House keeping loss is:
Finally the total power loss in the LM26400Y is:
Choose 1% resistors. R2 = 5.90kΩ.
PROGRAMMING OUTPUT VOLTAGE
INDUCTOR SELECTION
First make sure the required maximum duty cycle in steady
state is less than 80% so that the regulator will not lose reg-
ulation. The datasheet lower limit for maximum duty cycle is
about 90% over temperature (see Electrical Characteristics
table for the accurate value). The maximum duty cycle in
steady state happens at low line and full load.
An inductance value that gives a peak-to-peak ripple current
of 0.4A to 0.8A is recommended. Too large a ripple current
can reduce the maximum achievable DC load current be-
cause the peak current of the switch is limited to a typical of
3A. Too small a ripple current can cause the regulator to os-
cillate due to the lack of inductor current ramp signal, espe-
cially under high input voltages. Use the following equation to
determine inductance:
The output voltage is programmed through the feedback re-
sistors R1 and R2, as illustrated in Figure 6.
where VIN_MAX is the maximum input voltage of the applica-
tion.
The rated current of the inductor should be higher than the
maximum DC load current. Generally speaking, the lower the
DC resistance of the inductor winding, the higher the overall
regulator efficiency.
20200258
Ferrite core inductors are recommended for less AC loss and
less fringing magnetic flux. The drawback of ferrite core in-
ductors is their quick saturation characteristic. Once the in-
ductor gets saturated, its current can spike up very quickly if
the switch is not turned off immediately. The current limit cir-
cuit has a propagation delay and so is oftentimes not fast
enough to stop the saturated inductor from going above the
current limit. This has the potential to damage the internal
switch. So to prevent a ferrite core inductor from getting into
saturation, the inductor saturation current rating should be
higher than the switch current limit ICL. The LM26400Y is quite
robust in handling short pulses of current that is a few amps
above the current limit. When a compromise has to be made,
pick an inductor with a saturation current just above the lower
limit of the ICL. Be sure to validate the short-circuit protection
over the intended temperature range.
FIGURE 6. Programming Output Voltage
It is recommended that the lower feedback resistor R2 always
be 5.9kΩ. This simplifies the selection of the CFF value (For
an explanation of CFF, please refer to the section LOAD STEP
RESPONSE). The 5.9kΩ is also a suitable R2 value in appli-
cations that need to increase the output voltage on the fly by
paralleling another resistor with R2. Since the FB pin is 0.6V
during normal operation, the current through the feedback re-
sistors is normally 0.6V / 5.9kΩ = 0.1mA and the power
dissipation in R2 is 0.6V x 0.6V / 5.9kΩ = 61µW - low enough
for 0402 size or smaller resistors.
Use the following equation to determine the upper feedback
resistor R1.
To prevent the inductor from saturating over the entire -40°C
to 125°C range, pick one with a saturation current higher than
the upper limit of ICL in the Electrical Characteristics table.
Inductor saturation current is usually lower when hot. So con-
sult the inductor vendor if the saturation current rating is only
specified at room temperature.
To determine the maximum allowed resistor tolerance, use
the following equation:
Soft saturation inductors such as the iron powder types can
also be used. Such inductors do not saturate suddenly and
therefore are safer when there is a severe overload or even
shorted output. Their physical sizes are usually smaller than
the Ferrite core inductors. The downside is their fringing flux
and higher power dissipation due to relatively high AC loss,
especially at high frequencies.
where TOL is the set point accuracy of the regulator, Φ is the
tolerance of VFB
.
Example:
Example:
VOUT = 1.2V; VIN = 9V to 14V; IOUT = 2A max; Peak-to-peak
VOUT = 1.2V, with a set point accuracy of +/-3.5%.
Ripple Current ΔI = 0.6A.
15
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