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--- electrical_engineering_2:magnetic_circuits [2024/05/10 16:02]
mexleadmin [5.3 Mutual Induction and Coupling]
+++ electrical_engineering_2:magnetic_circuits [2024/07/11 18:54] (aktuell)
mexleadmin [Effects in the electric Circuits]
@@ Zeile 348: / Zeile 348: @@
 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*}
 \Psi_1 &= &\Psi_{11}        &+ \Psi_{12} \\
@@ Zeile 363: / Zeile 363: @@
 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 of 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.
 They are given by the formula $u_{{\rm ind},x} = -{\rm d}\Psi_x /{\rm d}t$:
@@ Zeile 391: / Zeile 386: @@
 <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 the windings, and therefore the sign of the generated flux. \\
+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. \\
 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.
@@ Zeile 426: / Zeile 421: @@
 \begin{align*} k_{21} = \pm {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*}
-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 oposite, the minus sign applies.
+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 second coil. \\
+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 opposite 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.
@@ Zeile 444: / Zeile 439: @@
 \end{align*}
-Note, that also $M_{21}$ and $M_{12}$ can be either positiv or negative, depending on the sign of the coupling coefficients.
+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:
@@ Zeile 454: / Zeile 449: @@
 \left( \begin{array}{c} i_1                                                                        \\ i_2                                                                        \end{array} \right)
 \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%>
 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 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>
-=== Step 1: Draw the problem as a network ===
+.  Simplify the configuration into three magnetic resistors and 2 voltage sources. Draw the problem as an equivalent circuit
-=== Step 2: Calculate the magnetic resistances ===
+#@HiddenBegin_HTML~5311,Result~@#
+<WRAP> <imgcaption ExImgNr11 | Equivalent Network> </imgcaption> {{drawio>CoreWithTwoCoils2network.svg}} </WRAP>
+#@HiddenEnd_HTML~5311,Result~@#
-The magnetic resistance is summed up by looking at the circuit from the source $1$:
+. 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~@#
+<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 $1$ (see <imgref ExImgNr13>):
 \begin{align*}
 R_{\rm m1} &= R_{\rm m,11} + R_{\rm m,ss} || R_{\rm m,22} \\
@@ Zeile 476: / Zeile 484: @@
 where the parts are given as
 \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*}
 With the given geometry this leads to
 \begin{align*}
-R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm 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_{\rm 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*}
 Similarly, the magnetic resistance $R_{m2}$ is
 \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*}
-== Step 3: Calculate the magnetic inductances ==
+#@HiddenEnd_HTML~5312,Path~@#
+. 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*}
 L_{11} &= {{N_1^2}\over{R_{\rm m1}}}                    &= 329 ~\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*}
+#@HiddenEnd_HTML~5313,Path~@#
+. 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~@#
+. 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>
-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.
+. 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 ~@#
+. 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 = \rm{sgn}(k_{12}) \sqrt{k_{12}\cdot k_{21}} \end{align*}
 ==== Effects in the electric Circuits ====
@@ Zeile 557: / Zeile 620: @@
 \end{align*}
-<panel type="info" title="Task 5.3.2 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 $A = 500 ~\rm mm^2$ and an average circumference of $l=280 ~\rm mm$.
@@ Zeile 579: / Zeile 642: @@
 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 $R_m$. \\
+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 $\theta_x$. \\
@@ Zeile 611: / Zeile 674: @@
 #@HiddenBegin_HTML~5_3_2r,Result~@#
-  - $0.10 ~\rm  mVs$
+  - $0.10 ~\rm mVs$
   - $0.40 ~\rm mVs$
 #@HiddenEnd_HTML~5_3_2r,Result~@#
@@ Zeile 769: / 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%>
-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?