Datasheet
LTC3863
24
3863f
For more information www.linear.com/3863
APPLICATIONS INFORMATION
The boundary output current for continuous/discontinuous
mode is calculated:
I
OUT(CDB)
=
55V
2
• 5V
2• 12µH• 320kHz • 55V + 5V
( )
2
= 0.55A
The maximum inductor peak current occurs at minimum
V
IN
of 4.5V and full load of 1.8A where LTC3863 operates
in continuous mode is:
I
L(PEAK _ MAX)
=
1.8A • 4.5V + 5V + 0.5V
( )
5V
+
∆I
L
2
= 3.6A +
0.644A
2
≈ 4.25A
Next, set the R
SENSE
resistor value to ensure that the
converter can deliver the maximum peak inductor current
of 4.25A with sufficient margin to account for component
variations and worst-case operating conditions. Using a
30% margin factor:
R
SENSE
=
95mV
1.3• 4.25A
= 17.2mΩ
Use a more readily available 16mΩ sense resistor. This
results in peak inductor current limit:
I
L(PEAK)
=
95mV
16mΩ
= 5.94A
Choose an inductor that has rated saturation current higher
than 5.94A with sufficient margin.
The output current limit can be calculated from the peak
inductor current limit and its minimum occurs at minimum
V
IN
of 5V:
I
LIMIT(MIN)
=
95mV
16mΩ
–
0.644A
2
•
5V
4.5V + 5V + 0.5V
( )
= 2.8A
In this example, 2.8A is the maximum output current the
switching regulator can support at V
IN
= 4.5V. It is larger
than the full load of 1.8A by a margin of 1A. If a larger
margin is needed, use a smaller R
SENSE
.
Next choose a P-channel MOSFET with the appropriate
BV
DSS
and I
D
rating. The BV
DSS
rating should be greater
than (55V + 5V + V
D
) with sufficient margin. In this ex-
ample, a good choice is the Vishay Si7469DP (BV
DSS
=
80V, I
D
= 10A, R
DS(ON)
= 30mΩ, ρ
100°C
= 1.8, V
MILLER
=
3.2V, C
MILLER
= 235pF, θ
JA
= 24°C/W). The highest power
dissipation and the resulting junction temperature for the
P-channel MOSFET occurs at the minimum V
IN
of 5V and
maximum output current of 1.8A. They can be calculated
at T
A
= 70°C as follows:
D=
5V
+
0.5V
4.5V + 5V + 0.5V
≈ 0.55
P
PMOS
= 0.55•
1.8A
1–0.55
2
• 1.8 • 30mΩ
+
320kHz • 235pF • 4.5V+ |–5V|+0.5V
( )
2
2
•
1.8A
1–0.55
( )
•
0.9Ω
4.5V – 3.2V
+
2Ω
3.2V
≈
0.475W
+
0.020W
≈
0.495W
T
J
= 70°C + 0.495W • 24°C/W = 82°C
The same calculations can be repeated for V
IN(MAX)
= 55V:
D=
5V + 0.5V
55V + 5V + 0.5V
≈ 0.091
P
PMOS
≈ 0.091•
1.8A
1–0.091
2
• 1.8 • 30mΩ
+
320kHz • 235pF • 55V+ |–5V|+0.5V
( )
2
2
•
1.8A
1–0.091
( )
•
0.9Ω
5V – 3.2V
+
2Ω
3.2V
≈ 0.019W + 0.39W ≈0.411W
T
J
= 70°C + 0.411W • 24°C/W = 80°C
Next choose an appropriate Schottky diode that will handle
the power requirements. The reverse voltage of the diode,
V
R
, should be greater than (55V + 5V). The Fairchild
S38 Schottky diode is selected (V
F
(3A,125°C) = 0.45V,