Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_2:magnetic_circuits [2023/03/17 10:44] – mexleadmin | electrical_engineering_2:magnetic_circuits [2025/05/27 07:56] (aktuell) – mexleadmin | ||
|---|---|---|---|
| Zeile 1: | Zeile 1: | ||
| - | ====== 5. Magnetic Circuits ====== | + | ====== 5 Magnetic Circuits ====== |
| < | < | ||
| Zeile 7: | Zeile 7: | ||
| < | < | ||
| - | In this chapter, we will investigate, how far we come with such an analogy and where it can be practically applied. | + | In this chapter, we will investigate how far we have come with such an analogy and where it can be practically applied. |
| - | ===== 5.1 Linear | + | ===== 5.1 Linear |
| For the upcoming calculations, | For the upcoming calculations, | ||
| Zeile 17: | Zeile 17: | ||
| - The fields inside of airgaps are homogeneous. This is true for small air gaps. | - The fields inside of airgaps are homogeneous. This is true for small air gaps. | ||
| - | One can calculate a lot of simple magnetic circuits when these assumptions and focusing on the average field line are applied. | + | One can calculate a lot of simple magnetic circuits when these assumptions |
| < | < | ||
| Zeile 25: | Zeile 25: | ||
| * a current-carrying coil | * a current-carrying coil | ||
| * a ferrite core | * a ferrite core | ||
| - | * an airgap (in the picture (2) ). | + | * an airgap (in picture (2) ). |
| < | < | ||
| Zeile 53: | Zeile 53: | ||
| * The field in the air gap can be used to generate (mechanical) effects within the air gap. \\ An example of this can be the force onto a permanent magnet (see <imgref ImgNr02> (3)). | * The field in the air gap can be used to generate (mechanical) effects within the air gap. \\ An example of this can be the force onto a permanent magnet (see <imgref ImgNr02> (3)). | ||
| - | With the above-mentioned assumptions the magnetic flux $\Phi$ must remain constant along the ferrite core, so $\Phi_{\rm core}=const.$. \\ | + | With the above-mentioned assumptions the magnetic flux $\Phi$ must remain constant along the ferrite core, so $\Phi_{\rm core}=\rm const.$. \\ |
| Since the magnetic field lines neither show sources nor sinks, also the flux passing over to the airgap must be $\Phi_{\rm airgap}=\Phi_{\rm core}=\rm const.$ This can also be seen in <imgref ImgNr04> (1) ). \\ | Since the magnetic field lines neither show sources nor sinks, also the flux passing over to the airgap must be $\Phi_{\rm airgap}=\Phi_{\rm core}=\rm const.$ This can also be seen in <imgref ImgNr04> (1) ). \\ | ||
| - | A different view onto this is the closed surface $\vec{A}$ (<imgref ImgNr04> (2)): Based on the examination in [[: | + | A different view of this is the closed surface $\vec{A}$ (<imgref ImgNr04> (2)): Based on the examination in [[: |
| The relationship $B=\mu \cdot H$, and $\mu_{\rm core}\gg\mu_{\rm airgap}$ lead to the fact that the ${H}$-Field must be much stronger within the airgap (<imgref ImgNr04> (3)). | The relationship $B=\mu \cdot H$, and $\mu_{\rm core}\gg\mu_{\rm airgap}$ lead to the fact that the ${H}$-Field must be much stronger within the airgap (<imgref ImgNr04> (3)). | ||
| Zeile 88: | Zeile 88: | ||
| With the prevoius formula $5.2.1$, this gets to: | With the prevoius formula $5.2.1$, this gets to: | ||
| \begin{align*} | \begin{align*} | ||
| - | \theta &= {{B}\over{\mu_0 \mu_{\rm r,core}}} l_{\rm core} + {{B}\over{\mu_0 \mu_{\rm r,airgap}}} \delta \\ | + | \theta &= {{B}\over{\mu_0 \mu_{\rm r,core}}} l_{\rm core} |
| - | & | + | & |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| - | Comparing the formula $5.2.3$ with the ohmic resistance and resistivity of two resistors in series, shows something interesting: | + | Comparing the formula $5.2.3$ with the ohmic resistance and resistivity of two resistors in series shows something interesting: |
| \begin{align*} | \begin{align*} | ||
| u &= &R_1 \cdot &i &+ &R_2 \cdot i \\ | u &= &R_1 \cdot &i &+ &R_2 \cdot i \\ | ||
| - | &= &\rho {{l_1}\over{A_1}} \cdot &i &+ &\rho {{l_2}\over{A_2}} \cdot I | + | &= &\rho {{l_1}\over{A_1}} \cdot &i &+ &\rho {{l_2}\over{A_2}} \cdot i |
| \end{align*} | \end{align*} | ||
| Zeile 140: | Zeile 140: | ||
| |Simplifications |The simplifications often work for good results \\ (small wire diameter, relatively constant resistivity) |The simplification is often too simple \\ (widespread beyond the mean magnetic path length, non-linearity of the permeability) | | |Simplifications |The simplifications often work for good results \\ (small wire diameter, relatively constant resistivity) |The simplification is often too simple \\ (widespread beyond the mean magnetic path length, non-linearity of the permeability) | | ||
| - | <panel type=" | + | <panel type=" |
