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| | | | | | | | | | | 把NC965_李斑竹在那个帖子的解说COPY 到这里
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一个脉冲来临第一时刻,是按串联等效电阻Rs和结电容C充电,因为Rs很小,充电瞬间完成,得到的(瞬时)电压是按C的容量分配给两个结电容的,如果C相等,则端电压相等,如果C不相等,则按C的大小比例分配,电容大者电压低。
然后考察并联等效电阻Rp的作用,它会有一个很小的电流,即漏电流。如果C和Rp都相等,这个电流和端电压不变,均压。如果C相等而Rp不相等,C上的端电压会在RC时间常数大致相当的时间后从最初的电容均压值逐渐过渡到按Rp的大小比例重新分配,电阻大者电压高。从波形上看就是电容按RpC放电。
于是;只要脉冲周期远远小于RpC时间常数,按电阻均压的效应就可以忽略,按电容均压的效应得以凸显。对于在一个脉宽内还叠加有(更短脉宽的)尖峰毛刺的一般情况更是如此。比如开关电源常见应用的二极管,尖峰毛刺的脉宽ns数量级,相当大的幅度且位居器件耐压的极限,这时的电阻均压就形同摆设,完全是结电容在左右局面。
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| | | | | | | | | | | 我个人的理解是:
电阻: 脉冲或者高频信号下,电阻并不是完全的集总模型,其分布电容的容抗,对响应起了主导作用 ~
请大师和王斑竹帮指点 ,谢谢 ~
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| | | | | | | | | | | | | | | | | 我讲的阻抗是R//C(并联于二极管),两组R//C 再串联。
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| | | | | | | | | | | | | | | | | | | ,
问题是:为什么要加电容并联在R 两端,不加可以吗?
为什么要加,基于什么思考?
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| | | | | | | | | | | | | | | | | | | | | 看得明6楼不? 那是 3 个不同漏电电阻和结电容的 diodes 串联,在不同频率下的表现,要在全频都均压的话,就要每个diode并联RC,用来mask out 本身的RC,尽量使每个diode看起来是相同的等效漏电电阻和结电容。
评分查看全部评分
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#34 (漏电电阻 结电容)=( 14.8 M ohm,58.6pF)
#40 (漏电电阻 结电容)=( 8.1M ohm ,77.6pF)
#51 (漏电电阻 结电容)=( 8.0 M ohm,79.4pF)
是这样吗?
另外, Q=2.35 nC 某一频率下,结电容的储存电荷?
结电容一定,告知了电荷相当于告知了频率?
观察了该频率好像是在1K , 如何由Q=2.35 nC 得到 频率值?
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| | | | | | | | | | | | | | | | | | | | | | | 嗯嗯 ,大师用mask out 太贴切了。 前面也隐约感觉这样,在8楼 (不敢肯定), 在高频下DIODE非集总因素就凸显出来了, 用需要外加RC 来mask out .
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| | | | | | | | | | | | | | | | | | | | | | | 6 楼图中 Q=2.35 nC什么意思? 该不是直流“偏置” 吧 ~
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| | | | | | | | | | | | | | | 欧姆定理(无论是直流电路欧姆定理还是交流电路欧姆定理),只适合稳态电路(方能使用阻抗的概念),不适合瞬态电路(不能使用阻抗的概念,即不能用欧姆定理求解)。
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| | | | | | | | | | | | | | | | | 反推如果欧姆定律都不能成立,
请教下李斑竹:瞬态电路 用什么数学工具来描述?
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| | | | | | | | | | | | | | | | | | | 与时间有关的函数,因为欧姆定理与时间无关,所以不能用。
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| | | | | | | | | | | | | | | | | 九天,李斑竹这个解释也能评分?? 你说说你的理由
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| | | | | | | | | | | | | | | | | | | 这位九天曾经“呛”过他几次,印象中尤其是版主的发言,不分好歹都会点赞(因为不懂,也分不出好歹 )。
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| | | | | | | | | | | | | | | | | | | | | 人家是外行,看热闹的,看不出门道,哪里热闹,哪里点赞,不必苛求
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| | | | | | | | | | | | | | | | | | | 此贴不是吗?讨论串联二极管电压应力,一旦有人用阻抗、容抗去分析,不用细看,就知道跑偏了题意。当然,同样的电路,用欧姆定理去进行稳态分析是可以的,比如你6楼的例子,适合直流、正弦波,也能得到电压应力,一旦遇到方波或者毛刺就必定傻眼(这需要瞬态分析)。
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| | | | | | | | | | | | | | | | | | | | | 看的有点晕。6楼你不用阻抗去分析,难道是凭感觉猜吗?
你说的时间常数只是时域的分析方法。
频域呢?频域也是有欧姆定律的 ~
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| | | | | | | | | | | | | | | | | | | | | | | 当然时域问题可以转换为频域问题进行分析,但对于串联二极管电压应力这样简单的问题,貌似没有必要。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | 7楼提出了最简单的分析模型,也能完全说明问题,何须更复杂的?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | 近来我意识到一个问题:反映一个人是否步入“老年化”的标志之一就是故步自封的程度。
李斑竹,这里没有故意冒犯您的意思。只是有感 ~
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 始终没见到你的见解,晒一晒? 最好有个具体案例得到一个具体值。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 李版,你这图想说什么啊?
