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 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.
  • 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

Polarity and the dot convention

Before we start with the transformer, we have to look at a common convention for the orientation of two coils to each other.

Ideal single-phase transformer

For an ideal transformer we assume:

• both windings are linked by the same magnetic flux \( \underline{\Phi} \),
• 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

\[ \begin{align*} \boxed{ \frac{\underline{U}_1^\phantom{O}}{\underline{U}_2^\phantom{O}} = \frac{N_1}{N_2} = n } \end{align*} \]

with

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

Linked fluxes and mutual inductance

For a single coil we already know that its flux linkage $\Psi = N\Phi$ is proportional to the current $i$ through the coil

\[ \begin{align*} \Psi=L i \end{align*} \]

\[ \begin{align*} \underline{\Psi}=L \underline{I} . \end{align*} \]

For two coupled coils $1$ and $2$, each flux linkage $\underline{\Psi}_1 = N\underline{\Phi}_1$ and $\underline{\Psi}_2 = N\underline{\Phi}_2$ depend on both currents $\underline{I}_1$ and $\underline{I}_2$.

Not only the current through the coil generates a part of the flux linkage, but also the other coil provides a part for the flux linkage.

\[ \begin{align*} \underline{\Psi}_1 &= \underbrace{{\color{green}{L_{11}\underline{I}_1}}}_{\text{self-linkage of coil 1}} + \underbrace{{\color{blue}{M_{12}\underline{I}_2}}}_{\text{mutual linkage from coil 2}}, \\[4pt] \underline{\Psi}_2 &= \underbrace{{\color{blue}{M_{21}\underline{I}_1}}}_{\text{mutual linkage from coil 1}} + \underbrace{{\color{green}{L_{22}\underline{I}_2}}}_{\text{self-linkage of coil 2}}. \end{align*} \]

For most transformer calculations we use the symmetric case of the mutual inductances.
(this is true for passive, stationary, and reciprocal situations, like transformers, but not necessarily for motors or complex setups)

\[ \begin{align*} {\color{blue}{M_{12}}}={\color{blue}{M_{21}}}={\color{blue}{M}}. \end{align*} \]

Often, the self-inductances are abbreviated: \[ \begin{align*} \color{green}{L_{11}} \rightarrow \color{green}{L_{1}} \\ \color{green}{L_{22}} \rightarrow \color{green}{L_{2}} \\ \end{align*} \]

Then

\[ \begin{align*} \boxed{ \begin{pmatrix} \underline{\Psi}_1\\ \underline{\Psi}_2 \end{pmatrix} = \begin{pmatrix} {\color{green}{L_{1}}} & {\color{blue}{M}}\\ {\color{blue}{M}} & {\color{green}{L_{2}}} \end{pmatrix} \begin{pmatrix} \underline{I}_1\\ \underline{I}_2 \end{pmatrix} } \end{align*} \]

and

\[ \begin{align*} M=k\sqrt{L_{1}L_{2}}. \end{align*} \]

Here \(k\) is the coupling coefficient. In the shown transformer \(k\) is 1 since all flux from $1$ flows through $2$ and vice versa.
In reality that is not the case as explained in the next chapters.

The mutual inductance \(M\) answers the question:

How much flux linkage appears in coil \(2\) when the current in coil \(1\) changes?

• A large \(M\) means strong interaction.
• A small \(M\) means weak interaction.

voltages by mutual inductances and resistances

For positive coupling, we get the following complex representation (since $u(t) = L\frac{{\rm d}i}{{\rm d}t} \: \longrightarrow \:\underline{U}=j\omega L \underline{I}$):

\[ \begin{align*} \underline{U}_1 &= R_1 \underline{I}_1 + {\color{green} {j\omega L_{1} \underline{I}_1 }} + {\color{blue}{j\omega M \underline{I}_2 }}, \\[4pt] \underline{U}_2 &= R_2 \underline{I}_2 + {\color{blue} {j\omega M \underline{I}_1 }} + {\color{green}{j\omega L_{2} \underline{I}_2 }}. \end{align*} \]
For negative coupling, the sign of the \(M\)-term changes in the chosen equation system, see figure 2 and figure 3.

Stray and Leakage

The image of the ideal transformer shall be consecutively developed to a more realistic transformer model.
To do so, we look at the situation of two coils near each other and expand this formula for the induced voltage.

The flux created by coil \(1\) can be split into

\[ \begin{align*} \Phi_{11} = {\color{blue}{\Phi_{21}}} + {\color{orange}{\Phi_{1 \rm \sigma}}}. \end{align*} \]

• \(\Phi_{11}\): total flux created by coil \(1\).
• \({\color{blue}{\Phi_{21}}}\): part of this flux that also links coil \(2\).
• \({\color{orange}{\Phi_{1 \rm \sigma}}}\): stray or leakage flux that does not link coil \(2\). The Greek letter sigma $\sigma$ is used to denote leakage or “stray” quantities.

For an example, we will have a look at the instantaneous voltage induced in coil \(2\):

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

The complex voltage induced in coil \(2\) is

\[ \begin{align*} \underline{U}_{2} = j\omega {\color{blue}{ \underline{\Psi}_{21} }} = j\omega N_2 {\color{blue}{ \underline{\Phi}_{21} }} . \end{align*} \]

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 'H' denotes the German word “Haupt” (sometimes also given as 'm' for “main”).
• 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 leakage flux can be interpreted as an additional virtual branch in the mechanical setup, see figure 5.

