Differences

This shows you the differences between two versions of the page.

--- dummy2 [2025/04/10 10:25] – tfischer
+++ dummy2 [2026/05/16 21:52] (current) – mexleadmin
@@ Line 1: / Line 1: @@
-asdasdasd
+====== Block 09/10 — Transformers and Magnetic Coupling ======
+===== Learning objectives =====
+<callout>
+After this 90-minute block, you can
+  * explain how two coils can exchange energy by a common magnetic flux \(\Phi\).
+  * use the ideal transformer equations
+\[
+\begin{align*}
+\frac{\underline{U}_1}{\underline{U}_2}=\frac{N_1}{N_2}=n,
+\qquad
+\frac{\underline{I}_1}{\underline{I}_2}=-\frac{1}{n}
+\end{align*}
+\]
+with a clear sign convention.
+  * explain mutual inductance \(M\) using flux linkage and magnetic reluctance \(R_{\rm m}\).
+  * distinguish **main flux**, **leakage flux**, **copper losses**, and **iron losses** in a real transformer.
+  * refer secondary-side quantities to the primary side using \( \underline{U}'_2=n\underline{U}_2\), \( \underline{I}'_2=\frac{1}{n}\underline{I}_2\), \(R'_2=n^2R_2\), and \(X'_{2\sigma}=n^2X_{2\sigma}\).
+  * interpret the no-load test and short-circuit test using the reduced equivalent circuit.
+  * calculate short-circuit voltage \(u_{\rm k}\), continuous short-circuit current \(I_{\rm 1k}\), and an estimated initial peak short-circuit current.
+  * connect transformer parameters to engineering applications in mechatronics and robotics, such as isolated power supplies, motor current measurement, welding transformers, and safety transformers.
+</callout>
+===== Preparation at Home =====
+Well, again
+  * read through the present chapter and write down anything you did not understand.
+  * Repeat the EEE1 ideas of [[:electrical_engineering_1:block18|magnetic flux and induction]], [[:electrical_engineering_1:block19|magnetic circuits]], and [[:electrical_engineering_1:block20|inductance and magnetic energy]].
+  * Repeat from EEE2 the use of [[block03|sinusoidal quantities]], [[block04|complex calculation]], and [[block06|complex power]].
+  * For a deeper field-theory view, see also [[:electrical_engineering_2:magnetic_circuits#mutual_induction_and_coupling|Mutual Induction and Coupling]].
+For checking your understanding please do the quick checks in the exercise section.
+===== 90-minute plan =====
+  * **Warm-up (10 min):**
+    * Where do transformers occur in robots and automation systems?
+    * Recall: Faraday induction from EEE1 — a changing magnetic flux induces a voltage.
+    * Recall: in AC analysis we use RMS phasors \(\underline{U}\), \(\underline{I}\), and impedances \(j\omega L\).
+  * **Core concepts and derivations (55 min):**
+    * Ideal transformer: common flux, voltage ratio, current ratio, power balance.
+    * Mutual inductance: how flux from one coil links another coil.
+    * Magnetic coupling with reluctance \(R_{\rm m}\).
+    * Real transformer: winding resistances, leakage inductances, iron-loss resistance.
+    * Reduced equivalent circuit: refer secondary quantities to the primary side.
+    * No-load and short-circuit operation: what can be measured, what can be neglected.
+  * **Practice (20 min):**
+    * Quick ratio calculations for step-up and step-down transformers.
+    * Unit checks for \(j\omega L\), \(j\omega N\Phi\), and \(u_{\rm k}\).
+    * Short-circuit current calculation for a transformer used in an actuator supply.
+  * **Wrap-up (5 min):**
+    * Summary box: ideal transformer, mutual inductance, real transformer, reduced circuit, short-circuit parameters.
+    * Common pitfalls checklist.
+===== Conceptual overview =====
+<callout icon="fa fa-lightbulb-o" color="blue">
+  * A transformer is **not** a DC component. It needs a changing magnetic flux. In normal operation this is usually a sinusoidal flux created by AC voltage.
+  * The transformer does not “create power”. Ideally, it trades voltage for current:
+\[
+\begin{align*}
+\text{higher voltage} \quad \Longleftrightarrow \quad \text{lower current}
+\end{align*}
+\]
+  * The link between the two windings is the magnetic field in the iron core. This continues directly from EEE1:
+    * [[:electrical_engineering_1:block18|induction]] explains why a changing flux induces voltage.
+    * [[:electrical_engineering_1:block19|magnetic circuits]] explains why the iron core guides the flux.
+    * [[:electrical_engineering_1:block20|inductance]] explains how flux linkage and current are connected.
+  * Mutual inductance \(M\) measures how strongly one coil “notices” the changing current in another coil.
+  * A real transformer is almost ideal, but not quite:
+    * \(R_1, R_2\): copper losses in the windings.
+    * \(L_{1\sigma}, L_{2\sigma}\): leakage flux that does not couple both windings.
+    * \(R_{\rm Fe}\): iron losses in the core.
+    * \(L_{\rm H}\): main magnetizing inductance needed to create the main flux.
+  * In engineering, transformer data such as \(u_{\rm k}\) are not abstract: they determine voltage drop, fault current, thermal stress, and protection design.
+</callout>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+===== Core content =====
+==== From EEE1 induction to an AC transformer ====
+<panel type="info" title="Transition from EEE1 to EEE2">
+In EEE1 we considered magnetic flux \(\Phi\), flux linkage \(\Psi\), and induction.
+For one coil with \(N\) turns the flux linkage is
+\[
+\begin{align*}
+\Psi = N\Phi .
+\end{align*}
+\]
+Faraday's law gives
+\[
+\begin{align*}
+u(t)=\frac{{\rm d}\Psi}{{\rm d}t}=N\frac{{\rm d}\Phi}{{\rm d}t}.
+\end{align*}
+\]
+In sinusoidal steady state this becomes the phasor equation
+\[
+\begin{align*}
+\underline{U}=j\omega\underline{\Psi}=j\omega N\underline{\Phi}.
+\end{align*}
+\]
+This is the starting point for the transformer.