从没有说你的时间常数法不正确。 我所反驳的是您认为别人用阻抗概念来认识这个问题是荒谬的 。
任何时域的表现 ,都能在频域给出对应关系。 你怎么能一口断定别人就“歇菜傻眼”了呢?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 电容峰值电压都相等,说明了高频下电路主要阻抗是容抗决定的,因为R>>>> 1/jwc (when frequnce is very high ) , 并联电阻不起作用。
您不认为您的仿真在帮助说明问题吗?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 39楼就是7楼的完全复制,看来我们的认识是一致的。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 估计你是想看另一个二极管的电压应力,可惜没有意外惊喜
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 这有什么意外?用阻抗和欧姆定律就可以计算出来。李版敝屣自珍了。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 我到觉得尘埃已经落地,可以结贴了。有人想用欧姆定理去算他就去算,友情提示:无需欧姆定理也能搞定。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 这么快就想结了?还是请李版先展示一下,如何无需欧姆定理也能搞定?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 7楼就是,其实上个贴一开始就已经搞定了,没欧姆定理什么事。
估计你是想秀一把用欧姆定理来描述RC放电曲线,我虽然不感兴趣(也许别人感兴趣呢?),但也不反对有人把简单问题复杂化。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 李版的金科玉律明确指出:“欧姆定理(无论是直流电路欧姆定理还是交流电路欧姆定理),只适合稳态电路(方能使用阻抗的概念),不适合瞬态电路(不能使用阻抗的概念,即不能用欧姆定理求解)。”
简单的例子有电容RC放电的方程:Vc=-R*C*dVc/dt,因为是瞬态,又用了欧姆,所以是不成立的,它的解 vc(t)=vco*exp(-t/RC),当然也是错的,可叹科学发达的今天,人类还没找到RC放电的正确数学描述。此其一。
本例中,妄想用 Vd1=Vpulse*Z1/(Z1+Z2) 的频域式子来算出管子压降,显然这式子是不对的,因为是瞬态,又用了欧姆2次,又用了阻抗的概念,但勉强计算下去,结果竟然跟仿真的一样,除非凑巧,式子是错的话,那仿真估计也是错的,不可靠的。这不成那不成,又一难题待解。此其二。
又指出 “与时间有关的函数,因为欧姆定理与时间无关,所以不能用。”
但凡类似 i(t)=Vpk*sin(ωt) /R 的欧姆算式是不能用的。大家注意。此其三。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 好像挺复杂的样子,即使如此复杂也没有看见V与 t 的对应关系,V=f(t),还没搞定,继续!
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | vc(t)=vco*exp(-t/RC),这个可以吗?亲
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Vc=-R*C*dVc/dt , 亲说说这是什么 ?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 设dvc/dt=∞,则Vc=-R*C*dVc/dt=-∞(V)?这就是你的答案?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 如果你的“设“”是成立的, “”则 “”也就能成立 ~
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 斑竹,你理解下你的“设”的物理含义,它代表的物理量是啥呢 ?再想想 ~
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | dvc/dt=∞,它代表方波的边缘(沿),由此得到无穷大均压(起始/终结)电压?
还是那句话,你能得到具体数值即可,V为方波幅度,R1//C1+R2//C2已知这4个元件共5个参数,如何得到一个唯一的均压值和均压值过程?其他都是空口白牙。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 李版的数学真是。vc(t) 是什么,dvc/dt 是什么,弄不清楚?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 知道你的意思,也知道蓝天兄弟的意思,但我关心的是均压值及其过程。这贴不是讨论均压吗? |
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 剩下的就是对李版的金句的肯定(50楼)和再肯定。李版自便。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 那不是金句,只是说了一个道理,正如你说(蓝天想说)的,虽然RC放电曲线方程是欧姆定理的微分形式的一个解,它已经表现为一个与时间有关的过程函数,正好与本帖均压问题的讨论完全契合,此时就无需再使用欧姆定理,也不需要引进阻抗的概念,直接可以求解。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 这不是需不需要用什么方法的问题,是可不可以用,能不能用的问题,或者我笨,我喜欢这样算呢,是李版自己斩钉截铁的说:“不适合瞬态电路,不能使用阻抗的概念,不能用欧姆定理求解)”,3个否定,这就是李版的道理?实在看不到道理何在。对好错好,知道便好。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 你喜欢这样算也可以,可是至今为止没见你算出一个数值。道理有很多,解决工程实际问题才是硬道理。39楼均压曲线的3个终了值尚未标识出,试试?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 我的方法不是已被否定了?还有什么可做? 我也没见李版算出一个数值,何妨演示一下,好让我们见识见识。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 我已经标识出起始电压550V,来至7楼的算法。正好还没标识出终了电压,你来,万勿推辞。
附,7楼算法用公式表达就是这样:
V(t)=V(0) - ΔV*(1-exp(-t/RC))
其中:
V1(0)/V2(0)=C2/C1, 即:V(0)=V2(0)=Vin/(C2/C1+1)
V1(∞)/V2(∞)=R1/R2,即:V(∞)=V2(∞)=Vin/(R1/R2+1)
ΔV=V(0)-V(∞)
R=R1//R2
C=C1//C2
