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MC33364D1R2 bảng dữ liệu(PDF) 11 Page - ON Semiconductor |
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MC33364D1R2 bảng dữ liệu(HTML) 11 Page - ON Semiconductor |
11 / 16 page MC33364 http://onsemi.com 11 The MOC8102 has a typical current transfer ratio (CTR) of 100% with 25% tolerance. When the TL431 is full--on, 5 mA will be drawn from the transistor within the MOC8102. The transistor should be in saturated state at that time, so its collector resistor must be Rcollector = Vref − Vsat ILED = 5.0 V − 0.3 V 5.0 mA = 940 Ω Since a resistor of 5.0 k is internally connected from the reference voltage to the feedback pin of the MC33364, the external resistor can have a higher value Rext = R3 = (Rint)(Rcollector) (Rint) − (Rcollector) = (5.0 k)(940) 5.0 k − 940 = 1157 Ω ≈ 1200 Ω This completes the design of the voltage feedback circuit. In no load condition there is only a current flowing through the optoisolator diode and the voltage sense divider on the secondary side. The load at that condition is given by: Rnoload = Vout (ILED + Idiv) = 6.0 V (5.0 mA + 0.25 mA) = 1143 Ω The output filter pole at no load is: fph = 1 (2π Rnoload Cout ) = 1 (2π)(1143)(300 mF) = 0.46 Hz In heavy load condition the ILED and Idiv is negligible. The heavy load resistance is given by: Rheavy = Vout Iout = 6.0 V 2.0 A = 3.0 Ω The output filter pole at heavy load of this output is fph = 1 (2π RheavyCout) = 1 (2π)(3)(300 mF) = 177 Hz The gain exhibited by the open loop power supply at the high input voltage will be: A = V in max − Vout 2 Ns (V in max)(Verror)(Np) = 382 V − 6.0 V 2 (7) (382 V)(1.2 V)(139) = 15.53 = 23.82 dB The maximum recommended bandwidth is approximately: fc = fs min 5 = 70 kHz 5 = 14 kHz The gain needed by the error amplifier to achieve this bandwidth is calculated at the rated load because that yields the bandwidth condition, which is: Gc = 20 log fc fph −A=20log14kHz 177 −23.82dB = 14.14 dB The gain in absolute terms is: Ac = 10(Gc∕20) = 10(14.14∕20) = 5.1 Now the compensation circuit elements can be calculated. The output resistance of the voltage sense divider is given by the parallel combination of resistors in the divider: Rin = Rupper || Rlower = 10 k || 14 k = 5833 Ω R9 = (Ac) (Rin) = 29.75 k ≈ 30 k C8 = 1 2π (Ac)(Rin)(fc) = 382 pF ≈ 390 pF The compensation zero must be placed at or below the light load filter pole: C7 = 1 2π (R9) (fpn) = 11.63 mF ≈ 10 mF |
Số phần tương tự - MC33364D1R2 |
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Mô tả tương tự - MC33364D1R2 |
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