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SC9301 Datasheet(PDF) 20 Page - Semtech Corporation |
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SC9301 Datasheet(HTML) 20 Page - Semtech Corporation |
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20 / 29 page ![]() SC9301 20 Applications Information (continued) SW INMAX OUT ON f V V T T ON = 379 ns at 13.2VIN, 1.5VOUT, 300kHz H 27 . 1 A 5 . 3 ns 379 ) 5 . 1 2 . 13 ( L A slightly smaller value of 1.2μH is selected. This will increase the maximum I RIPPLE to 3.7A. 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 451 V V R pF 25 T INMIN OUT TON VINMIN _ ON L T ) V V ( I ON OUT IN RIPPLE A 5 . 3 H 2 . 1 ns 451 ) 5 . 1 8 . 10 ( I VINMIN _ RIPPLE Capacitor Selection The output capacitors are chosen based upon required ESR and capacitance. The maximum ESR requirement is controlled by the output ripple requirement and the DC tolerance. 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. A change in the output ripple voltage will lead to a change in DC voltage at the output. The design goal for output voltage ripple is 2% of 1.5V or 30mV. The maximum ESR value allowed is shown by the following equations. A 7 . 3 mV 30 I V ESR RIPPLEMAX RIPPLE MAX ESR MAX = 8.1 mΩ The output capacitance is usually chosen to meet tran- sient 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 capacitance. 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 1.65V (150mV rise upon load release), and a 10A load release, the required capaci- tance is shown by the next equation. F 357 2 5 . 1 2 65 . 1 2 7 . 3 2 1 10 H 2 . 1 MIN COUT During the load release time, the voltage cross the induc- tor is approximately -V OUT . This causes a down-slope or falling di/dt in the inductor. If the load di/dt is not much faster than the di/dt of the inductor, then the inductor current will tend to track the falling load current. This will reduce the excess inductive energy that must be absorbed by the output capacitor, therefore a smaller capacitance can be used. The following can be used to calculate the needed capaci- tance 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 3.7 = 11.85A dt dl Current Load of change of Rate LOAD I MAX = maximum load release = 10A OUT PK LOAD MAX OUT LPK LPK OUT V V 2 dt dl I V I L I C |
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