SC424 View Datasheet(PDF) - Semtech Corporation
Part Name
Description
MFG CO.
'SC424' PDF : 29 Pages View PDF
SC414/SC424
Applications Information (continued)
There are two values of load current to evaluate — con-
tinuous load current and peak load current. Continuous
load current relates to thermal stresses which drive the
selection of the inductor and input capacitors. Peak load
current determines instantaneous component stresses and
filtering requirements such as inductor saturation, output
capacitors, and design of the current limit circuit.
The following values are used in this design.
• V = 12V + 10%
• IN
V = 1V + 4%
• OUT
f = 250kHz
• SW
Load = 6A maximum
Frequency Selection
Selection of the switching frequency requires making a
trade-off between the size and cost of the external filter
components (inductor and output capacitor) and the
power conversion efficiency.
The desired switching frequency is 250kHz which results
from using components selected for optimum size and
cost .
A resistor (RTON) is used to program the on-time (indirectly
setting the frequency) using the following equation.
RTON
1
400: u VIN
25pF u fSW
VOUT
To select R , use the maximum value for V , and for T
TON
IN
ON
use the value associated with maximum V .
IN
TON
V OUT
VINMAXu f SW
TON = 303 ns at 13.2VIN, 1VOUT, 250kHz
Substituting for R results in the following solution.
TON
RTON = 130.9kΩ, use RTON = 130kΩ
Inductor Selection
In order to determine the inductance, the ripple current
must first be defined. Low inductor values result in smaller
size but create higher ripple current which can reduce
efficiency. Higher inductor values will reduce the ripple
current/voltage and for a given DC resistance are more
efficient. However, larger inductance translates directly
into larger packages and higher cost. Cost, size, output
ripple, and efficiency are all used in the selection process.
The ripple current will also set the boundary for power-
save operation. The switching will typically enter power-
save mode when the load current decreases to 1/2 of the
ripple current. For example, if ripple current is 4A then
Power-save operation will typically start for loads less than
2A. If ripple current is set at 40% of maximum load current,
then power-save will start for loads less than 20% of
maximum current.
The inductor value is typically selected to provide a ripple
current that is between 25% to 50% of the maximum load
current. This provides an optimal trade-off between cost,
efficiency, and transient performance.
During the DH on-time, voltage across the inductor is
(V - V ). The equation for determining inductance is
IN OUT
shown next.
L (VIN VOUT ) u TON
IRIPPLE
Example
In this example, the inductor ripple current is set equal to
50% of the maximum load current. Therefore ripple
current will be 50% x 6A or 3A. To find the minimum
inductance needed, use the V and T values that corre-
IN
ON
spond to V .
INMAX
L (13.2V 1V) u 318ns 1.26PH
3A
A slightly larger value of 1.5μH is selected. This will
decrease the maximum I to 2.53A.
RIPPLE
Note that the inductor must be rated for the maximum DC
load current plus 1/2 of the ripple current.
The ripple current under minimum V conditions is also
IN
checked using the following equations.
TON _ VINMIN
25pF u RTON u VOUT 10ns 311ns
VINMIN
22
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