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LTC2913 датащи(PDF) 17 Page - Linear Technology |
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LTC2913 датащи(HTML) 17 Page - Linear Technology |
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17 / 24 page ![]() LT4363 17 4363fa APPLICATIONS INFORMATION highcurrentapplications.Noiseimmunitywillbeimproved significantly by locating resistive dividers close to the pins with short VCC and GND traces. Design Example As a design example, take an application with the follow- ing specifications: VCC = 8V to 14V DC with a transient of 150V and decay time constant ( τ) of 400ms, VOUT ≤ 27V, current limit (ILIM) at 5A, low battery detection of 6V, input overvoltage level at 60V, and 1ms of overvoltage early warning (Figure 8). Selection of SMAJ58A for D1 will limit the voltage at the VCC pin to less than 71V during 150V surge. The minimum required voltage at the VCC pin is 4V when VIN is at 8V; the supply current for LT4363 is 1.5mA. The maximum value for R7 to ensure proper operation is: R7 = 8V – 4V 1.5mA = 2.67kΩ Select 1kΩ for R7 to accommodate all conditions. The maximum current through R7 into D1 is then calcu- lated as: ID1= 150V – 64V 1k Ω = 86mA which is easily handled by the SMAJ58A for more than 500ms. With 0.1µF of bypass capacitance, C1, along with 1k of R7, high voltage transients up to 200V with a pulse width less than 10µs are filtered out at the VCC pin. Next, calculate the resistive divider value to limit VOUT to 27V during an overvoltage event: VREG = 1.275V • R1 +R2 ( ) R2 = 27V Set the current through R1 and R2 during the overvoltage condition to 250µA. R2 = 1.275V 250µA = 5kΩ Choose 4.99kΩ for R2. R1 = 27V – 1.275V ( )•R2 1.275V = 100.7kΩ The nearest standard value for R1 is 100kΩ. Next calculate the sense resistor, RSNS, value: RSNS = 50mV ILIM = 50mV 5A = 10mΩ CTMR is then chosen for 1ms of early warning time: CTMR = 1ms • 6µA 100mV = 60nF The nearest standard value for CTMR is 47nF. Finally, calculate R4, R5, and R6 for 6V low battery detec- tion and 60V input overvoltage level: 6V • R5 +R6 R4 +R5+R6 = 1.275V 60V • R6 R4 +R5+R6 = 1.275V Choose 10kΩ for R6. R4 +R5= 60V • 10k Ω 1.275V – 10k Ω = 460.6kΩ R5 = 1.275V • 460.6k Ω + 10kΩ 6V – 10k Ω = 90kΩ R4 = 460.6kΩ – 90kΩ = 370.6kΩ Select 90.9kΩ for R5 and 374kΩ for R4. The pass transistor, Q1, should be chosen to withstand a short-circuit with VCC = 14V. In the case of a severe output short where VOUT = 0V, the total overcurrent fault time is: tOC = 47nF • 0.875V 45.5µA = 0.904ms |
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