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electrical_engineering_2:magnetic_circuits [2023/03/17 10:44]
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mexleadmin [Effects in the electric Circuits]
Zeile 1: Zeile 1:
-====== 5. Magnetic Circuits ======+====== 5 Magnetic Circuits ======
  
 <callout> For this and the following chapter the online Book 'DC Electrical Circuit Analysis - A Practical Approach' is strongly recommended as a reference. In detail this is chapter [[https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/DC_Electrical_Circuit_Analysis_-_A_Practical_Approach_(Fiore)/10%3A_Magnetic_Circuits_and_Transformers/10.3%3A_Magnetic_Circuits|10.3 Magnetic Circuits]] </callout> <callout> For this and the following chapter the online Book 'DC Electrical Circuit Analysis - A Practical Approach' is strongly recommended as a reference. In detail this is chapter [[https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/DC_Electrical_Circuit_Analysis_-_A_Practical_Approach_(Fiore)/10%3A_Magnetic_Circuits_and_Transformers/10.3%3A_Magnetic_Circuits|10.3 Magnetic Circuits]] </callout>
Zeile 9: Zeile 9:
 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 come with such an analogy and where it can be practically applied.
  
-===== 5.1 Linear magnetic Circuits =====+===== 5.1 Linear Magnetic Circuits =====
  
 For the upcoming calculations, the following assumptions are made For the upcoming calculations, the following assumptions are made
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) ).
  
 <WRAP> <imgcaption ImgNr02 | A simple magnetic Circuit> </imgcaption> {{drawio>SimpleMagnCircuit.svg}} </WRAP> <WRAP> <imgcaption ImgNr02 | A simple magnetic Circuit> </imgcaption> {{drawio>SimpleMagnCircuit.svg}} </WRAP>
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 [[:electrical_engineering_2:the_time-dependent_magnetic_field#recap_of_magnetic_field|Recap of magnetic Field]] we know that the flux into the volume must be equal the flux out of the volume, or $\Phi_{\rm m} = \iint_{A} \vec{B} \cdot d \vec{A} = 0$.+A different view of this is the closed surface $\vec{A}$ (<imgref ImgNr04> (2)): Based on the examination in [[:electrical_engineering_2:the_time-dependent_magnetic_field#recap_of_magnetic_field|Recap of magnetic Field]] we know that the flux into the volume must be equal the flux out of the volume, or $\Phi_{\rm m} = \iint_{A} \vec{B} \cdot d \vec{A} = 0$.
  
 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}                       + {{B}\over{\mu_0 \mu_{\rm r,airgap}}} \delta \\  
-       &= {{\Phi \cdot l_{\rm core}}\over{A \cdot \mu_0 \mu_{\rm r,core}}}  + {{\Phi \cdot \delta}\over{A \cdot \mu_0 \mu_{\rm r,airgap}}} \\  +       &= {{\Phi \cdot l_{\rm core}}\over{A \cdot \mu_0 \mu_{\rm r,core}}}      + {{\Phi \cdot \delta}\over{A \cdot \mu_0 \mu_{\rm r,airgap}}} \\  
-       &= {{1}\over{\mu_0 \mu_{\rm r,core}}}{{l_{core}}\over{A}} \cdot \Phi + {{1}\over{\mu_0 \mu_{\rm r,airgap}}}{{\delta}\over{A}}   \cdot \Phi \tag{5.2.3} +       &= {{1}\over{\mu_0 \mu_{\rm r,core}}}{{l_{\rm core}}\over{A}} \cdot \Phi + {{1}\over{\mu_0 \mu_{\rm r,airgap}}}{{\delta}\over{A}}   \cdot \Phi \tag{5.2.3} 
 \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 168: Zeile 168:
 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="xs" type="link" collapse="Solution_5_1_2_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_2_1_Result" collapsed="true">+#@HiddenBegin_HTML~5_1_2s,Solution~@#
  
