LT1976EFE#TR Linear Technology, LT1976EFE#TR Datasheet - Page 16
![IC REG SW STEP DWN 1.5A 16-TSSOP](/photos/7/34/73402/161-16-tssop_sml.jpg)
LT1976EFE#TR
Manufacturer Part Number
LT1976EFE#TR
Description
IC REG SW STEP DWN 1.5A 16-TSSOP
Manufacturer
Linear Technology
Type
Step-Down (Buck)r
Datasheet
1.LT1976IFEPBF.pdf
(28 pages)
Specifications of LT1976EFE#TR
Internal Switch(s)
Yes
Synchronous Rectifier
No
Number Of Outputs
1
Voltage - Output
1.2 ~ 54 V
Current - Output
1.5A
Frequency - Switching
200kHz
Voltage - Input
3.3 ~ 60 V
Operating Temperature
-40°C ~ 125°C
Mounting Type
Surface Mount
Package / Case
16-TSSOP Exposed Pad, 16-eTSSOP, 16-HTSSOP
Lead Free Status / RoHS Status
Contains lead / RoHS non-compliant
Power - Output
-
Available stocks
Company
Part Number
Manufacturer
Quantity
Price
LT1976/LT1976B
APPLICATIO S I FOR ATIO
1. Choose a value in microhenries from the graph of
Table 4. Inductor Selection Criteria
VENDOR/
PART NUMBER
Coiltronics
UP2B-150
UP2B-330
UP2B-470
UP2B-680
UP2B-101
UP3B-150
UP3B-330
UP3B-470
UP3B-680
UP3B-101
Sumida
CDRH8D28-150M
CDRH124-150M
CDRH127-150M
CDRH8D28-330M
CDRH124-330M
CDRH127-330M
CDRH8D28-470M
CDRH125-470M
CDRH127-470M
CDRH124-680M
CDRH127-680M
CDRH124-101M
CDRH127-101M
Coilcraft
DT3308P-153
DT3308P-333
DT3308P-473
16
maximum load current. Choosing a small inductor with
lighter loads may result in discontinuous mode of
operation, but the LT1976 is designed to work well in
either mode.
VALUE (μH)
U
100
100
100
100
15
33
47
68
15
33
47
68
15
15
15
33
33
33
47
47
47
68
68
15
33
47
U
I
RMS
0.95
1.25
2.4
1.7
1.4
1.2
3.9
2.4
1.9
1.6
1.4
2.2
3.2
4.5
1.4
2.7
3.0
1.8
2.5
1.5
2.1
1.2
1.7
2.0
1.4
1
(A)
W
DCR (Ω) HEIGHT (mm)
0.041
0.062
0.139
0.179
0.271
0.032
0.069
0.101
0.156
0.205
0.053
0.122
0.048
0.150
0.058
0.076
0.228
0.05
0.02
0.97
0.30
0.17
0.47
0.1
0.1
0.3
U
6.8
6.8
6.8
6.8
6.8
4.5
4.5
4.5
4.5
6
6
6
6
6
3
8
3
8
3
6
8
8
8
3
3
3
2. Calculate peak inductor current at full load current to
3. Decide if the design can tolerate an “open” core geom-
4. After making an initial choice, consider the secondary
Assume that the average inductor current is equal to
load current and decide whether or not the inductor
must withstand continuous fault conditions. If maxi-
mum load current is 0.5A, for instance, a 0.5A inductor
may not survive a continuous 2A overload condition.
For applications with a duty cycle above 50%, the
inductor value should be chosen to obtain an inductor
ripple current of less than 40% of the peak switch
current.
ensure that the inductor will not saturate. Peak current
can be significantly higher than output current, especially
with smaller inductors and lighter loads, so don’t omit
this step. Powdered iron cores are forgiving because they
saturate softly, whereas ferrite cores saturate abruptly.
Other core materials fall somewhere in between. The
following formula assumes continuous mode of opera-
tion, but it errs only slightly on the high side for discon-
tinuous mode, so it can be used for all conditions.
etry like a rod or barrel, which have high magnetic field
radiation, or whether it needs a closed core like a toroid
to prevent EMI problems. This is a tough decision
because the rods or barrels are temptingly cheap and
small and there are no helpful guidelines to calculate
when the magnetic field radiation will be a problem.
things like output voltage ripple, second sourcing, etc.
Use the experts in the Linear Technology’s applications
department if you feel uncertain about the final choice.
They have experience with a wide range of inductor
types and can tell you about the latest developments in
low profile, surface mounting, etc.
V
f = switching frequency, 200kHz
I
PEAK
IN
= maximum input voltage
=
I
OUT
+
V
OUT
2
( )( )( )
(
f L V
V
IN
–
V
IN
OUT
)
1976bfg