电容的计算公式_范文大全网

范文一：电容的计算公式

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电容计算公式

1、电容的基本公式

电容器的种类虽多，但就其构成原理来说，基本上是相同的。任意两个被绝缘材料隔开的金属导体，就构成一个电容器。电容器的结构及其在电路中的一般表示符号，如图1所示。

图一

电容器的两个金属导体，叫做电极或极板，用导线引出。极板中间可以是空气或其它绝缘材料，称为介质。

电容器具有在短时间内贮存电能的特性。例如，在电容器的两个极板间接入直流电源后，电源两极上的正、负电荷将分别向电容器的两个极板上移动。因为电容器的极板间被绝缘材料隔开，电荷不能通过，所以电荷就在极介质板上积存起来。实验证明，电容器极板上积存的电量q，与极板间所加电压成正比。也就是说，电容器极板上积存的电量q与极板间的电压U的比值是一个常量;这个比例常数，1.其它资源见:江上清风http://www.docin.com/hcyylhl 2.欢迎转载。转载请包含链接全文转载

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q C=...................................(1) U

电容器的电容C，是一个衡量电容器积存电荷能力大小的物理量。它的大小只与电容器的极板面积、极间距离以及中间绝缘材料的性质有关，而与外加电压无关。在式(1)中，q的单位是库仑，U的单位是伏特，电容C的单位是法拉，简称法，用符号F表示。一般地，电容器的电容要比1F小得多，故通常用微法(pF)或皮法(pF)作为电容器的单位，它们的换算关系是

-6 lμF=10F

-6-121pF=10μF=10F

电容器上(或铭牌上)除标出电容量外，一般标有一个电压值，这个电压叫电容器的最大工作电压，也叫电容器的耐压，使用时要注意电容上的标记。

2、介质的电介系数

上式(1)是电容的理论公式，实际情况并非如此，上面也提到了，一般情况下，电容的两个极板间往往充入介质材料，这是因为充入介质材料以后，在电压不变的情况下，在电容尺寸不变的情况下，电容量往往可以增加很多。

当电极间为真空时，在电场作用下，极板上的电荷量为q0，如图二所示，极板间的电容由下式表示:

qεs00C==...............................................(2) 0dU

式中C------------真空中的电容 0

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?14ε------------真空中介电常数，ε=8.86×10F/cm; 00

2S--------极板面积(cm)

d------极板距离(cm)。

图二

当电极间放入电介质后，在靠近电极的电介质表面形成束缚电荷q’，它将从电源吸引一部分额外电荷来“中和”，使极板上储存的电荷增加，因此极板间的电容为:

′q+qεS0C==...................................(3) Ud

ε-----------------------介质的介电系数，又称电容率。用式(3)除以式(2)，则有

Cε==ε.......................................(4) rCε00

这里的ε称为介质相对介电常数，通常用来表征介质的介电特性。 r

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所以，在一定的几何尺寸下，为了获得更大的电容量，就要选用相对介电常数(ε)大的电介质。例如，在电力电容器的制造中，以合r

成液体(ε约为3,5)代替由石油制成的电容器油(ε=2.2)，这rr样就可增大电容量或减小电容器的体积和质量。

由(4)式得到

ε=εε ............................................. (5) 0r

式中:ε ------------介质的介电系数又称电容率(F/m);

?12ε -----------------真空的介电系数，ε=8.85×10F/m; 00

ε ------------------相对介电系数，是个纯数，见表一所示。 r

表一:几种材料的相对介电系数ε r

改性聚苯乙烯材料空气聚苯乙烯聚四氟乙烯聚丙烯 (204)

约1.0 2.5 3.12 2.1 2.2 ε r

材料聚酯(涤纶) 聚碳酸酯聚砜聚酰亚胺聚苯醚

3.1 3.0 3.1 3.4 2.58 ε r

酚醛玻璃纤材料尼龙66(增高频酚醛塑料氧化铝氧化钽维塑料

(FQBD-12) 强) (塑14-6)

4.0,4.6 7 8 10 27 ε r

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结构形式电容量(F) 平板电容(略去其边缘效应的影响) εSC= d

