The transformer

The transformer is a stationary electrical machine and serves to convert electrical energy through the magnetic circuit back into electrical energy, particularly changing the voltage level. Transformers are available in various designs and power ratings, from a few watts (Figure 2) up to the high megawatt range (Figure 1).

In public power supply, the energy generated by power plants is transformed to a higher voltage to enable long-distance transmission. Transmission losses decrease with higher voltage, and thus, for the same power, the current decreases quadratically (Equation 1 in three-phase alternating current).

\begin {equation} P_\mathrm {v} = 3\cdot R_\mathrm {L}\cdot I_\mathrm {L}^2\label {GlVerlustleistung} \end {equation}

The transmitted power in three-phase alternating current is:

\begin {equation} \underline {S}_\mathrm {N} = \sqrt {3}\cdot \underline {U}_\mathrm {N} \cdot \underline {I}_\mathrm {L}^\ast \end {equation}

The losses are therefore proportional to \(\frac {1}{\underline {U}_\mathrm {N}^2}\)

A transformer makes it possible to provide comprehensive energy supply, as power stations can also be located far away from consumers.

In power supply networks (see Figure 3), there are several voltage levels. The extra-high voltage level, which in Europe is predominantly \(380\,\)kV, serves as the transmission network for transporting electrical power over long distances, i.e., several hundred kilometers, even across national borders. Large power plants are directly connected at this level. The high-voltage level (\(110\,\)kV or \(220\,\)kV) is referred to as the distribution network and is used to connect cities and large industrial consumers. The medium-voltage level (\(5\)–\(35\,\)kV) provides the connection of local distribution transformers for supplying individual streets. While the higher voltage levels are usually implemented using overhead lines, the medium-voltage network is predominantly realized with underground cables. In rural areas, however, overhead lines are also used at this level.

The power grid is divided into several voltage levels that enable the transmission and distribution of energy.

1 Principle of the transformer

To discuss the principle of the transformer, we start with a completely lossless, ideal transformer. A sinusoidal (actually cosinusoidal, but this is the same shape and only corresponds to a phase shift of \(90^\circ \)) voltage \(\underline {U}_1\) is applied to the primary coil. The time curve of \(\underline {U}_1\) corresponds to \(u\) in Equation 3.

\begin {equation} u = \hat {U}\cdot \cos (\omega t)\label {GlSpannung1} \end {equation}

According to the inverse of the law of induction (equation 4), this voltage causes a changing magnetic flux. \begin {equation} u = -N\cdot \frac {\mathrm {d}\varPhi }{\mathrm {d}t}\label {GlInduktionsgesetz} \end {equation}

The shape of the magnetic flux is also sinusoidal for a sinusoidal voltage waveform, since the integral of a cosine is a sine. The effective value of the magnetic flux can therefore be represented by equation 5 by solving the induction law for \(\varPhi \) and forming the effective value. This eliminates the sine and thus the time dependence. \begin {equation} \varPhi = \frac {\sqrt {2}\cdot \underline {U}_1}{2\pi \cdot N_1 \cdot f}\label {GlMagnetischerFluss} \end {equation}

The direction of the magnetic field always moves according to the right-hand rule (see Module 4), in this example counterclockwise. The magnetic flux generated in this way induces a voltage \(\underline {U}_2\) in the secondary coil of the transformer, which is optimally penetrated by the same flux \(\varPhi \). This voltage can also be calculated using the law of induction (equation 4). \begin {equation} u_2 = -N_2\cdot \frac {\mathrm {d}\varPhi }{\mathrm {d}t}\tag {\ref {GlInduktionsgesetz}} \end {equation}

