LTC3862 Datasheet, PDF (31/40 Pages) Linear Technology – Multi-Phase Current Mode Step-Up DC/DC Controller

LTC3862
APPLICATIONS INFORMATION
The value of the input capacitor is a function of the
source impedance, and in general, the higher the source
impedance, the higher the required input capacitance.
The required amount of input capacitance is also greatly
affected by the duty cycle. High output current applica-
tions that also experience high duty cycles can place great
demands on the input supply, both in terms of DC current
and ripple current.
The input ripple current in a multi-phase boost converter
is relatively low (compared with the output ripple current),
because this current is continuous and is being divided
between two or more inductors. Nonetheless, significant
stress can be placed on the input capacitor, especially
in high duty cycle applications. Figure 24 illustrates the
normalized input ripple current, where:
I_NORM
=
V_IN
L•f
A Design Example
Consider the LTC3862 application circuit is shown in Fig-
ure 25a. The output voltage is 48V and the input voltage
range is 5V to 36V. The maximum output current is 5A
when the input voltage is 24V to 36V. Below 24V, current
limit will linearly reduce the maximum load to 1A at 5V
in (see Figure 25b).
1. The duty cycle range (where 5A is available at the
output) is:
D_MAX
=
⎛
⎝⎜
V_O + V_F
V_O +
– V_IN ⎞
V_{F ⎠⎟}
=
⎛
⎝⎜
48V + 0.5V – 24V ⎞
48V + 0.5V ⎠⎟
=
50.5%
D_MIN
=
⎛
⎝⎜
48V + 0.5V – 36V ⎞
48V + 0.5V ⎠⎟
=
25.8%
1.00
0.90
0.80
0.70
0.60
1-PHASE
0.50
0.40
2-PHASE
0.30
0.20
0.10
0
0
0.2 0.4 0.6 0.8 1.0
DUTY CYCLE
3862 F24
Figure 24. Normalized Input Peak-to-Peak Ripple Current
2. The operating frequency is chosen to be 300kHz so
the period is 3.33μs. From Figure 6, the resistor from
the FREQ pin to ground is 45.3k.
3. The minimum on-time for this application operating
in CCM is:
t_ON(MIN)
=
1•
f
⎛
⎝⎜
V_O
+
V_F – V_IN(MAX) ⎞
V_O + V_F
⎠⎟
=
1
300kHz
•
⎛
⎝⎜
48V + 0.5V – 36V
48V + 0.5V
⎞
⎠⎟
=
859ns
The maximum DC input current is:
I_IN(MAX) =
I_O(MAX)
1– D_MAX
=
5A
1– 0.505
= 10.1A
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