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-====== Block 09/10 — Magnetically Coupled Coils and Transformers ======
-===== 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.
-  * 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 basic voltage drop under load.
-  * 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 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.
-    * Magnetic coupling: reluctance model, main inductance, mutual inductance.
-    * Real transformer: winding resistances, leakage inductances, iron-loss resistance.
-    * Reduced equivalent circuit: refer secondary quantities to 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, 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.
-  * 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>
-==== 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~~
-==== Magnetically coupled coils and mutual inductance ====
-The iron core is often described as a **magnetic circuit**. The magnetic flux is caused by the magnetic voltage, also called magnetomotive force,
-\[
-\begin{align*}
-\Theta = N_1\underline{I}_1+N_2\underline{I}_2.
-\end{align*}
-\]
-With magnetic reluctance \(R_{\rm mFe}\) of the iron path and permeance \(\Lambda_{\rm H}=\frac{1}{R_{\rm mFe}}\), the main flux is
-\[
-\begin{align*}
-\underline{\Phi}
-=
-\frac{\Theta}{R_{\rm mFe}}
-=
-\Lambda_{\rm H}\Theta
-=
-\Lambda_{\rm H}
-\left(
-N_1\underline{I}_1+N_2\underline{I}_2
-\right).
-\end{align*}
-\]
-The flux linkages are then
-\[
-\begin{align*}
-\underline{\Psi}_1
-&=
-N_1\underline{\Phi}
-=
-L_{1{\rm H}}\underline{I}_1
-+
-M\underline{I}_2,
-\\
-\underline{\Psi}_2
-&=
-N_2\underline{\Phi}
-=
-M\underline{I}_1
-+
-L_{2{\rm H}}\underline{I}_2.
-\end{align*}
-\]
-The inductances are
-\[
-\begin{align*}
-L_{1{\rm H}}&=N_1^2\Lambda_{\rm H},
-&
-L_{2{\rm H}}&=N_2^2\Lambda_{\rm H},
-&
-M&=N_1N_2\Lambda_{\rm H}.
-\end{align*}
-\]
-<callout>
-**Meaning of the symbols**
-  * \(L_{1{\rm H}}\), \(L_{2{\rm H}}\): main self-inductances caused by the main flux path.
-  * \(M\): mutual inductance. It describes how current in one winding contributes to flux linkage in the other winding.
-  * \(\Lambda_{\rm H}\): permeance of the main magnetic path. A high-permeability iron core gives large \(\Lambda_{\rm H}\).
-</callout>
-<panel type="info" title="Engineering picture">
-A current in the primary winding creates a magnetic flux in the core.
-The secondary winding is placed around the same flux path. Therefore, the secondary voltage is not caused by an electrical wire connection, but by **magnetic coupling**. This is why transformers provide galvanic isolation.
-</panel>
-==== 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="Real transformer equivalent with magnetic coupling">
-Using phasors, the voltage equations are
-\[
-\begin{align*}
-\underline{U}_1
-&=
-R_1\underline{I}_1+j\omega\underline{\Psi}_1
-\\
-&=
-R_1\underline{I}_1
-+j\omega L_{1\sigma}\underline{I}_1
-+j\omega L_{1{\rm H}}\underline{I}_1
-+j\omega M\underline{I}_2,
-\\[4pt]
-\underline{U}_2
-&=
-R_2\underline{I}_2+j\omega\underline{\Psi}_2
-\\
-&=
-R_2\underline{I}_2
-+j\omega L_{2\sigma}\underline{I}_2
-+j\omega L_{2{\rm H}}\underline{I}_2
-+j\omega M\underline{I}_1.
-\end{align*}
-\]
-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.
-</panel>
-<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>
-~~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. In the lecture slides 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.
-The initial peak short-circuit current from the lecture slides is
-\[
-\begin{align*}
-\boxed{
-I_{\rm sk}=2.54\cdot I_{1{\rm k}}
-}
-\end{align*}
-\]
-This is a maximum current value shortly after the fault occurs.
-<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>
-~~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: 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~Exercise2~@#
-\[
-\begin{align*}
-R'_2
-=
-n^2R_2
-=
-^2\cdot 0.20~\Omega
-=
-\cdot 0.20~\Omega
-=
-.0~\Omega.
-\end{align*}
-\]
-\[
-\begin{align*}
-X'_{2\sigma}
-=
-n^2X_{2\sigma}
-=
-^2\cdot 0.35~\Omega
-=
-\cdot 0.35~\Omega
-=
-.75~\Omega.
-\end{align*}
-\]
-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.
-  * Calculate the initial peak short-circuit current \(I_{\rm sk}\) using \(I_{\rm sk}=2.54 I_{1{\rm k}}\).
-#@ResultBegin_HTML~Exercise3~@#
-\[
-\begin{align*}
-I_{1{\rm k}}
-=
-I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}}
-=
-~{\rm A}\cdot \frac{100~\%}{5~\%}
-=
-~{\rm A}.
-\end{align*}
-\]
-\[
-\begin{align*}
-I_{\rm sk}
-=
-.54\cdot I_{1{\rm k}}
-=
-.54\cdot 200~{\rm A}
-=
-~{\rm A}.
-\end{align*}
-\]
-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~Exercise4~@#
-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 \(u(t)\), \(i(t)\).
-  * **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.
-  * **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>
-~~PAGEBREAK~~ ~~CLEARFIX~~