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Applications Information (continued)
There are two values of load current to consider: con-
tinuous load current and peak load current. Continuous
load current is concerned with thermal stresses which
drive the selection of input capacitors, MOSFETs and
diodes. Peak load current determines instantaneous
component stresses and filtering requirements such as
inductor saturation, output capacitors and design of the
current limit circuit.
Design example: Side1 will be used.
VIN = 10V min, 20V max
VOUT1 = 1.8V +/- 4%
Load = 10A maximum
Inductor Selection
Low inductor values result in smaller size but create high-
er ripple current. Higher inductor values will reduce the
ripple current but are larger and more costly. Because
wire resistance varies widely for different inductors and
because magnetic core losses vary widely with operating
conditions, it is often difficult to choose which inductor
will optimize efficiency. The general rule is that higher
inductor values have better efficiency at light loads due
to lower core losses and lower peak currents, but at high
load the smaller inductors are better because of lower re-
sistance. The inductor selection is generally based on the
ripple current which is typically set between 20% to 50%
of the maximum load current. Cost, size, output ripple
and efficiency all play a part in the selection process.
The first step is to select the switching frequency. In this
case VOUT1 will be used at a nominal 270kHz.
For 15V input and 1.8V output, the typical on-time should
be:
TONtyp = VOUT/VIN/Frequency
TONtyp = 444nsec.
The timing resistor RTON is selected to provide TONtyp:
RTON = (TONtyp – 35) × (VIN / (3.3 × VOUT) – 37
RTON = 976K.
We will use RTON = 1Meg.
Note that side2 will run typically 20% faster than side1, in
this case 320kHz.
During the DH on-time, voltage across the inductor is (VIN
- VOUT). To determine the inductance, the ripple current
must be defined. Smaller ripple current will give smaller
output ripple and but will lead to larger inductors. The
ripple current will also set the boundary for Psave opera-
tion. The switcher will typically enter Psave operation
when the load current decreases to of the ripple cur-
rent; (i.e. if ripple current is 4A then Psave operation will
typically start for loads less than 2A. If ripple current is
set at 40% of maximum load current, then Psave will oc-
cur for loads less than 20% of maximum current).
The equation for inductance is:
L = (VIN - VOUT) × TON / I
RIPPLE
Use the maximum value for VIN, and for TON use the val-
ue associated with maximum VIN, and that side’s TON us-
ing the RTON value selected. For selecting the inductor,
we start with the highest VOUT setting and a maximum
ripple current of 4A.
TON1 = 343 nsec at 20VIN, 1.8VOUT
L = (20V - 1.8V) × 343 nsec / 4A = 1.56μH
We will use 1.5μH which will slightly increase the maxi-
mum I
RIPPLE
to 4.2A.
Note: the inductor must be rated for the maximum DC
load current plus of the ripple current.
The minimum ripple current is also calculated. This oc-
curs when VIN is at the minimum value of 10V.
TON
VINMIN
= 3.3 × (RTON + 37) × (VOUT/VIN) + 35
TON
VINMIN
= 651nsec
I
RIPPLE
= (VIN - VOUT) × TON / L
I
RIPPLE_VINMIN
= (10 – 1.8) × 651nsec / 1.5μH = 3.55A