| A coil is set up onto a toroidal plastic ring ($\mu_{\rm r}=1$) with an average circumference of $l_R = 300 ~\rm mm$. | A coil is set up onto a toroidal plastic ring ($\mu_{\rm r}=1$) with an average circumference of $l_R = 300 ~\rm mm$. | ||
| Zeile 164: | Zeile 164: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| Calculate the magnetic resistances of cylindrical coreless (=ironless) coils with the following dimensions: | Calculate the magnetic resistances of cylindrical coreless (=ironless) coils with the following dimensions: | ||
| - | - $l=35.8~\rm cm$, $d=1.9~\rm cm$ | + | - $l=35.8~\rm cm$, $d=1.90~\rm cm$ |
| - | - $l=22.5~\rm cm$, $d=1.5~\rm cm$ | + | - $l=11.1~\rm cm$, $d=1.50~\rm cm$ |
| - | <button size=" | + | # |
| - | - $1.5\cdot 10^5 ~\rm {{1}\over{H}}$ | + | The magnetic resistance is given by: |
| - | - $3.0\cdot 10^5 ~\rm {{1}\over{H}}$ | + | \begin{align*} |
| + | \ R_{\rm m} & | ||
| + | \end{align*} | ||
| - | </collapse> | + | With |
| + | * the area $ A = \left({{d}\over{2}}\right)^2 \cdot \pi $ | ||
| + | * the vacuum magnetic permeability $\mu_{0}=4\pi\cdot 10^{-7} ~\rm H/m$, and | ||
| + | * the relative permeability $\mu_{\rm r}=1$. | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | - $1.00\cdot 10^9 ~\rm {{1}\over{H}}$ | ||
| + | - $0.50\cdot 10^9 ~\rm {{1}\over{H}}$ | ||
| + | # | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| Calculate the magnetic resistances of an airgap with the following dimensions: | Calculate the magnetic resistances of an airgap with the following dimensions: | ||
| Zeile 196: | Zeile 208: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| - | Calculate the magnetic voltage necessary | + | Calculate the magnetic voltage necessary to create a flux of $\Phi=0.5 ~\rm mVs$ in an airgap with the following dimensions: |
| - $\delta=1.7~\rm mm$, $A=4.5~\rm cm^2$ | - $\delta=1.7~\rm mm$, $A=4.5~\rm cm^2$ | ||
| Zeile 212: | Zeile 224: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| Calculate the magnetic flux created on a magnetic resistance of $R_m = 2.5 \cdot 10^6 ~\rm {{1}\over{H}}$ with the following magnetic voltages: | Calculate the magnetic flux created on a magnetic resistance of $R_m = 2.5 \cdot 10^6 ~\rm {{1}\over{H}}$ with the following magnetic voltages: | ||
| Zeile 230: | Zeile 242: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| - | A core shall consist of two parts as seen in <imgref ImgExNr08> | + | A core shall consist of two parts, as seen in <imgref ImgExNr08> |
| - | In the coil with $600$ windings shall pass the current $I=1.30 ~\rm A$. | + | In the coil, with $600$ windings shall pass the current $I=1.30 ~\rm A$. |
| The cross sections are $A_1=530 ~\rm mm^2$ and $A_2=460 ~\rm mm^2$. | The cross sections are $A_1=530 ~\rm mm^2$ and $A_2=460 ~\rm mm^2$. | ||
| The mean magnetic path lengths are $l_1 = 200 ~\rm mm$ and $l_2 = 130 ~\rm mm$. | The mean magnetic path lengths are $l_1 = 200 ~\rm mm$ and $l_2 = 130 ~\rm mm$. | ||
| - | The air gaps on the coupling joint between both parts have the length $\delta=0.23 ~\rm mm$ each. | + | The air gaps on the coupling joint between both parts have the length $\delta = 0.23 ~\rm mm$ each. |
| The permeability of the ferrite is $\mu_r = 3000$. | The permeability of the ferrite is $\mu_r = 3000$. | ||
| The cross-section area $A_{\delta}$ of the airgap can be considered the same as $A_2$ | The cross-section area $A_{\delta}$ of the airgap can be considered the same as $A_2$ | ||
| Zeile 258: | Zeile 270: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| The magnetic circuit in <imgref ImgExNr09> | The magnetic circuit in <imgref ImgExNr09> | ||
| Zeile 277: | Zeile 289: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| - | The choke coil shown in < | + | The choke coil shown in < |
| The number of windings shall be $N$ and the current through a single winding $I$. | The number of windings shall be $N$ and the current through a single winding $I$. | ||
| - | < | + | < |
| - Draw the lumped circuit of the magnetic system | - Draw the lumped circuit of the magnetic system | ||
| Zeile 296: | Zeile 308: | ||
| ===== 5.3 Mutual Induction and Coupling ===== | ===== 5.3 Mutual Induction and Coupling ===== | ||
| - | Situation: Two coils $1$ and $2$ near each other. \\ Questions: | + | Imagine charging your phone wirelessly by simply placing it on a charging pad. |
| + | This seamless experience is made possible by the fascinating phenomenon of mutual induction and coupling between two coils. | ||
| + | |||
| + | This situation is depicted in <imgref ImgNr09>: | ||
| + | When an alternating current flows through one coil (Coil $1$), it creates a time-varying magnetic field that induces a voltage in the nearby coil (Coil $2$), even though they are not physically connected. | ||
