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1、For ensure two learn a do learning education made effectiveness, promote all members consciously respected Constitution, and comply with party rules, with XI General Secretary series important speech spirit armed mind, and guide practice, and promoted work, do qualified Communist, according to Commu
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12、rmation guidance and example, and take this opportunity to point to gradually expand the scope of education application of information technology. Formed Department personnel grasp business and educational information development and application; application and一种有效的反激钳位电路设计方法0 引言 单端反激式开关电源具有结构简单、输入
13、输出电气隔离、电压升降范围宽、易于多路输出、可靠性高、造价低等优点,广泛应用于小功率场合。然而,由于漏感影响,反激变换器功率开关管关断时将引起电压尖峰,必须用钳位电路加以抑制。由于RCD钳位电路比有源钳位电路更简洁且易实现,因而在小功率变换场合RCD钳位更有实用价值。 1 漏感抑制 变压器的漏感是不可消除的,但可以通过合理的电路设计和绕制使之减小。设计和绕制是否合理,对漏感的影响是很明显的。采用合理的方法,可将漏感控制在初级电感的2左右。 设计时应综合变压器磁芯的选择和初级匝数的确定,尽量使初级绕组可紧密绕满磁芯骨架一层或多层。绕制时绕线要尽量分布得紧凑、均匀,这样线圈和磁路空间上更接近垂直关
14、系,耦合效果更好。初级和次级绕线也要尽量靠得紧密。2 RCD钳位电路参数设计2.1 变压器等效模型 图1为实际变压器的等效电路,励磁电感同理想变压器并联,漏感同励磁电感串联。励磁电感能量可通过理想变压器耦合到副边,而漏感因为不耦合,能量不能传递到副边,如果不采取措施,漏感将通过寄生电容释放能量,引起电路电压过冲和振荡,影响电路工作性能,还会引起EMI问题,严重时会烧毁器件,为抑制其影响,可在变压器初级并联无源RCD钳位电路,其拓扑如图2所示。2.2 钳位电路工作原理 引入RCD钳位电路,目的是消耗漏感能量,但不能消耗主励磁电感能量,否则会降低电路效率。要做到这点必须对RC参数进行优化设计,下面
15、分析其工作原理: 当S1关断时,漏感Lk释能,D导通,C上电压瞬间充上去,然后D截止,C通过R放电。实验表明R或C值越小就会这样,R太小,放电就快,C太小很快充满,小到一定程度就会这样回到零。实验表明,C越大,这儿就越平滑均是将反射电压吸收了部分就是反射电压 1)若C值较大,C上电压缓慢上升,副边反激过冲小,变压器能量不能迅速传递到副边,见图3(a);2)若C值特别大,电压峰值小于副边反射电压,则钳位电容上电压将一直保持在副边反射电压附近,即钳位电阻变为死负载,一直在消耗磁芯能量,见图3(b); 3)若RC值太小,C上电压很快会降到副边反射电压,故在St开通前,钳位电阻只将成为反激变换器的死负
16、载,消耗变压器的能量,降低效率,见图3(c): 4)如果RC值取得比较合适,使到S1开通时,C上电压放到接近副边反射电压,到下次导通时,C上能量恰好可以释放完,见图3(d),这种情况钳位效果较好,但电容峰值电压大,器件应力高。 第2)和第3)种方式是不允许的,而第1)种方式电压变化缓慢,能量不能被迅速传递,第4)种方式电压峰值大,器件应力大。可折衷处理,在第4)种方式基础上增大电容,降低电压峰值,同时调节R,使到S1开通时,C上电压放到接近副边反射电压,之后RC继续放电至S1下次开通,如图3(e)所示。本人认为此分析清楚地说明RC放电时间常数要大于开关周期,至少要大于截止时间,也就是RC振荡频
17、率小于开关频率。2.3 参数设计 S1关断时,Lk释能给C充电,R阻值较大,可近似认为Lk与C发生串联谐振,谐振周期为TLC=2,经过14谐振周期,电感电流反向,D截止,这段时间很短。由于D存在反向恢复,电路还会有一个衰减振荡过程,而且是低损的,时间极为短暂,因此叮以忽略其影响。总之,C充电时间是很短的,相对于整个开关周期,可以不考虑。本人认为这讲的极有道理,开关管截止时间里充电过后就要放电,所以在实际实验中如果R太小还没到开关管导通C的电已放完了,故出现了一个平台,这时会消耗反射电压的能量,所以R的取值一定要使C的放电电压在开关管导通时不小于反射电压。在进入到导通时间后C的电压为负值,千万不
