參數(shù)資料
型號: LT1374IT7
廠商: LINEAR TECHNOLOGY CORP
元件分類: 穩(wěn)壓器
英文描述: RADIATION HARDENED HIGH EFFICIENCY, 5 AMP SWITCHING REGULATORS
中文描述: 8.5 A SWITCHING REGULATOR, 560 kHz SWITCHING FREQ-MAX, PSFM7
封裝: PLASTIC, TO-220, 7 PIN
文件頁數(shù): 24/28頁
文件大?。?/td> 261K
代理商: LT1374IT7
24
LT1374
APPLICATIO
S I
FOR
ATIO
U
Operating duty cycle:
W
U
U
DC
V
0 3
.
V
V
V
V
OUT
F
IN
OUT
F
=
+
+
+
(This formula uses an average value for switch loss, so it
may be several percent in error.)
With the conditions above:
DC
=
+
.
+ +
5
=
5
0 3
0 5
.
5 5
.
0 5
.
51
%
This duty cycle is close enough to 50% that I
P
can be
assumed to be 4.5A.
OUTPUT DIVIDER
If the adjustable part is used, the resistor connected to
V
OUT
(R2) should be set to approximately 5k. R1 is
calculated from:
(
.
R
R V
OUT
2
1
–2 42
2 42
=
)
INDUCTOR VALUE
Unlike buck converters, positive-to-negative converters
cannot use large inductor values to reduce output ripple
voltage. At 500kHz, values larger than 25
μ
H make almost
no change in output ripple. The graph in Figure 16 shows
peak-to-peak output ripple voltage for a 5V to –5V con-
verter versus inductor value. The criteria for choosing the
inductor is therefore typically based on ensuring that peak
switch current rating is not exceeded. This gives the
lowest value of inductance that can be used, but in some
cases (lower output load currents) it may give a value that
creates unnecessarily high output ripple voltage. A com-
promise value is often chosen that reduces output ripple.
As you can see from the graph, largeinductors will not
Figure 16. Ripple Voltage on Positive-to-Negative Converter
INDUCTOR SIZE (
μ
H)
0
O
P
)
250
200
150
100
50
0
20
1374 F16
5
10
15
5V TO –5V CONVERTER
OUTPUT CAPACITOR’S
ESR = 0.05
I
LOAD
= 0.25A
I
LOAD
= 1A
give arbitrarily low ripple, but smallinductors can give
high ripple.
The difficulty in calculating the minimum inductor size
needed is that you must first know whether the switcher
will be in continuous or discontinuous mode at the critical
point where switch current is 4.5A. The first step is to use
the following formula to calculate the load current where
the switcher must use continuous mode. If your load
current is less than this, use the discontinuous mode
formula to calculate minimum inductor needed. If load
current is higher, use the continuous mode formula.
Output current where continuous mode is needed:
I
V
I
V
V
V
V
V
CONT
IN
)
IN
OUT
IN
OUT
F
=
( ) ( )
(
+
(
+
+
)
2
2
4
Minimum inductor discontinuous mode:
=
(
( )( )
L
V
I
f I
MIN
OUT
OUT
2
P
)(
)
2
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