ε ----介质的介电系

数又称电容率(F/m)

下同

S---电极有效面积

2(m)

d---介质厚度(m)

2εS卷绕型电容 C= d

S---电极(铝箔)有效

2面积(m)

d---元件极间介质厚

度(m)

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圆管型电容 2πεl

C= dln(1+)r

l-----电极有效长度(m)

d-----介质厚度

r-----圆管内半径，与d同单位。

球形电容 r4πεrabC= r?rba

若r ?，则半径为r的孤立导ba

体球的电容为:C=4πεr a

r、r-------内球外表面与外球内表ab

面的半径(m)

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结构形式电容量(F)

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结构形式电容量(F)

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范文二：电容补偿的计算公式

电容补偿的计算公式

未补偿前的负载功率因数为COS ∮1。负载消耗的电流值为I1。

负载功率（KW ）*1000

则I1＝----------------------

√3*380*COS∮1

负载功率（KW ）*1000

则I2＝----------------------

√3*380*COS∮2

补偿后的负载功率因数为COS ∮2，负载消耗的电流值为I2 则所需补偿的电流值为：I ＝I1－I2 所需采用的电容容量参照如下：

得到所需COS ∮2每KW 负荷所需电容量（KVA R ）

例：

现有的负载功率为1500KW ，未补偿前的功率因数为COS ∮1＝0.60，现需将功率因数提高到COS ∮2=0.96。则

1500*1000

则I1＝-----------------＝3802（安培）

√3*380*0.60

1500*1000

则I2＝------------------＝2376（安培）

√3*380*0.96

即未进行电容补偿的情况下，功率因数COS ∮1＝0.60，在此功率因数的状况下，1500KW 负载所需消耗的电流值为I1＝3802安培。

进行电容补偿后功率因数上升到COS ∮2＝0.95，在此功率因数的状况下，1500KW 负载所需消耗的电流值为I2＝2376安培。

所以功率因数从0.60升到0.96。所需补偿的电流值为I1－I2＝1426安培

查表COS ∮1=0.60，COS ∮2=0.96时每KW 负载所需的电容量为1.04KVA R ，现负载为1500KW ，则需采用的电容量为1500*1.04＝1560KVA R 。现每个电容柜的容量为180KVAR ，则需电容柜的数量为 1500÷180＝8.67个即需9个容量为180KVA R 电容柜。

范文三：电容补偿的计算公式[新版]

电容补偿的计算公式

未补偿前的负载功率因数为COS?1。负载消耗的电流值为I1。

负载功率(KW)*1000

则I1,----------------------

?3*380*COS?1

负载功率(KW)*1000

则I2,----------------------

?3*380*COS?2

补偿后的负载功率因数为COS?2，负载消耗的电流值为I2 则所需补偿的电流值为:I,I1,I2

所需采用的电容容量参照如下:

得到所需COS?2每KW负荷所需电容量(KVAR)

例:

现有的负载功率为1500KW，未补偿前的功率因数为COS?1,0.60，现需将功率因数提高到COS?

2=0.96。则

1500*1000

3802(安培) 则I1,-----------------,

?3*380*0.60

1500*1000

则I2,------------------,2376(安培)

?3*380*0.96

即未进行电容补偿的情况下，功率因数COS?1,0.60，在此功率因数的状况下，1500KW负载所需消

耗的电流值为I1,3802安培。

进行电容补偿后功率因数上升到COS?2,0.95，在此功率因数的状况下，1500KW负载所需消耗的电流值为I2,2376安培。

所以功率因数从0.60升到0.96。所需补偿的电流值为I1,I2,1426安培

查表COS?1=0.60，COS?2=0.96时每KW负载所需的电容量为1.04KVAR，现负载为1500KW，则需采用的电容量为1500*1.04,1560KVAR。现每个电容柜的容量为180KVAR，则需电容柜的数量为 1500?180,8.67个即需9个容量为180KVAR电容柜。

范文四：平板计算公式

Design Formulas

Stress Formulas

Strength of Materials

Beam Formulas, Bending Moments Properties of Sections, Moments of Inertia Flat Plate Formulas

Designing for Equal Stiffness Designing for Impact Resistance Designing for Thermal Stress

74 Whether you are designing in metals or

plastics, it is necessary to choose the specific

structural property values for use in stan-

dard design equations. With metals, such

property values are relatively constant over

a wide range of temperatures and time.

