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ISL78845ASEHVF Datasheet(PDF) 10 Page - Intersil Corporation |
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ISL78845ASEHVF Datasheet(HTML) 10 Page - Intersil Corporation |
10 / 15 page ISL78840ASEH, ISL78841ASEH, ISL78843ASEH, ISL78845ASEH 10 FN7952.0 May 29, 2012 For a flyback converter, Vn can be solved in terms of input voltage, current transducer components, and primary inductance, yielding Equation 12: where RCS is the current sense resistor, Tsw is the switching period, Lp is the primary inductance, VIN is the minimum input voltage, and D is the maximum duty cycle. The current sense signal at the end of the ON time for CCM operation is Equation 13: where VCS is the voltage across the current sense resistor, Ls is the secondary winding inductance, and IO is the output current at current limit. Equation 13 assumes the voltage drop across the output rectifier is negligible. Since the peak current limit threshold is 1.00V, the total current feedback signal plus the external ramp voltage must sum to this value when the output load is at the current limit threshold as: shown in Equation 14. Substituting Equations 12 and 13 into Equation 14 and solving for RCS yields Equation 15: Adding slope compensation is accomplished in the ISL7884xASEH using an external buffer transistor and the RTCT signal. A typical application sums the buffered RTCT signal with the current sense feedback and applies the result to the CS pin as shown in Figure 6. Assuming the designer has selected values for the RC filter (R6 and C4) placed on the CS pin, the value of R9 required to add the appropriate external ramp can be found by superposition. The factor of 2.05 in Equation 16 arises from the peak amplitude of the sawtooth waveform on RTCT minus a base-emitter junction drop. That voltage multiplied by the maximum duty cycle is the voltage source for the slope compensation. Rearranging to solve for R9 yields Equation 17: The value of RCS determined in Equation 15 must be rescaled so that the current sense signal presented at the CS pin is that predicted by Equation 13. The divider created by R6 and R9 makes this necessary. Example: VIN = 12V VO = 48V Ls = 800µH Ns/Np = 10 Lp = 8.0µH IO = 200mA Switching Frequency, fsw = 200kHz Duty Cycle, D = 28.6% R6 = 499Ω Solve for the current sense resistor, RCS, using Equation 15. RCS = 295mΩ Determine the amount of voltage, Ve, that must be added to the current feedback signal using Equation 12. Ve = 92.4mV Using Equation 17, solve for the summing resistor, R9, from CT to CS. R9 = 2.67kΩ Determine the new value of RCS (R’CS) using Equation 18. R’CS = 350mΩ Additional slope compensation may be considered for design margin. The above discussion determines the minimum external ramp that is required. The buffer transistor used to create the external ramp from RTCT should have a sufficiently high gain (>200) so as to minimize the required base current. Whatever base current is required reduces the charging current into RTCT and will reduce the oscillator frequency. V e DT ⋅ SW VIN RCS ⋅⋅ L p ---------------------------------------------------- 1 π --- 0.5 + ⎝⎠ ⎛⎞ 1 1D – ------------- 1 – ⎝⎠ ⎛⎞ = V (EQ. 12) V CS N S RCS ⋅ N P ------------------------ I O 1D – () V O T ⋅⋅ sw 2L s ---------------------------------------------- + ⎝⎠ ⎜⎟ ⎛⎞ = V (EQ. 13) V e V CS + 1V = (EQ. 14) R CS 1 DT sw VIN ⋅⋅ L p --------------------------------- 1 π --- 0.5 + 1D – ------------------ 1 – ⎝⎠ ⎜⎟ ⎜⎟ ⎛⎞ ⋅ N s N p ------- I O 1D – () V O Tsw ⋅⋅ 2L s ---------------------------------------------- + ⎝⎠ ⎜⎟ ⎛⎞ ⋅ + --------------------------------------------------------------------------------------------------------------------------------------------------------- = (EQ. 15) CS RTCT R6 C4 R9 VREF FIGURE 6. SLOPE COMPENSATION V e 2.05D R 6 ⋅ R 6 R 9 + ---------------------------- = V (EQ. 16) R 9 2.05D V e – () R 6 ⋅ V e ---------------------------------------------- = Ω (EQ. 17) R ′ CS R 6 R 9 + R 9 --------------------- R CS ⋅ = (EQ. 18) |
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