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LTC1265I Ver la hoja de datos (PDF) - Linear Technology

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LTC1265I Datasheet PDF : 16 Pages
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LTC1265/LTC1265-3.3/LTC1265-5
APPLICATIONS INFORMATION
Design Example
As a design example, assume VIN = 5V, VOUT = 3.3V, IMAX
= 0.8A and f = 250kHz. With this information we can easily
calculate all the important components.
From (1),
RSENSE = 100mV/0.8 = 0.125
From (2) and assuming VD = 0.4V,
CT 100pF
Using (3), the value of the inductor is:
L 5.2(105)(0.125)(100pF)3.3V = 22µH
For the catch diode, a MBRS130LT3 or 1N5818 will be
sufficient in this application.
CIN will require an RMS current rating of at least 0.4A at
temperature, and COUT will require an ESR of (from 5):
ESRCOUT < 0.25
The inductor ripple current is given by:
) IRIPPLE =
VOUT + VD
L
tOFF = 0.22A
At light loads the peak inductor current is at:
IPEAK = 25mV/0.125 = 0.2A
Therefore, at load current less than 0.1A the LTC1265 will
be in Burst Mode operation. Figure 8 shows the complete
circuit and Figure 9 shows the efficiency curve with the
above calculated component values.
VIN
5V
+
CIN
0.1µF PWR VIN
SHDN
VIN
SW
1k
3900pF 100pF
LTC1265-3.3
ITH
PGND
SENSE+
CT
SENSE
SGND
22µH
D1
0.125
+
VOUT
3.3V
0.8A
COUT
1000pF
LTC1265 F08
Figure 8. Design Example Circuit
100
L = DALE LPT4545-220 (22µH)
VOUT = 3.3V
95 CT = 100pF
90
85
80
75
70
0.01
0.1
LOAD CURRENT (mA)
1.0
1265 G11
Figure 9. Design Example Efficiency Curve
12

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