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SC419 Datasheet(PDF) 17 Page - Semtech Corporation |
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SC419 Datasheet(HTML) 17 Page - Semtech Corporation |
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17 / 24 page ![]() SC419 17 To find the minimum inductance needed, use the V IN and T ON values that correspond to VINMAX. H 77 . 0 A 5 ns 318 ) 05 . 1 2 . 13 ( L A slightly larger value of 0.88µH is selected. This will decrease the maximum I RIPPLE to 4.4A. 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 V IN conditions is also checked using the following equations. ns 384 ns 10 V V R pF 25 T INMIN OUT TON VINMIN _ ON L T ) V V ( I ON OUT IN RIPPLE A 25 . 4 H 088 ns 384 ) 05 . 1 8 . 10 ( I VINMIN _ RIPPLE 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 toler- ance. 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 500mV refer- ence tolerance is 1%. Allowing 1% tolerance from the FB resistor divider, this allows 2% tolerance due to V OUT ripple. Since this 2% error comes from 1/2 of the ripple voltage, the allowable ripple is 4%, or 42mV for a 1.05V output. The maximum ripple current of 4.4A creates a ripple voltage across the ESR. The maximum ESR value allowed is shown by the following equations. A 4 . 4 mV 42 I V ESR RIPPLEMAX RIPPLE MAX ESR MAX = 9.5 mΩ The output capacitance is chosen to meet transient requirements. A worst-case load release, from maximum load to no load at the exact moment when inductor current is at the peak, determines the required capaci- tance. 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. 2 OUT 2 PEAK 2 RIPPLEMAX OUT MIN V V I 2 1 I L COUT Assuming a peak voltage V PEAK of 1.150 (100mV rise upon load release), and a 10A load release, the required capaci- tance is shown by the next equation. 2 2 2 MIN 05 . 1 15 . 1 4 . 4 2 1 10 H 88 . 0 COUT COUT MIN = 595µF 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 500mV reference, the DL output is high and the low-side MOSFET is on. During this time, the voltage across the inductor is approximately -V OUT. This causes a down-slope or falling di/dt in the inductor. If the load di/dt is not 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 induc- tive energy that must be absorbed by the output capaci- tor, therefore a smaller capacitance can be used. The following five equations can be used to calculate the needed capacitance for a given dI LOAD/dt. Peak inductor current is shown by the next equation. I LPK = IMAX + 1/2 x IRIPPLEMAX I LPK = 10 + 1/2 x 4.4 = 12.2A dt dl Current Load of change of Rate LOAD Applications Information (continued) |
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