标签: lc

You sure have had your own A-Ha moment. The Merriam-Webster dictionary defines it as "a moment of sudden realisation, inspiration, insight, recognition, or comprehension." But I'm taking it one step further— a-ha! with an exclamation point. This is a more dramatic realisation. It's the moment when you discover a great truth, when something that was complicated or unpredictable suddenly becomes clear. As engineers, I'm sure we've had many. One of my first a-ha!s came in grade school. I was a hobbyist, and I enjoyed creating things from the local Radio Shack, even though I didn't know why they worked. I had hoarded quite the collection of resistors, capacitors, tubes, and speakers. I could read the colour bands on a resistor to get its value, and I had rudimentary soldering skills, but I didn't know how to design anything. One day, my older brother, who had been a Navy technician, explained Ohm's Law to me. A lightning bolt ignited in my head. You mean, there's a relationship between voltage, current, and resistance? That makes a lot of sense. And the world became a little bit more understandable. Building a circuit was a little less about pleasing the electron gods and a little more like engineering. I've had several of these moments since then. Calculus was a key ingredient to many. Though I knew formulas from high school physics, calculus enabled me to derive the formulas. As a ham radio operator, I knew how to calculate the resonate frequency of an LC network.   But I didn't know why that was the resonant frequency. With calculus, I was able to derive it myself, and I discovered why it also explained the resonant frequency of a pendulum or a spring and mass. Calculus explained that the current through a capacitor was proportional to the first derivative of the voltage. Suddenly, first-order differential equations explained time-domain and frequency-domain phenomena. This was another a-ha! moment. I've had many since then. Boolean logic explained digital circuits to me—no longer a mystery. A more recent a-ha! moment came when I learned how WCDMA worked. In retrospect, all these things seem obvious, but I can recall the very day that each of these a-ha! moments came. Science and engineering aren't the only subjects that have created these moments. An Economist article about international trade led the reader through a simplified two-party, two-industry model, where one party had a productivity advantage for both industries, and the other party had inferior productivity for both. Much to my surprise, simple arithmetic in the example showed that both parties produced and acquired more goods than either one could have done by itself if trading could occur between them. Until I had done the math, I had assumed trade was only advantageous if each had an absolute advantage in some industry. This was the principle of comparative advantage—and another a-ha! moment for me. I remember the day I did the surprising arithmetic. What about you? I'd like to hear about your a-ha! moments. Larry Desjardin Consultant  

LC 电路相关属性计算来由,这些困惑小弟很久,还请各位帮忙引导,L 的阻抗大小跟什么有关系,及感抗,C的大小确定.

by CQ.Lee RFStory.com已经改为 MiniIC.com 原文链接： http://www.rfstory.com/2011/06/24/lowpass-filter-lc/ 在射频里面，用串联L再并联C的方法做低通滤波器非常普遍。有个疑问，为什么不只用一个串联的L或者并联的C呢？在高频，L表现为开路，和负载串联分得到大部分电压，C和表现为短路，和负载并联分得到大部分电流，都可以达到滤波效果啊。   原因是滤波器的负载在不通频率表现出不同的阻抗特性，一般低频电路中的负载均为较高的阻抗，而传输线的阻抗并不是很高，这就造成了传输线特性阻抗与负载的失配，那么在LC滤波器的负载阻抗在不通频率时就不一样，规律是绕着史密斯圆图旋转。举一个例子，比如在某一频率点负载呈低阻特性阻抗为1欧姆，那么只串联一个并联个电容显然分不到多少电流，达不到滤波效果。相应的在某频率也会达到高阻，这时如果只用串联L也分不到多少电压。   那么就有人会问了，单一的串联L虽然L没有分得到大部分电压，但是L和负载加起来还是表现为高阻不会吸收多少能量啊。错了，因为源端也有可能是高阻！这要注意的一点，源端的阻抗也是随着频率变化的，如果源端和负载端达到了共轭匹配，那么引入的滤波的效果就很差了。   所以建议大家在设计电源供电电路的时候，最好采用LC的滤波方式，而不要简单的只用L或者C，那样造成的恶果就是，在某一频点有个杂散泄漏，不论你换成一个什么样的电容都滤不掉，很悲惨！   本文为原创文章，转载请注明：转载自 射频那些事儿 。