+</panel>
+<WRAP>
+<panel type="default">
+<imgcaption fig_transformer_principle|Idealized single-phase transformer with primary winding, secondary winding, iron core, and common main flux \(\Phi\).></imgcaption>
+{{drawio>block09_transformer_principle.svg}}
+</panel>
+</WRAP>
+<callout type="info" icon="true">
+**Unit check for induced voltage**
+\[
+\begin{align*}
+\underline{U}=j\omega N\underline{\Phi}
+\end{align*}
+\]
+with
+\[
+\begin{align*}
+[\omega\Phi]={\rm s}^{-1}\cdot {\rm Wb}
+={\rm s}^{-1}\cdot {\rm V\,s}
+={\rm V}.
+\end{align*}
+\]
+The number of turns \(N\) is dimensionless.
+</callout>
+==== Mutual induction: the key idea before the transformer ====
+Before we look at the transformer, we need one important idea:
+<callout icon="fa fa-lightbulb-o" color="blue">
+A changing current in coil \(1\) creates a changing magnetic flux.
+If part of this flux passes through coil \(2\), a voltage is induced in coil \(2\).
+This is called **mutual induction**.
+</callout>
+<WRAP>
+<panel type="default">
+<imgcaption fig_mutual_induction_two_coils|Mutual induction of two coils: only part of the flux created by coil \(1\) links coil \(2\).></imgcaption>
+{{:electrical_engineering_2:mutualinductiontwocoils1.svg?650}}
+</panel>
+</WRAP>
+The flux created by coil \(1\) can be split into
+\[
+\begin{align*}
+\Phi_{11}
+=
+\Phi_{21}
++
+\Phi_{\rm S1}.
+\end{align*}
+\]
+  * \(\Phi_{11}\): total flux created by coil \(1\).
+  * \(\Phi_{21}\): part of this flux that also links coil \(2\).
+  * \(\Phi_{\rm S1}\): stray or leakage flux that does **not** link coil \(2\).
+The voltage induced in coil \(2\) is
+\[
+\begin{align*}
+u_{{\rm ind},2}(t)
+=
+-N_2\frac{{\rm d}\Phi_{21}}{{\rm d}t}.
+\end{align*}
+\]
+<panel type="info" title="First-year analogy: two coils and a leaky air duct">
+Imagine two pistons connected by an air duct.
+  * If piston \(1\) moves, pressure travels through the duct and can move piston \(2\).
+  * If the duct is tight, piston \(2\) strongly feels piston \(1\).
+  * If the duct leaks, part of the pressure escapes and piston \(2\) feels less.
+For coupled coils:
+  * the moving piston corresponds to changing current,
+  * the air duct corresponds to the magnetic path,
+  * leakage air corresponds to leakage flux,
+  * the strength of the coupling is described by mutual inductance \(M\).
+The analogy is not perfect, but it helps to remember: **coupling is strong when much of the field of one coil also passes through the other coil.**
+</panel>
+==== Linked fluxes and mutual inductance ====
+For a single coil we already know
+\[
+\begin{align*}
+\Psi=L i .
+\end{align*}
+\]
+For two coupled coils, each flux linkage can depend on both currents:
+\[
+\begin{align*}
+\Psi_1
+&=
+L_{11}i_1
++
+M_{12}i_2,
+\\
+\Psi_2
+&=
+M_{21}i_1
++
+L_{22}i_2.
+\end{align*}
+\]
+For most transformer calculations we use the symmetric case
+\[
+\begin{align*}
+M_{12}=M_{21}=M.
+\end{align*}
+\]
+Then
+\[
+\begin{align*}
+\boxed{
+\begin{pmatrix}
+\Psi_1\\
+\Psi_2
+\end{pmatrix}
+=
+\begin{pmatrix}
+L_{11} & M\\
+M & L_{22}
+\end{pmatrix}
+\begin{pmatrix}
+i_1\\
+i_2
+\end{pmatrix}
+}
+\end{align*}
+\]
+and
+\[
+\begin{align*}
+M=k\sqrt{L_{11}L_{22}}.
+\end{align*}
+\]
+Here \(k\) is the coupling coefficient.
+<tabcaption tab_coupling_coefficient|Meaning of the coupling coefficient \(k\)>
+^ Coupling coefficient ^ Interpretation ^ Typical example ^
+| \(k=0\) | no useful flux from one coil links the other coil | coils far apart |
+| \(0<k<1\) | partial coupling | wireless charger with air gap or misalignment |
+| \(k\approx 1\) | almost all useful flux links both coils | transformer with iron core |
+| sign of \(k\) | depends on winding direction and reference arrows | dot convention |
+==== Polarity and the dot convention ====
+The sign of the mutual term depends on the winding direction and on the chosen current reference arrows.
+<WRAP>
+<panel type="default">
+<imgcaption fig_direction_of_coupling|Dot convention: the dots indicate corresponding winding ends.></imgcaption>
+{{:electrical_engineering_2:directionofcoupling.svg?650}}
+</panel>
+</WRAP>
+<callout>
+**Rule of thumb**
+  * If both currents enter dotted terminals, the mutual fluxes support each other.
+  * If one current enters a dotted terminal and the other current leaves a dotted terminal, the mutual fluxes oppose each other.
+</callout>
+<WRAP group>
+<WRAP column half>
+<panel type="default">
+<imgcaption fig_positive_coupling|Positive coupling: currents enter corresponding dotted terminals.></imgcaption>
+{{:electrical_engineering_2:poscoupling.svg?500}}
+</panel>
+</WRAP>
+<WRAP column half>
+<panel type="default">
+<imgcaption fig_negative_coupling|Negative coupling: only one current enters a dotted terminal.></imgcaption>
+{{:electrical_engineering_2:negcoupling.svg?500}}
+</panel>
+</WRAP>
+</WRAP>
+For positive coupling:
+\[
+\begin{align*}
+u_1
+&=
+L_{11}\frac{{\rm d}i_1}{{\rm d}t}
++
+M\frac{{\rm d}i_2}{{\rm d}t},
+\\
+u_2
+&=
+M\frac{{\rm d}i_1}{{\rm d}t}
++
+L_{22}\frac{{\rm d}i_2}{{\rm d}t}.