18、要认为是某个电压对C反向充电,本人认为是开关管导通后呈现的低电位。 对于理想的钳位电路工作方式,见图3(e)。S1关断时,漏感释能,电容快速充电至峰值Vcmax,之后RC放电。由于充电过程非常短,可假设RC放电过程持续整个开关周期。 RC值的确定需按最小输入电压(但有的书上说是按最大值,实际测试表明似乎应是最大值),最大负载,即最大占空比条件工作选取,否则,随着D的增大,副边导通时间也会增加,钳位电容电压波形会出现平台,钳位电路将消耗主励磁电感能量。 对图3(c)工作方式,峰值电压太大,现考虑降低Vcmax。Vcmax只有最小值限制,必须大于副边反射电压 可做线性化处理来设定Vcmax,如图4
19、所示,由几何关系得 为保证S1开通时,C上电压刚好放到需满足 将(1)式代入(2)式可得 对整个周期RC放电过程分析,有 根据能量关系有 式中:IpkLk释能给C的电流峰值将式(1)和式(4)代人式(5),得同理这公式有错误应是除以LnDon. 结合式(3),得应是 电阻功率选取依据 式中:fs为变换器的工作频率。3 实验分析 输入直流电压30(12)v,输出12VlA,最大占空比Dmax=0.45,采用UC3842控制,工作于DCM方式,变压器选用CER28A型磁芯,原边匝数为24匝,副边取13匝。有关实验波形如图5图8所示。 图7显示在副边反射电压点没有出现平台,说明结果与理论分析吻合。4
20、 结语 按照文中介绍的方法设计的钳位电路,可以较好地吸收漏感能量,同时不消耗主励磁电感能量。经折衷优化处理,既抑制了电容电压峰值,减轻了功率器件的开关应力,又保证了足够电压脉动量,磁芯能量可以快速、高效地传递,为反激变换器的设计提供了很好的依据。网上相关人员讨论:1关于吸收电路的问题,很有分析的必要,我也曾对此仔细分析过。我再分析一下,你可以按照这个思路自己进行计算。 开关管漏极上的电压由三部分组成:电源电压,反击感应电压(等于输出电压除以杂比),漏感冲击电压。 吸收电路,一定要让他只吸收漏感冲击电压,而不要对另外电压起作用,那样不仅会增大吸收电阻的负担,还会降低开关电源的效率。 首先计算吸收
21、电阻的功耗,如果能做到只对漏感能量吸收,那么他的功率容量应该是漏感功率的1.5-2倍。 漏感的量能为0.5*Ls*Ip*Ip*f,f=工作频率,Ls=漏感,Ip关断时的开关管峰值电流,这样算出来的结果是很准确的。 由于吸收电容的另一端是接在正电源上的,所以它的电压只有两部分:反击感应电压(等于输出电压除以杂比),漏感冲击电压。电压是一个微分波形,也就是电容放电波形,随着放电,电压会越来越低,当开关管的截止期结束时,一定不要让电压下降到反激感应电压以下,否则就会损耗“本体”能量。 再计算吸收元件的数值,电容太小时,漏感能量灌入后,电压会突升的太高,有可能击穿开关管,可以根据你的开关管耐压,和你希
22、望的振铃高度,确定一个峰值电压,比如100伏,截止期结束时,我们给他定一个终止电压,比如50伏,这样,就可以计算出吸收电容的数值来: 原理是,电容电压变化量所导致的能量差 = 一个周期的漏感能量。(上面的公式5) 假设反激感应电压为U,那么电容电压的最大值就是(U+100),最小值就是(U+50),电容中的能量有一个计算公式,Ec=0.5*C*U*U, 所以,能量差就是:Ech-Ecl=0.5*C*(U+100)*(U+100)-(U+50)*(U+50),U是已知的,能量差也是已知的,电容还算不出来吗? 最后计算吸收电阻。电容放电公式:u=Uo*exp(-t/),t/=-ln(U+50)/(
23、U+100)经本人推算应是t/=-ln(U+100)/(U+50),或-t/=-ln(U+50)/(U+100),掉了个负号原文作者在发贴时可能笔误,t=截止期时间(按正常工作时的截止时间计算),可以算出,=RC吸收时间常数,那么吸收电阻不也就出来吗?本人认为这个讲的有道理.2.按上述理论进行计算: 变压器初级电感L=632uH,漏感Llou= 29uH。 先算Ip: 假定最大输出功率时是DCM模式. 则Pin = 0.5*Ls*Ip*Ip*f Ip= (Pin/0.5*Ls*f)(0.5) = (P0/*0.5*Ls*f)(0.5) = (150/0.85*0.5*623*10(-6)*70
24、*10(3) = 2.7A 漏感的能量为0.5*Ls*Ip*Ip*f,f=工作频率,Ls=漏感,Ip关断时的开关管峰值电流 Wlou= 0.5*Ls*Ip*Ip*f = 0.5 * 29*10(-6) * 2.7 * 2.7 * 70*10(3) = 7.3 W 由上面漏感能量数值可看出,漏感能量太大了,如果此能量全都由电阻来消耗,按两倍功率计算,要15W的电阻。这是无法办到的。 这么大的功耗,从上面计算可以看出,是由于初级Ip太大造成的。如果是几十W的电源,那么功耗就可以接受了。 对以上结果,请问计算有问题没有?有什么办法?3是的,这个功耗是太大,漏感功耗没有别的去处,只能消耗在吸收电阻上。
25、像这种功率较大的开关电源,一般都是工作在连续状态,否则,开关管的功率容量和磁芯的功率容量都得不到充分利用,还有一个问题,就是工作在不连续或者临界状态的变压器,由于其磁通变化量太大,变压器的发热量也是个不容忽视的问题。我上面没说,你的初级电感量太小,变压器可能工作在非连续状态。增大电感量,初级电流自然就降下来了。你可以这样计算:让磁通的变化量(p-p)/磁通平均值=0.3左右。 另外,如果电源的安全系数要求不是太高(医疗仪器要求高),可以适当减小初次级之间的绝缘厚度,以减小漏感,你的漏感量在正常的数值范围内,但不是特别的小,大功率的电源,漏感就是个很麻烦的问题4.你好,非常感谢。 初级电感和漏感