| + | This mutual influence is governed by the principle of electromagnetic induction. | ||
| + | |||
| + | <WRAP center 35%> < | ||
| + | |||
| + | The key factor determining the strength of mutual induction is the mutual inductance ($M$) between the coils. | ||
| + | It quantifies the magnetic flux linkage and depends on factors like the number of turns, current, and relative orientation of the coils. | ||
| + | |||
| + | While geometric properties play a role, the fundamental principle can be described using electric properties alone, making mutual induction a versatile concept with numerous applications, | ||
| + | |||
| + | * Wireless power transfer | ||
| + | * Transformers | ||
| + | * Inductive coupling in communication systems | ||
| + | * Inductive sensors | ||
| + | |||
| + | As we explore this chapter, we'll delve into the mathematical models, equations, and practical considerations of mutual induction and coupling, unlocking a world of innovative technologies that shape our modern lives. | ||
| + | We explicitly try to answer the following questions: | ||
| * Which effect do the coils have on each other? | * Which effect do the coils have on each other? | ||
| * Can we describe the effects with mainly electric properties (i.e. no geometric properties) | * Can we describe the effects with mainly electric properties (i.e. no geometric properties) | ||
| - | <WRAP center 35%> < | ||
| ==== Effect of Coils on each other ==== | ==== Effect of Coils on each other ==== | ||
| Zeile 317: | Zeile 348: | ||
| For the single coil, we got the relationship between the linked flux $\Psi$ and the current $i$ as: $\Psi = L \cdot i$. \\ | For the single coil, we got the relationship between the linked flux $\Psi$ and the current $i$ as: $\Psi = L \cdot i$. \\ | ||
| - | Now the coils also are interacting | + | Now the coils also interact |
| \begin{align*} | \begin{align*} | ||
| \Psi_1 &= & | \Psi_1 &= & | ||
| Zeile 332: | Zeile 363: | ||
| The formula can also be described as: | The formula can also be described as: | ||
| - | \begin{align*} | + | {{drawio> |
| - | \left( \begin{array}{c} \Psi_1 \\ \Psi_2 \end{array} | + | |
| - | \left( \begin{array}{c} L_{11} & M_{12} \\ M_{21} & L_{22} \end{array} \right) | + | |
| - | \cdot | + | |
| - | \left( \begin{array}{c} i_1 \\ i_2 \end{array} | + | |
| - | \end{align*} | + | |
| - | The view onto the magnetic flux is sometimes good when effects like an acting Lorentz force in of interest. | + | The view of the magnetic flux is advantageous |
| - | More often the coils are coupling two electric circuits | + | However, more often the coils couple |
| Here, the effect on the circuits is of interest. This can be calculated with the induced electric voltages $u_{\rm ind,1}$ and $u_{\rm ind,2}$ in each circuit. | Here, the effect on the circuits is of interest. This can be calculated with the induced electric voltages $u_{\rm ind,1}$ and $u_{\rm ind,2}$ in each circuit. | ||
| They are given by the formula $u_{{\rm ind},x} = -{\rm d}\Psi_x /{\rm d}t$: | They are given by the formula $u_{{\rm ind},x} = -{\rm d}\Psi_x /{\rm d}t$: | ||
| Zeile 353: | Zeile 379: | ||
| The main question is now: How do we get $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$? | The main question is now: How do we get $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$? | ||
| - | ==== Magnetic Circuit with 2 Sources ==== | + | ==== Magnetic Circuit with two Sources ==== |
| - | In order to get the self-induction and mutual induction of two interacting coils, we are going to investigate two coils on an iron core with a middle leg (see <imgref ImgNr08> | + | To get the self-induction and mutual induction of two interacting coils, we are going to investigate two coils on an iron core with a middle leg (see <imgref ImgNr08> |
| There the stray flux of the previous situation is only located in the middle leg. This also means, that there is no stray flux outside of the iron core. | There the stray flux of the previous situation is only located in the middle leg. This also means, that there is no stray flux outside of the iron core. | ||
| < | < | ||
| - | The <imgref ImgNr08> shows the fluxes on each part. The black dots nearby | + | The <imgref ImgNr08> shows the fluxes on each part. The black dots near the windings mark the direction of the windings, and therefore the sign of the generated flux. \\ |
| - | When there is one current | + | All the fluxes caused by currents flowing into the __marked pins__ are summed up __positively__ in the core. \\ |
| + | When there is a current | ||