+\end{align*}
+\]
+For negative coupling, the sign of \(M\) is negative in the chosen equation system.
+<panel type="info" title="Engineering example: wireless charging">
+In wireless charging, the transmitter coil and receiver coil are separated by an air gap.
+The coupling coefficient \(k\) is much smaller than in a transformer with an iron core.
+If the receiver is misaligned, less flux from the transmitter passes through it.
+Then \(M\) decreases, the induced voltage decreases, and the transmitted power decreases.
+</panel>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+==== Magnetic coupling with reluctance only ====
+We now use the magnetic circuit model from EEE1, but only with **magnetic reluctance** \(R_{\rm m}\).
+No inverse magnetic quantity is needed here.
+The magnetic voltage, also called magnetomotive force, is
+\[
+\begin{align*}
+\Theta=N i .
+\end{align*}
+\]
+For a magnetic path with reluctance \(R_{\rm m}\), Hopkinson's law is
+\[
+\begin{align*}
+\Theta=R_{\rm m}\Phi
+\qquad \Longleftrightarrow \qquad
+\Phi=\frac{\Theta}{R_{\rm m}}.
+\end{align*}
+\]
+<panel type="info" title="Analogy to an electric circuit">
+\[
+\begin{align*}
+\text{electric:}\quad U=R I
+\qquad
+\text{magnetic:}\quad \Theta=R_{\rm m}\Phi
+\end{align*}
+\]
+  * Electric voltage \(U\) pushes electric current \(I\) through resistance \(R\).
+  * Magnetic voltage \(\Theta=N i\) pushes magnetic flux \(\Phi\) through reluctance \(R_{\rm m}\).
+</panel>
+For two windings on the same main magnetic path, the total magnetic voltage is
+\[
+\begin{align*}
+\Theta
+=
+N_1\underline{I}_1
++
+N_2\underline{I}_2.
+\end{align*}
+\]
+The main flux is
+\[
+\begin{align*}
+\underline{\Phi}
+=
+\frac{\Theta}{R_{\rm mH}}
+=
+\frac{N_1\underline{I}_1+N_2\underline{I}_2}{R_{\rm mH}}.
+\end{align*}
+\]
+The flux linkages are
+\[
+\begin{align*}
+\underline{\Psi}_1
+&=
+N_1\underline{\Phi}
+=
+\frac{N_1^2}{R_{\rm mH}}\underline{I}_1
++
+\frac{N_1N_2}{R_{\rm mH}}\underline{I}_2,
+\\
+\underline{\Psi}_2
+&=
+N_2\underline{\Phi}
+=
+\frac{N_1N_2}{R_{\rm mH}}\underline{I}_1
++
+\frac{N_2^2}{R_{\rm mH}}\underline{I}_2.
+\end{align*}
+\]
+Therefore
+\[
+\begin{align*}
+\boxed{
+L_{1{\rm H}}=\frac{N_1^2}{R_{\rm mH}}
+}
+\qquad
+\boxed{
+L_{2{\rm H}}=\frac{N_2^2}{R_{\rm mH}}
+}
+\qquad
+\boxed{
+M=\frac{N_1N_2}{R_{\rm mH}}
+}
+\end{align*}
+\]
+for an ideal common magnetic path.
+<callout type="info" icon="true">
+**Unit check for mutual inductance**
+\[
+\begin{align*}
+M=\frac{N_1N_2}{R_{\rm mH}}
+\end{align*}
+\]
+with
+\[
+\begin{align*}
+[R_{\rm mH}]
+=
+\frac{{\rm A}}{{\rm Vs}}
+=
+\frac{1}{{\rm H}}.
+\end{align*}
+\]
+Thus
+\[
+\begin{align*}
+[M]=\frac{1}{1/{\rm H}}={\rm H}.
+\end{align*}
+\]
+</callout>
+==== Ideal single-phase transformer ====
+For an ideal transformer we assume:
+  * both windings are linked by the same magnetic flux \(\Phi\),
+  * there is no leakage flux,
+  * there are no winding resistances,
+  * there are no iron losses,
+  * the transformer stores no net energy over one period.
+Let \(N_1\) be the number of turns of the primary winding and \(N_2\) the number of turns of the secondary winding.
+\[
+\begin{align*}
+\underline{\Psi}_1 &= N_1\underline{\Phi},
+&
+\underline{U}_1 &= j\omega\underline{\Psi}_1
+= j\omega N_1\underline{\Phi},
+\\
+\underline{\Psi}_2 &= N_2\underline{\Phi},
+&
+\underline{U}_2 &= j\omega\underline{\Psi}_2
+= j\omega N_2\underline{\Phi}.
+\end{align*}
+\]
+Dividing the two voltage equations gives the **turns ratio**
+\[
+\begin{align*}
+\boxed{
+\frac{\underline{U}_1}{\underline{U}_2}
+=
+\frac{N_1}{N_2}
+=
+n
+}
+\end{align*}
+\]
+with
+\[
+\begin{align*}
+n=\frac{N_1}{N_2}.
+\end{align*}
+\]
+<WRAP column 100%>
+<panel type="danger" title="Remember: ideal transformer ratios">
+With the indicated reference arrows and a lossless transformer:
+\[
+\begin{align*}
+\underline{U}_1\underline{I}_1+\underline{U}_2\underline{I}_2=0
+\end{align*}
+\]
+and therefore
+\[
+\begin{align*}
+\boxed{
+\frac{\underline{I}_1}{\underline{I}_2}
+=
+-\frac{\underline{U}_2}{\underline{U}_1}
+=
+-\frac{N_2}{N_1}
+=
+-\frac{1}{n}
+}
+\end{align*}
+\]
+The minus sign is not a “loss”. It is caused by the chosen current arrows. The primary side absorbs power while the secondary side delivers power to the load.
+</panel>
+</WRAP>
+<panel type="info" title="Physical interpretation">
+  * If \(N_2<N_1\), the transformer steps the voltage down: \(U_2<U_1\).
+  * At the same time, the secondary current can be higher: \(I_2>I_1\).