-  - $1.5\cdot 10^5 ~\rm {{1}\over{H}}$ +The magnetic resistance is given by: 
-  - $3.0\cdot 10^5 ~\rm {{1}\over{H}}$+\begin{align*} 
 +\ R_{\rm m} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}  
 +\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 $\mu_{\rm r}=1$. 
 + 
 +#@HiddenEnd_HTML~5_1_2s,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~5_1_2r,Result~@# 
 +  - $1.00\cdot 10^9 ~\rm {{1}\over{H}}$ 
 +  - $0.50\cdot 10^9 ~\rm {{1}\over{H}}$ 
 +#@HiddenEnd_HTML~5_1_2r,Result~@#
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 198: Zeile 210:
 <panel type="info" title="Task 5.1.4 Magnetic Voltage"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Task 5.1.4 Magnetic Voltage"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-Calculate the magnetic voltage necessary in order to create a flux of $\Phi=0.5 ~\rm mVs$ in an airgap with the following dimensions:+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 233: Zeile 245:
  
 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 $A_2=460 ~\rm mm^2$. 
Zeile 279: Zeile 291:
 <panel type="info" title="Task 5.1.8 Coil on a ferrite Core with airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Task 5.1.8 Coil on a ferrite Core with airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-The choke coil shown in <imgref ImgNr08> shall be given, with a constant cross-section in all legs $l_0$, $l_1$, $l_2$. +The choke coil shown in <imgref ImgExNr10> shall be given, with a constant cross-section in all legs $l_0$, $l_1$, $l_2$. 
 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 $I$.
  
-<WRAP> <imgcaption ImgNr08 | Example for a Choke Coil> </imgcaption> {{drawio>ChokeCoilEx1.svg}} </WRAP>+<WRAP> <imgcaption ImgExNr10 | Example for a Choke Coil> </imgcaption> {{drawio>ChokeCoilEx1.svg}} </WRAP>
  
   - 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%> <imgcaption ImgNr09 | Mutual Induction of two Coils> </imgcaption> {{drawio>MutualInductionTwoCoils1.svg}} </WRAP> 
 + 
 +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, including: 
 + 
 +  * 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%> <imgcaption ImgNr09 | Mutual Induction of two Coils> </imgcaption> {{drawio>MutualInductionTwoCoils1.svg}} </WRAP> 
  
 ==== 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 with each other. This must also be reflected in the relationship $\Psi_1 = f(i_1, i_2)$, $\Psi_2 = f(i_1, i_2)$: +Now the coils also interact with each other. This must also be reflected in the relationship $\Psi_1 = f(i_1, i_2)$, $\Psi_2 = f(i_1, i_2)$: 
 \begin{align*}  \begin{align*} 
 \Psi_1 &= &\Psi_{11}        &+ \Psi_{12} \\  \Psi_1 &= &\Psi_{11}        &+ \Psi_{12} \\ 
Zeile 332: Zeile 363:
  
 The formula can also be described as:  The formula can also be described as: 
-\begin{align*}  +{{drawio>VectorialformulaOfInduction.svg}}
-\left( \begin{array}{c} \Psi_1 \\          \Psi_2 \end{array}           \right) =  +
-\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}             \right)  +
-\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 when effects like an acting Lorentz force is of interest.  
-More often the coils are coupling two electric circuits linked in a transformer or a wireless charger. +However, more often the coils couple electric circuits, like in a transformer or a wireless charger. 
 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.
  
 <WRAP> <imgcaption ImgNr08 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils.svg}} </WRAP> <WRAP> <imgcaption ImgNr08 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils.svg}} </WRAP>
  
-The <imgref ImgNr08> shows the fluxes on each part. The black dots nearby the windings mark the direction of winding: \\  +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 for each winding ingoing into the marked pin, the fluxes will get added up positively.+All the fluxes caused by currents flowing into the __marked pins__ are summed up __positively__ in the core. \\ 
 +When there is a current flowing into a __non-marked pin__, its flux has to be __subtracted__ from the others.
  
-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 [[:electrical_engineering_2:the_time-dependent_magnetic_field#self-induction1|Self-Induction]] as This was given in the chapter [[:electrical_engineering_2:the_time-dependent_magnetic_field#self-induction1|Self-Induction]] as
  
