====== Block 12 - Capacitors and Capacitance ====== ===== Learning objectives ===== After this 90-minute block, you can - define a **capacitor** and **capacitance** $C$ and use it for an ideal plate capacitor, including unit checks $[C]={\rm F}$. - relate fields and material. - compute $C$ for key geometries (parallel plates, coaxial, spherical) and explain how $A$, $d$, $\varepsilon_{\rm r}$ scale $C$. ===== Preparation at Home ===== Well, again * read through the present chapter and write down anything you did not understand. * Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting). For checking your understanding please do the following exercises: * ... ===== 90-minute plan ===== - Warm-up (8 min): - Quick recall quiz: $C=\dfrac{Q}{U}$, $\vec{D}=\varepsilon\vec{E}$, units of $E$ (${\rm V/m}$), $D$ (${\rm C/m^2}$). - Estimate: how $C$ changes when $A$ doubles or $d$ halves (plate model). - Core concepts & derivations (60 min): - From fields to $C$ (plate capacitor): $U=\int\vec{E}\cdot{\rm d}\vec{s}=E\,d$, $Q=\oint\vec{D}\cdot{\rm d}\vec{A}=D\,A$, $\Rightarrow C=\varepsilon_0\varepsilon_{\rm r}\dfrac{A}{d}$. Dimensional check: $[\varepsilon_0\varepsilon_{\rm r}A/d]={\rm F}$. - Other geometries: coaxial and spherical capacitor formulas; where fields are highest (edge intuition kept qualitative). - Practice (20 min): - Mini-calcs: (i) $C$ of given $A,d,\varepsilon_{\rm r}$; (ii) coaxial $C$ per length; (iii) allowable $U$ from $E_0$ and $d$; (iv) energy at given $U$. - Discuss the provided “glass plate in capacitor” task. - Wrap-up (2 min): Summary box + pitfalls checklist; connect to next block (capacitor circuits). ===== Conceptual overview ===== - A **capacitor** is two conductors separated by a dielectric. It stores **charge** and **energy** in the electric field; no conduction current flows through the ideal dielectric. :contentReference[oaicite:15]{index=15} - **Capacitance** measures how much charge per volt: $C=\dfrac{Q}{U}$. For parallel plates, $C=\varepsilon_0\varepsilon_{\rm r}\dfrac{A}{d}$ → increase $A$ or $\varepsilon_{\rm r}$, decrease $d$ to raise $C$. :contentReference[oaicite:16]{index=16} - **Other geometries:** Closed forms exist for coaxial and spherical capacitors; useful as building blocks and for cables/sensors. :contentReference[oaicite:18]{index=18} ===== Core content ===== ==== Capacitor ==== * A capacitor can "store" charges. The total charge of a two-plate capacitor is in general 0. * From the mechanical point of view a capacitor has two electrodes (= conductive areas), which are separated by a dielectric (= non-conductor). * In a capacitor an electric field can be established without charge carriers moving internally. * The characteristic of the capacitor is the capacitance $C$. * In addition to the capacitance, every capacitor also has resistance and inductance. However, both of these are usually very small and are neglected for an ideal capacitor. * Examples of capacitor are * the electrical component "capacitor", * an open switch, * a wire related to ground, * a human being $\rightarrow$ Thus, for any arrangement of two conductors separated by an insulating material, a capacitance can be specified. ==== Capacitance $C$ ==== The capacitance $C$ can be derived as follows: - A plate capacitor has a nearly homogenious field. \\ Therefore, it is given for the voltage: \\ $$U = \int \vec{E} {\rm d} \vec{s} = E \cdot l$$ \\ and hence \\ $$E= {{U}\over{l}}$$ or with $D = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot E $ $$D= \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot {{U}\over{l}}$$ - Furthermore, the charge $Q$ can be given as $${Q= \rlap{\Large \rlap{\int_A} \int} \, \LARGE \circ} \; \vec{D} \cdot {\rm d} \vec{A} $$ The idealizion for the plate capacitor leads to: $$Q=D \cdot A$$. - Thus, the charge $Q$ is given by: \begin{align*} Q = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot {{U}\over{l}} \cdot A \end{align*} - This means that $Q \sim U$, given the geometry (i.e., $A$ and $d$) and the dielectric ($\varepsilon_{ \rm r} $). - So it is reasonable to determine a proportionality factor ${{Q}\over{U}}$. The capacitance $C$ of an idealized plate capacitor is defined as \begin{align*} \boxed{C = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot {{A}\over{l}} = {{Q}\over{U}}} \end{align*} Some of the main results here are: * The capacity can be increased by increasing the dielectric constant $\varepsilon_{ \rm r} $, given the the same geometry. * As near together the plates are as higher the capacity will be. * As larger the area as higher the capacity will be. This relationship can be examined in more detail in the following simulation: --> Capacitor lab# If the simulation is not displayed optimally, [[https://phet.colorado.edu/sims/cheerpj/capacitor-lab/latest/capacitor-lab.html?simulation=capacitor-lab&locale=de|this link]] can be used. {{url>https://phet.colorado.edu/sims/cheerpj/capacitor-lab/latest/capacitor-lab.html?simulation=capacitor-lab&locale=de 900,800 noborder}} <-- The shows the topology of the electric field inside a plate capacitor. {{url>https://www.falstad.com/vector2de/vector2de.html?f=ChargedPlateDipole&fc=Floor%3A%20field%20magnitude&fl=Overlay%3A%20equipotentials&d=vectors&m=Mouse%20%3D%20Adjust%20Angle&st=20&vd=32&a1=63&a2=16&rx=77&ry=5&rz=107&zm=1.165 600,400 noborder}} ==== Symbols ==== * The symbol of a general capacitor is given be two parallel lines nearby each other. \\ * Since **electrolytic capacitors** can only withstand voltage in one direction, the **polarisation** is often shown by a curved electrode (US) or a unfilled one (EU). \\ Be aware that electrolytic capacitors can explode, once used in the wrong direction. {{drawio>electrical_engineering_and_electronics_1:CapSymbols01.svg}} ==== Designs and types of capacitors ==== To calculate the capacitance of different designs, the definition equations of $\vec{D}$ and $\vec{E}$ are used. This can be viewed in detail, e.g., in [[https://www.youtube.com/watch?v=kAXg1xMkR_4&ab_channel=PatrickKaploo|this video]]. \\ Based on the geometry, different equations result (see also ). {{drawio>GeometryCapacitors.svg}} ^Shape of the Capacitor^ Parameter ^ Equation for the Capacity ^ |plate capacitor | area $A$ of plate \\ distance $l$ between plates | \begin{align*}C = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot {{A}\over{l}} \end{align*}| |cylinder capacitor |radius of outer conductor $R_{ \rm o}$ \\ radius of inner conductor $R_{ \rm i}$ \\ length $l$ | \begin{align*}C = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot 2\pi {{l}\over{{\rm ln} \left({{R_{ \rm o}}\over{R_{ \rm i}}}\right)}} \end{align*}| |spherical capacitor |radius of outer spherical conductor $R_{ \rm o}$ \\ radius of inner spherical conductor $R_{ \rm i}$ | \begin{align*}C = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot 4 \pi {{R_{ \rm i} \cdot R_{ \rm o}}\over{R_{ \rm o} - R_{ \rm i}}} \end{align*} | ~~PAGEBREAK~~ ~~CLEARFIX~~ {{drawio>DesignsCapacitors.svg}} In different designs of capacitors can be seen: - **{{wp>variable_capacitor|rotary variable capacitor}}** (also variable capacitor or trim capacitor). - A variable capacitor consists of two sets of plates: a fixed set and a movable set (stator and rotor). These represent the two electrodes. - The movable set can be rotated radially into the fixed set. This covers a certain area of $A$. - The size of the area is increased by the number of plates. Nevertheless, only small capacities are possible because of the necessary distance. - Air is usually used as the dielectric; occasionally, small plastic or ceramic plates are used to increase the dielectric