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SC104 датащи(PDF) 6 Page - Semtech Corporation |
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SC104 датащи(HTML) 6 Page - Semtech Corporation |
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6 / 12 page ![]() 6 2005 Semtech Corp. www.semtech.com SC104 POWER MANAGEMENT Applications Information Component Selection - Introduction Referring to the 6 LED typical schematic below, there are three components that depend upon the application that need to be determined: RSET - this resistor sets the output current for the device RLIM - this resistor sets the peak inductor current L - the output inductor All the other components can be mostly generalized and are addressed below the following design steps. ENABLE RSET 23.2R LED2 D1 IOUT = 15mA U1 SC104 1 2 3 4 5 6 7 8 ADJ FB GND OUT LX IN LIM EN LED1 LED3 VIN = 3V to 5V LED5 RLIM 7.50k LED4 CIN 4.7uF IOUT ADJUST L1 12uH LED6 COUT 0.47uF Step 1: Continuous or Discontinuous? The first thing to do when designing with the SC104 is to determine whether the output inductor will be operating in continuous mode (where the inductor current does not drop to zero while the device is switching) or discontinuous mode (where the inductor current drops to zero while switching). This determination can be made simply by calculating the required duty cycle needed for the target output voltage, and comparing it to the guaranteed minimum value for the maximum duty cycle from the Electrical Characteristics on Page 3. % DC(MIN) = 70% (or 0.7 duty). If DC is greater than 0.7 then discontinuous mode is required. The required duty cycle is calculated as follows: () () f ) SAT ( CE OUT f IN OUT V V V V V V DC + − + − = Where: V OUT = output voltage, the sum of the total LED (max.) forward voltage drop at the required output voltage plus the feedback voltage, 0.35V. V IN = minimum input voltage V f = Schottky diode (D1) forward voltage drop V CE(SAT) = power switch saturation voltage Using the 6 LED example above: V OUT = (6 * 3.475) + 0.35 = 21.2V V IN = 3V V f = 0.35V V CE(SAT) = 0.25V thus DC = 0.87 Since this value is greater than the guaranteed minimum value for maximum duty cycle, the device will be operating in discontinuous mode to provide the desired output. Note that the duty cycle does not depend upon the output current, and that unless the output to input ratio is low, the device will usually need to be in discontinuous mode, so we will cover that first (Step 1 through Step 5). Continuous mode calculations start at Step 6. Step 2: Calculating the Inductor for Discontinuous Mode Having determined that we need to be operating in discontinuous mode, we next need to calculate the maximum inductor value allowed that will permit the part to output the correct power. The maximum discontinuous inductor value, L (D) is given by: () () () () f IN OUT f ) SAT ( CE OUT ) MIN ( OFF ) MIN ( ON OUT OUT ) SAT ( CE IN IN 2 ) MIN ( ON ) D ( V V V V V V t t I V 4 . 1 2 V V V t L + − + − • + • • • • − • • = Where: t ON(MIN) = minimum switch on-time = 1.8µs I OUT = required output current t OFF(MIN) = minimum switch off-time = 0.6µs Using our 6 LED example: I OUT = 15mA thus L (D) = 14.4µH Selecting the next lower standard value gives us L (D) = 12µH. Of course a lower value inductor may be used if desired, but may not necessarily be the most efficient choice. Step 3: Calculating the Current Limit Required with this Inductor for Discontinuous Mode Having determined the inductor value we are going to use, we next need to calculate the current limit required to meet the necessary output power. The discontinuous mode current limit, I LIM(D), is given by: ( ) ) D ( ) MIN ( ON ) SAT ( CE IN ) D ( LIM L t V V I • − = |
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