The applied primary voltage \(\underline {U}_1\) and the secondary voltage \(\underline {U}_2\) can be put into a ratio (Equation 7). The rate of change of the magnetic flux \(\varPhi \) is thereby eliminated, leaving only the number of turns on the primary side \(N_1\) and the secondary side \(N_2\). This ratio is called the transformation ratio ü (equation 7). If the transformation ratio ü is greater than one, the voltage is reduced (step-down transformer). If the transformation ratio ü is less than one, the voltage is increased (step-up transformer). Since the transformer can be used in both directions, a transformation ratio greater than or equal to one is always specified in the technical data. \begin {equation} \frac {\underline {U}_1}{\underline {U}_2} = \frac {-N_1\cdot \frac {\mathrm {d}\varPhi }{\mathrm {d}t}}{-N_2\cdot \frac {\mathrm {d}\varPhi }{\mathrm {d}t}} = \frac {N_1}{N_2} = \textit {ü} \label {Gluebertragung} \\ \end {equation}

\begin {align} \textit {ü} < 1 &\rightarrow \text {Voltage goes up}\nonumber \\ \textit {ü} > 1 &\rightarrow \text {Voltage goes down}\nonumber \end {align}

In a real transformer, not all of the magnetic flux generated by the primary winding penetrates the secondary side (shown as a red dotted line in Figure 4). These losses are called stray flux and are represented in the equivalent circuit diagram by the stray inductance . Therefore, the real secondary voltage is smaller than the ideal secondary voltage calculated here.

2 Transformer structure

In simple terms, a transformer consists of two coils that are coupled together by means of a magnetically conductive core (see Figure 5). The core is made of a ferromagnetic material. Due to its high permeability, the ferromagnetic material has the ability to conduct the magnetic flux between two coils as efficiently as possible (see Module 6). However, effects such as hysteresis and eddy current losses interfere with the resistance-free flow. These effects are referred to as iron losses.

The coils are wound around the magnetic conductor. As mentioned above, the number of turns determines the voltage transformation ratio (equation 7). They consist of electrically conductive wires such as copper. In reality, these also exhibit resistance losses, which are also referred to as copper losses.

In the circuit diagram of the ideal transformer, the magnetically coupled coils are represented as adjacent black rectangles (see Figure 6). The two lines between the coils indicate that they are connected by a continuous magnetically conductive core. The dots on the respective coils provide information about the winding direction. They can be located at the bottom or at the top. If they are at the same height, the phase position of the two coils is the same. If they are at different heights, this means that the winding direction is reversed, resulting in a phase shift of \(180^\circ \).

Ideal three-phase mains transformer An ideal three-phase power transformer has an apparent power of \(\underline {S}_\mathrm {N} = 100\,\rm {kVA}\), an upper-side voltage of \(\underline {U}_1 = 20\,\rm {kV}\) (delta connection) and a lower-side voltage of \(\underline {U}_2 = 400\,\rm {V}\) (delta connection).

a)

What is the gear ratio?\begin {align} \textit {ü} &= \frac {\underline {U}_1}{\underline {U}_2} = \frac {N_1}{N_2} \nonumber \\ &= \frac {20\,\rm {kV}}{400\,\rm {V}} = 50\nonumber \end {align}

b)

What are the primary and secondary currents in nominal operation at a power factor of \(\cos (\varphi )=1\)?\begin {align} \cos (\varphi )=1 &\rightarrow |\underline {S}| = P\nonumber \\ P &= U \cdot I\nonumber \\ I_1 &= \frac {P}{U} = \frac {100\,\rm {kW}}{20\,\rm {kV}} = 5\,\rm {A}\nonumber \\ I_2 &= \frac {P}{U} = \frac {100\,\rm {kW}}{400\,\rm {V}} = 250\,\rm {A}\nonumber \end {align}

3 Euivalent circuit diagram

For better understanding, the equivalent circuit of the real transformer (Figure 7) is first represented functionally as a galvanically isolated system with a primary and a secondary circuit. Compared to the ideal transformer, this equivalent circuit is extended by the losses described previously. The magnetic circuit is transformed into an electrical circuit using the main inductance \(L_h\) and the leakage inductances \(L_{\sigma 1}\) and \(L_{\sigma 2}\). The copper losses in the primary and secondary windings described earlier are represented by the resistances \(R_1\) and \(R_2\). The quantity \(\underline {I}\mu \) denotes the magnetizing current and describes the component of the no-load current required to excite the magnetic core of the transformer. For simplicity, iron losses are neglected in this example. If they are to be taken into account, the iron losses are represented by a resistor connected in parallel with the main inductance.