constant. - **{{wp>Ceramic_capacitor#Multi-layer_ceramic_capacitors_(MLCC)|multilayer capacitor}}** - In the multilayer capacitor, there are again two electrodes. Here, too, the area $A$ (and thus the capacitance $C$) is multiplied by the finger-shaped interlocking. - Ceramic is used here as the dielectric. - The multilayer ceramic capacitor is also referred to as KerKo or MLCC. - The variant shown in (2) is an SMD variant (surface mound device). - Disk capacitor - A ceramic is also used as a dielectric for the disk capacitor. This is positioned as a round disc between two electrodes. - Disc capacitors are designed for higher voltages, but have a low capacitance (in the microfarad range). - **{{wp>Electrolytic capacitor}}**, in German also referred to as //Elko// for //__El__ektrolyt__ko__ndensator// - In electrolytic capacitors, the dielectric is an oxide layer formed on the metallic electrode. The second electrode is the liquid or solid electrolyte. - Different metals can be used as the oxidized electrode, e.g., aluminum, tantalum, or niobium. - Because the oxide layer is very thin, a very high capacitance results (depending on the size: up to a few millifarads). - Important for the application is that it is a polarized capacitor. I.e., it may only be operated in one direction with DC voltage. Otherwise, a current can flow through the capacitor, which destroys it and is usually accompanied by an explosive expansion of the electrolyte. To avoid reverse polarity, the negative pole is marked with a dash. - The electrolytic capacitor is built up wrapped, and often has a cross-shaped predetermined breaking point at the top for gas leakage. - **{{wp>film capacitor}}**, in German also referred to as //Folko//, for //__Fol__ien__ko__ndensator//. - A material similar to a "chip bag" is used as an insulator: a plastic film with a thin, metalized layer. - The construction shows a high pulse load capacitance and low internal ohmic losses. - In the event of an electrical breakdown, the foil enables "self-healing": the metal coating evaporates locally around the breakdown. Thus the short-circuit is canceled again. - With some manufacturers, this type is referred to as MKS (__M__metallized foil__c__capacitor, Polye__s__ter). - **{{wp>Supercapacitor}}** (engl. Super-Caps) - As a dielectric is - similar to the electrolytic capacitor - very thin. In the actual sense, there is no dielectric at all. - The charges are not only stored in the electrode, but - similar to a battery - the charges are transferred into the electrolyte. Due to the polarization of the charges, they surround themselves with a thin (atomic) electrolyte layer. The charges then accumulate at the other electrode. - Supercapacitors can achieve very large capacitance values (up to the Kilofarad range), but only have a low maximum voltage ~~PAGEBREAK~~ ~~CLEARFIX~~ {{elektrotechnik_1:kondensatorensmd.jpg}}{{elektrotechnik_1:kondensatorentht.jpg}} In are shown different capacitors: - The above two SMD capacitors - On the left a $100~{ \rm µF}$ electrolytic capacitor - On the right a $100~{ \rm nF}$ MLCC in the commonly used {{wp>Surface-mount_technology#Packages}} 0603 ($1.6~{ \rm mm}$ x $0.8~{ \rm mm}$) - below different THT capacitors (__T__hrough __H__ole __T__echnology) - A big electrolytic capacitor with $10~{ \rm mF}$ in blue, the positive terminal is marked with $+$ - In the second row is a Kerko with $33~{ \rm pF}$ and two Folkos with $1,5~{ \rm µF}$ each - In the bottom row, you can see a trim capacitor with about $30~{ \rm pF}$ and a tantalum electrolytic capacitor and another electrolytic capacitor Various conventions have been established for designating the capacitance value of a capacitor [[https://www.eit.edu.au/resources/different-types-of-capacitors/|various conventions]]. - There are polarized capacitors. With these, the installation direction and current flow must