+  * This is useful in robotics power supplies: a mains-side transformer or isolated converter stage may reduce voltage while increasing available current for actuators.
+</panel>
+==== Example: step-down transformer for a robot controller ====
+A transformer has \(N_1=800\) turns and \(N_2=80\) turns.
+The primary RMS voltage is \(U_1=230~{\rm V}\).
+\[
+\begin{align*}
+n &= \frac{N_1}{N_2}
+= \frac{800}{80}
+=10,
+\\
+U_2 &= \frac{U_1}{n}
+= \frac{230~{\rm V}}{10}
+=23~{\rm V}.
+\end{align*}
+\]
+If the secondary side supplies \(I_2=4~{\rm A}\), the ideal primary current magnitude is
+\[
+\begin{align*}
+I_1 &= \frac{I_2}{n}
+= \frac{4~{\rm A}}{10}
+=0.4~{\rm A}.
+\end{align*}
+\]
+The apparent power is equal on both sides:
+\[
+\begin{align*}
+S_1 &= U_1I_1=230~{\rm V}\cdot 0.4~{\rm A}=92~{\rm VA},
+\\
+S_2 &= U_2I_2=23~{\rm V}\cdot 4~{\rm A}=92~{\rm VA}.
+\end{align*}
+\]
+~~PAGEBREAK~~ ~~CLEARFIX~~
+==== Real transformer: leakage and losses ====
+In a real transformer, not all flux links both windings.
+  * The **main flux** \(\Phi_{\rm H}\) links primary and secondary winding.
+  * The **primary leakage flux** \(\Phi_{1\sigma}\) mainly links only the primary winding.
+  * The **secondary leakage flux** \(\Phi_{2\sigma}\) mainly links only the secondary winding.
+<WRAP>
+<panel type="default">
+<imgcaption fig_main_and_leakage_flux|Main flux and leakage fluxes in a real transformer.></imgcaption>
+{{drawio>block09_main_and_leakage_flux.svg}}
+</panel>
+</WRAP>
+The real flux linkage equations become
+\[
+\begin{align*}
+\underline{\Psi}_1
+&=
+L_1\underline{I}_1+M\underline{I}_2,
+&
+L_1&=L_{1{\rm H}}+L_{1\sigma},
+\\
+\underline{\Psi}_2
+&=
+L_2\underline{I}_2+M\underline{I}_1,
+&
+L_2&=L_{2{\rm H}}+L_{2\sigma}.
+\end{align*}
+\]
+The winding resistances \(R_1\) and \(R_2\) cause copper losses:
+\[
+\begin{align*}
+P_{\rm Cu,1}=R_1I_1^2,
+\qquad
+P_{\rm Cu,2}=R_2I_2^2.
+\end{align*}
+\]
+<panel type="info" title="Color scheme for the equivalent equations">
+In the following formulas:
+  * blue terms: useful main magnetic coupling,
+  * orange terms: leakage flux,
+  * red terms: winding resistance and copper loss.
+</panel>
+\[
+\begin{align*}
+\underline{U}_1
+&=
+\underbrace{{\color{red}{R_1\underline{I}_1}}}_{\text{primary copper drop}}
++
+\underbrace{{\color{orange}{j\omega L_{1\sigma}\underline{I}_1}}}_{\text{primary leakage drop}}
++
+\underbrace{{\color{blue}{j\omega L_{1{\rm H}}\underline{I}_1+j\omega M\underline{I}_2}}}_{\text{main magnetic coupling}},
+\\[6pt]
+\underline{U}_2
+&=
+\underbrace{{\color{red}{R_2\underline{I}_2}}}_{\text{secondary copper drop}}
++
+\underbrace{{\color{orange}{j\omega L_{2\sigma}\underline{I}_2}}}_{\text{secondary leakage drop}}
++
+\underbrace{{\color{blue}{j\omega L_{2{\rm H}}\underline{I}_2+j\omega M\underline{I}_1}}}_{\text{main magnetic coupling}}.
+\end{align*}
+\]
+<callout>
+The blue terms are responsible for transformer action.
+The orange terms are unwanted but unavoidable.
+The red terms convert electrical energy into heat.
+</callout>
+With the leakage reactances
+\[
+\begin{align*}
+X_{1\sigma}=\omega L_{1\sigma},
+\qquad
+X_{2\sigma}=\omega L_{2\sigma},
+\qquad
+X_{\rm M}=\omega M
+\end{align*}
+\]
+these equations can be represented by an equivalent circuit.
+<callout type="info" icon="true">
+**Unit check for leakage reactance**
+\[
+\begin{align*}
+X_{1\sigma}=\omega L_{1\sigma}
+\end{align*}
+\]
+with
+\[
+\begin{align*}
+[\omega L]={\rm s}^{-1}\cdot{\rm H}
+=
+{\rm s}^{-1}\cdot \frac{{\rm V\,s}}{{\rm A}}
+=
+\Omega .
+\end{align*}
+\]
+So \(jX_{1\sigma}\) is an impedance.
+</callout>
+==== Why leakage flux matters in engineering ====
+<panel type="info" title="Mechatronics example: motor start-up">
+A robot axis may draw a high current during acceleration.
+This high current produces a voltage drop at the leakage reactance and winding resistance of the transformer.
+Possible effects:
+  * the secondary voltage decreases,
+  * a DC link after a rectifier may sag,
+  * controllers may reset due to undervoltage,
+  * cables and protective devices must withstand the fault current.
+So leakage is not only a “field theory detail”. It directly affects the electrical behavior of the machine.
+</panel>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+==== Reduced equivalent circuit referred to the primary side ====
+For calculations it is convenient to move all secondary-side quantities to the primary side. This is called **referring** or **transforming** the secondary side to the primary side.
+\[
+\begin{align*}
+\boxed{
+\underline{U}'_2=n\underline{U}_2
+}
+\qquad
+\boxed{
+\underline{I}'_2=\frac{1}{n}\underline{I}_2
+}
+\end{align*}
+\]
+The secondary resistance and leakage reactance are transformed by \(n^2\):
+\[
+\begin{align*}
+\boxed{
+R'_2=n^2R_2
+}
+\qquad
+\boxed{
+X'_{2\sigma}=n^2X_{2\sigma}
+}
+\end{align*}
+\]
+<callout type="info" icon="true">
+**Unit check for referred resistance**
+The turns ratio \(n\) is dimensionless. Therefore
+\[
+\begin{align*}
+[R'_2]=[n^2R_2]=\Omega .