26、的数值在上面第十贴中写出来了,我是刚测的数据。 测时发现,初次级间不加铜皮屏蔽漏感小。这应是正常的吧。也可能是漏感加大的缘故,加了屏蔽后尖峰反而大了。5. 初次级间不加铜皮屏蔽漏感小,是正常的。所谓漏感是通过本线圈的磁力线没有完全通过另一线圈所产生的,增加铜皮屏蔽,相当于线圈之间的耦合难度增大,故漏感增大,分布电容减少。 想减少尖峰,最好的办法是减少变压器漏感,其次是在MOS管漏极加磁珠,这样都会减少损耗,还有就是无损吸收,最后就是用RCD这种有损吸收的方式。6. 是的,铜箔不是磁性材料,它只对电场起作用,对磁场而言,它和绝缘材料差不多. 网上有人这样讲: rcd的rc时间常数必须长于开关周期
27、,也就是rc震荡频率要小于开关频率,这样子防止在管子未开通前放电完毕而导致二极管再次开通,造成系统的震荡.本人仔细分析了一下,这样讲有一定的道理, 但P126里讲RC时间常数等于第一个尖峰和第二个尖峰时间的3倍就够了,这个我认为有点错, 因为有人讲振荡频率是指第一个脉冲以后的,从图上看基本差不多,第一个脉冲是漏感往C里面充电的过程,然后根据回复时间D有一个关断过程,当然认为是一个振荡也可以,只是时间和后面的振荡相比就太长了。所以一般认为一两个脉冲之后的才算振荡(前几个脉冲由于单向导电也不象正旋波),因为后面的振荡和RC基本无关了, 只有第一个尖峰脉冲的能量被吸收,后面的脉冲电压都达不到吸收的门
28、槛电平,所以是在自己震荡,与R,C无关了. 只有第一个尖峰脉冲的能量被吸收,后面的脉冲电压都达不到吸收的门槛电平,所以是在自己震荡(本人认为是漏感和分布电容),与R,C无关了.如果RC时间常数太小了,在截止时间内C的电放完了,那反激电压岂不是又要向C充电而形成振荡?本人认为至少要大于开关管截止时间.这是电源网上另一个人讲的:RCD是给变压器漏感提供泻放回路的。反激变换同正激有所不同,反激变压器可以看作是一个耦合电感,所以不像正激一样有磁芯复位的要求。但是,由于反激变压器大都开了气隙,所以漏感比较大,漏感能量也比较大需要一个泻放的通路。上面接法的RCD的作用就是当开关管关断时漏感电流通过D对C充
29、电,由于C的存在形成LC谐振。一般来讲是需要限制这个谐振电压的峰值的。应该是输入直流电压最大值+次级按变比折算过来的反射电压+LC谐振峰值电压MOS的额定电压。R的作用是降低LC的Q值,使震荡衰减。一般控制LC频率在开关频率三倍以上,这不是绝对。频率越高则电容越小,但是谐振峰值大,频率低了谐振的时间就比较长,影响能量传递。R大了衰减的比较慢,增加管子的损耗和干扰(谐振能量)。小了则增加损耗降低效率。定期分享海量电源工程师实用学习资料,不打广告,不做推销,致力于打造一个纯净的电源工程师交流与进步的平台。真心想学习的请进,微信号搜索“电源研发精英圈”或“dianyuankaifa”加关注,也可扫描
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43、rise signed cooperation agreement, low cost built cover City School of fiber Education City domain network and achieved broadband access, to achieved has regional within all schools, and all classroom of broadband network access as typical case in national informed. Three XX municipal party Committe
44、e municipal education information development, policy measures included in the citys development plan, formulated education plan, and guaranteed on the funds. According to the Ministry of education documents and XX actual, Education Bureau of XX municipality formulated XX City Informatization of edu
45、cation three-year development plan for the XX teaching information application of the three-year planning document, guide the education information construction. Early education work priorities, specifically the inclusion of educational information construction areas, and application of information technology and construction in special steering steering room, and at the time of the annual assessment as an important indicator of school assessment. Since 2008, the citys education
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