-MAX, PDIP8 |
| 封裝： | PLASTIC, DIP-8 |
| 文件頁數(shù)： | 7/16頁 |
| 文件大?。?/td> | 442K |
| 代理商： | ADP1073AN-5 |
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 I
PEAK
value calculated for the minimum value of
V
IN
. This will improve efficiency by producing a constant I
PEAK
as V
IN
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
I
LIM
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:
I
PEAK
=
2
×
I
OUT
DC
V
IN
–
V
SW
+
V
D
V
OUT
+
V
D
(6)
where
DC
= duty cycle (0.72 for the ADP1073)
V
SW
= voltage drop across the switch
V
D
= diode drop (0.5 V for a 1N5818)
I
OUT
= output current
V
OUT
= the output voltage
V
IN
= the minimum input voltage
As previously mentioned, the switch voltage is higher in step-
down mode than in step-up mode. V
SW
is a function of switch
current and is therefore a function of V
IN
, L, time and V
OUT
.
For most applications, a V
SW
value of 1.5 V is recommended.
The inductor value can now be calculated:
L
=
V
IN
(
MIN
)
–
V
SW
–
V
OUT
I
PEAK
×
t
ON
(7)
where
t
ON
= switch ON time (38
μ
s)
If the input voltage will vary (such as an application which must
operate from a battery), an R
LIM
resistor should be selected
from Figure 4. The R
LIM
resistor will keep switch current con-
stant as the input voltage rises. Note that there are separate
R
LIM
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:
I
PEAK
=
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
μ
s
=
144
μ
H
Since 144
μ
H is not a standard value, the next lower standard
value of 100
μ
H would be specified.
To avoid exceeding the maximum switch current when the
input voltage is at +9 V, an R
LIM
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:
P
L
=
|
V
OUT
|
+
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:
I
L
(
t
)
=
V
L
R
'
1–
e
–
R
'
t
L
(9)
where
R
' = 0.65
+ R
L(DC)
V
L
=
V
IN
– 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:
P
L
=
|
5
V
|
+
0.5
V
During each switching cycle, the inductor must supply the fol-
lowing energy:
(
)
×
(75
mA
)
=
413
mW
P
L
f
OSC
=
413
mW
19
kHz
=
21.7
μ
J
Using a standard inductor value of 330
μ
H, with 1
dc resis-
tance, will produce a peak switch current of:
I
PEAK
=
4.5
V
–0.75
V
0.65
+
1
1–
e
–1.65
×
38
μ
s
330
μ
H
=
393
mA
Once the peak current is known, the inductor energy can be
calculated from Equation 9:
E
L
=
1
2(330
μ
H
)
×
(393
mA
)
2
=
25.5
μ
J
The inductor energy of 25.5
μ
J is greater than the P
L
/f
OSC
requirement of 21.7
μ
J, so the 330
μ
H 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 R
LIM
resistor and the I
LIM
pin can be connected di-
rectly to V
IN
. Care should be taken, of course, to ensure that the
peak current does not exceed 800 mA.
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