+\end{align*}
+\]
+The value changes, but the unit does not.
+</callout>
+<WRAP>
+<panel type="default">
+<imgcaption fig_reduced_equivalent_circuit|Reduced transformer equivalent circuit with secondary quantities referred to the primary side.></imgcaption>
+{{drawio>block09_reduced_transformer_equivalent_circuit.svg}}
+</panel>
+</WRAP>
+In the reduced equivalent circuit:
+  * \(R_1\) and \(R'_2\) model copper losses.
+  * \(jX_{1\sigma}\) and \(jX'_{2\sigma}\) model leakage flux.
+  * \(jX_{1{\rm H}}\) models the magnetizing branch.
+  * \(R_{\rm Fe}\) is placed parallel to \(jX_{1{\rm H}}\) to model iron losses.
+<panel type="info" title="Why the reduced circuit is useful">
+Once all quantities are referred to one side, the transformer can be calculated like an AC network with impedances.
+This uses the same method as [[block04|complex network calculation]]: replace components by impedances and apply Kirchhoff's laws.
+</panel>
+==== No-load operation of the real transformer ====
+No-load operation means that the secondary side is open:
+\[
+\begin{align*}
+\underline{I}_2=0.
+\end{align*}
+\]
+The primary side still draws a small no-load current \(\underline{I}_{10}\). This current has two parts:
+\[
+\begin{align*}
+\underline{I}_{10}
+=
+\underline{I}_{\rm Fe}
++
+\underline{I}_{\rm m}.
+\end{align*}
+\]
+  * \(\underline{I}_{\rm Fe}\): current through \(R_{\rm Fe}\), in phase with voltage, represents iron losses.
+  * \(\underline{I}_{\rm m}\): magnetizing current through \(jX_{1{\rm H}}\), approximately \(90^\circ\) lagging.
+<WRAP>
+<panel type="default">
+<imgcaption fig_no_load_phasor|No-load phasor diagram: the no-load current is the sum of iron-loss current and magnetizing current.></imgcaption>
+{{drawio>block09_no_load_phasor_diagram.svg}}
+</panel>
+</WRAP>
+The technical voltage ratio is often defined from the no-load voltages. Here it is denoted by \(\ddot{u}\):
+\[
+\begin{align*}
+\ddot{u}
+=
+\frac{\text{higher voltage}}{\text{lower voltage}}
+\bigg|_{\rm no~load}.
+\end{align*}
+\]
+For a step-down transformer:
+\[
+\begin{align*}
+\ddot{u}
+=
+\frac{U_{1{\rm N}}}{U_{20}}.
+\end{align*}
+\]
+Here \(U_{1{\rm N}}\) is the rated primary voltage and \(U_{20}\) is the open-circuit secondary voltage.
+<callout>
+Because of real voltage drops and magnetizing effects,
+\[
+\begin{align*}
+\ddot{u}\neq n,
+\end{align*}
+\]
+but for many practical transformers
+\[
+\begin{align*}
+\ddot{u}\approx n.
+\end{align*}
+\]
+</callout>
+==== Short-circuit operation of the real transformer ====
+In the short-circuit test, the secondary side is shorted:
+\[
+\begin{align*}
+\underline{U}_2=0.
+\end{align*}
+\]
+Because the required primary voltage is small, the magnetizing branch is often neglected:
+\[
+\begin{align*}
+X_{1{\rm H}},\;R_{\rm Fe}
+\gg
+X'_{2\sigma},\;R'_2.
+\end{align*}
+\]
+This gives the short-circuit equivalent circuit with
+\[
+\begin{align*}
+\boxed{
+R_{\rm k}=R_1+R'_2
+}
+\qquad
+\boxed{
+X_{\rm k}=X_{1\sigma}+X'_{2\sigma}
+}
+\end{align*}
+\]
+and
+\[
+\begin{align*}
+\underline{Z}_{\rm k}
+=
+R_{\rm k}+jX_{\rm k}.
+\end{align*}
+\]
+<WRAP>
+<panel type="default">
+<imgcaption fig_short_circuit_equivalent|Short-circuit equivalent circuit of a real transformer.></imgcaption>
+{{drawio>block09_short_circuit_equivalent_circuit.svg}}
+</panel>
+</WRAP>
+<panel type="info" title="Definition: rated short-circuit voltage">
+The **rated short-circuit voltage** \(U_{1{\rm k}}\) is the primary voltage that must be applied while the secondary side is shorted so that rated primary current \(I_{1{\rm N}}\) flows.
+As a relative value:
+\[
+\begin{align*}
+\boxed{
+u_{\rm k}
+=
+\frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
+}
+\end{align*}
+\]
+Small \(u_{\rm k}\) means: small internal impedance and high possible fault current.
+Large \(u_{\rm k}\) means: stronger current limitation and larger voltage drop under load.
+</panel>
+The continuous short-circuit current for rated primary voltage is
+\[
+\begin{align*}
+\boxed{
+I_{1{\rm k}}
+=
+\frac{U_{1{\rm N}}}{U_{1{\rm k}}}\cdot I_{1{\rm N}}
+=
+I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}}
+}
+\end{align*}
+\]
+where \(u_{\rm k}\) is inserted as a percentage value.
+<callout type="info" icon="true">
+**Unit check for \(u_{\rm k}\)**
+\[
+\begin{align*}
+u_{\rm k}
+=
+\frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
+\end{align*}
+\]
+is dimensionless. It is usually stated in percent.
+</callout>
+==== Why the first short-circuit peak can be about \(2.54 I_{1{\rm k}}\) ====
+The RMS short-circuit current \(I_{1{\rm k}}\) does not describe the highest instantaneous current.
+When a short circuit starts, the current can contain
+  * a sinusoidal AC component, and
+  * a decaying DC offset.
