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MCR03EZPFX1002 Datenblatt(PDF) 20 Page - International Rectifier |
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MCR03EZPFX1002 Datenblatt(HTML) 20 Page - International Rectifier |
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20 / 32 page ![]() Rev 16.0 20 PD-97660 IR3831WMPbF The transfer function (V e/Vo) is given by: The (s) indicates that the transfer function varies as a function of frequency. This configuration introduces a gain and zero, expressed by: First select the desired zero-crossover frequency (F o): Use the following equation to calculate R3: VOUT VREF R9 R8 C POLE C4 R3 Ve FZ F POLE E/A Zf Frequency Gain(dB) H(s) dB Fb Comp Z IN Fig. 13. Type II compensation network and its asymptotic gain plot (18) ..... ) ( 4 8 4 3 1 C sR C sR Z Z s H V V IN f o e + − = − = = () (20) ........ .......... .......... * * (19) ......... .......... .......... ......... 4 3 8 3 2 1 C R F R R s H z π = = () s ESR o F F F * 1/10 ~ 1/5 F and o ≤ > (21) ....... .......... .......... * * * * 2 8 3 LC in ESR o osc F V R F F V R = Where: V in = Maximum Input Voltage V osc = Oscillator Ramp Voltage F o = Crossover Frequency F ESR = Zero Frequency of the Output Capacitor F LC = Resonant Frequency of the Output Filter R 8 = Feedback Resistor To cancel one of the LC filter poles, place the zero before the LC filter resonant frequency pole: Use equations (20), (21) and (22) to calculate C4. One more capacitor is sometimes added in parallel with C4 and R3. This introduces one more pole which is mainly used to suppress the switching noise. The additional pole is given by: The pole sets to one half of the switching frequency which results in the capacitor C POLE: For a general solution for unconditional stability for any type of output capacitors, and a wide range of ESR values, we should implement local feedback with a type III compensation network. The typically used compensation network for voltage-mode controller is shown in figure 14. Again, the transfer function is given by: By replacing Z in and Zf according to figure 14, the transfer function can be expressed as: (22) ....... .......... .......... .......... * * . % o o z LC z C L F F F π 2 1 75 0 75 = = ...(23) .......... .......... .......... * * * POLE POLE P C C C C R F + = 4 4 3 2 1 π .(24) 1 1 1 3 4 3 . .......... .......... *F *R C *F *R C s s POLE π π ≅ − = IN f o e Z Z s H V V − = = ) ( () [ ] (25) .... ) ( * ) ( ) ( ) ( 7 10 3 4 3 4 3 3 4 8 10 8 7 4 3 1 1 1 1 C sR C C C C sR C C sR R R sC C sR s H + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + + + + + + − = |
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