Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
| electrical_engineering_2:magnetic_circuits [2024/05/10 20:20] – mexleadmin | electrical_engineering_2:magnetic_circuits [2025/05/27 07:56] (current) – mexleadmin | ||
|---|---|---|---|
| Line 7: | Line 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 Magnetic Circuits ===== | ===== 5.1 Linear Magnetic Circuits ===== | ||
| Line 17: | Line 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 |
| < | < | ||
| Line 140: | Line 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 $l_R = 300 ~\rm mm$. | ||
| Line 164: | Line 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: | ||
| Line 192: | Line 192: | ||
| </ | </ | ||
| - | <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: | ||
| Line 208: | Line 208: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| 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: | ||
| Line 224: | Line 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: | ||
| Line 242: | Line 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 $I=1.30 ~\rm A$. | ||
| Line 250: | Line 250: | ||
| 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 $A_2$ | ||
| Line 270: | Line 270: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| The magnetic circuit in <imgref ImgExNr09> | The magnetic circuit in <imgref ImgExNr09> | ||
| Line 289: | Line 289: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| The choke coil shown in <imgref ImgExNr10> | The choke coil shown in <imgref ImgExNr10> | ||
| Line 450: | Line 450: | ||
| \end{align*} | \end{align*} | ||
| - | Based on energy considerations, | + | For most of the applications |
| - | * the mutual inductances are equal: $M_{12} = M_{21} = M$ | + | * In General: the mutual inductance $M$ is: $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$ |
| - | * 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*} | * 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=" | + | <panel type=" |
| The magnetical configuration in <imgref ExImgNr01> | 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 area of the cross-section is $A=9 ~\rm cm^2$ in all parts, the permeability is $\mu_r=800$, | ||
| - | |||
| - | Calculate | ||
| - | * the self inductions $L_{11}$, $L_{22}$, | ||
| - | * the mutual inductions $M_{12}$, and $M_{21}$, | ||
| - | * the coupling factors $k_{12}$ and $k_{21}$. | ||
| < | < | ||
| - | === 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> < | <WRAP right> < | ||
| Line 492: | Line 489: | ||
| \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) \\ | ||
| Line 503: | Line 500: | ||
| \end{align*} | \end{align*} | ||
| - | == Step 3: Calculate the self-induction == | + | # |
| - | \\ | + | |
| + | 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. | 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*} | ||
| Line 510: | Line 510: | ||
| L_{22} &= {{N_2^2}\over{R_{\rm m2}}} &= 247 ~\rm mH\\ \\ | L_{22} &= {{N_2^2}\over{R_{\rm m2}}} &= 247 ~\rm mH\\ \\ | ||
| \end{align*} | \end{align*} | ||
| + | # | ||
| + | |||
| + | 4. Calculate the coupling factors $k_{12}$ and $k_{21}$. | ||
| - | == Step 4: Calculate the coupling factors == | + | # |
| - | \\ | + | |
| <WRAP right> < | <WRAP right> < | ||
| Line 526: | Line 528: | ||
| A similar approach leads to $k_{12}$ with $k_{12}= 1/4$. | A similar approach leads to $k_{12}$ with $k_{12}= 1/4$. | ||
| + | # | ||
| + | 5. Calculate the mutual inductions $M_{12}$, and $M_{21}$, | ||
| + | |||
| + | # | ||
| \begin{align*} | \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_{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\\ \\ | 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*} | \end{align*} | ||
| + | # | ||
| </ | </ | ||
| + | |||
| + | # | ||
| + | |||
| + | 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 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*} | ||
| + | |||
| + | # | ||
| + | |||
| + | 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. | ||
| + | |||
| + | # | ||
| + | |||
| + | |||
| + | |||
| ==== Effects in the electric Circuits ==== | ==== Effects in the electric Circuits ==== | ||
| Line 548: | Line 588: | ||
| * the direction of the windings, and | * the direction of the windings, and | ||
| - | * the orientation/ | + | * The orientation/ |
| <WRAP center 50%> < | <WRAP center 50%> < | ||
| Line 558: | Line 598: | ||
| < | < | ||
| - | In this case, the **mutual induction is positiv | + | In this case, the **mutual induction is positive |
| The formula of the shown circuitry is then: | The formula of the shown circuitry is then: | ||
| Line 566: | Line 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 is negativ | + | In this case, the **mutual induction is negative |
| The formula of the shown circuitry is then: | The formula of the shown circuitry is then: | ||
| Line 580: | Line 620: | ||
| \end{align*} | \end{align*} | ||
| - | <panel type=" | + | <panel type=" |
| 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> | - The coils shall pass the currents with positive polarity (see the image **A** in <imgref ImgEx14> | ||
| Line 596: | Line 636: | ||
| **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 | + | Since there are no branches, all of the core can be lumped |
| < | < | ||
| Line 602: | Line 642: | ||
| Hopkinson' | Hopkinson' | ||
| - | It connects the magnetic flux $\Phi$ and the magnetic voltage $\theta$ on the single magnetic resistor $R_m$. \\ | + | 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 $\theta_x$. \\ | It also connects the single magnetic fluxes $\Phi_x$ (with $x = {1,2}$) and the single magnetic voltages $\theta_x$. \\ | ||
| Line 634: | Line 674: | ||
| # | # | ||
| - | - $0.10 ~\rm mVs$ | + | - $0.10 ~\rm mVs$ |
| - $0.40 ~\rm mVs$ | - $0.40 ~\rm mVs$ | ||
| # | # | ||
| Line 643: | Line 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: | ||
| Line 669: | Line 709: | ||
| ==== 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: | ||
| Line 695: | Line 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_{\rm m} {\rm d}V} \\ | W_{\rm m} &= \iiint_V{w_{\rm m} {\rm d}V} \\ | ||
| Line 715: | Line 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*} | ||
| Line 725: | Line 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 \\ | ||
| - | | + | |
| & | & | ||
| & | & | ||
| Line 747: | Line 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 $B=\mu H$. | ||
| Line 754: | Line 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 | ||
| Line 777: | Line 815: | ||
| 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. | ||
| Line 788: | Line 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? | ||
| Line 805: | Line 843: | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| - | The book [[https:// | + | The book [[https:// |
| </ | </ | ||
| Line 814: | Line 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 ==== | ==== Moving a Plate into an Air Gap ==== | ||