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 $k_{21}=100~\%$, there is no flux in the middle leg but only in the second coil. \\ +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 $k_{21}=-100~\%$, there is no flux in the middle leg but only in the second coil and in the opposite direction as the flux that originates from the second coil. \\ 
 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:
        &= k_{21}                  \cdot {{N_1 \cdot N_2 }\over {R_{\rm m1}}} \\         &= k_{21}                  \cdot {{N_1 \cdot N_2 }\over {R_{\rm m1}}} \\ 
 \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                                                             \\ \Psi_2                                                             \end{array} \right) +\left( \begin{array}{c} \Psi_1                                                                     \\ \Psi_2                                                                     \end{array} \right) 
 =  = 
-\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*}
 +
 +For most of the applications the induction matrix has to be symmetric((This can be derived from energy considerations, when only electric circuits are coupled without additional flow of mechanical energy. This is, for example  not the case for motors with a mechanical load.)). \\ Therefore, the following applies:
 +
 +  * In General: the mutual inductance $M$ is:     $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$
 +  * For symmetric induction matrix: The mutual inductances are equal: $M_{12} = M_{21} = M$
 +  * The resulting **total coupling** $k$ is given as \begin{align*} k = \rm{sgn}(k_{12}) \sqrt{k_{12}\cdot k_{21}} \end{align*}
 +
 +
  
 <panel type="info" title="Task 5.3.1 Example for magnetic Circuit with two Sources"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Task 5.3.1 Example for magnetic Circuit with two Sources"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 The magnetical configuration in <imgref ExImgNr01> shall be given. \\  The magnetical configuration in <imgref ExImgNr01> shall be given. \\ 
-The area of the cross-section is $A=9 ~\rm cm^2$ in all parts, the permeability is $\mu_r=800$, the length $l=12 ~\rm cm$ and the number of windings $N_1 = 400$, $N_2=300$.  +The area of the cross-section is $A=9 ~\rm cm^2$ in all parts, the permeability is $\mu_r=800$, the length $l=12 ~\rm cm$ and the number of windings $N_1 = 400$, $N_2=300$.
-The coupling factors are $k_{12}=0.6$ and $k_{21}=0.8$. +
- +
-Calculate $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$.+
  
 <WRAP> <imgcaption ExImgNr01 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils2.svg}} </WRAP> <WRAP> <imgcaption ExImgNr01 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils2.svg}} </WRAP>
  
-=== Step 1: Draw the problem as a network ===+1.  Simplify the configuration into three magnetic resistors and 2 voltage sources. Draw the problem as an equivalent circuit 
 + 
 +#@HiddenBegin_HTML~5311,Result~@# 
 +<WRAP> <imgcaption ExImgNr11 | Equivalent Network> </imgcaption> {{drawio>CoreWithTwoCoils2network.svg}} </WRAP> 
 +#@HiddenEnd_HTML~5311,Result~@# 
 + 
 +2. Calculate all magnetic resistances. Additionally, calculate the magnetic resistances $R_{\rm m1}$  and $R_{\rm m2}$ seen from the magnetic voltage source $1$ and $2$. 
 + 
 +#@HiddenBegin_HTML~5312,Path~@#
  
-=== Step 2: Calculate the magnetic resistances ===+<WRAP right> <imgcaption ExImgNr13 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkSingleVolt.svg}} </WRAP>
  
-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 \left(3 + {{1\cdot 2}\over{1 + 2}}\right) \\  
-           &= {{1}\over{\mu_0 \mu_{r}}}{{l}\over{A}}\cdot {{11}\over{3}} \\  +           &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{3}} &&= 486 \cdot 10^{3} ~\rm {{1}\over{H}}\\ 
-           &= 133 \cdot 10^{3}                      \cdot {{11}\over{3}} ~\rm {{1}\over{H}}\\ +
 \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}} &&= 365 \cdot 10^{3} ~\rm {{1}\over{H}}\\ 
-           &= 133 \cdot 10^{3}                          \cdot {{11}\over{4}} ~\rm {{1}\over{H}}\\ +
 \end{align*} \end{align*}
  
