RT6218A/B
Application Information
inductor current (and the output current) while ensuring
that IL(PEAK) does not exceed the upper current limit
level.
Inductor Selection
Selecting an inductor involves specifying its inductance
and also its required peak current. The exact inductor
value is generally flexible and is ultimately chosen to
obtain the best mix of cost, physical size, and circuit
efficiency. Lower inductor values benefit from reduced
size and cost and they can improve the circuit's
transient response, but they increase the inductor
ripple current and output voltage ripple and reduce the
efficiency due to the resulting higher peak currents.
Conversely, higher inductor values increase efficiency,
but the inductor will either be physically larger or have
higher resistance since more turns of wire are required
and transient response will be slower since more time
is required to change current (up or down) in the
inductor. A good compromise between size, efficiency,
and transient response is to use a ripple current (IL)
about 20% to 50% of the desired full output load
current. Calculate the approximate inductor value by
selecting the input and output voltages, the switching
frequency (fSW), the maximum output current
(IOUT(MAX)) and estimating a IL as some percentage of
that current.
For best efficiency, choose an inductor with a low DC
resistance that meets the cost and size requirements.
For low inductor core losses some type of ferrite core is
usually best and a shielded core type, although
possibly larger or more expensive, will probably give
fewer EMI and other noise problems.
Considering the Typical Operating Circuit for 1.2V
output at 2A and an input voltage of 12V, using an
inductor ripple of 0.6A (30%), the calculated inductance
value is :
1.2 121.2
12650kHz0.6A
L
2.77μH
The ripple current was selected at 0.6A and, as long as
we use the calculated 2.77H inductance, that should
be the actual ripple current amount. The ripple current
and required peak current as below :
1.2 121.2
12650kHz2.77μH
I =
= 0.6A
= 2.3A
L
0.6A
and I
= 2A +
L(PEAK)
2
VOUT VIN VOUT
L =
V fSW IL
IN
For the 2.77H value, the inductor's saturation and
thermal rating should exceed 2.3A. Since the actual
value used was 2.77H and the ripple current exactly
0.6A, the required peak current is 2.3A.
Once an inductor value is chosen, the ripple current
(IL) is calculated to determine the required peak
inductor current.
VOUT VIN VOUT
and IL(PEAK) = IOUT(MAX)
IL
2
Input Capacitor Selection
IL=
V fSW L
IN
Input capacitors are needed to smooth out the RMS
ripple current (IRMS) imposed by the switching currents
and drawn from the input power source, by reducing
the ripple voltage amplitude seen at the input of the
converters. The voltage rating of the input filter
capacitors must be greater than the maximum input
voltage. It’s also important to consider the ripple current
capabilities of capacitors.
To guarantee the required output current, the inductor
needs a saturation current rating and a thermal rating
that exceeds IL(PEAK). These are minimum requirements.
To maintain control of inductor current in overload and
short circuit conditions, some applications may desire
current ratings up to the current limit value. However,
the IC's output under-voltage shutdown feature make
this unnecessary for most applications.
The RMS ripple current (IRMS) of the regulator can be
determined by the input voltage (VIN), output voltage
(VOUT), and rated output current (IOUT) as the following
equation :
IL(PEAK) should not exceed the minimum value of IC's
upper current limit level or the IC may not be able to
meet the desired output current. If needed, reduce the
inductor ripple current (IL) to increase the average
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12
DS6218A/B-01 October 2018
RICHTEK [ RICHTEK TECHNOLOGY CORPORATION ]
