LTC3863_15 Datasheet, PDF (24/38 Pages) Linear Technology – 60V Low IQ Inverting DC/DC Controller

LTC3863
Applications Information
If RFB1 is chosen to be 187k, then RFB2 needs to be 30.1k.
The FREQ pin is tied to signal ground in order to program
the switching frequency to 350kHz. The on-time required
to generate –5V output from 55V V_IN in continuous mode
can be calculated as:
t_ON(CCM)
=
320kHz
5V + 0.5V
•(55V + 5V
+
0.5V
)
=
260ns
This on-time, t_ON, is larger than LTC3863’s minimum on-
time with sufficient margin to prevent cycle skipping. Use
a lower frequency if a larger on-time margin is needed to
account for variations from minimum on-time and switch-
ing frequency. As load current decreases, the converter
will eventually start cycle skipping.
Next, set the inductor value such that the inductor ripple
current is 60% of the average inductor current at maximum
V_IN = 55V and full load = 1.8A:
L
=
0.6
• 1.8A
55V² •(5V + 0.5V)
• 320kHz •(55V + 5V
+ 0.5V)²
≈
13.1µH
Select a standard value of 12μH.
The resulting ripple current at minimum V_IN of 4.5V is:
∆I_L
=
12µH
•
4.5V •(5V + 0.5V)
320kHz •(4.5V + 5V
+
0.5V)
=
0.644A
The boundary output current for continuous/discontinuous
mode is calculated:
I_OUT(CDB)
=
55V² •(5V + 0.5V)
2 •12µH• 320kHz •(55V + 5V
+
0.5V)²
=
0.59A
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
4.5V
+
0.5V)
+
∆I_L
2
=
3.6A
+
0.644A
2
≈
3.92A
Next, set the R_SENSE resistor value to ensure that the
converter can deliver the maximum peak inductor current
of 3.92A with sufficient margin to account for component
variations and worst-case operating conditions. Using a
30% margin factor:
R_SENSE
=
95mV
1.3 • 3.92A
=
18.6mΩ
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


•
(4.5V
+
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.
3863fa
24
For more information www.linear.com/3863