be observed, as otherwise, an explosion can occur. - Depending on the application - and the required size, dielectric strength, and capacitance - different types of capacitors are used. - The calculation of the capacitance is usually __not__ via $C = \varepsilon_0 \cdot \varepsilon_{ \rm r} \cdot {{A}\over{l}} $ . The capacitance value is given. - The capacitance value often varies by more than $\pm 10~{ \rm \%}$. I.e., a calculation accurate to several decimal places is rarely necessary/possible. - The charge current seems to be able to flow through the capacitor because the charges added to one side induce correspondingly opposite charges on the other side. ~~PAGEBREAK~~ ~~CLEARFIX~~ ===== Common pitfalls ===== * **Mixing up geometry symbols.** Use $d$ (or $l$) strictly as **plate spacing** and $A$ as **active area** in $C=\varepsilon_0\varepsilon_{\rm r}\dfrac{A}{d}$. Check units to catch mistakes. * **Forgetting the field relations.** $U=\int \vec{E}\cdot{\rm d}\vec{s}$ and $Q=\oint \vec{D}\cdot{\rm d}\vec{A}$; without them, layered-dielectric problems are guessed instead of solved. * **Assuming conduction through the dielectric.** The apparent “current through a capacitor” is displacement-related; no charge carriers traverse the ideal dielectric. * **Real-part issues.** Ignoring polarity of electrolytics and tolerance spreads ($\pm 10~\%$ and more) causes design errors; pick suitable component types. ===== Exercises ===== {{fa>pencil?32}} A plate capacitor with a distance of $d = 2 ~{ \rm cm}$ between the plates and with air as dielectric ($\varepsilon_{ \rm r}=1$) gets charged up to $U = 5~{ \rm kV}$. In between the plates, a thin metal foil with the area $A = 45~{ \rm cm^2}$ is introduced parallel to the plates. Calculate the amount of the displaced charges in the thin metal foil. * What is the strength of the electric field $E$ in the capacitor? * Calculate the displacement flux density $D$ * How can the charge $Q$ be derived from $D$? $Q = 10 ~{ \rm nC}$ {{fa>pencil?32}} An ideal plate capacitor with a distance of $d_0 = 7 ~{ \rm mm}$ between the plates gets charged up to $U_0 = 190~{ \rm V}$ by an external source. The source gets disconnected. After this, the distance between the plates gets enlarged to $d_1 = 7 ~{ \rm cm}$. - What happens to the electric field and the voltage? - How does the situation change (electric field/voltage), when the source is not disconnected? * Consider the displacement flux through a surface around a plate - $U_1 = 1.9~{ \rm kV}$, $E_1 = 27~{ \rm kV/m}$ - $U_1 = 190~{ \rm V}$, $E_1 = 2.7~{ \rm kV/m}$ {{fa>pencil?32}} An ideal plate capacitor with a distance of $d_0 = 6 ~{ \rm mm}$ between the plates and with air as dielectric ($\varepsilon_0=1$) is charged to a voltage of $U_0 = 5~{ \rm kV}$. The source remains connected to the capacitor. In the air gap between the plates, a glass plate with $d_{ \rm g} = 4 ~{ \rm mm}$ and $\varepsilon_{ \rm r} = 8$ is introduced parallel to the capacitor plates. 1. Calculate the partial voltages on the glas $U_{ \rm g}$ and on the air gap $U_{ \rm a}$. #@HiddenBegin_HTML~1531T,Tipps~@# * Build a formula for the sum of the voltages first * How is the voltage related to the electric field of a capacitor? #@HiddenEnd_HTML~1531T,Tipps~@# #@HiddenBegin_HTML~1531P,Path~@# The sum of the voltages across the glass and the air gap gives the total voltage $U_0$, and each individual voltage is given by the $E$-field in the individual material by $E = {{U}\over{d}}$: \begin{align*} U_0 &= U_{\rm g} + U_{\rm a} \\ &= E_{\rm g} \cdot d_{\rm g} + E_{\rm a} \cdot d_{\rm a} \end{align*} The displacement field $D$ must be continuous across the different materials since it is only based on the charge $Q$ on the plates. \begin{align*} D_{\rm g} &= D_{\rm a} \\ \varepsilon_0 \varepsilon_{\rm r, g} \cdot E_{\rm g} &= \varepsilon_0 \cdot