Subsequently, in the T-equivalent circuit of the transformer (Figure 8), it is assumed that there is no galvanic isolation between the primary and secondary sides, although in reality this is of course the case. This step allows the quantities on the secondary side to be referred to the primary side and the calculation rules known from Module 4 to be applied. The referred quantities are denoted as primed quantities, in this example as \(\underline {U}^\prime _2\), \(\underline {I}^\prime _2\), \({R}^\prime _2\), and \({L}^\prime _{\sigma 2}\).

To convert the voltage to the primary side, the voltage \(\underline {U}^\prime _2\) is multiplied by the transformation ratio. To calculate the incoming current \(\underline {I}_1\), the current \(\underline {I}_2\) is multiplied by the inverse transformation ratio and then added to the magnetising current \(\underline {I}_\mu \) (Equation ??). \begin {align} \underline {U}_1 = \underline {U}^\prime _2 &= \frac {N_1}{N_2}\cdot \underline {U}_2 \\ \underline {I}_1 = \underline {I}^\prime _2 + \underline {I}_\mu &= \frac {N_2}{N_1}\cdot \underline {I}_2 + \underline {I}_\mu \label {Glstromumrechnung} \end {align}

The losses occurring on the secondary side \(L_{\sigma 2}^\prime \) and \(R_2^\prime \) are converted to the primary side using the squared transmission ratio (Equations ?? and ??). \begin {align} L_{\sigma 2}^\prime &= \left (\frac {N_1}{N_2}\right )^2\cdot L_{\sigma 2} \label {Glstreuinduktivitätumrechnung} \\ R_2^\prime &= \left (\frac {N_1}{N_2}\right )^2\cdot R_2\label {Glwiderstandumrechnung} \end {align}

Real three-phase mains transformer A real three-phase mains transformer (\(f=\,50\,\rm {Hz}\)) has the following data:

• Power: \(\underline {S}_\mathrm {N} = 100\,\rm {kVA}\)
• Overvoltage \(\underline {U}_1 = 20\,\rm {kV}\)
• Transmission ratio: \(50\)

• \(L_\mathrm {h} = 500\,\rm {H}\) \(\Rightarrow \underline {Z}_\mathrm {Lh} = j 157,08\,\rm {k}\Omega \)
• \(L_{\sigma 1} = L_{\sigma 2}^\prime = 190\,\rm {mH}\) \(\Rightarrow \underline {Z}_{\rm {L}\sigma } = j 59,69\,\Omega \)
• \(R_1 = R_2^\prime = 30\,\Omega \)

a)

What is the open-circuit voltage \(\underline {U}_2\) for this transmission ratio? \begin {align} \underline {U}_2^\prime &= \underline {U}_1\cdot \frac {\underline {Z}_\mathrm {h}}{\underline {Z}_\mathrm {h} + \underline {Z}_{\sigma 1} + R_1}\nonumber \\ &= 19,992\,\text {kV}\cdot e^{j0,011^\circ }\nonumber \\ \underline {U}_2 &= 399,85\,\text {V}\cdot e^{j0,011^\circ }\nonumber \end {align}

b)

How high is the no-load current (on the primary side)? \begin {align} \underline {I}_1 &= \frac {\underline {U}_1}{\underline {Z}_\mathrm {h} + \underline {Z}_{\sigma 1} + R_1}\nonumber \\ &= \frac {20\,\text {kV}}{157,139\,\text {k}\Omega \cdot e^{j 89,989^\circ }}\nonumber \\ &= 127,28\,\text {mA}\cdot e^{-j 89,989^\circ }\nonumber \end {align}

c)

How significant are the idle losses? \begin {align} \underline {S}_\mathrm {v} &= \underline {U}\cdot \underline {I}^\ast \nonumber \\ &= 20\,\text {kV} \cdot (127,28\,\text {mA}\cdot e^{-j 89,989^\circ })^\ast \nonumber \\ &= 2,546\,\text {kVA}\cdot e^{j 89,989^\circ }\nonumber \end {align}