But for plastics, the appropriate values are

dependent on temperature, stress level,

and life expectancy of the part.

As far as design practices are involved,

the principles defined in many good engi-

neering handbooks are applicable. However,

the nature of high polymer materials requires

even more attention to appropriate safety

factors.

The information and formulas provided

in this chapter can help you solve many of

the design problems commonly met in the

structural design of plastic parts.

However, it is important that designers

and design engineers understand that the

formulas and the data expressed in this

brochure are given only as guides. They

may not be pertinent to the design of a

particular part, with its own special require-

ments and end-use environments.

Generally, the symbols used in this man-

ual’s various figures, formulas, and text

have the definitions shown in the boxed

column on this page.

Our customers can expect efficient design

assistance and aid from the technical sup-

port services at Dow Plastics. We invite

you to discuss your needs with us.

Above all, there is an aspect of profes-

sional and competent design engineering

that holds true throughout. That is the fact

that, after all the science, mathematics, and

experience have been properly used in

“solving” the design needs of a part, it is

strongly recommended that prototype parts

be produced and thoroughly tested in the

expected end-use conditions and environ-

ments before committing the design to

full-scale production.

Partial list of engineering symbols and letters used, and meanings.

? =Angle

A =Area . . . cross-sectional

? =Coefficient of linear thermal expansion y =Deflection of cantilever; height of undercut

? =Density

D =Diameter

MD =Diameter, major

PD =Diameter, pitch

c =Distance from neutral axis to outer fiber, centroid

z =Distance from q to neutral axis

P =Force . . . P = deflection force

? =Friction, coefficient

a,b,h,t=Height or thickness

I =Inertia, moment of (neutral axis) =Length

? =Length, change

E =Modulus (Young’s)

q =Point within a beam or internal

pressure

R =Radius

r =Radius

=Secant modulus

M =Sectional bending moment

τ=Shear stress

=Strain

? =Stress

TCF =Thickness conversion factor

T =Temperature

? T =Temperature, change

v =Poisson’s ratio

? =Velocity, constant angular,

radius/second

=Width at base

a =Width . . . wall thickness

Stress Formulas

Tensile or Compressive Stress

Tensile or compressive stress ? is the forcecarried per unit of area and is expressedby the equation:

? = P = P

Where:? =stress P =force

A =cross-sectional areaa =width b

=height

The force (P) produces stresses normal(i.e., perpendicular) to the cross section ofthe part. If the stress tends to lengthen thepart, it is called tensile stress. If the stresstends to shorten the part, it is called com-pressive stress. (For compression loading,the part should be relatively short, or it mustbe constrained against lateral bucking.)

Strain

Strain is the ratio of the change in thepart’s length, over the original length. It isexpressed as the percentage of change inlength, or percent elongation.

In direct tension and compression loading,the force is assumed to act along a linethrough the center of gravity of membershaving uniform cross-sections, calledcentroids.

Within the elastic limits of the materials,design formulas developed for metals canalso be applied to plastics. Stress levels aredetermined only by load and part geometry,so standard equations can be used. Deflec-tion is determined by two other materialproperty values: the elastic, or Young’s modulus (E); and Poisson’s ratio (v). Sincethe modulus of a plastic material varies withtemperature and duration of the stress, thismodulus may need replacement in deflec-tion equations by the appropriate creepmodulus. It may be helpful to review vari-ous sections of Chapter 4 for assistance inchoosing modulus values appropriate tothe specific stress level, temperature, anddesign life of the part.

Poisson’s ratio varies with temperature,strain level, and strain rate. These differ-ences are too small to significantly affect acalculation. For example, Poisson’s ratio atroom temperature for CALIBRE polycarbon-ate resin is 0.37, and it ranges from 0.35 to0.40 over the operational temperature range.By selecting the correct modulus and assum-ing the value of Poisson’s ratio to be constant,standard equations can be used to designa part for fabrication in thermoplastics.