E_{\rm a} \end{align*} Therefore, we can put $E_\rm a= \varepsilon_{\rm r, g} \cdot E_\rm g $ into the formula of the total voltage and rearrange to get $E_\rm g$: \begin{align*} U_0 &= E_{\rm g} \cdot d_{\rm g} + \varepsilon_{\rm r, g} \cdot E_{\rm g} \cdot d_{\rm a} \\ &= E_{\rm g} \cdot ( d_{\rm g} + \varepsilon_{\rm r, g} \cdot d_{\rm a}) \\ \rightarrow E_{\rm g} &= {{U_0}\over{d_{\rm g} + \varepsilon_{\rm r, g} \cdot d_{\rm a}}} \end{align*} Since we know that the distance of the air gap is $d_{\rm a} = d_0 - d_{\rm a}$ we can calculate: \begin{align*} E_{\rm g} &= {{5'000 ~\rm V}\over{0.004 ~{\rm m} + 8 \cdot 0.002 ~{\rm m}}} \\ &= 250 ~\rm{{kV}\over{m}} \end{align*} By this, the individual voltages can be calculated: \begin{align*} U_{ \rm g} &= E_{\rm g} \cdot d_\rm g &&= 250 ~\rm{{kV}\over{m}} \cdot 0.004~\rm m &= 1 ~{\rm kV}\\ U_{ \rm a} &= U_0 - U_{ \rm g} &&= 5 ~{\rm kV} - 1 ~{\rm kV} &= 4 ~{\rm kV}\\ \end{align*} #@HiddenEnd_HTML~1531P,Path~@# #@HiddenBegin_HTML~1531R,Results~@# $U_{ \rm a} = 4~{ \rm kV}$, $U_{ \rm g} = 1 ~{ \rm kV}$ #@HiddenEnd_HTML~1531R,Results~@# 2. What would be the maximum allowed thickness of a glass plate, when the electric field in the air-gap shall not exceed $E_{ \rm max}=12~{ \rm kV/cm}$? #@HiddenBegin_HTML~1532P,Path~@# Again, we can start with the sum of the voltages across the glass and the air gap, such as the formula we got from the displacement field: $D = \varepsilon_0 \varepsilon_{\rm r, g} \cdot E_{\rm g} = \varepsilon_0 \cdot E_\rm a $. \\ Now we shall eliminate $E_\rm g$, since $E_\rm a$ is given in the question. \begin{align*} U_0 &= E_{\rm g} \cdot d_{\rm g} + E_{\rm a} \cdot d_{\rm a} \\ &= {{E_\rm a}\over{\varepsilon_{\rm r,g}}} \cdot d_{\rm g} + E_{\rm a} \cdot d_{\rm a} \\ \end{align*} The distance $d_\rm a$ for the air is given by the overall distance $d_0$ and the distance for glass $d_\rm g$: \begin{align*} d_{\rm a} = d_0 - d_{\rm g} \end{align*} This results in: \begin{align*} U_0 &= {{E_{\rm a}}\over{\varepsilon_{\rm r,g}}} \cdot d_{\rm g} + E_{\rm a} \cdot (d_0 - d_{\rm g}) \\ {{U_0}\over{E_{\rm a} }} &= {{1}\over{\varepsilon_{\rm r,g}}} \cdot d_{\rm g} + d_0 - d_{\rm g} \\ &= d_{\rm g} \cdot ({{1}\over{\varepsilon_{\rm r,g}}} - 1) + d_0 \\ d_{\rm g} &= { { {{U_0}\over{E_{\rm a} }} - d_0 } \over { {{1}\over{\varepsilon_{\rm r,g}}} - 1 } } &= { { d_0 - {{U_0}\over{E_{\rm a} }} } \over { 1 - {{1}\over{\varepsilon_{\rm r,g}}} } } \end{align*} With the given values: \begin{align*} d_{\rm g} &= { { 0.006 {~\rm m} - {{5 {~\rm kV} }\over{ 12 {~\rm kV/cm}}} } \over { 1 - {{1}\over{8}} } } &= { {{8}\over{7}} } \left( { 0.006 - {{5 }\over{ 1200}} } \right) {~\rm m} \end{align*} #@HiddenEnd_HTML~1532P,Path~@# #@HiddenBegin_HTML~1532R,Results~@# $d_{ \rm g} = 2.10~{ \rm mm}$ #@HiddenEnd_HTML~1532R,Results~@# {{fa>pencil?32}} Two concentric spherical conducting plates set up a spherical capacitor. The radius of the inner sphere is $r_{ \rm i} = 3~{ \rm mm}$, and the inner radius from the outer sphere is $r_{ \rm o} = 9~{ \rm mm}$. - What is the capacity of this capacitor, given that air is used as a dielectric? - What would be the limit value of the capacity when the inner radius of the outer sphere goes to infinity ($r_{ \rm o} \rightarrow \infty$)? * What is the displacement flux density of the inner sphere? * Out of this derive the strength of the electric field $E$ * What ist the general relationship between $U$ and $\vec{E}$? Derive from this the voltage between the spheres. - $C = 0.5~{ \rm pF}$ - $C_{\infty} = 0.33~{ \rm pF}$ {{fa>pencil?32}} {{youtube>NyRjHj2uy6k}} ===== Embedded resources ===== The background behind the dielectric constant $\varepsilon_{ \rm r} $ and the field is explained in the following video \\ {{youtube>rkntp3_cZl4}} Electrolytic capacitors can explode! \\ {{youtube>sW0a9d_vWoc}} ~~PAGEBREAK~~ ~~CLEARFIX~~