| - | In order to get $L_{11}$ and $L_{22}$, we look back to the inductance $L$ of a long coil with the length $l$. \\ | + | To get $L_{11}$ and $L_{22}$, we look back to the inductance $L$ of a long coil with the length $l$. \\ |
| This was given in the chapter [[: | This was given in the chapter [[: | ||
| Zeile 389: | Zeile 416: | ||
| \end{align*} | \end{align*} | ||
| - | In order to get the effect of the mutual induction, a coupling coefficient $k$ is introduced. | + | To get the effect of the mutual induction, a coupling coefficient $k$ is introduced. |
| $k_{21}$ describes how much of the flux from coil $1$ is acting on coil $2$ (similar for $k_{12}$): | $k_{21}$ describes how much of the flux from coil $1$ is acting on coil $2$ (similar for $k_{12}$): | ||
| - | \begin{align*} k_{21} = {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*} | + | \begin{align*} k_{21} = \pm {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*} |
| - | When $k_{21}=100~\%$, | + | The sign of $k_{21}$ depends on the direction of $\Phi_{21}$ relative to $\Phi_{22}$! If the directions are the same, the positive sign applies, if the directions are opposite, the minus sign applies. |
| + | |||
| + | When $k_{21}=+100~\%$, there is no flux in the middle leg but only in the second coil and in the same direction as the flux that originates from the second coil. \\ | ||
| + | When $k_{21}=-100~\%$, | ||
| For $k_{21}=0~\%$ all the flux is in the middle leg circumventing the second coil, i.e. there is no coupling. | For $k_{21}=0~\%$ all the flux is in the middle leg circumventing the second coil, i.e. there is no coupling. | ||
| Zeile 408: | Zeile 438: | ||
| & | & | ||
| \end{align*} | \end{align*} | ||
| + | |||
| + | Note, that also $M_{21}$ and $M_{12}$ can be either positive or negative, depending on the sign of the coupling coefficients. | ||
| The formula is finally: | The formula is finally: | ||
| \begin{align*} | \begin{align*} | ||
| - | \left( \begin{array}{c} \Psi_1 | + | \left( \begin{array}{c} \Psi_1 |
| = | = | ||
| - | \left( \begin{array}{c} {{N_1^2}\over{R_{m1}}} & k_{12}\cdot{{N_1 \cdot N_2}\over{R_{m2}}} \\ k_{21}\cdot{{N_1 \cdot N_2}\over{R_{m1}}} & {{N_2^2}\over{R_{m2}}} \end{array} \right) | + | \left( \begin{array}{c} {{N_1^2}\over{R_{\rm m1}}} & k_{12}\cdot{{N_1 \cdot N_2}\over{R_{\rm m2}}} \\ k_{21}\cdot{{N_1 \cdot N_2}\over{R_{\rm m1}}} & {{N_2^2}\over{R_{\rm m2}}} \end{array} \right) |
| \cdot | \cdot | ||
| - | \left( \begin{array}{c} i_1 \\ i_2 \end{array} \right) | + | \left( \begin{array}{c} i_1 \\ i_2 \end{array} \right) |
| \end{align*} | \end{align*} | ||
| - | <panel type=" | + | For most of the applications the induction matrix has to be symmetric((This can be derived from energy considerations, |
| - | The magnetical configuration in <imgref ExImgNr01> | + | * In General: |
| - | The area of the cross-section is $A=9 ~\rm cm^2$ in all parts, the permeability | + | * For symmetric induction matrix: The mutual inductances are equal: |
| - | The coupling | + | |
| - | Calculate $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$. | + | |
| + | |||
| + | <panel type=" | ||
| + | |||
| + | The magnetical configuration in <imgref ExImgNr01> | ||
| + | The area of the cross-section is $A=9 ~\rm cm^2$ in all parts, the permeability is $\mu_r=800$, the length | ||
| < | < | ||
| - | === Step 1: Draw the problem as a network === | + | 1. Simplify the configuration into three magnetic resistors and 2 voltage sources. |
| + | |||
| + | # | ||
| + | < | ||
| + | # | ||
| + | |||
| + | 2. Calculate all magnetic resistances. Additionally, | ||
| + | |||
| + | # | ||
| - | === Step 2: Calculate the magnetic resistances === | + | <WRAP right> < |
| - | The magnetic resistance is summed up by looking at the circuit from the source $1$: | + | The magnetic resistance is summed up by looking at the circuit from the source $1$ (see <imgref ExImgNr13> |
| \begin{align*} | \begin{align*} | ||
| R_{\rm m1} &= R_{\rm m,11} + R_{\rm m,ss} || R_{\rm m,22} \\ | R_{\rm m1} &= R_{\rm m,11} + R_{\rm m,ss} || R_{\rm m,22} \\ | ||
| Zeile 439: | Zeile 484: | ||
| where the parts are given as | where the parts are given as | ||
| \begin{align*} | \begin{align*} | ||
| - | R_{\rm m,11} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{3\cdot l}\over{A}} \\ | + | R_{\rm m,11} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{3\cdot l}\over{A}} &&= 398 \cdot 10^{3} ~\rm {{1}\over{H}} \\ |
| - | R_{\rm m,ss} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{1\cdot l}\over{A}} \\ | + | R_{\rm m,ss} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{1\cdot l}\over{A}} &&= 133 \cdot 10^{3} ~\rm {{1}\over{H}} \\ |
| - | R_{\rm m,22} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{2\cdot l}\over{A}} \\ | + | R_{\rm m,22} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{2\cdot l}\over{A}} &&= 265 \cdot 10^{3} ~\rm {{1}\over{H}} \\ |
