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[求助]
超級(jí)電容器電容的計(jì)算
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各位蟲(chóng)友,大家好。 1 將電極材料組裝成超級(jí)電容器器件,測(cè)定器件阻抗,可根據(jù) C=-1/(2*3.14*f*Z'') 得到電容,這個(gè)電容值C是器件總電容么?從阻抗譜中計(jì)算得到的C',C''及其它直接數(shù)據(jù)都是整個(gè)器件的數(shù)據(jù)吧?器件的面積電容CA=C/2S , S為單個(gè)電極的面積;單個(gè)電極的面積電容CA=2C/S, 這里的C都是指整個(gè)器件的電容.因此,整個(gè)器件的面積電容是單個(gè)電極面積電容的1/4? 2 質(zhì)量電容和體積電容的計(jì)算方法與上述類(lèi)似,都是整個(gè)器件是單個(gè)電極的1/4? 3 如果測(cè)試整個(gè)器件的充放電曲線,計(jì)算電容C=IΔt/ΔU,這里直接得到的也是整個(gè)器件的電容吧?而單個(gè)電極的面積電容應(yīng)該是CA=2*IΔt/S*ΔU,S為單個(gè)電極的面積? 4 另外看文獻(xiàn),有一句話“電極材料的質(zhì)量比電容Cwt等于總電容C除以電極材料的質(zhì)量m,公式為:Cwt=C/m”,這句話對(duì)么?我認(rèn)為應(yīng)該是Cwt=2C/m,大家覺(jué)著呢? |
| 超級(jí)電容器計(jì)算容量一般都是通過(guò)CV曲線的面積積分或者恒流充電曲線的斜率計(jì)算的,至于通過(guò)阻抗計(jì)算的很少,這個(gè)針對(duì)于物理平板電容器的吧,因?yàn)槌?jí)電容器電極有效利用面積是很難界定的。容量四分之一的問(wèn)題,主要是針對(duì)單電極和雙電極而言的,雙電極由兩個(gè)單電極串聯(lián)而成,串聯(lián)后容量就是二分之一,兩個(gè)單電極質(zhì)量就是2m,所以雙電極的容量就是單電極容量的四分之一了。文獻(xiàn)里面如果是用三電極測(cè)試的,這里的容量計(jì)算的容量就是單電極的容量,如果是正負(fù)極組裝的電容器,這里計(jì)算的容量是雙電極的容量。 |
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謝謝您,您看文獻(xiàn)里的公式:the special capacitances of the ECs were evaluated in area (μF cm−2) and volume (F cm−3) units. The specific areal capacitance of EC devices, CA (μF cm−2), was calculated from EIS spectra by CA=-1/(2*π*f*Z''*s) ,where f is frequency (Hz), Z” is the imaginary resistance (Ω) and s is the area of the electrode (cm2).The specific volumetric capacitance, CV (F cm−3), was calculated by equation S2: CV =1/ (*π*f*Z'' 𝑣 , where 𝑣 is the volume of the two electrodes; 𝑣 = 2d s and d is the thickness of a single electrode. 這里的CA和CV明顯是器件的面積和體積電容,但前者是C/S,S是單個(gè)電極的面積,后者是C/V,V是兩個(gè)電極的體積。為什么求器件體積電容,要除以兩個(gè)電極的體積? |
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謝謝您,您看文獻(xiàn)里的公式:the special capacitances of the ECs were evaluated in area (μF cm−2) and volume (F cm−3) units. The specific areal capacitance of EC devices, CA (μF cm−2), was calculated from EIS spectra by CA=-1/(2*π*f*Z''*s) ,where f is frequency (Hz), Z” is the imaginary resistance (Ω) and s is the area of the electrode (cm2). The specific volumetric capacitance of EC devices, CV (F cm−3), was calculated from EIS spectra by CV=-1/(2*π*f*Z''*v) ,where f is frequency (Hz), Z” is the imaginary resistance (Ω) and v is the volume of two electrodes, v=2ds, d is the thickness of a single electrode . |
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