如何理解電容器的靜電容量

1、多層片式陶瓷電容器（MLCC）技 術(shù) 交 流如何理解電容器的靜電容量HOW TO UNDERSTAND THE CAPACITANCE 江門(mén)市新會(huì)三巨電子科技有限公司 JIANGMEN CITY XINHUI SANJV ELECTRONIC CO，LTD地址：廣東省江門(mén)市新會(huì)區(qū)中心南路37號(hào)廣源大廈B座郵政編碼：529100聯(lián)系電話：07508686169 傳真：07506331711 E-Mail: 公司網(wǎng)址：如何理解電容器的靜電容量A.電容量電容器的基本特性是能夠儲(chǔ)存電荷（Q），而Q值與電容量（C）和外加電壓（V）成正比。Q = CV因此充電電流被定義為：= dQ/dt = CdV/dt

2、當(dāng)外加在電容器上的電壓為1伏特，充電電流為1安培，充電時(shí)間為1秒時(shí)，我們將電容量定義為1法拉。C = Q/V = 庫(kù)侖/伏特 = 法拉由于法拉是一個(gè)很大的測(cè)量單位，在實(shí)際使用中很難達(dá)到，因此通常采用的是法拉的分?jǐn)?shù)，即：皮法（pF） = 10-12F納法（nF） = 10-9F微法（mF）= 10-6FB.電容量影響因素對(duì)于任何給定的電壓，單層電容器的電容量正比于器件的幾何尺寸和介電常數(shù)：C = KA/f(t)K = 介電常數(shù)A = 電極面積t = 介質(zhì)層厚度f(wàn) = 換算因子在英制單位體系中，f = 4.452，尺寸A和t的單位用英寸，電容量用皮法表示。單層電容器為例，電極面積1.01.0，介質(zhì)

3、層厚度0.56，介電常數(shù)2500，C = 2500（1.0）（1.0）/4.452（0.56）= 10027 pF如果采用公制體系，換算因子f = 11.31，尺寸單位改為cm，C = 2500（2.54）（2.54）/11.31（0.1422）= 10028 pF正如前面討論的電容量與幾何尺寸關(guān)系，增大電極面積和減小介質(zhì)層厚度均可獲得更大的電容量。然而，對(duì)于單層電容器來(lái)說(shuō)，無(wú)休止地增大電極面積或減小介質(zhì)層厚度是不切實(shí)際的。因此，平行列陣迭片電容器的概念被提出，用以制造具有更大比體積電容的完整器件，如下圖所示。在這種“多層”結(jié)構(gòu)中，由于多層電極的平行排列以及在相對(duì)電極間的介質(zhì)層非常薄，電極面積

4、A得以大大增加，因此電容量C會(huì)隨著因子N（介質(zhì)層數(shù)）的增加和介質(zhì)層厚度t的減小而增大。這里A指的是交迭電極的重合面積。C = KAN/4.452（t）以前在1.01.00.56的單片電容器上所獲得的容量，現(xiàn)在如果采用相同的介質(zhì)材料，以厚度為0.001的30層介質(zhì)相迭加成尺寸僅為0.0500.0400.040的多層元件即可獲得（這里重合電極面積A為0.0300.020）。C = 2500（0.030）（0.020）30/4.452（0.01）= 10107 pF上面的實(shí)例表明在多層結(jié)構(gòu)電容器尺寸相對(duì)于單層電容器小700倍的情況下仍能提供相同的電容量。因此通過(guò)優(yōu)化幾何尺寸，選擇有很高介電常數(shù)和良好

5、電性能（能在形成薄層結(jié)構(gòu)后保持良好的絕緣電阻和介質(zhì)強(qiáng)度）的介質(zhì)材料即可設(shè)計(jì)和制造出具有最大電容量體積系數(shù)的元件。HOW TO UNDERSTAND THE CAPACITANCEA. CAPACITANCEThe principal characteristic of a capacitor is that it can store an electric charge (Q), which is directly proportional to the capacitance value (C) and the voltage applied (V).Q = CVThe charging c

6、urrent I is therefore defined asI = dQ/dt = CdV/dt.The value of capacitance is defined as one Farad when the voltage across the capacitor is one volt, and a charging current of one ampere flows for one second.C = Q/V = Coulomb/Volt = FaradBecause the Farad is a very large unit of measurement, and is

7、 not encountered in practical applications, fractions of the Farad are commonly used, namely:picofarad (pF) = 10-12 Faradnanofarad (nF) = 10-9 Faradmicrofarad (mF) = 10-6 FaradB.FACTORS AFFECTING CAPACITANCEFor any given voltage the capacitance value of the single plate device is directly proportion

8、al to the geometry and dielectric constant of the device:C = KA/f(t)K = dielectric constantA = area of electrodet = thickness of dielectricf = conversion factorIn the English system of units, f=4.452, and using dimensions in inches for A and t, thecapacitance value is expressed in picofarads (pF). F

9、or example: for a single plate device, with a 1.0 X 1.0 area, .056 dielectric thickness, and a dielectric constant of 2500,C = 2500 (1.0)(1.0)/4.452 (.056) = 10,027 pFutilizing the Metric System, the conversion factor is f= 11.31, and dimensions are incentimeters.C = 2500 (2.54)(2.54)/11.31 (.1422)

10、= 10,028 pF.As is evident from the above relationship of capacitance to geometry, greater capacitance canbe achieved by increasing the electrode area while decreasing the dielectric thickness. As it is physically impractical to increase area in a single plate device with thinner dielectric, the conc

11、ept of stacking capacitors in a parallel array was conceived to produce a physically sound device with more capacitance per unit volume, as illustrated in below Figure.In this “multilayer” configuration, the area A is increased by virtue of many electrodes in parallel arrangement, in a construction permitting very thin dielectric thickness between opposing electrodes, suc