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SC403B датащи(PDF) 23 Page - Semtech Corporation |
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SC403B датащи(HTML) 23 Page - Semtech Corporation |
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23 / 32 page ![]() 23 SC403B Applications Information (continued) Frequency Selection Selection of the switching frequency requires making a trade-off between the size and cost of the external filter components (inductor and output capacitor) and the power conversion efficiency. The desired switching frequency is 300kHz which results from using components selected for optimum size and cost. A resistor (R TON) is used to program the on-time (indirectly setting the frequency) using the following equation. OUT IN ON TON V pF 25 V ) ns 10 t( R To select R TON, use the maximum value for VIN, and for tON use the value associated with maximum V IN. SW INMAX OUT ON f V V t t ON = 379 ns at 13.2VIN, 1.5VOUT, 300kHz Substituting for R TON results in the following solution. R TON = 129.9kΩ, use RTON = 130kΩ Inductor Selection In order to determine the inductance, the ripple current must first be defined. Low inductor values result in smaller size but create higher ripple current which can reduce effi- ciency. Higher inductor values will reduce the ripple current and ripple voltage and for a given DC resistance are more efficient. However, larger inductance translates directly into larger packages and higher cost. Cost, size, output ripple, and efficiency are all used in the selection process. The ripple current will also set the boundary for PSAVE operation. The switching will typically enter PSAVE mode when the load current decreases to 1/2 of the ripple current. For example, if ripple current is 4A then PSAVE operation will typically start for loads less than 2A. If ripple current is set at 40% of maximum load current, then PSAVE will start for loads less than 20% of maximum current. The inductor value is typically selected to provide a ripple current that is between 25% to 50% of the maximum load current. This provides an optimal trade-off between cost, efficiency, and transient performance. During the DH on-time, voltage across the inductor is (V IN - VOUT). The equation for determining inductance is shown next. RIPPLE ON OUT IN I t ) V V ( L Example In this example, the inductor ripple current is set equal to 50% of the maximum load current. Therefore ripple current will be 50% x 6A or 3A. To find the minimum inductance needed, use the V IN and tON values that corre- spond to V INMAX. H 48 . 1 A 3 ns 379 ) 5 . 1 2 . 13 ( L A slightly larger value of 1.5µH is selected. This will decrease the typical I RIPPLE to 2.7A. 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 461 ns 10 V V R pF 25 t INMIN OUT TON VINMIN _ ON L t ) V V ( I ON OUT IN RIPPLE A 38 . 2 ) 2 . 0 1 ( H 5 . 1 ns 461 ) 5 . 1 8 . 10 ( I MIN _ RIPPLE A 7 . 3 ) 2 . 0 1 ( H 5 . 1 ns 379 ) 5 . 1 8 . 10 ( I MAX _ RIPPLE The value of L has been adjusted by +20% for the equa- tions above assuming an inductor tolerance of +20%. |
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