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## LM2575T-5.0 Datasheet(PDF) 15 Page - Texas Instruments

 Part No. LM2575T-5.0 Description LM1575/LM2575/LM2575HV SIMPLE SWITCHER 1A Step-Down Voltage Regulator Download 43 Pages Scroll/Zoom 100% Maker TI1 [Texas Instruments] Homepage http://www.ti.com Logo

## LM2575T-5.0 Datasheet(HTML) 15 Page - Texas Instruments

 15 / 43 page LM1575, LM2575-N, LM2575HVwww.ti.comSNVS106E – MAY 1999 – REVISED APRIL 2013PROCEDURE (Adjustable Output Voltage Versions)EXAMPLE (Adjustable Output Voltage Versions)Given:Given:VOUT = Regulated Output VoltageVOUT = 10VVIN(Max) = Maximum Input VoltageVIN(Max) = 25VILOAD(Max) = Maximum Load CurrentILOAD(Max) = 1AF = Switching Frequency (Fixed at 52 kHz)F = 52 kHz1. Programming Output Voltage (Selecting R1 and R2, as shown 1. Programming Output Voltage (Selecting R1 and R2)in Figure 25 and Figure 26)Use the following formula to select the appropriate resistor values.(1)(3)R1 can be between 1k and 5k. (For best temperature coefficient andR2 = 1k (8.13− 1) = 7.13k, closest 1% value is 7.15kstability with time, use 1% metal film resistors)(2)2. Inductor Selection (L1)2. Inductor Selection (L1)A. Calculate the inductor Volt • microsecond constant,A. Calculate E • T (V •μs)E • T (V •μs), from the following formula:(5)(4) B. E • T = 115 V • μsB. Use the E • T value from the previous formula and match it with C. ILOAD(Max) = 1Athe E • T number on the vertical axis of the Inductor ValueD. Inductance Region = H470Selection Guide shown in Figure 31.E. Inductor Value = 470μH Choose from AIE part #430-0634, PulseC. On the horizontal axis, select the maximum load current.Engineering part #PE-53118, or Renco part #RL-1961.D. Identify the inductance region intersected by the E • T value andthe maximum load current value, and note the inductor code for thatregion.E. Identify the inductor value from the inductor code, and select anappropriate inductor from the table shown in Table 2. Part numbersare listed for three inductor manufacturers. The inductor chosenmust be rated for operation at the LM2575 switching frequency (52kHz) and for a current rating of 1.15 × ILOAD. For additional inductorinformation, see INDUCTOR SELECTION.3. Output Capacitor Selection (COUT)3. Output Capacitor Selection (COUT)A. The value of the output capacitor together with the inductor A.defines the dominate pole-pair of the switching regulator loop. Forstableoperation,thecapacitormustsatisfythefollowing(7)requirement:However, for acceptable output ripple voltage selectCOUT ≥ 220 μF(6)COUT = 220 μF electrolytic capacitorThe above formula yields capacitor values between 10μF and 2000μF that will satisfy the loop requirements for stable operation. But toachieve an acceptable output ripple voltage, (approximately 1% ofthe output voltage) and transient response, the output capacitor mayneed to be several times larger than the above formula yields.B. The capacitor's voltage rating should be at last 1.5 times greaterthan the output voltage. For a 10V regulator, a rating of at least 15Vor more is recommended.Higher voltage electrolytic capacitors generally have lower ESRnumbers, and for this reason it may be necessary to select acapacitor rate for a higher voltage than would normally be needed.(Continued)(Continued)Copyright © 1999–2013, Texas Instruments IncorporatedSubmit Documentation Feedback15Product Folder Links: LM1575 LM2575-N LM2575HV