Block 09/10 — Transformers and Magnetic Coupling

Learning objectives

Preparation at Home

Well, again

• read through the present chapter and write down anything you did not understand.
• Repeat the EEE1 ideas of magnetic flux and induction, magnetic circuits, and inductance and magnetic energy.
• Repeat from EEE2 the use of sinusoidal quantities, complex calculation, and complex power.
• For a deeper field-theory view, see also 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

Core content

From EEE1 induction to an AC transformer

Mutual induction: the key idea before the transformer

Before we look at the transformer, we need one important idea:

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

\[ \begin{align*} u_{{\rm ind},2}(t) = -N_2\frac{{\rm d}\Phi_{21}}{{\rm d}t}. \end{align*} \]

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 \(k\)>

Polarity and the dot convention

The sign of the mutual term depends on the winding direction and on the chosen current reference arrows.

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.

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

\[ \begin{align*} \Theta=R_{\rm m}\Phi \qquad \Longleftrightarrow \qquad \Phi=\frac{\Theta}{R_{\rm m}}. \end{align*} \]

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.

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

\[ \begin{align*} n=\frac{N_1}{N_2}. \end{align*} \]

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:

\[ \begin{align*} S_1 &= U_1I_1=230~{\rm V}\cdot 0.4~{\rm A}=92~{\rm VA}, \\ S_2 &= U_2I_2=23~{\rm V}\cdot 4~{\rm A}=92~{\rm VA}. \end{align*} \]

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.

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:

\[ \begin{align*} P_{\rm Cu,1}=R_1I_1^2, \qquad P_{\rm Cu,2}=R_2I_2^2. \end{align*} \]

\[ \begin{align*} \underline{U}_1 &= \underbrace{{\color{red}{R_1\underline{I}_1}}}_{\text{primary copper drop}} + \underbrace{{\color{orange}{j\omega L_{1\sigma}\underline{I}_1}}}_{\text{primary leakage drop}} + \underbrace{{\color{blue}{j\omega L_{1{\rm H}}\underline{I}_1+j\omega M\underline{I}_2}}}_{\text{main magnetic coupling}}, \\[6pt] \underline{U}_2 &= \underbrace{{\color{red}{R_2\underline{I}_2}}}_{\text{secondary copper drop}} + \underbrace{{\color{orange}{j\omega L_{2\sigma}\underline{I}_2}}}_{\text{secondary leakage drop}} + \underbrace{{\color{blue}{j\omega L_{2{\rm H}}\underline{I}_2+j\omega M\underline{I}_1}}}_{\text{main magnetic coupling}}. \end{align*} \]

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.

Why leakage flux matters in engineering

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 \(n^2\):

\[ \begin{align*} \boxed{ R'_2=n^2R_2 } \qquad \boxed{ X'_{2\sigma}=n^2X_{2\sigma} } \end{align*} \]

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.

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.

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.

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

\[ \begin{align*} \underline{Z}_{\rm k} = R_{\rm k}+jX_{\rm k}. \end{align*} \]

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.

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

\[ \begin{align*} i_{\rm p} \approx 1.8\cdot\sqrt{2}\cdot I_{1{\rm k}} = 2.55\,I_{1{\rm k}} \approx 2.54\,I_{1{\rm k}}. \end{align*} \]

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.

Construction types and cooling

Transformer behavior is influenced by construction.

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.

Typical technical transformer data

<tabcaption tab_transformer_types|Typical transformer types, short-circuit voltage, and secondary voltage>

Exercises

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.

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