76 Stress Acting at an Angle

The standard stress equation is valid when

the cross-section being considered is

perpendicular to the force. However, when

the cross-section is at an angle other than

90° to the force, as shown in Figure 57, the

equation must be adapted. These stresses

are always less than the standard case, i.e.,

maximum normal stress occurs when ? = 0.

Shear Stress

In addition to the normal stress calculated

in the previous section, a plane at an angle

to the force has a shear stress component.

Here, unlike tensile and compressive stress,

the force produces stress in the plane of the

cross-section, i.e., the shear stresses are

perpendicular to tensile or compressive

stresses. The equations for calculating

planar shear stress, based on Figure 58 are:

τ? = P sin ? cos ?

Max ? =P

2A (when ? = 45° or 135°)

Torsional Stress

When a stress acts to twist a component, it produces torsional stress. If a solid circular shaft, or shaft-like component, is subject to a twisting moment, or torsion, the resulting shear stress (q) is calculated by:

G ? r

where:

q = shear stress

G = modulus of rigidity

(see Chapter 4, page 35)

? = angle of twist, in radians

r = radius of shaft

= length of shaft

The torque (T) carried by the shaft is given by G ?

where I

is the polar second moment of

? d4

A useful rearrangement of the formula is

Figure 57 – Diagram of Stress Acting at

an Angle ? Figure 58 – Representation of Shear Stress q =

T = I

area =

? =

Strength of Materials

Beams

When a straight beam of uniform cross-sectional area is subjected to a perpendicular load, the beam bends. If shear is negligible, the vertical deflection is largely due to bend-ing. Fibers on the convex side of the beam lengthen, and fibers on the concave side compress.

There is a neutral surface within any beam that contains the centroids of all sections and is perpendicular to the plane of the load for such deflections. In a uniform, symmetri-cal beam, the neutral axis of the beam is the horizontal, central axis. Tensile or compres-sive stress and strain on the neutral axis are essentially zero. At all other points within the beam, the stress is a tensile stress if the point lies between the neutral axis and convex surfaces of the beam, and is a compressive stress if the point lies between the neutral axis and concave surfaces of the beam, see Figure 59. The fiber stress ? for any point (q) within the beam is calculated using the equation:? = Mz

where:

M =bending moment of the section containing q (values can be taken from the appro-priate beam formula, Figures 60 to 68). z =the distance from q to the neutral axis I =the moment of inertia with respect to the neutral axis (values can be taken from the appropriate cross-sectional area formula, Figures 69 to 91).

The maximum fiber stress in any section occurs at the points farthest from the neu-tral surface and at the section of greatest bending moment, i.e., when z = Max z, and M = Max M. Maximum fiber stress is given by the equation:

Max ? = Mc

where:

c =the distance from the neutral axis to the

extreme outermost fiber.

Such equations are valid if:

? The beam is of homogeneous material, so that it has the same modulus of elasticity in tension and compression.? Plane sections remain planar.

If several loads are applied at the same time, the total stress and deflection at any point are found by superimposition. Compute the stress and deflection for each load acting on the point, and add them together. Figure 59 – Bending of a Beam

Figure 60 – Cantilever Beam, concen-trated load at free end

Beam Formulas, Bending Moments

Figure 63 – Simple Beam, concentratedload off center

Figure 61 – Cantilever Beam, uniform

load, w per unit length, total load WFigure 62 – Simple Beam, concentratedload at center

Figure 64 – Simple Beam, two equal, concentrated loads, symmetrically placed Figure 66 – Beam fixed at both ends, concentrated load at center