-== Step 3: Calculate the magnetic inductances ==+#@HiddenEnd_HTML~5312,Path~@#
  
 +3. Calculate the self-inductions $L_{11}$ and $L_{22}$
 +
 +#@HiddenBegin_HTML~5313,Path~@#
 +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*}
 +#@HiddenEnd_HTML~5313,Path~@#
 +
 +4. Calculate the coupling factors $k_{12}$ and $k_{21}$.
 +
 +#@HiddenBegin_HTML~5314,Path~@#
 +<WRAP right> <imgcaption ExImgNr12 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkCoupling.svg}} </WRAP>
 +
 +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 ("magnetic voltage source"), see <imgref ExImgNr12>. Based on the image, we see that the flux $\Phi_{11}$ divides into $\Phi_{S1}$ and $\Phi_{21}$ based on the magnetic resistance $\Phi_{\rm m,SS}$ and $\Phi_{\rm m,22}$. 
 +In step 2, we have calculated that $R_{\rm m,22}$ is twice $R_{\rm m,SS}$.  So from the incoming flux, only $1/3$ reaches $R_{\rm m,22}$ and - by this - "crosses the other coil". 
 +
 +Therefore, the coupling factor $k_{21}$ is: $k_{21}= 1/3$.
 +
 +A similar approach leads to $k_{12}$ with $k_{12}= 1/4$.
 +#@HiddenEnd_HTML~5314,Path~@#
 +
 +5. Calculate the mutual inductions $M_{12}$, and $M_{21}$,
 +
 +#@HiddenBegin_HTML~5315,Path~@#
 +\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*}
 +#@HiddenEnd_HTML~5315,Path~@#
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-For symmetrical magnetic structures and $\mu_{\rm r} = \rm const.$ the following applies:+#@TaskTitle_HTML@# 5.3.2 Wireles Charging #@TaskText_HTML@# 
 + 
 +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's battery pack. \\ 
 +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. 
 + 
 +#@HiddenBegin_HTML~53211,Path ~@# 
 + 
 +The given self-inductances are $L_{\rm T} = L_{11}$, $L_{\rm R} = L_{22}$. \\ 
 +By this, the following formula can be applied: 
 + 
 +\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*} 
 + 
 +#@HiddenEnd_HTML~53211,Path ~@# 
 + 
 +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. 
 + 
 +#@TaskEnd_HTML@# 
  
-  * 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 498: Zeile 598:
 <imgcaption ImgNr12 | Example Circuits with positive Polarity> </imgcaption> {{drawio>posCoupling.svg}} <imgcaption ImgNr12 | Example Circuits with positive Polarity> </imgcaption> {{drawio>posCoupling.svg}}
  
-In this case, the **mutual induction added positively**.+In this case, the **mutual induction is positiv $(M>0)$**.
  
 The formula of the shown circuitry is then:  The formula of the shown circuitry is then: 
Zeile 512: Zeile 612:
 <WRAP> <imgcaption ImgNr13 | Example Circuits with negative Polarity> </imgcaption> {{drawio>negCoupling.svg}} </WRAP> <WRAP> <imgcaption ImgNr13 | Example Circuits with negative Polarity> </imgcaption> {{drawio>negCoupling.svg}} </WRAP>
  
-In this case, the **mutual induction added negatively**.+In this case, the **mutual induction is negativ $(M<0)$***.
  
 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="info" title="Task 5.3.1 toroidal Core with two Coils"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Task 5.3.3 toroidal Core with two Coils"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-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 $I_2=300 ~\rm mA$. 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 $I_2=300 ~\rm mA$.
  
-  - The coils shall pass the currents with positive polarity (see the top image in <imgref ImgEx14>).    What is the magnetic flux in the coil? +  - The coils shall pass the currents with positive polarity (see the image **A** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm A}$ in the coil? 
-  - The coils shall pass the currents with negative polarity (see the bottom image in <imgref ImgEx14>). What is the magnetic flux in the coil?+  - The coils shall pass the currents with negative polarity (see the image **B** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm B}$ in the coil?
  
 <WRAP> <imgcaption ImgEx14 | toroidal core with two coils in positive and negative polarity> </imgcaption> {{drawio>torCoilPosNeg.svg}} </WRAP> <WRAP> <imgcaption ImgEx14 | toroidal core with two coils in positive and negative polarity> </imgcaption> {{drawio>torCoilPosNeg.svg}} </WRAP>
  
-<button size="xs" type="link" collapse="Solution_5_3_1_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_3_1_1_Result" collapsed="true">+#@HiddenBegin_HTML~5_3_2s,Solution~@#
  
-  - $0.40 ~\rm  mVs$ +The resulting flux can be derived from a superposition of the individual fluxes $\Phi_1(I_1)$ and $\Phi_2(I_2)$, or alternatively by summing the magnetic voltages in the loop ($\sum_x \theta_x = 0$).
-  - $0.10 ~\rm mVs$+
  