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 何以得到此值?可曾用到欧姆定理?可曾用了阻抗的概念?谅你也用不上。
我不用欧姆定理,不用阻抗也能得到此值,不捞你费心。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 谅你也看不明白:谅你不用欧姆定理,不用阻抗一定不能得到此值,这我不担心。
(既然有人看不明白,删了 Mathcad )
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 7楼你有个P算法,只是个口水描述。上面用阻抗、欧姆的方法我演示过了,现轮到你了,有本事就算来看看,不用啰嗦。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 7楼算法已经在68楼用函数表达:
v(t)=V(0) - ΔV*(1-exp(-t/RC)),将39楼有关参数(不计二极管内阻)代入后,不用欧姆定理和阻抗,可得到:
v(t)=550 - 235.7143*(1-exp(-35000t)),于是
v(250us)=314.323V(貌似这里的误差是你被阻抗坑了一下)
v(25us)=412.546V
v(0)=550V
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 好热闹啊, 李斑竹姑且不论您的结果是否正确,
您这个式子,本质就是欧姆定律的解啊。没有欧姆定律,哪来的下面这个式子 ??
v(t)=V(0) - ΔV*(1-exp(-t/RC)),
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 我的意思,有了这个瞬态函数,就无需再用欧姆定理,也就无需再用阻抗的概念,用了就会出错,比如314V就会变成317V。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 你这个瞬态函数,从哪里来的?怎么来的? 它还是来自欧姆定理啊 。
它是欧姆定理的解!
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 算法来至49楼(欧姆定理和RC放电二选一),RC放电曲线是众所周知的算法,用起来也简单明快,就不必再回过头去追究它与欧姆定理(抑或麦克斯韦方程)的关系。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 你就直接告诉我 RC放电曲线是怎么来的 ? 一问一答 !
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 世界上所有事物(包括所有物理定理),(追根寻源的话)都是从宇宙大爆炸原点来的。 |
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 李版自己斩钉截铁的说:“不适合瞬态电路,不能使用阻抗的概念,不能用欧姆定理求解“”
忘记了,这是你说的吗?
能不能用,和想不想用,是两个概念吧 ~
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 是我说的,但此话有个前提:瞬态函数方程已经成立
你5楼说心里有个答案,说出来听听?有点好奇
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 317V 是125us 时的值,脉宽不是125us么 ?
当初对原理的错误认识,提出此金句,后来知道有漏洞,只好圆场,还自打嘴巴,现在又补上一个前提,自己也不觉得好笑。 瞬态函数方程是凭空而来的?试试输入是锯齿波,三角波,馒头波的?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | v(t)=550 - 235.7143*(1-exp(-35000t))
v(12.5uS)=466.473V
v(125uS)=317.253V
果然搞错,想必我可能还有其他错误,比如标点语法什么的,麻烦找出来。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 估计也不复杂,但确实没那闲心,工程上遇到了再议,谢谢。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 估计就84楼的,李版也说不出个所以然来。 真怀疑是请qiang的。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 晕! 真没想到你还有这种心思。
你不防把7楼(上贴21楼)的狗屁算法逐字逐句理解一遍,相信你的理解能力还不会差到连口水话都理解不了的程度,它与68楼的算法完全对应,84楼只不过是这个算法的一个具体案例而已。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 理由:如果任何人都不能证明它是错误的,那它必然就是正确的。 |
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 这表达式却是对的,待李版说理由。
接97楼:
7楼的 狗屁说什么? 有RC自然是有充有放,电压自然是按元件数值分配,就这样的一个一般性说法,就等于说了个瞬态函数,得出答案? 这么容易的话,请试给个3管子串联的来看看,求中间管子压降。也许真是这么容易。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | V1(0)/V2(0)=C2/C1 ,这个式子很关键啊。你怎么来的?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 李版看这个式子对否?N 个串联
Vj= Vin /[ Cj*(1/C1+1/C2+...+1/Cn)]
Vj : 第j个电容两端电压
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 你想考察一个器件,只看它的结电容即可(无论它处于什么位置),其他器件的结电容是串联关系(可等效为1个电容)。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 呵呵, 你咋就知道是两头呢,如果考察的是中间的呢?
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 你给30个具体数值,然后算出其中任意29个的串联值,然后代入68楼公式里
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 我问的是中间那节电容,亲。
你那个能算“中间”的那节电容吗? 能明白您的等效意思。
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | 不会因为中间或者边缘而有什么不同,什么叫中间?4个串哪个是中间?100个串哪个是边缘?你不觉得问题太低级? |
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