4KW PFC相关电容电感计算1.输入电容计算参阅IR1153应用规格书2000W PFC计算如下:誓 High Frequency Input Capacitor (C()The purpose of the high-frequency capacitor is to supply thqhigh-frequency component of the inductor current (the ripple component) via the shortest possible loop. This has the advaof acting like an EMI filter,itminimizes the high-frequency current nequirennent from the AC line. Typically a high-frequency, film type capacitor with low ESL mnd high-voltage rating (630V) is 謂胡聚酣类麻也嚮Equivalent Series IntJudance)等対磊联l瞄High-frequency input capacitodesign is essentially a tra&-off between: sizing it big 巳 no ugh to minimize the noise injected back into the AC line sizing it small enough to avoid line current zero-crcfsjnig .distortion (flattening) The high-frequency input capacitor is determined as follows:5 =2.1where:k胡i = inductor current ripple factor, of 35% as mentioned earlierr- maximum high frequency input voltage ripple factor (aVin/Vin)s assumed 9% A standard 2.2pFt 630V capacitor is selected for Cin for this converter.，所以需要先求I，参阅IR1153应用IN (RMS) MAX因为C = kIN (RMS) MAXINAIl 2兀 X f X r X VswIN (RMS) MIN规格书2000W PFC计算如下：lit Is necessary to determine the maximum input currents & peak) from the specifications in Table 1 before proceeding with detailed design of the PFC boost converter. The maximum input current is typically enBun怕red at highest load & lowest input line situation (2000W, 170VAC)L Assuming a nominal efficiency of 92% at this situationh the maximum input power can be calculated：ihS _20(W_?|74w/“MNFro mi this, the maximumi RIMS AC line current is then calculated: 可AiW (”艸Rh霑、亦)PF0.92(170/)0.0.92当 PouT=4000W 时，P一 Pq(MAX)OUTIN (MAX)耳MIN因为一般需要对市电220VAC (- 10%, +15%)变动范围内的PFC运行情况进行确 认是否存在异常，即198V254VAC，所以V 198V。假设当PFC在4000WIN (RMS) MIN负载情况下运行功率因数cos 0为0.998,贝y：PO (MAX)IN (RMS) MAX 耳(V) PF0.92 X198V x 0.998MIN IN (RMS) MIN4000WI-J 2I-迈 x 22 A 31.1A；IN (PEAK) MAXIN (RMS) MAX综上所述，高频输入电容计算如下所示：22 AC kIN (RMS) MAXinail 2兀 x f x r x VswIN (RMS) MIN所以一个标准的3.3uF或者2.2uF，630V的聚酯(薄膜)电容可以选用。2.输出电容计算参阅IR1153应用规格书2000W PFC计算如下：35% X 2kx 22.2kHzx9% x198V _ 3.1uF ； Output Capacitor (Cojt)Output Capacitor design is based on hold-up time requirement For 20ms hold-up time and capacitance is first calculated:minimum output vollage of 285V the output_ 2 - 2000.205 丽w二殛m函=1194炒Minimum capa citor value must be de-rated for capacitor lol era nee (20%) to guarantee minimum hold-uptime.3 standard 470uFr 450V capacitors connected in parallleL which yields about 1410uF total can be seise ted for this converter. The hold-up time will be slightly less than 20ms in the worst case where the DC bus capacitances are at 80% of their rated value,2 x P x At由计算公式：CO ，当POUT=4000W时，对于50Hz的市电来讲,OUT(MIN) V2 V2UU 丄OO (MIN)At 20ms, V 380V, V 285V ?，将各个参数代入得：OO (MIN)增加20%余量:2 x 4000W x 20ms160160C 2533uFOUT (MIN)-(285V )2144400 81225 63175COUT3166.25uFC2533uFOUT( MIN) 1 AC1 0.2TOL所以4个680uF /450V的电容并联使用达2720uF可以满足4000W PFC的需要。3.升压电感(Lbst)计算IR1153 IC is an average current mode conlrqPer. An on-chip RC filter is sized to effectively filter the boost inductor current ripple to generate a clean average current signal for the IC. The averaging function in the IC can accommodate a maximum limit of 40% inductor current ripple factor at maximum input current. The boost inductance has to be sized so tiat the inductor r ipple current factor is not more th自n 40% 曲 maximLim input current condition (at peak of AC sinusoid). This is because: Higher ripple current factors will interfere with the Average Current Mode operation of One Cycle Control algorithm in IR1153 leading to duty cycle instabilities-pulse skipping which results in current distortion Eind sometimes even audible noise失真 power devices are stressed more with higher ripple currents as the peak in ductor curre nt (Iufkjmax also in creases proportionately注：由以上叙述可知，升压电感的纹波电流应该小于最大电流的40%；参阅IR1153 应用规格书2000W PFC计算如下：In this calculation, an inductor current ripple factor of 35% is selected (typical ripple factdris -20% for most PFC designs). The ripple current at peak of AC sinusoid at maximum input current is:AZ. =035x IMAnd, peak Inductor current is:AFKMAX=18.M +21.3/!In order to det ermine the boost induct 自 nc& the power switch duty cycle at peak of AC sinusoid (at lowest input line of 17 VAC) is required.=2W嘉J愜曲3S5K-240F 朋5矿Based on the boost converter voltage conversion ratio,电乐转扭比*-0.38The boost inductance is then given by:L _ Vfw.FEAf：imy * _240FxO.JH亢52妙A convenient value of 700uH is selected for LBST for this 亡 on verier当Put=4000W时，选择典型的纹波因数n = 20%计算，因为一般需要对市电 220VAC (-10%, +15%)变动范围内的PFC运行情况进行确认是否存在异常，即 198V254VAC,所以 V二 198V。IN (RMS) MIN假设当PFC在4000W负载情况下运行功率因数cos 0为0.998，贝V：=22 AP4000W