+The worst case occurs when the fault starts at an unfavorable phase angle. Then the first current peak is larger than the normal sinusoidal peak \(\sqrt{2}I_{1{\rm k}}\).
+A common engineering form is
+\[
+\begin{align*}
+i_{\rm p}
+=
+\kappa\sqrt{2}\,I''_{\rm k}.
+\end{align*}
+\]
+Here
+  * \(i_{\rm p}\) is the instantaneous peak short-circuit current,
+  * \(I''_{\rm k}\) is the initial symmetrical RMS short-circuit current,
+  * \(\kappa\) is a peak factor depending mainly on the \(R/X\) ratio of the short-circuit impedance.
+For a strongly inductive transformer short circuit, a practical approximation is
+\[
+\begin{align*}
+\kappa\approx 1.8.
+\end{align*}
+\]
+Then
+\[
+\begin{align*}
+i_{\rm p}
+\approx
+.8\cdot\sqrt{2}\cdot I_{1{\rm k}}
+=
+.55\,I_{1{\rm k}}
+\approx
+.54\,I_{1{\rm k}}.
+\end{align*}
+\]
+<WRAP column 100%>
+<panel type="danger" title="Do not overinterpret the factor 2.54">
+The factor \(2.54\) is an engineering approximation for the first peak current.
+It is **not** a universal transformer law.
+For real protection design, use the applicable standard, manufacturer data, and the actual \(R/X\) ratio of the installation.
+</panel>
+</WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+==== Real transformer under load ====
+Under load, the short-circuit equivalent circuit is often sufficient for engineering estimates.
+\[
+\begin{align*}
+\underline{U}_{\rm k}
+=
+\left(R_{\rm k}+jX_{\rm k}\right)\underline{I}_1.
+\end{align*}
+\]
+This voltage drop is subtracted vectorially from the primary-side voltage relation. Therefore the secondary voltage depends on
+  * load current magnitude,
+  * load power factor,
+  * winding resistance,
+  * leakage reactance.
+<WRAP>
+<panel type="default">
+<imgcaption fig_kapp_triangle|Kapp triangle: approximate voltage drop under load using \(R_{\rm k}I\) and \(X_{\rm k}I\).></imgcaption>
+{{drawio>block09_kapp_triangle.svg}}
+</panel>
+</WRAP>
+<panel type="info" title="Engineering use: voltage regulation">
+In a robot with motors, the supply transformer may show a lower output voltage during high acceleration because the motor currents increase.
+The Kapp triangle helps estimate this voltage drop. This is important for:
+  * selecting transformer size,
+  * checking whether the DC link after a rectifier remains high enough,
+  * designing fuses and protective devices,
+  * avoiding undervoltage resets in control electronics.
+</panel>
+==== Construction types and cooling ====
+Transformer behavior is influenced by construction.
+<WRAP group>
+<WRAP column half>
+<panel type="default">
+<imgcaption fig_core_transformer|Core-type transformer: usually smaller short-circuit voltage \(u_{\rm k}\).></imgcaption>
+{{drawio>block09_core_type_transformer.svg}}
+</panel>
+</WRAP>
+<WRAP column half>
+<panel type="default">
+<imgcaption fig_shell_transformer|Shell-type transformer: usually larger short-circuit voltage \(u_{\rm k}\).></imgcaption>
+{{drawio>block09_shell_type_transformer.svg}}
+</panel>
+</WRAP>
+</WRAP>
+Cooling types:
+  * **Dry-type transformer:** air cooling, often used inside machines or buildings at lower and medium power.
+  * **Oil transformer:** oil provides insulation and heat transfer, typical for higher power.
+<panel type="info" title="Mechatronics examples">
+  * **Isolating transformer:** safe diagnostic supply for laboratory setups.
+  * **Control transformer:** supplies \(24~{\rm V}\) or similar low-voltage control circuits.
+  * **Current transformer:** measures large motor currents with galvanic isolation.
+  * **Welding transformer:** intentionally high short-circuit voltage and current limitation for welding processes.
+</panel>
+==== Typical technical transformer data ====
+<tabcaption tab_transformer_types|Typical transformer types, short-circuit voltage, and secondary voltage>
+^ Name / use ^ Typical \(u_{\rm k}\) ^ Secondary voltage \(U_2\) ^ Important note ^
+| Power transformer | \(4\ldots 12~\%\) | application-dependent | low voltage drop, high fault currents possible |
+| Isolating transformer | \(\approx 10~\%\) | max. \(250~{\rm V}\) | galvanic isolation for safety and measurement |
+| Toy transformer | \(\approx 20~\%\) | max. \(24~{\rm V}\) | current limitation is desired |
+| Doorbell transformer | \(\approx 40~\%\) | max. \(12~{\rm V}\), often several taps | simple robust low-voltage supply |
+| Ignition transformer | \(\approx 100~\%\) | \(\leq 14~{\rm kV}\) | high voltage, limited current |
+| Welding transformer | \(\approx 100~\%\) | max. \(70~{\rm V}\) | large current, strong current limitation |
+| Voltage transformer | \(<1~\%\) | \(100~{\rm V}\) | operate with high load resistance, approximately no-load |
+| Current transformer | \(100~\%\) | \(0~{\rm V}\) ideal secondary voltage | operate with low burden, approximately short-circuit |
+<WRAP column 100%>
+<panel type="danger" title="Important safety note: current transformers">
+A current transformer secondary must not be opened while primary current flows.
+If the secondary circuit is open, the transformer tries to maintain the magnetic balance and can generate dangerous high voltages.
+</panel>
+</WRAP>
+===== Exercises =====
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Quick check: ideal transformer voltage and current ratio
+#@TaskText_HTML@#
+A transformer has \(N_1=1200\) turns and \(N_2=300\) turns.
+The primary RMS voltage is \(U_1=230~{\rm V}\).
+The secondary side supplies a load current \(I_2=2.0~{\rm A}\).
+  * Calculate the turns ratio \(n\).
+  * Calculate the ideal secondary voltage \(U_2\).
+  * Calculate the magnitude of the ideal primary current \(I_1\).