| \end{align*} | \end{align*} | ||
| - | With the given geometry this leads to | + | With the given geometry, this leads to |
| \begin{align*} | \begin{align*} | ||
| - | R_{\rm m1} &= {{1}\over{\mu_0 \mu_{r}}}{{l}\over{A}}\cdot (3 + {{1\cdot 2}\over{1 + 2}}) \\ | + | R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot |
| - | & | + | & |
| - | & | + | |
| \end{align*} | \end{align*} | ||
| Similarly, the magnetic resistance $R_{m2}$ is | Similarly, the magnetic resistance $R_{m2}$ is | ||
| \begin{align*} | \begin{align*} | ||
| - | R_{\rm m2} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{4}} | + | R_{\rm m2} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{4}} |
| - | & | + | |
| \end{align*} | \end{align*} | ||
| - | == Step 3: Calculate the magnetic inductances == | + | # |
| + | 3. Calculate the self-inductions $L_{11}$ and $L_{22}$ | ||
| + | |||
| + | # | ||
| + | For the self-induction the effect on the electrical circuit is relevant. That is why the number of windings has to be considered. | ||
| \begin{align*} | \begin{align*} | ||
| L_{11} &= {{N_1^2}\over{R_{\rm m1}}} &= 329 ~\rm mH\\ \\ | L_{11} &= {{N_1^2}\over{R_{\rm m1}}} &= 329 ~\rm mH\\ \\ | ||
| L_{22} &= {{N_2^2}\over{R_{\rm m2}}} &= 247 ~\rm mH\\ \\ | L_{22} &= {{N_2^2}\over{R_{\rm m2}}} &= 247 ~\rm mH\\ \\ | ||
| - | M_{21} &= k_{21}\cdot{{N_1 \cdot N_2}\over{R_{\rm m1}}} &= 197 ~\rm mH\\ \\ | ||
| - | M_{12} &= k_{12}\cdot{{N_1 \cdot N_2}\over{R_{\rm m2}}} &= 197 ~\rm mH\\ | ||
| \end{align*} | \end{align*} | ||
| + | # | ||
| + | |||
| + | 4. Calculate the coupling factors $k_{12}$ and $k_{21}$. | ||
| + | |||
| + | # | ||
| + | <WRAP right> < | ||
| + | |||
| + | The coupling factor $k_{21}$ is defined as "how much of the flux created by one coil ($\Phi_{11}$) crosses the other coil ($\Phi_{21}$) ": | ||
| + | \begin{align*} | ||
| + | k_{21} &= {{\Phi_{21}}\over{\Phi_{11}}} | ||
| + | \end{align*} | ||
| + | |||
| + | For this, we look at the circuit considering only one coil (" | ||
| + | In step 2, we have calculated that $R_{\rm m,22}$ is twice $R_{\rm m, | ||
| + | |||
| + | Therefore, the coupling factor $k_{21}$ is: $k_{21}= 1/3$. | ||
| + | |||
| + | A similar approach leads to $k_{12}$ with $k_{12}= 1/4$. | ||
| + | # | ||
| + | |||
| + | 5. Calculate the mutual inductions $M_{12}$, and $M_{21}$, | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | M_{21} &= k_{21}\cdot{{N_1 \cdot N_2}\over{R_{\rm m1}}} &&= {{1}\over{3}}\cdot{{400 \cdot 300}\over{ 486 \cdot 10^{3} ~\rm {{1}\over{H}} }} &&= 82.2 ~\rm mH\\ \\ | ||
| + | M_{12} &= k_{12}\cdot{{N_1 \cdot N_2}\over{R_{\rm m2}}} &&= {{1}\over{4}}\cdot{{400 \cdot 300}\over{ 365 \cdot 10^{3} ~\rm {{1}\over{H}} }} &&= 82.2 ~\rm mH\\ \\ | ||
| + | \end{align*} | ||
| + | # | ||
| </ | </ | ||
| - | For symmetrical magnetic structures | + | # |
| + | |||
| + | For Electric vehicles, sometimes wireless charging systems are employed. These use the principle of mutual inductance to transfer power from a charging pad on the ground to the vehicle' | ||
| + | This system consists of two coils: a transmitter coil embedded in the charging pad and a receiver coil mounted on the underside of the vehicle. | ||
| + | |||
| + | * The transmitter coil has a self-inductance of $L_{\rm T} = 200 ~\rm \mu H$. | ||
| + | * The receiver coil has a self-inductance of $L_{\rm R} = 150 ~\rm \mu H$. | ||
| + | * The mutual inductance between the coils at this distance is measured to be $M = 20 ~\rm \mu H$ - when the vehicle is properly aligned over the charging pad. | ||
| + | |||
| + | 1. Calculate the coupling coefficient $k$ between the transmitter and receiver coils when the vehicle is properly aligned over the charging pad. | ||
| + | |||
| + | # | ||
| + | |||
| + | The given self-inductances are $L_{\rm T} = L_{11}$, $L_{\rm R} = L_{22}$. \\ | ||
| + | By this, the following | ||
| + | |||
| + | \begin{align*} | ||
| + | M = k \cdot \sqrt{L_{\rm T} \cdot L_{\rm R}} | ||
| + | \end{align*} | ||
| + | |||
| + | Therefore, $k$ is given as: | ||
| + | \begin{align*} | ||
| + | k = {{M}\over{ \sqrt{ L_{\rm T} \cdot L_{\rm R} } }} | ||
| + | \end{align*} | ||
| + | |||
| + | # | ||
| + | |||
| + | 2. If the vehicle is misaligned by 10 cm from the center of the charging pad, the mutual inductance drops to $M = 12 ~\rm \mu H$. Calculate the new coupling coefficient in this misaligned position. | ||
| + | |||
| + | # | ||
| - | * the mutual inductances are equal: $M_{12} = M_{21} = M$ | ||
| - | * the mutual inductance $M$ is: $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$ | ||
| - | * The resulting *total coupling* $k$ is given as \begin{align*} k = \sqrt{k_{12}\cdot k_{21}} \end{align*} | ||
