LT1374IR#PBF Linear Technology, LT1374IR#PBF Datasheet - Page 26
![IC SW REG 4.5A ADJ STEP-DWN 7-DD](/photos/7/59/75907/lt1374ir_pbf_sml.jpg)
LT1374IR#PBF
Manufacturer Part Number
LT1374IR#PBF
Description
IC SW REG 4.5A ADJ STEP-DWN 7-DD
Manufacturer
Linear Technology
Type
Step-Down (Buck)r
Datasheet
1.LT1374CFEPBF.pdf
(32 pages)
Specifications of LT1374IR#PBF
Internal Switch(s)
Yes
Synchronous Rectifier
No
Number Of Outputs
1
Voltage - Output
2.42 ~ 21.5 V
Current - Output
4.5A
Frequency - Switching
500kHz
Voltage - Input
5 ~ 25 V
Operating Temperature
-40°C ~ 125°C
Mounting Type
Surface Mount
Package / Case
D²Pak, TO-263 (7 leads + tab)
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Power - Output
-
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
LT1374
APPLICATIONS
With the conditions above:
This duty cycle is close enough to 50% that I
assumed to be 4.5A.
OUTPUT DIVIDER
If the adjustable part is used, the resistor connected to
V
calculated from:
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 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
26
OUT
Figure 16. Ripple Voltage on Positive-to-Negative Converter
(R2) should be set to approximately 5k. R1 is
250
200
150
100
50
0
0
U
5
INDUCTOR SIZE (µH)
INFORMATION
U
5V TO – 5V CONVERTER
OUTPUT CAPACITOR’S
ESR = 0.05Ω
10
I
I
LOAD
LOAD
15
W
= 1A
= 0.25A
1374 F16
20
U
P
can be
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.
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:
Minimum inductor discontinuous mode:
Minimum inductor continuous mode:
For the example above, with maximum load current of 1A:
1374fd