+  * State whether this is a step-up or step-down transformer.
+#@ResultBegin_HTML~Exercise1~@#
+\[
+\begin{align*}
+n=\frac{N_1}{N_2}
+=
+\frac{1200}{300}
+=
+.
+\end{align*}
+\]
+The secondary voltage is
+\[
+\begin{align*}
+U_2
+=
+\frac{U_1}{n}
+=
+\frac{230~{\rm V}}{4}
+=
+.5~{\rm V}.
+\end{align*}
+\]
+The primary current magnitude is
+\[
+\begin{align*}
+I_1
+=
+\frac{I_2}{n}
+=
+\frac{2.0~{\rm A}}{4}
+=
+.50~{\rm A}.
+\end{align*}
+\]
+Because \(U_2<U_1\), it is a step-down transformer.
+#@ResultEnd_HTML@#
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Quick check: mutual inductance from reluctance
+#@TaskText_HTML@#
+Two coils are wound on the same ideal magnetic core.
+The main magnetic reluctance is
+\[
+\begin{align*}
+R_{\rm mH}=2.0\cdot 10^6~\frac{1}{\rm H}.
+\end{align*}
+\]
+The number of turns is \(N_1=500\) and \(N_2=100\).
+  * Calculate \(L_{1{\rm H}}\).
+  * Calculate \(L_{2{\rm H}}\).
+  * Calculate \(M\).
+  * Check whether the units are correct.
+#@ResultBegin_HTML~Exercise2~@#
+\[
+\begin{align*}
+L_{1{\rm H}}
+=
+\frac{N_1^2}{R_{\rm mH}}
+=
+\frac{500^2}{2.0\cdot 10^6~1/{\rm H}}
+=
+.125~{\rm H}.
+\end{align*}
+\]
+\[
+\begin{align*}
+L_{2{\rm H}}
+=
+\frac{N_2^2}{R_{\rm mH}}
+=
+\frac{100^2}{2.0\cdot 10^6~1/{\rm H}}
+=
+.0050~{\rm H}
+=
+.0~{\rm mH}.
+\end{align*}
+\]
+\[
+\begin{align*}
+M
+=
+\frac{N_1N_2}{R_{\rm mH}}
+=
+\frac{500\cdot 100}{2.0\cdot 10^6~1/{\rm H}}
+=
+.025~{\rm H}
+=
+~{\rm mH}.
+\end{align*}
+\]
+The unit is correct because \(1/(1/{\rm H})={\rm H}\).
+#@ResultEnd_HTML@#
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Quick check: referring secondary quantities to the primary side
+#@TaskText_HTML@#
+A transformer has \(n=5\).
+The secondary winding resistance is \(R_2=0.20~\Omega\) and the secondary leakage reactance is \(X_{2\sigma}=0.35~\Omega\).
+Calculate the values \(R'_2\) and \(X'_{2\sigma}\) referred to the primary side.
+#@ResultBegin_HTML~Exercise3~@#
+\[
+\begin{align*}
+R'_2
+=
+n^2R_2
+=
+^2\cdot 0.20~\Omega
+=
+\cdot 0.20~\Omega
+=
+.0~\Omega.
+\end{align*}
+\]
+\[
+\begin{align*}
+X'_{2\sigma}
+=
+n^2X_{2\sigma}
+=
+^2\cdot 0.35~\Omega
+=
+\cdot 0.35~\Omega
+=
+.75~\Omega.
+\end{align*}
+\]
+The unit remains \(\Omega\), because \(n\) is dimensionless.
+#@ResultEnd_HTML@#
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Quick check: short-circuit voltage and fault current
+#@TaskText_HTML@#
+A transformer has a rated primary current \(I_{1{\rm N}}=10~{\rm A}\) and a short-circuit voltage \(u_{\rm k}=5~\%\).
+  * Calculate the continuous short-circuit current \(I_{1{\rm k}}\) when rated primary voltage is applied.
+  * Estimate the initial peak short-circuit current \(i_{\rm p}\) using \(i_{\rm p}\approx 2.54 I_{1{\rm k}}\).
+#@ResultBegin_HTML~Exercise4~@#
+\[
+\begin{align*}
+I_{1{\rm k}}
+=
+I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}}
+=
+~{\rm A}\cdot \frac{100~\%}{5~\%}
+=
+~{\rm A}.
+\end{align*}
+\]
+\[
+\begin{align*}
+i_{\rm p}
+\approx
+.54\cdot I_{1{\rm k}}
+=
+.54\cdot 200~{\rm A}
+=
+~{\rm A}.
+\end{align*}
+\]
+The short-circuit current is much larger than the rated current. Protection devices must be selected accordingly.
+#@ResultEnd_HTML@#
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Longer exercise: transformer equivalent circuit for an actuator supply
+#@TaskText_HTML@#
+A single-phase transformer supplies an actuator driver.
+Rated data and equivalent circuit data are:
+\[
+\begin{align*}
+U_{1{\rm N}}&=230~{\rm V},
+&
+U_{2{\rm N}}&=23~{\rm V},
+&
+I_{2{\rm N}}&=5.0~{\rm A},
+\\
+R_1&=1.2~\Omega,
+&
+X_{1\sigma}&=1.8~\Omega,
+\\
+R_2&=0.012~\Omega,
+&
+X_{2\sigma}&=0.018~\Omega.
+\end{align*}
+\]
+Assume \(n=\frac{U_{1{\rm N}}}{U_{2{\rm N}}}\). The magnetizing branch is neglected for the loaded operating point.
+  * Calculate \(n\).
+  * Refer \(R_2\) and \(X_{2\sigma}\) to the primary side.
+  * Calculate \(R_{\rm k}\) and \(X_{\rm k}\).
+  * Calculate the primary rated current magnitude \(I_{1{\rm N}}\) using the ideal current ratio.
+  * Estimate the magnitude of the internal voltage drop \(U_{\rm k}\approx |\underline{Z}_{\rm k}|I_{1{\rm N}}\).
+#@ResultBegin_HTML~Exercise5~@#
+The turns ratio is
+\[
+\begin{align*}
+n
+=
+\frac{U_{1{\rm N}}}{U_{2{\rm N}}}
+=
+\frac{230~{\rm V}}{23~{\rm V}}
+=
+.