80 Figure 67 – Beam fixed at both ends, concentrated load at any point

Properties of Sections, Moments of Inertia

Figure 71Figure 74Figure 77

Figure 79Figure 81Figure 84

NOTICE: Dow believes the information and recommendations contained herein to be accurate and reliable as of March 2001. However, since any assistance furni shed by Dow with reference to the proper use and disposal of its products is provided without charge, and since use conditi ons and disposal are not within its control, Dow assumes no obligation or liability for such assistance and does not guarantee results from use of such products or other infor-mation contained herein. No warranty, express or implied, is given nor is freedom from any patent owned by Dow or others to be inferred. Information contained herei n concerning laws and regulations is based on U.S. federal laws and regulations except where specifi c reference is made to those of other juri sdi cti ons. Since use conditi ons and governmental regulations may differ from one location to another and may change with time, it is the Buyer’s responsi bi li ty to determine whether Dow’s products are appropriate for Buyer’s use, and to assure Buyer’s workplace and disposal practices are in compliance with laws, regulations, ordinances, and other governmental enactments applicable in the jurisdiction(s) having authority over Buyer’s operations.

范文五：电容计算公式

电容定义式

C=Q/U

Q=I*T 电容放电时间计算：C=(Vwork+

Vmin)*I*t/( Vwork -Vmin ) 2 2电容计算公式.xlsx

电压(V) = 电流(I) x 电阻(R)

电荷量(Q) = 电流(I) x 时间(T)

功率(P) = V x I (I=P/U; P=Q*U/T)

能量(W) = P x T = Q x V

容量 F= 库伦（C ） / 电压（V ）

将容量、电压转为等效电量

电量=电压（V) x 电荷量（C ）

实例估算：电压5.5V 1F（1法拉电容）的电量为5.5C （库伦），电压下限是3.8V ，电容放电的有效电压差为

5.5-3.8=1.7V，所以有效电量为1.7C 。

1.7C=1.7A*S（安秒）=1700mAS（毫安时）=0.472mAh（安时）

若电流消耗以10mA 计算，1700mAS/10mA=170S=2.83min（维持时间分钟）

电容放电时间的计算

在超级电容的应用中，很多用户都遇到相同的问题，就是怎样计算一定容量的超级电容在以一定电流放电时的放电时间，或者根据放电电流及放电时间，怎么选择超级电容的容量，下面我们给出简单的计算公司，用户根据这个公式，就可以简单地进行电容容量、放电电流、放电时间的推算，十分地方便。

C(F)：超电容的标称容量；

R(Ohms)：超电容的标称内阻；

ESR(Ohms)：1KZ 下等效串联电阻；

Vwork(V)：正常工作电压

Vmin(V)：截止工作电压；

t(s)：在电路中要求持续工作时间；

Vdrop(V)：在放电或大电流脉冲结束时，总的电压降；

I(A)：负载电流；

超电容容量的近似计算公式，

保持所需能量=超级电容减少的能量。

保持期间所需能量=1/2I(Vwork+ Vmin)t；

超电容减少能量=1/2C(Vwork -Vmin) ，

因而，可得其容量(忽略由IR 引起的压降) 22

C=(Vwork+ Vmin)*I*t/( Vwork2 -Vmin2)

举例如下：

如单片机应用系统中，应用超级电容作为后备电源，在掉电后需要用超级电容维持100mA 的电流，持续时间为10s, 单片机系统截止工作电压为4.2V ，那么需要多大容量的超级电容能够保证系统正常工作？

由以上公式可知：

工作起始电压Vwork ＝5V

工作截止电压Vmin ＝4.2V

工作时间t=10s

工作电源I ＝0.1A

那么所需的电容容量为：

C=(Vwork+ Vmin)*I*t/( Vwork2 -Vmin2)

=(5+4.2)*0.1*10/(52 -4.22)

=1.25F

根据计算结果，可以选择5.5V 1.5F 电容就可以满足需要了。

实例：

假设磁带驱动的工作电压5V ，安全工作电压3V 。如果直流马达要求0.5A 保持2秒（可以安全工作），问需要选用多大容量的超级电容？

解： C=(Uwork+ Umin)It/(Uwork*Uwork -Umin*Umin)

＝（5+3）*0.2*2／（5*5-3*3）

＝0.5F

因为5V 的电压超过了单体电容器的标称工作电压。因而，可以将两电容器串联。如两相同的电容器串联的话，那每只的电压即是其标称电压2.5V 。

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