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TPS65320QPWPRQ1 Datasheet(PDF) 25 Page - Texas Instruments |
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TPS65320QPWPRQ1 Datasheet(HTML) 25 Page - Texas Instruments |
25 / 34 page ![]() SW 1 C8 R4 f = p ´ ´ out ESR C R C8 R4 ´ = P _ mod 1 C6 2 R4 f = p ´ ´ co out out ps ref ea 2 f C V R4 gm V gm æ ö æ ö p ´ ´ = ´ ç ÷ ç ÷ ç ÷ ´ è ø è ø SW co P _ mod f f f 2 = ´ co P _ mod Z _ mod f f f = ´ Z _ mod ESR out 1 f 2 R C = p ´ ´ m ax P _ mod L out out out I 1 f 2 R C 2 V C = = p ´ ´ p ´ ´ TPS65320-Q1 www.ti.com SLVSAY9A – DECEMBER 2012 – REVISED APRIL 2013 To get started, calculate the modulator pole, ƒP_mod, and the ESR zero, ƒz_mod using Equation 32 and Equation 33. For COUT, use a derated value of 40 μF. Use Equation 34 and Equation 35 to estimate a starting point for the crossover frequency, ƒco, to design the compensation. For the example design, ƒP_mod is 2.39 kHz and ƒZ_mod is 1.33 MHz. Equation 34 is the geometric mean of the modulator pole and the ESR zero and Equation 35 is the mean of the modulator pole and the switching frequency. Equation 34 yields 56.4 kHz and Equation 35 gives 51.3 kHz. Use the lower value of Equation 34 or Equation 35 for an initial crossover frequency. For this example, the target ƒco is 51.3 kHz. Next, calculate the compensation components. Use a resistor in series with a capacitor to create a compensating zero. A capacitor in parallel to these two components forms the compensating pole. (32) (33) (34) (35) The total loop gain, which consists of the product of the modulator gain, the feedback voltage-divider gain, and the error amplifier gain at ƒco should be equal to 1. One can juse Equation 36 to determine the compensation resistor, R4 (see schematics in Figure 23). Assume the power-stage transconductance, gmps, is 10.5 S. The output voltage, VOUT, reference voltage, VREF, and amplifier transconductance, gmea, are 5 V, 0.8 V and 310 μS, respectively. The calculated value for R4 is 24.74 k Ω; use 27 kΩ in this design. Use Equation 37 to set the compensation zero to the modulator pole frequency. Equation 35 yields 2468 pF for compensating capacitor C6 (see schematics in Figure 23); use 2700 pF for this design. (36) (37) One can implement a compensation pole if desired using an additional capacitor C8 in parallel with the series combination of R4 and C6. Use the larger value of Equation 38 and Equation 39 to calculate the C8, to set the compensation pole. Type 2B compensation does not use C8. This would demand a low ESR of the output capacitor. (38) (39) LDO Depending upon an end application, one may use different values of external components. In order to program the output voltage, select the feedback resistors (R5 and R6) carefully. Using smaller resistors results in higher current consumption, whereas using very large resistors impacts the sensitivity of the regulator. TI therefore recommends selecting feedback resistors such that the sum of R5 and R6 is between 20 k Ω and 200 kΩ. If the desired regulated output voltage is 3.3 V on selecting R6, one can calculate R5. Knowing VREF = 0.8 V (typical), VOUT = 3.3 V, and selecting R6 = 20 kΩ, the calculated value of R5 is 62 kΩ. There may be a requirement for a larger output capacitor during fast load steps to prevent the output from temporarily dropping down. TI recommends a low-ESR ceramic capacitor with dielectric of type X5R or X7R. Additionally, one can connect a bypass capacitor at the output to decouple high-frequency noise, per the end application. Copyright © 2012–2013, Texas Instruments Incorporated Submit Documentation Feedback 25 Product Folder Links: TPS65320-Q1 |
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