SC173AMLTRT View Datasheet(PDF) - Semtech Corporation
Part Name
Description
MFG CO.
SC173AMLTRT
Semtech Corporation
'SC173AMLTRT' PDF : 26 Pages View PDF
SC173A
Applications Information (continued)
Example
In this example, the inductor ripple current is set equal
to 30% of the maximum load current. Therefore ripple
current will be 30% x 3A or 0.9A. To find the minimum
inductance needed, use the VIN and TON values that cor-
respond to VINMAX.
TON_VINMAX
=
1V
5.5V ⋅ 800kHz
=
227ns
L = (5.5V -1V) • 227ns = 1.14mH
0.9A
A larger value of 2µH is selected. This will decrease the
maximum IRIPPLE to 0.511A.
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 VIN conditions is also
checked using the following equations.
TON_VINMIN
=
4.5V
1V
× 800kHz
=
277ns
IRIPPLE
=
(VIN
-
VOUT) ×
L
TON
IRIPPLE_VINMIN
=
(4.5V - 1V)×
2mH
277ns
=
0.485A
Capacitor Selection
The output capacitors are chosen based on required ESR
and capacitance. The maximum ESR requirement is con-
trolled by the output ripple requirement and the DC tol-
erance. The output voltage has a DC value that is equal
to the valley of the output ripple plus 1/2 of the peak-
to-peak ripple. Change in the output ripple voltage will
lead to a change in DC voltage at the output.
The design goal is for the output voltage regulation to be
±4% under static conditions. The internal 750mV refer-
ence tolerance is 1%. Assuming a 1% tolerance from the
FB resistor divider, this allows 2% tolerance due to VOUT
ripple. Since this 2% error comes from 1/2 of the ripple
voltage, the allowable ripple is 4%, or 40mV for a 1V out-
put.
© 2010 Semtech Corporation
The maximum ripple current of 0.511A creates a ripple
voltage across the ESR. The maximum ESR value allowed
is shown by the following equations.
ESRMAX
= VRIPPLE
IRIPPLEMAX
= 40mV
0.51A
ESRMAX = 78.3 mΩ
The output capacitance is chosen to meet transient re-
quirements. A worst-case load release, from maximum
load to no load at the exact moment when inductor cur-
rent is at the peak, determines the required capacitance.
If the load release is instantaneous (load changes from
maximum to zero in < 1µs), the output capacitor must
absorb all the inductor’s stored energy. This will cause a
peak voltage on the capacitor according to the following
equation.
COUTMIN
=
L
× (IOUT
+
1
2
(VPEAK )2
× IRIPPLEMAX )2
- (VOUT )2
Assuming a peak voltage VPEAK of 1.050V (50mV rise upon
load release), and a 3A load release, the required capaci-
tance is shown by the next equation.
2mH × (3A + 1 × 0.511A) 2
COUT MIN =
2
(1.05V) 2 - (1.0V) 2
= 207 mF
If the load release is relatively slow, the output capacitance
can be reduced. At heavy loads during normal switching,
when the FB pin is above the 750mV reference, the DL
output is high and the low-side MOSFET is on. During this
time, the voltage across the inductor is approximately
-VOUT. This causes a down-slope or falling di/dt in the
inductor. If the load di/dt is not much faster than the
-di/dt in the inductor, then the inductor current will tend
to track the falling load current. This will reduce the
excess inductive energy that must be absorbed by the
output capacitor, therefore a smaller capacitance can be
used.
The following can be used to calculate the needed ca-
pacitance for a given dILOAD/dt. Peak inductor current is
shown by the next equation.
15
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