| ==== Effects in the electric Circuits ==== | ==== Effects in the electric Circuits ==== | ||
| - | * Whenever two coils are magnetically coupled, not only the self-induction $L$, but also the mutual induction $M$ applies. | + | * Whenever two coils are magnetically coupled, not only the self-induction $L$ but also the mutual induction $M$ applies. |
| * Based on the currents $i_1$, $i_2$ in the two circuits, the induced voltages are given by: | * Based on the currents $i_1$, $i_2$ in the two circuits, the induced voltages are given by: | ||
| Zeile 488: | Zeile 588: | ||
| * the direction of the windings, and | * the direction of the windings, and | ||
| - | * the orientation/ | + | * The orientation/ |
| <WRAP center 50%> < | <WRAP center 50%> < | ||
| Zeile 498: | Zeile 598: | ||
| < | < | ||
| - | In this case, the **mutual induction | + | In this case, the **mutual induction |
| The formula of the shown circuitry is then: | The formula of the shown circuitry is then: | ||
| Zeile 506: | Zeile 606: | ||
| \end{align*} | \end{align*} | ||
| - | === negative | + | === Negative |
| - | The polarity is negative when only one current | + | The polarity is negative when only one current flows into the dotted pin and the other one out of the dotted pin (see <imgref ImgNr13> |
| < | < | ||
| - | In this case, the **mutual induction | + | In this case, the **mutual induction |
| The formula of the shown circuitry is then: | The formula of the shown circuitry is then: | ||
| \begin{align*} | \begin{align*} | ||
| - | u_1 &= R_1 \cdot i_1 &+ L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} &- M \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\ | + | u_1 &= R_1 \cdot i_1 &+ L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} & + M \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\ |
| - | u_2 &= R_2 \cdot i_2 &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} &- M \cdot {{{\rm d}i_1}\over{{\rm d}t}} & \\ | + | u_2 &= R_2 \cdot i_2 &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} & + M \cdot {{{\rm d}i_1}\over{{\rm d}t}} & \\ |
| \end{align*} | \end{align*} | ||
| - | <panel type=" | + | <panel type=" |
| - | A toroidal core (ferrite, $\mu_{\rm r} = 900$) has a cross-sectional area of $A = 500mm^2$ and an average circumference of $l=280 ~\rm mm$. | + | A toroidal core (ferrite, $\mu_{\rm r} = 900$) has a cross-sectional area of $A = 500 ~\rm mm^2$ and an average circumference of $l=280 ~\rm mm$. |
| - | On the core, there are two coils $N_1=500$ and $N_2=250$ wound. The currents on the coils are $I_1 = 250 ~\rm mA$ and $I_2=300 ~\rm mA$. | + | At the core, there are two coils $N_1=500$ and $N_2=250$ wound. The currents on the coils are $I_1 = 250 ~\rm mA$ and $I_2=300 ~\rm mA$. |
| - | - The coils shall pass the currents with positive polarity (see the top image in <imgref ImgEx14> | + | - The coils shall pass the currents with positive polarity (see the image **A** in <imgref ImgEx14> |
| - | - The coils shall pass the currents with negative polarity (see the bottom | + | - The coils shall pass the currents with negative polarity (see the image **B** in <imgref ImgEx14> |
| < | < | ||
| - | <button size=" | + | # |
| - | - $0.40 ~\rm mVs$ | + | The resulting flux can be derived from a superposition of the individual fluxes |
| - | - $0.10 ~\rm mVs$ | + | |
| - | </collapse> | + | **Step 1 - Draw an equivalent magnetic circuit** |
| + | |||
| + | Since there are no branches, all of the core can be lumped into a single magnetic resistance (see <imgref ImgEx14circ> | ||
| + | < | ||
| + | |||
| + | **Step 2 - Get the absolute values of the individual fluxes** | ||
| + | |||
| + | Hopkinson' | ||
| + | It connects the magnetic flux $\Phi$ and the magnetic voltage $\theta$ on the single magnetic resistor $R_\rm m$. \\ | ||
| + | It also connects the single magnetic fluxes $\Phi_x$ (with $x = {1,2}$) and the single magnetic voltages $\theta_x$. \\ | ||
| + | |||
| + | \begin{align*} | ||
| + | \theta_x | ||
| + | N_x \cdot I_x &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}} \cdot \Phi_x \\ | ||
| + | \rightarrow \Phi_x | ||
| + | = {{1}\over{R_{\rm m} }} \cdot N_x \cdot I_x \\ | ||
| + | \end{align*} | ||
| + | |||
| + | With the given values we get: $R_{\rm m} = 495 {\rm {kA}\over{Vs}}$ | ||
| + | |||
| + | **Step 3 - Get the signs/ | ||
| + | |||
| + | The < | ||
| + | The fluxes have to be added regarding these directions and the given direction of the flux in question. | ||
| + | < | ||
| + | |||
| + | Therefore, the formulas are | ||
| + | \begin{align*} | ||
| + | \Phi_{\rm A} & | ||
| + | & | ||
| + | & = 0.25 ~{\rm mVs} - 0.15 ~{\rm mVs} \\ | ||
| + | \Phi_{\rm B} & | ||