The total main flux linking is given by (see figure 5) \[ \begin{align*} \underline{\Phi}_{1 \rm H} = \underline{\Phi}_{21} + \underline{\Phi}_{12} = \underline{\Phi}_{\rm H} \\[4pt] \underline{\Phi}_{2 \rm H} = \underline{\Phi}_{21} + \underline{\Phi}_{12} = \underline{\Phi}_{\rm H} \end{align*} \]

For the flux linkage also the leakage flux has to be considered: \[ \begin{align*} \underline{\Psi}_{1} =N_1 ( \underline{\Phi}_{21} + \underline{\Phi}_{12} + \underline{\Phi}_{1\sigma}) \\[4pt] \underline{\Psi}_{2} =N_2 ( \underline{\Phi}_{21} + \underline{\Phi}_{12} + \underline{\Phi}_{2\sigma}) \end{align*} \]

The real flux linkage equations become

\[ \begin{align*} \underline{\Psi}_1 &= \underbrace{{\color{blue }{L_{1{\rm H}}\underline{I}_1 + M\underline{I}_2}} }_{\text{main magnetic path}} + \underbrace{{\color{orange}{L_{1\sigma }\underline{I}_1}} }_{\text{primary leakage} }, \\[4pt] \underline{\Psi}_2 &= \underbrace{{\color{blue }{M\underline{I}_1 + L_{2{\rm H}}\underline{I}_2 }} }_{\text{main magnetic path}} + \underbrace{{\color{orange}{L_{2\sigma }\underline{I}_2}} }_{\text{secondary leakage} }. \end{align*} \]

Equivalently,

\[ \begin{align*} \underline{\Psi}_1 &= L_1\underline{I}_1+M\underline{I}_2, & L_1&=L_{1{\rm H}}+L_{1\sigma}, \\ \underline{\Psi}_2 &= L_2\underline{I}_2+M\underline{I}_1, & L_2&=L_{2{\rm H}}+L_{2\sigma}. \end{align*} \]

The winding resistances \(R_1\) and \(R_2\) cause copper losses:

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

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

Based, on the block19 and block20 of last semester, the inductances can be calculated by the reluctance $R_{\rm mFe}$ of the iron core and the number of turns $N_1$, $N_2$:

\[ \begin{align*} \boxed{ L_{1 \rm H} = \frac{N_1^2}{R_{\rm mFe} } \\ L_{2 \rm H} = \frac{N_2^2}{R_{\rm mFe} } \\ M = \frac{N_1 N_2}{R_{\rm mFe} } \\ } \end{align*} \]

Reduced equivalent circuit referred to the primary side

Since we know, that we can transform the current and voltage by the transformer, we can use this also to simplify the circuit.

Fig. 6: equivalent circuit of a real transformer

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 \(n^2\) since $R = U / I$:

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

Fig. 7: reduced equivalent circuit of a real transformer

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.
• \(R_{\rm L}\) is a load resistor for the phasor diagram.

The phasor diagram is shown in figure 8.

• The upper one only shows the main voltages and currents.
  For an ohmic load current is parallel to $\underline{U}'_2$. However, $\underline{I}'_2$ is drawn as a current entering the transformer secondary port. Since the transformer delivers power to the load, $\underline{I}'_2$ is antiparallel to the load current and therefore antiparallel $\underline{U}'_2$.
• The lower one shows all voltages and currents.

Fig. 8: phasor diagram of a real transformer

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 \(\ddot{\rm u}\):

\[ \boxed{ \begin{align*} \ddot{\rm u} = \frac{\text{higher voltage}}{\text{lower voltage}} \bigg|_{\rm no~load}. \end{align*} } \]

For a step-down transformer:

\[ \begin{align*} \ddot{\rm u} = \frac{U_{1{\rm N}}}{U_{20}}. \end{align*} \]

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, so $\underline{U}_2=0$.
The primary voltage $\underline{U}_1$ is increased only until rated current flows.
Because the short-circuit impedance is small, this requires only a small fraction of the rated primary voltage.
Therefore the main flux and the magnetizing current are small, so the magnetizing branch can usually be neglected.

\[ \begin{align*} \Big|\; jX_{1{\rm H}} \; || \;R_{\rm Fe} \;\Big| \;\; \gg \;\; \Big|\; jX_{1\sigma} + \;R_1 + jX'_{2\sigma} + \;R'_2 \;\Big| . \end{align*} \]

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

\[ \begin{align*} \boxed{ I_{1{\rm k}} = \frac{U_{1{\rm N}}}{U_{1{\rm k}}}\cdot I_{1{\rm N}} = I_{1{\rm N}}\cdot \frac{100~\%}{{\rm u_k}} } \end{align*} \]

where \(\rm u_k\) is inserted as a percentage value.

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.

\[ \begin{align*} \boxed{ \underline{U}_1 - \underline{U}_{\rm k} = \underline{U}'_2 = n \underline{U}_2 } \end{align*} \]

If the magnetizing branch is neglected for the load calculation, then the series current relation is approximately $\underline{I}'_2 \approx -\underline{I}_1$. Therefore the ideal current transformation can still be used as an approximation.

\[ \begin{align*} \boxed{ \frac{\underline{I}_1}{\underline{I}_2} = -\frac{N_2}{N_1} = -\frac{1}{n} } \end{align*} \]

The Kapp triangle (see yellow triangle in figure 11) represents the formula for $\underline{U}_{\rm k}$.

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

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 \(\ddot{\rm u}\): 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 \(\rm u_k\) as a voltage in volts: \(\rm u_k\) 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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