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LTC3548 Datasheet(PDF) 13 Page - Linear Technology |
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LTC3548 Datasheet(HTML) 13 Page - Linear Technology |
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13 / 30 page ![]() LT3510 13 3510fe maximum duty cycle. The duty cycle is the fraction of time that the internal switch is on during a clock cycle. Unlike most fixed frequency regulators, the LT3510 will not switch off at the end of each clock cycle if there is sufficient volt- age across the boost capacitor (C3 in Figure 1) to fully saturate the output switch. Forced switch off for a minimum time will only occur at the end of a clock cycle when the boost capacitor needs to be recharged. This operation has the same effect as lowering the clock frequency for a fixed off time, resulting in a higher duty cycle and lower minimum input voltage. The resultant duty cycle depends on the charging times of the boost capacitor and can be approximated by the following equation: DCMAX = 1 1 + 1 B where B is 2A divided by the typical boost current from the Electrical Characteristics. This leads to a minimum input voltage of: VIN(MIN) = VOUT + VD DCMAX –VD + VSW where VSW is the voltage drop of the internal switch. Figure 3 shows a typical graph of minimum input voltage vs load current for the 3.3V and 1.8V application on the first page of this data sheet. The maximum input voltage is determined by the absolute maximum ratings of the VIN and BST pins and by the frequency and minimum duty cycle. The minimum duty cycle is defined as : DCMIN = tON(MIN) • Frequency Maximum input voltage as: VIN(MAX) = VOUT + VD DCMIN –VD + VSW Note that the LT3510 will regulate if the input voltage is taken above the calculated maximum voltage as long as maximum ratings of the VIN and BST pins are not violated. However operation in this region of input voltage will exhibit pulse skipping behavior. Example: VOUT = 3.3V, IOUT = 1A, Frequency = 1MHz, Temperature = 25°C VSW = 0.1V, B = 40 (from boost characteristics specifica- tion), VD = 0.4V, tON(MIN) = 200ns DCMAX = 1 1 + 1 40 = 98% VIN(MIN) = 3.3 + 0.4 0.98 – 0.4 + 0.1= 3.48V DCMIN = tMIN(ON) •f = 0.200 VIN(MAX) = 3.3 + 0.4 0.200 – 0.4 + 0.1= 18.2V Inductor Selection and Maximum Output Current A good first choice for the inductor value is: L = VIN –VOUT ()•VOUT VIN •f where f is frequency in MHz and L is in μH. With this value the maximum load current will be ~2A, independent of input voltage. The inductor’s RMS current Figure 3. Minimum Input Voltage vs Load Current CURRENT (mA) 1 3.0 3.5 4.0 4.5 5.0 5.5 6.0 10 100 1000 10000 3510 F03 VOUT = 3.3V START-UP RUNNING APPLICATIONS INFORMATION |
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