| + | & | ||
| + | & = 0.25 ~{\rm mVs} + 0.15 ~{\rm mVs} | ||
| + | \end{align*} | ||
| + | |||
| + | # | ||
| + | |||
| + | |||
| + | # | ||
| + | - $0.10 ~\rm mVs$ | ||
| + | - $0.40 ~\rm mVs$ | ||
| + | # | ||
| </ | </ | ||
| Zeile 542: | Zeile 683: | ||
| The magnetic field of a coil stores magnetic energy. | The magnetic field of a coil stores magnetic energy. | ||
| - | The energy transfer from the electric circuit to the magnetic field is also the cause of the " | + | The energy transfer from the electric circuit to the magnetic field is also the cause of the " |
| The energetic turnover for charging an conductor from $i(t_0=0)=0$ to $i(t_1)=I$ is given by: | The energetic turnover for charging an conductor from $i(t_0=0)=0$ to $i(t_1)=I$ is given by: | ||
| Zeile 557: | Zeile 698: | ||
| \boxed{W_m = {{1}\over{2}}L\cdot I^2 } | \boxed{W_m = {{1}\over{2}}L\cdot I^2 } | ||
| \end{align*} | \end{align*} | ||
| + | |||
| + | |||
| + | |||
| + | |||
| ==== magnetic Energy of a magnetic Circuit ==== | ==== magnetic Energy of a magnetic Circuit ==== | ||
| - | With this formula also the stored energy in a magnetic circuit can be calculated. | + | With this formula also the stored energy in a magnetic circuit can be calculated. For this, the formula be rewritten by the properties linked flux $\Psi = N \cdot \Phi = L \cdot I$ and magnetic voltage $\theta=N \cdot I = \Phi \cdot R_{\rm m}$ of the magnetic circuit: \begin{align*} \boxed{W_{\rm m} = {{1}\over{2}} \Psi \cdot I = {{1}\over{2}} {{\Psi^2}\over{L}}= {{1}\over{2}}{{\Phi^2 }\over{N^2 \cdot L}} = {{1}\over{2}} \Phi^2 \cdot R_{\rm m} = {{1}\over{2}}{{\theta^2 }\over{R_{\rm m}}}} \end{align*} |
| - | For this, the formula be rewritten by the properties linked flux $\Psi = N \cdot \Phi = L \cdot I$ and magnetic voltage $\theta=N \cdot I = \Phi \cdot R_{\rm m}$ of the magnetic circuit: | + | |
| - | \begin{align*} | + | |
| - | \boxed{W_{\rm m} = {{1}\over{2}} | + | |
| - | = {{1}\over{2}}{{\Psi^2 }\over{L}}} | + | |
| - | \end{align*} | + | |
| ==== magnetic Energy of a toroid Coil ==== | ==== magnetic Energy of a toroid Coil ==== | ||
| - | The formula can also be used for calculating the stored energy of a toroid coil with $N$ windings, the cross-section $A$, and an average length $l$ of a field line. | + | The formula can also be used for calculating the stored energy of a toroid coil with $N$ windings, the cross-section $A$, and an average length $l$ of a field line. \\ |
| By this, the following formulas can be used: | By this, the following formulas can be used: | ||
| - | \begin{align*} | + | - For the magnetic voltage: $\theta = H \cdot l = N \cdot I $ \\ |
| - | \theta = H \cdot l = N \cdot I \\ | + | - For the magnetic flux: $\Phi = B \cdot A $ |
| - | \Phi = B \cdot A | + | |
| - | \end{align*} | + | |
| With the above-mentioned formulas of the magnetic circuit, we get: | With the above-mentioned formulas of the magnetic circuit, we get: | ||
| Zeile 595: | Zeile 733: | ||
| ==== generalized magnetic Energy ==== | ==== generalized magnetic Energy ==== | ||
| - | The general term to find the magnetic energy (e.g. for inhomogeneous magnetic fields) is given by | + | The general term to find the magnetic energy (e.g., for inhomogeneous magnetic fields) is given by |
| \begin{align*} | \begin{align*} | ||
| - | W_{\rm m} &= \iiint_V{w_m | + | W_{\rm m} &= \iiint_V{w_{\rm m} |
| &= \iiint_V{\vec{B}\cdot \vec{H} {\rm d}V} | &= \iiint_V{\vec{B}\cdot \vec{H} {\rm d}V} | ||
| \end{align*} | \end{align*} | ||
| Zeile 615: | Zeile 753: | ||
| \end{align*} | \end{align*} | ||
| - | Multiplying with $i$ and with $dt$ we get the principle of conservation of energy $dw = u \cdot i \cdot {\rm d}t$ for each small time step. | + | Multiplying with $i$ and with $dt$, we get the principle of conservation of energy $dw = u \cdot i \cdot {\rm d}t$ for each small time step. |
| \begin{align*} | \begin{align*} | ||
| Zeile 625: | Zeile 763: | ||
| \begin{align*} | \begin{align*} | ||
| dW_{\rm m} &= N {{{\rm d}\Phi}\over{{\rm d}t}} \cdot i \cdot {\rm d}t \\ | dW_{\rm m} &= N {{{\rm d}\Phi}\over{{\rm d}t}} \cdot i \cdot {\rm d}t \\ | ||
| - | | + | |
| & | & | ||
| & | & | ||
| Zeile 641: | Zeile 779: | ||
| We can conclude that the magnetic energy $W_{\rm m}$: | We can conclude that the magnetic energy $W_{\rm m}$: | ||
| - | $W_{\rm m}$ can be calculated from the $H$-$B$ curve by integrating the external magnetic field strength $H$ for each small step of the flux density $dB$. | + | $W_{\rm m}$ can be calculated from the $H$-$B$ curve by integrating the external magnetic field strength $H$ for each small step of the flux density ${\rm d}B$. |