-</collapse>+**Step 1 - Draw an equivalent magnetic circuit** 
 + 
 +Since there are no branches all of the core can be lumped to a single magnetic resistance (see <imgref ImgEx14circ>). 
 +<WRAP> <imgcaption ImgEx14circ | equivalent magnetic circuit> </imgcaption> {{drawio>torCoilPosNegCirc.svg}} </WRAP> 
 + 
 +**Step 2 - Get the absolute values of the individual fluxes** 
 + 
 +Hopkinson's Law can be used here as a starting point. \\ 
 +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             &= R_{\rm m}                                  \cdot \Phi_x \\ 
 +N_x \cdot I_x        &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}} \cdot \Phi_x \\ 
 +\rightarrow \Phi_x   &= \mu_0 \mu_{\rm r} {{A}\over{l}} \cdot N_x \cdot I_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/directions of the fluxes** 
 + 
 +The <imgref5_3_2_Solution> shows how to get the correct direction for every single flux by use of the right-hand rule. \\ 
 +The fluxes have to be added regarding these directions and the given direction of the flux in question. 
 +<WRAP> <imgcaption 5_3_2_Solution| toroidal core with two coils in positive and negative polarity> </imgcaption> {{drawio>torCoilPosNeg_solution.svg}} </WRAP> 
 + 
 +Therefore, the formulas are 
 +\begin{align*}  
 +\Phi_{\rm A}   &= \Phi_{1} - \Phi_{2} \\ 
 +               &={{1}\over{R_{\rm m} }} \cdot \left( N_1 \cdot I_1  -  N_2 \cdot I_2 \right) \\ 
 +               & = 0.25 ~{\rm mVs} - 0.15 ~{\rm mVs} \\ 
 +\Phi_{\rm B}   &= \Phi_{1} + \Phi_{2} \\ 
 +               &={{1}\over{R_{\rm m} }} \cdot \left( N_1 \cdot I_1  +  N_2 \cdot I_2 \right) \\ 
 +               & = 0.25 ~{\rm mVs} + 0.15 ~{\rm mVs}  
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~5_3_2s,Solution~@# 
 + 
 + 
 +#@HiddenBegin_HTML~5_3_2r,Result~@# 
 +  - $0.10 ~\rm mVs$ 
 +  - $0.40 ~\rm mVs$ 
 +#@HiddenEnd_HTML~5_3_2r,Result~@#
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
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}}  \Psi\cdot I  +
-                 = {{1}\over{2}}{{\Psi^2 }\over{L}}}  +
-\end{align*}+
  
 ==== magnetic Energy of a toroid Coil ==== ==== magnetic Energy of a toroid Coil ====
Zeile 597: Zeile 737:
 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                  {\rm d}V} \\ +W_{\rm m} &= \iiint_V{w_{\rm m}            {\rm d}V} \\ 
           &= \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 641: Zeile 781:
  
 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 675: Zeile 815:
 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 to the magnetic field and the demagnetization to an energy outtake. The magnetization corresponds to an energy intake to the magnetic field and the demagnetization to an energy outtake.
Zeile 692: Zeile 832:
 <panel type="info" title="Task 5.1.9 Application: Shaded Pole Motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Task 5.1.9 Application: Shaded Pole Motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-The <imgref ImgTask01> and <imgref ImgTask01> show a shaded pole motor of a commercial oven.+The <imgref ImgTask01> and <imgref ImgTask02> show a shaded pole motor of a commercial oven.
  
   * 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 716: Zeile 856:
 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 there 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://en.wikipedia.org/wiki/Magnetic_circuit#Resistance–reluctance_model|Magnetic_circuit}}, which was also presented in this chapter. This model can solve more questions, however, is a bit less intuitive based on this course and less commonly used compared to the {{https://en.wikipedia.org/wiki/Magnetic_circuit#Resistance–reluctance_model|Magnetic_circuit}}, which was also presented in this chapter.
 +
 +==== Moving a Plate into an Air Gap ====
 +
 +<WRAP> <imgcaption ImgNr81 | a magnetic circuit with a moving plate> {{electrical_engineering_2:plate_in_airgap_50_.gif}} {{electrical_engineering_2:stiftt_in_luftspalt_2.3.gif}} </imgcaption> </WRAP>
  
 ==== Switch Reluctance Motor ==== ==== Switch Reluctance Motor ====