LT1375 查看數據表（PDF） - Linear Technology
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LT1375 Datasheet PDF : 28 Pages
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LT1375/LT1376
APPLICATIONS INFORMATION
INDUCTOR VALUE
Unlike buck converters, positive-to-negative converters
cannot use large inductor values to reduce output ripple
voltage. At 500kHz, values larger than 25µH make almost
no change in output ripple. The graph in Figure 19 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, large inductors will not
give arbitrarily low ripple, but small inductors can give
high ripple.
150
5V TO –5V CONVERTER
OUTPUT CAPACITOR
120
ESR = 0.1Ω
90
ILOAD = 0.25A
60
ILOAD = 0.1A
30
0
0
5
10
15
20
25
INDUCTOR SIZE (µH)
1375/76 F19
Figure 19. Ripple Voltage on Positive-to-Negative Converter
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 1.5A. 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:
( ( )() ( ) ) ICONT =
22
VIN IP
4 VIN + VOUT VIN + VOUT + VF
Minimum inductor discontinuous mode:
( )( ) LMIN = 2 VOUT
IOUT
2
( )( )f IP
Minimum inductor continuous mode:
( )( ) LMIN =
( )( ) ( ) 2 f
VIN VOUT
VIN + VOUT
IP
−
IOUT
1+
VOUT + VF
VIN
For the example above, with maximum load current of
0.25A:
(5)2(1.5)2
( )( ) ICONT =
= 0.37A
4 5 + 5 5 + 5 + 0.5
This says that discontinuous mode can be used and the
minimum inductor needed is found from:
2(5)(0.25)
( ) LMIN
=
500
•
103
1.5
2
=
2.2µH
In practice, the inductor should be increased by about
30% over the calculated minimum to handle losses and
variations in value. This suggests a minimum inductor of
3µH for this application, but looking at the ripple voltage
chart shows that output ripple voltage could be reduced by
a factor of two by using a 15µH inductor. There is no rule
25
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