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ISL6752AAZA Datasheet(PDF) 11 Page - Intersil Corporation |
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ISL6752AAZA Datasheet(HTML) 11 Page - Intersil Corporation |
11 / 16 page 11 FN9181.3 October 31, 2008 Since the peak current limit threshold is 1.00V, the total current feedback signal plus the external ramp voltage must sum to this value. Substituting Equations 13 and 14 into Equation 15 and solving for RCS yields Equation 16: For simplicity, idealized components have been used for this discussion, but the effect of magnetizing inductance must be considered when determining the amount of external ramp to add. Magnetizing inductance provides a degree of slope compensation to the current feedback signal and reduces the amount of external ramp required. The magnetizing inductance adds primary current in excess of what is reflected from the inductor current in the secondary. where VIN is the input voltage that corresponds to the duty cycle D and Lm is the primary magnetizing inductance. The effect of the magnetizing current at the current sense resistor, RCS, is expressed in Equation 18: If ΔVCS is greater than or equal to Ve, then no additional slope compensation is needed and RCS becomes Equation 19: If ΔVCS is less than Ve, then Equation 16 is still valid for the value of RCS, but the amount of slope compensation added by the external ramp must be reduced by ΔVCS. Adding slope compensation may be accomplished in the ISL6752 using the CTBUF signal. CTBUF is an amplified representation of the sawtooth signal that appears on the CT pin. It is offset from ground by 0.4V and is 2x the peak-to-peak amplitude of CT (0.4V to 4.4V). A typical application sums this 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 placed on the CS pin, the value of R9 required to add the appropriate external ramp can be found by superposition. Rearranging to solve for R9 yields Equation 21: The value of RCS determined in Equation 16 must be rescaled so that the current sense signal presented at the CS pin is that predicted by Equation 14. The divider created by R6 and R9 makes this necessary. Example: VIN = 280V VO = 12V LO = 2.0µH Np/Ns = 20 Lm = 2mH IO = 55A Oscillator Frequency, FSW = 400kHz Duty Cycle, D = 85.7% NCT = 50 R6 = 499 Ω Solve for the current sense resistor, RCS, using Equation 16. RCS = 15.1Ω. V e V CS + 1 = (EQ. 15) R CS N P NCT ⋅ N S ------------------------ 1 I O V O L O -------- t SW 1 π --- D 2 ---- + ⎝⎠ ⎛⎞ + ---------------------------------------------------- ⋅ = Ω (EQ. 16) I P Δ V IN DtSW ⋅ L m ----------------------------- = A (EQ. 17) ΔV CS ΔI P RCS ⋅ N CT -------------------------- = V (EQ. 18) R CS N CT N S N P -------- I O Dt SW 2L O --------------- V IN N S N P -------- ⋅ V O – ⎝⎠ ⎜⎟ ⎛⎞ ⋅ + ⎝⎠ ⎜⎟ ⎛⎞ ⋅ V IN DtSW ⋅ L m ----------------------------- + ---------------------------------------------------------------------------------------------------------------------------------- = (EQ. 19) FIGURE 6. ADDING SLOPE COMPENSATION R6 C4 R9 CTBUF CS 1 2 4 3 5 6 7 89 10 11 12 13 14 15 16 RCS ISL6752 V e ΔV CS – DV CTBUF 0.4 – () 0.4 + () R6 ⋅ R6 R9 + ------------------------------------------------------------------------------- = V (EQ. 20) R9 DV CTBUF 0.4 – () V e ΔV CS 0.4 ++ – () R6 ⋅ V e ΔV CS – ------------------------------------------------------------------------------------------------------------------- = Ω (EQ. 21) R ′ CS R6 R9 + R9 ---------------------- R CS ⋅ = (EQ. 22) ISL6752 |
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