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SC3303 Datasheet(PDF) 17 Page - Semtech Corporation |
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SC3303 Datasheet(HTML) 17 Page - Semtech Corporation |
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17 / 26 page ![]() 17 SC3303 17 Applications Information (continued) The following values are used in this design example. V IN = 24V + 10% V OUT = 3.3V + 4% f SW = 300kHz Load = 3A maximum 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 500kHz 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. 2 . 1 V V 400 f 25pF 1 R OUT IN SW TON To select R TON , use the maximum value for V IN , and for T ON use the value associated with maximum V IN . SW INMAX OUT ON f V V t t ON = 417 ns at 26.4VIN, 3.3VOUT, 300kHz Substituting for R TON results in the following solution. R TON = 157kΩ, use RTON = 154kΩ 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 efficiency. Higher inductor values will reduce the ripple current/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 power- save operation. The switching will typically enter power- save mode when the load current decreases to 1/2 of the • • • • ripple current. For example, if ripple current is 2A then power-save operation will typically start for loads less than 1A. If ripple current is set at 40% of maximum load current, then power-save will start for loads less than 20% of maximum current. During the DH on-time, voltage across the inductor is (V IN - V OUT ). 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 33% of the maximum load current. Therefore ripple current will be 33% x 3A or 1A. To find the minimum inductance needed, use the V IN and T ON values that corre- spond to V INMAX . H A ns V V L 7 . 8 1 417 ) 3 . 3 4 . 26 ( A standard value of 10μH is selected. This gives a maximum I RIPPLE of 963mA. The peak ripple can be calculated by the equation, below where L TOL is assumed to be an inductor tolerance of 20%. I RIPPLE_PEAK = IRIPPLE_MAX / (1 - LTOL) = 1.203A Note that the inductor must be rated for the maximum DC load current plus 1/2 of the ripple current. I LSAT_MIN = IRIPPLE_PEAK X 0.5 + IOUT = 3.602A The ripple current under minimum V IN conditions is also checked using the following equations. L t ) V V ( I ON OUT IN RIPPLE mA H ns V V I VINMIN RIPPLE 763 10 417 ) 3 . 3 6 . 21 ( _ |
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