| This will be shown for the case of a linear magnetic behavior, a nonlinear behavior, and the situation with magnetic hysteresis shortly. | This will be shown for the case of a linear magnetic behavior, a nonlinear behavior, and the situation with magnetic hysteresis shortly. | ||
| Zeile 647: | Zeile 785: | ||
| In <imgref ImgNr15> the situation for a magnetic material with a linear relationship between $B$ and $H$ is shown. | In <imgref ImgNr15> the situation for a magnetic material with a linear relationship between $B$ and $H$ is shown. | ||
| - | Given by the maximum current $I_{\rm max}$ the maximum field strength $H_{\rm max}$ can be derived. | + | Given the maximum current $I_{\rm max}$ the maximum field strength $H_{\rm max}$ can be derived. |
| In the circuit in <imgref ImgNr14>, | In the circuit in <imgref ImgNr14>, | ||
| Therefore, the $B$-$H$-curve gets passed through positive and negative values of $H$ and $H$ along the line of $B=\mu H$. | Therefore, the $B$-$H$-curve gets passed through positive and negative values of $H$ and $H$ along the line of $B=\mu H$. | ||
| Zeile 654: | Zeile 792: | ||
| The situation for integrating the area in the graph is also shown: | The situation for integrating the area in the graph is also shown: | ||
| - | For each step ${\rm d}B$ the corresponding value of the field strength $H$ has to be integrated. | + | For each step ${\rm d}B$, the corresponding value of the field strength $H$ has to be integrated. |
| For $B_0=0$ to $B=B_{\rm max}$ the magnetic energy is | For $B_0=0$ to $B=B_{\rm max}$ the magnetic energy is | ||
| Zeile 675: | Zeile 813: | ||
| In this case, the permeability $\mu_{\rm r}$ is not a constant but can be represented as a function: $\mu_{\rm r}= f(B)$. | In this case, the permeability $\mu_{\rm r}$ is not a constant but can be represented as a function: $\mu_{\rm r}= f(B)$. | ||
| Here, the formula $W_{\rm m} = V\int_0^{B} H(B) \cdot {\rm d}B$ also applies - so the magnetic energy is again the area between the curve and the $B$-axis. | Here, the formula $W_{\rm m} = V\int_0^{B} H(B) \cdot {\rm d}B$ also applies - so the magnetic energy is again the area between the curve and the $B$-axis. | ||
| - | As an example the situation of the field strength $H(t_1)=H_1$ is shown. | + | As an example, the situation of the field strength $H(t_1)=H_1$ is shown. |
| This shall be the field strength after magnetizing the ferrite material to $H_{\rm max}$ (yellow arrows) and then partly demagnetizing the material again (blue arrow). | This shall be the field strength after magnetizing the ferrite material to $H_{\rm max}$ (yellow arrows) and then partly demagnetizing the material again (blue arrow). | ||
| - | The magnetization corresponds to an energy intake | + | The magnetization corresponds to an energy intake |
| - | Moving along the $H$-$B$-curve, | + | Moving along the $H$-$B$-curve, |
| This means that the magnetization and demagnetization take place lossless in this example. | This means that the magnetization and demagnetization take place lossless in this example. | ||
| - | This is a good approximation for magnetically soft materials, however, does not work for magnetically hard materials like a permanent magnet. | + | This is a good approximation for magnetically soft materials; however, |
| Here, hysteresis also has to be considered. | Here, hysteresis also has to be considered. | ||
| Zeile 688: | Zeile 826: | ||
| < | < | ||
| - | ===== Tasks ===== | + | ===== Exercise |
| - | <panel type=" | + | <panel type=" |
| - | The <imgref ImgTask01> | + | The <imgref ImgTask01> |
| * Find out how this motor works - explicitly: why is there a preferred direction of the motor? | * Find out how this motor works - explicitly: why is there a preferred direction of the motor? | ||
| Zeile 705: | Zeile 843: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| - | The book [[https:// | + | The book [[https:// |
| </ | </ | ||
| Zeile 714: | Zeile 852: | ||
| An alternative interpretation of the magnetic circuits is the {{https:// | An alternative interpretation of the magnetic circuits is the {{https:// | ||
| - | The big difference there is, that there the magnetic flux $\Phi$ is not interpreted as an analogy to the electric current $I$ but to the electric charge $Q$. | + | The big difference there is that the magnetic flux $\Phi$ is not interpreted as an analogy to the electric current $I$ but to the electric charge $Q$. |
| - | This model can solve more questions, however, is a bit less intuitive based on this course and less commonly used compared to the {{https:// | + | This model can solve more questions; however, |
| + | |||
| + | ==== Moving a Plate into an Air Gap ==== | ||
| + | |||
| + | < | ||
| ==== Switch Reluctance Motor ==== | ==== Switch Reluctance Motor ==== | ||