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TK65015MTL Datasheet(PDF) 6 Page - TOKO, Inc |
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TK65015MTL Datasheet(HTML) 6 Page - TOKO, Inc |
6 / 8 page Page 6 2-2-96 February, 1996 TOKO, Inc. 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 “V I” to represent the converter input voltage in the equations, one may not know what “V I” 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 2 Ω 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 V IN pin (i.e., the normal case), then the IC may become inoperable due to insufficient V IN. This is why the application test circuit features an RC filter into the V IN 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 “V I” 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 1 Ω 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 2 Ω (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. |
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