+\end{align*}
+\]
+The secondary quantities referred to the primary side are
+\[
+\begin{align*}
+R'_2
+&=
+n^2R_2
+=
+^2\cdot 0.012~\Omega
+=
+.2~\Omega,
+\\
+X'_{2\sigma}
+&=
+n^2X_{2\sigma}
+=
+^2\cdot 0.018~\Omega
+=
+.8~\Omega.
+\end{align*}
+\]
+Therefore
+\[
+\begin{align*}
+R_{\rm k}
+&=
+R_1+R'_2
+=
+.2~\Omega+1.2~\Omega
+=
+.4~\Omega,
+\\
+X_{\rm k}
+&=
+X_{1\sigma}+X'_{2\sigma}
+=
+.8~\Omega+1.8~\Omega
+=
+.6~\Omega.
+\end{align*}
+\]
+The primary current magnitude is
+\[
+\begin{align*}
+I_{1{\rm N}}
+=
+\frac{I_{2{\rm N}}}{n}
+=
+\frac{5.0~{\rm A}}{10}
+=
+.50~{\rm A}.
+\end{align*}
+\]
+The magnitude of the short-circuit impedance is
+\[
+\begin{align*}
+|\underline{Z}_{\rm k}|
+=
+\sqrt{R_{\rm k}^2+X_{\rm k}^2}
+=
+\sqrt{(2.4~\Omega)^2+(3.6~\Omega)^2}
+=
+.33~\Omega.
+\end{align*}
+\]
+Thus the internal voltage drop estimate is
+\[
+\begin{align*}
+U_{\rm k}
+\approx
+|\underline{Z}_{\rm k}|I_{1{\rm N}}
+=
+.33~\Omega\cdot 0.50~{\rm A}
+=
+.17~{\rm V}.
+\end{align*}
+\]
+This is a primary-side voltage drop. On the secondary side it corresponds approximately to
+\[
+\begin{align*}
+\frac{2.17~{\rm V}}{10}=0.217~{\rm V}.
+\end{align*}
+\]
+For a \(23~{\rm V}\) actuator supply this is small but not zero.
+#@ResultEnd_HTML@#
+#@TaskEnd_HTML@#
+===== Common pitfalls =====
+  * **Using a transformer with DC:** A transformer needs changing flux. With DC, after the switching transient, an ideal transformer no longer transfers voltage. A real transformer may overheat because the winding resistance limits the current only weakly.
+  * **Forgetting the current ratio sign:** The minus sign in \(\frac{\underline{I}_1}{\underline{I}_2}=-\frac{1}{n}\) comes from reference arrows. Do not interpret it as negative power loss.
+  * **Mixing peak values and RMS values:** In AC power and transformer ratings, \(U\) and \(I\) usually mean RMS values. Time functions are written \(u(t)\), \(i(t)\). Instantaneous short-circuit peaks are written here as \(i_{\rm p}\).
+  * **Confusing reluctance and resistance:** Magnetic reluctance \(R_{\rm m}\) has the unit \(1/{\rm H}\), not \(\Omega\).
+  * **Confusing \(n\) and the technical no-load voltage ratio \(\ddot{u}\):** The ideal ratio is \(n=\frac{N_1}{N_2}\). The measured no-load voltage ratio is close to \(n\), but not exactly equal for a real transformer.
+  * **Forgetting the square when referring impedances:** Voltages transform with \(n\), currents with \(\frac{1}{n}\), but impedances transform with \(n^2\).
+  * **Ignoring leakage reactance:** Leakage reactance is often the dominant part of short-circuit impedance. It strongly affects fault current and voltage drop.
+  * **Treating \(u_{\rm k}\) as a voltage in volts:** \(u_{\rm k}\) is normally given in percent. Insert it consistently in formulas.
+  * **Using \(2.54\) as a universal law:** The first short-circuit peak depends on the \(R/X\) ratio and on the switching instant. The factor \(2.54\) is an approximation.
+  * **Opening a current transformer secondary:** This can create dangerous voltages. Current transformers are operated with a low burden, approximately as a short-circuit.
+  * **Assuming ideal isolation at every frequency:** Real transformers have parasitic capacitances between windings. For high-frequency noise and EMC, the “isolated” sides can still be capacitively coupled.
+===== Embedded resources =====
+<WRAP group>
+<WRAP column half>
+<panel type="info" title="PhET: Faraday's Law">
+Use this simulation to revisit the EEE1 idea that a changing magnetic flux induces voltage.
+This is the physical basis of the transformer equations in this block.
+{{url>https://phet.colorado.edu/sims/html/faradays-law/latest/faradays-law_en.html 700,500 noborder}}
+</panel>
+</WRAP>
+<WRAP column half>
+<panel type="info" title="Falstad / CircuitJS: transformer circuits">
+Use CircuitJS for qualitative experiments with coupled inductors and transformer circuits.
+Suggested activity: search the example circuits for “transformer”, change the turns ratio, and observe voltage and current.
+{{url>https://www.falstad.com/circuit/circuitjs.html 700,500 noborder}}
+</panel>
+</WRAP>
+</WRAP>
+<WRAP group>
+<WRAP column half>
+<panel type="info" title="Falstad / CircuitJS: coupled inductors">
+Suggested experiment:
+  * build two coupled inductors,
+  * change the coupling coefficient,
+  * observe how the secondary voltage changes.
+This is especially useful for understanding the difference between an iron-core transformer and loosely coupled wireless charging coils.
+</panel>
+</WRAP>
+<WRAP column half>
+<panel type="info" title="Further reading in this wiki">
+Relevant continuity pages:
+  * [[:electrical_engineering_1:block18|EEE1: Induction]]
+  * [[:electrical_engineering_1:block19|EEE1: Magnetic circuits]]
+  * [[:electrical_engineering_1:block20|EEE1: Inductance]]
+  * [[:electrical_engineering_2:magnetic_circuits#mutual_induction_and_coupling|Mutual Induction and Coupling]]
+</panel>
+</WRAP>
+</WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~