TK65015 データシートの表示（PDF） - Toko America Inc
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TK65015 Datasheet PDF : 8 Pages
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TK65015
cally, the ripple voltage cannot be made smooth enough by
this capacitor alone, so an output filter is used. In any case,
to minimize the dissipation required by the output filter, the
output capacitor should still be chosen with consideration
to smoothing the voltage ripple. This implies that its ESR
(equivalent series resistance) should be low. This usually
means choosing a larger size than the smallest available
for a given capacitance. To determine the peak ripple
voltage on the output capacitor for a single switching cycle,
multiply the ESR by the peak current which was calculated
in Eq. (4). ESR can be a strong function of temperature,
being worst case when cold. The capacitance should be
capable of integrating a current pulse with little ripple.
Typically, if a capacitor is chosen with reasonably low ESR
and if the capacitor is the right type of capacitor for the
application (typically aluminum electrolytic or tantalum),
then the capacitance will be sufficient.
Higher-Order Considerations
In practice, it may be that the peak current (calculated in
Eq. (4)) flowing out of the battery and into the converter will
cause a substantial input ripple voltage dropped across the
resistance inside the battery. This becomes a more likely
case for cold temperature (when battery series resistance
is higher), higher load rating converters (whose inductor’s
must draw higher peak currents), and when the battery is
undersized for the peak current application.
While the simple analysis used a parameter “VI” to
represent the converter input voltage in the equations, one
may not know what “VI” value to use if it is delivered by a
battery that allows high ripple to occur. For example,
assuming that the converter draws a peak current of
100mA for a 1V input, and assuming that the input is
powered by a AAA battery which might have a series
resistance of 2Ω at 0°C, then if the battery measures in at
1V without load, in the converter the battery voltage will sag
to about 0.8V during the on-time. This can cause two
problems: (1) with the effective input voltage to the con-
verter reduced in this way, the converter output current
capability will decrease, (2) if the same battery is powering
the TK65015 at the VIN pin (i.e., the normal case), then the
IC may become inoperable due to insufficient VIN. This is
why the application test circuit features an RC filter into the
VIN pin. The current draw is very small, so the voltage drop
across this filter resistor is negligible. The filter serves to
average out the input ripple caused by the battery resis-
tance.
A more power-efficient method comes at the price of a
large capacitor. This can be placed in parallel with the
battery to help channel the converter current pulses away
from the battery. The capacitor must have low ESR
compared to the battery resistance in order to accomplish
this effectively.
Still another solution is to filter the DC input with an RC
or LC filter. However, it is more likely that the filter will either
be too large or too lossy. It is of questionable benefit to
smooth the input if the DC loss through the filter is large.
Assuming that input ripple voltage at the battery terminal
and converter input is large, and that we filter the VIN pin of
the IC as in the test circuit, then the parameter “VI” in the
previous equations is not usable, and we will need to use
parameters to represent both the source voltage and the
source resistance.
The on-resistance of the TK65015’s internal switch is
about 1Ω maximum. Using the previously stated example
of 100mA peak current, the voltage drop across the switch
would reach 100mV during the on-time. This subtracts
from the voltage which is impressed across the inductor to
store energy during the on-time, so less energy is delivered
to the output during the off-time.
It is quite possible for the inductor winding resistance to
meet or exceed 1W, also. Voltage drop across the winding
resistance of the inductor also subtracts from the voltage
used to store energy in the core. So it also degrades
efficiency.
As the inductor delivers energy into the output capacitor
during the off-time, its current decays at a rate proportional
to the voltage drop across it. The idealized equations
assume that the voltage at the switching node is clamped
at a diode drop above the output voltage. However, the
ESR of the output capacitor can increase the voltage drop
across the inductor by the additional voltage dropped
across the ESR when the peak current flows in it. For
example, the voltage across a capacitor with an ESR of 2Ω
(not unusual at cold temperature) would jump by 200mV
when 100mA peak current began to flow in it. This extra
voltage drop would cause the inductor current to ramp
down more quickly, thus, depleting the available output
current.
Page 6
2-2-96
February, 1996 TOKO, Inc.
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