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electrical_engineering_2:magnetic_circuits [2024/07/11 18:54] – [Effects in the electric Circuits] mexleadminelectrical_engineering_2:magnetic_circuits [2025/05/27 07:56] (aktuell) – mexleadmin
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 <WRAP> <imgcaption ImgNr01 | Similarities magnetic Circuit vs electric Circuit> </imgcaption> {{drawio>CompMagElCircuit.svg}} </WRAP> <WRAP> <imgcaption ImgNr01 | Similarities magnetic Circuit vs electric Circuit> </imgcaption> {{drawio>CompMagElCircuit.svg}} </WRAP>
  
-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 Magnetic Circuits ===== ===== 5.1 Linear Magnetic Circuits =====
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   - 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 are applied and focusing on the average field line are applied.
  
 <WRAP> <imgcaption ImgNr03 | Simplifications and Linearization> </imgcaption> {{drawio>SimplificationLin.svg}} </WRAP> <WRAP> <imgcaption ImgNr03 | Simplifications and Linearization> </imgcaption> {{drawio>SimplificationLin.svg}} </WRAP>
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 |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="info" title="Task 5.1.1 Coil on a plastic Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.1 Coil on a plastic Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 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 $l_R = 300 ~\rm mm$. 
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.2 magnetic Resistance of a cylindrical coil"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.2 magnetic Resistance of a cylindrical coil"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 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:
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.3 magnetic Resistance of an airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.3 magnetic Resistance of an airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 Calculate the magnetic resistances of an airgap with the following dimensions: Calculate the magnetic resistances of an airgap with the following dimensions:
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<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="Exercise 5.1.4 Magnetic Voltage"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 Calculate the magnetic voltage necessary 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:
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.5 Magnetic Flux"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.5 Magnetic Flux"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 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:
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.6 Two-parted ferrite Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.6 Two-parted ferrite Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-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 $I=1.30 ~\rm A$.
  
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 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 $A_2$
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.7 Comparison with simplified Calculation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.7 Comparison with simplified Calculation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 The magnetic circuit in <imgref ImgExNr09> passes a magnetic flux density of $0.4 ~\rm T$ given by an excitation current of $0.50 ~\rm A$ in $400$ windings.  The magnetic circuit in <imgref ImgExNr09> passes a magnetic flux density of $0.4 ~\rm T$ given by an excitation current of $0.50 ~\rm A$ in $400$ windings. 
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<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="Exercise 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 ImgExNr10> 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 $l_0$, $l_1$, $l_2$. 
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-<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="Exercise 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. \\ 
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 \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_{\rm r}}}{{l}\over{A}}\cdot \left(3 + {{1\cdot 2}\over{1 + 2}}\right) \\  R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot \left(3 + {{1\cdot 2}\over{1 + 2}}\right) \\ 
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 #@TaskTitle_HTML@# 5.3.2 Wireles Charging #@TaskText_HTML@# #@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. \\+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. 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.
  
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   * the direction of the windings, and   * the direction of the windings, and
-  * the orientation/counting of the current in the circuit.+  * The orientation/counting of the current in the circuit.
  
 <WRAP center 50%> <imgcaption ImgNr11 | Polarity of Coupling> </imgcaption> {{drawio>DirectionOfCoupling.svg}} </WRAP> <WRAP center 50%> <imgcaption ImgNr11 | Polarity of Coupling> </imgcaption> {{drawio>DirectionOfCoupling.svg}} </WRAP>
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 <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 is positiv $(M>0)$**.+In this case, the **mutual induction is positive $(M>0)$**.
  
 The formula of the shown circuitry is then:  The formula of the shown circuitry is then: 
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 \end{align*} \end{align*}
  
-=== negative Polarity ===+=== Negative Polarity ===
  
-The polarity is negative when only one current either flows into the dotted pin and the other one out of the dotted pin (see <imgref ImgNr13>).+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>).
  
 <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 is negativ $(M<0)$***.+In this case, the **mutual induction is negative $(M<0)$***.
  
 The formula of the shown circuitry is then:  The formula of the shown circuitry is then: 
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 \end{align*} \end{align*}
  
-<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%>+<panel type="info" title="Exercise 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 $A = 500 ~\rm mm^2$ 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 image **A** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm A}$ 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?
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 **Step 1 - Draw an equivalent magnetic circuit** **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>).+Since there are no branches, all of the core can be lumped into a single magnetic resistance (see <imgref ImgEx14circ>).
 <WRAP> <imgcaption ImgEx14circ | equivalent magnetic circuit> </imgcaption> {{drawio>torCoilPosNegCirc.svg}} </WRAP> <WRAP> <imgcaption ImgEx14circ | equivalent magnetic circuit> </imgcaption> {{drawio>torCoilPosNegCirc.svg}} </WRAP>
  
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 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 "current dampening" effect of the inductor. +The energy transfer from the electric circuit to the magnetic field is also the cause of the "current-damping" effect of the inductor. 
 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:
  
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 ==== 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: 
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 ==== 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_{\rm m}            {\rm d}V} \\  W_{\rm m} &= \iiint_V{w_{\rm m}            {\rm d}V} \\ 
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 \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*} 
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 \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 \\ 
- W_{\rm d} &=   \int                                                  {\rm d}W_{\rm m} \\ + W_{\rm m} &=   \int                                                  {\rm d}W_{\rm m} \\ 
            &= N \int_0^t {{{\rm d}\Phi}\over{{\rm d}t}} \cdot i \cdot {\rm d}t \\             &= N \int_0^t {{{\rm d}\Phi}\over{{\rm d}t}} \cdot i \cdot {\rm d}t \\ 
            &= N \int_0^                                {\Phi} i \cdot {\rm d}\Phi \\             &= N \int_0^                                {\Phi} i \cdot {\rm d}\Phi \\ 
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 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>, the inductor will experience increasing and decreasing current.  In the circuit in <imgref ImgNr14>, the inductor will experience increasing and decreasing current. 
 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 $B=\mu H$.
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 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
  
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 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 into the magnetic field, and the demagnetization to an energy outtake.
  
-Moving along the $H$-$B$-curve, one can see, that the energy intake and outtake are the same when coming back to a start point. +Moving along the $H$-$B$-curve, one can see that the energy intake and outtake are the same when coming back to a start point. 
 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, it does not work for magnetically hard materials like a permanent magnet. 
 Here, hysteresis also has to be considered. Here, hysteresis also has to be considered.
  
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 <WRAP> <imgcaption ImgNr17 | H-B-Curve material with Hysteresis> </imgcaption> {{drawio>HBcurvehyst.svg}} </WRAP> <WRAP> <imgcaption ImgNr17 | H-B-Curve material with Hysteresis> </imgcaption> {{drawio>HBcurvehyst.svg}} </WRAP>
  
-===== Tasks =====+===== Exercise =====
  
-<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="Exercise 5.1.9 Application: Shaded Pole Motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
 The <imgref ImgTask01> and <imgref ImgTask02> show a shaded pole motor of a commercial oven. The <imgref ImgTask01> and <imgref ImgTask02> show a shaded pole motor of a commercial oven.
Zeile 845: Zeile 843:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.10 Further exercise"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.10 Further exercise"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-The book [[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|DC Electrical Circuit Analysis - A Practical Approach (Fiore)]] has some nice exercise for beginning in the topic of magnetic circuits.+The book [[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|DC Electrical Circuit Analysis - A Practical Approach (Fiore)]] has some nice exercises for beginning in the topic of magnetic circuits.
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
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 An alternative interpretation of the magnetic circuits is the {{https://en.wikipedia.org/wiki/Gyrator–capacitor model|Gyrator–capacitor model}}.  An alternative interpretation of the magnetic circuits is the {{https://en.wikipedia.org/wiki/Gyrator–capacitor model|Gyrator–capacitor model}}. 
-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://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, it 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 ==== ==== Moving a Plate into an Air Gap ====