Home / Transformer / Autotransformer: Working, Advantages, Disadvantages

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## Autotransformer Definition

The autotransformer has a single winding on an iron core. One of the coil terminals is common to both input and output, and the other output terminal is movable so that it can make contact with any turn on the winding.

## Step-Up/ Step-Down Autotransformer

An autotransformer can be used as a step-up or a step-down transformer. As a step-up, it is often referred to as a boost, and as a step-down, it is called a buck connection. Figure 1 shows the schematic representation of buck connection while figure 2 shows the boost connection.

## Autotransformer Working Principle

The autotransformer serves a function similar to that of the ordinary transformer to raise or lower voltage. It consists of a single continuous winding with a tap brought out at some intermediate point as shown in Fig.1. Because the primary and secondary windings of the autotransformer are physically connected, the supply and output voltage are not insulated from each other.

Fig.1: Autotransformer Diagram

When a voltage V1 is applied to the primary of the autotransformer, the induced voltages are related by

$\begin{matrix} \frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{E}_{ac}}}{{{E}_{bc}}}=\frac{{{N}_{1}}}{{{N}_{2}}}=a & {} & \left( 1 \right) \\\end{matrix}$

Neglecting voltage drops in the windings

$\begin{matrix} \frac{{{V}_{1}}}{{{V}_{2}}}=a & {} & \left( 2 \right) \\\end{matrix}$

When a load is connected to the secondary of the autotransformer, a current I2 flows in the direction shown in Fig.1. By Kirchhoff’s current law,

$\begin{matrix} {{I}_{2}}={{I}_{1}}+\text{ }{{I}_{3}} & {} & \left( 3 \right) \\\end{matrix}$

As in the ordinary transformer, the primary and secondary ampere-turns balance each other, except for the small current required for core magnetization:

$\begin{matrix} {{N}_{1}}{{I}_{1}}={{N}_{2}}{{I}_{2}} & {} & (4) \\\end{matrix}$

Equation 4 may also be written as

$\begin{matrix} \frac{{{I}_{2}}}{{{I}_{2}}}~=\frac{{{N}_{1}}}{{{N}_{2}}}~=a & {} & \left( 5 \right) \\\end{matrix}$

Substituting Equation (5) into Equation (3), the ratio of the winding current is found as

$\begin{matrix} \frac{{{I}_{3}}}{{{I}_{1}}}~=a-1 & {} & \left( 6 \right) \\\end{matrix}$

In an autotransformer, the total power transmitted from the primary to the secondary does not actually pass through the whole winding. This means that a greater amount of power can be transferred without exceeding the current rating of the windings of the transformer.

$\begin{matrix} {{S}_{1}}=\text{ }{{V}_{1}}{{I}_{1}} & {} & \left( 7 \right) \\\end{matrix}$

Similarly, the output apparent power is given by

$\begin{matrix} {{S}_{2}}=\text{ }{{V}_{2}}{{I}_{2}} & {} & \left( 8 \right) \\\end{matrix}$

However, the apparent power in the transformer windings is

$\begin{matrix} {{S}_{w}}=\text{ }{{V}_{2}}{{I}_{3}}=({{V}_{1}}-{{V}_{2}}){{I}_{2}}=\text{ }{{S}_{ind}} & {} & \left( 9 \right) \\\end{matrix}$

This power is the component of the power transferred by transformer action or by electromagnetic induction.

The difference (S2 –Sw) between the output apparent power and the apparent power in the windings is the component of the output transferred by electrical conduction. This is equal to

$\begin{matrix} {{S}_{cond}}=\text{ }{{V}_{2}}{{I}_{2}}-\text{ }{{V}_{2}}{{I}_{3}}=\text{ }{{V}_{2}}{{I}_{1~~~~~~~~~~~~~~~~~~~}} & {} & \left( 10 \right) \\\end{matrix}$

## Advantages of an Autotransformer

Note that if the turns ratio of the transformer is large, the power rating as an autotransformer will be much larger than the rating as a conventional transformer. In a conventional transformer, all of the power is transformed, whereas in an autotransformer, most of the power is conducted at an elevated potential. Asa result, an autotransformer is much smaller than a conventional transformer of the same rating.

Other advantages of an autotransformer over a two winding transformer are:

1. Cheaper
2. More efficient, because losses stay the same while the rating goes up compared to a conventional transformer
3. Lower exciting current
4. Better voltage regulation

## Disadvantages of an Autotransformer

Some of the disadvantages of an autotransformer are:

1. Larger short-circuit current
2. No isolation exists between the primary and secondary windings
3. Only useful for moderately smaller voltage changes

Applications of Autotransformer

Practical applications of autotransformers include:

1. They are generally used to connect transmission lines of slightly different voltages ( e.g., 115 kV and 138 kV or 138 kV and 161 kV)
2. They are employed to compensate for voltage drops on long feeder circuits where it is important that each load device receives the same voltage ( e.g., on airfield lighting circuits to ensure uniform lamp intensity)
3. They offer variable voltage control in the laboratory setup: as we move the sliding contact, virtually all of the coil can become the series coil. Therefore, the entire coil must be sized for maximum current.
4. They are used to adjust the transformer output voltage in order to keep the system voltage constant with varying load.

## Autotransformer Example

A single-phase, 10-kVA, 440/110-V, two- winding transformer is connected as an autotransformer to supply a load at 550 V from a 440 V supply as shown below. Calculate the following.

1. kVA rating as an autotransformer
2. apparent power transferred by conduction
3. apparent power transferred by electromagnetic induction

Solution

The single –phase, a two-winding transformer is reconnected as an autotransformer as shown in Fig.2. The current ratings of the windings are given by

Fig.2: Autotransformer Example

$\begin{array}{*{35}{l}} {{I}_{ab}}=10,000/110\text{ }=\text{ }90.9\text{ }A \\ {{I}_{ab}}=10,000/440\text{ }=\text{ }22.7\text{ }A \\\end{array}$

At full or rated load, the primary and secondary terminal currents are

$\begin{array}{*{35}{l}} {{I}_{2}}=\text{ }90.9\text{ }A \\ {{I}_{1~}}={{I}_{2}}+{{I}_{3}}~=90.9+22.7~=~113.6\text{ }A \\\end{array}$

Therefore, the kVA rating of the autotransformer is

$\begin{array}{*{35}{l}} kV{{A}_{1}}~=\left( 440 \right)\left( 113.6 \right)/1000~=50\text{ }kVA \\ kV{{A}_{2}}~=\left( 550 \right)\left( 90.9 \right)/1000~=50\text{ }kVA \\\end{array}$

Note that this transformer, whose rating as an ordinary two-winding transformer is only 10 kVA, is capable of handling 50 kVA as an autotransformer. However, not all of the 50 kVA is transformed by electromagnetic induction. A large part is merely transferred electrically by conduction.

The apparent power transformed by induction is

${{S}_{ind}}={{V}_{1}}{{I}_{3}}=\left( 440 \right)\left( 22.7 \right)VA=10kVA$

The apparent power transformed by conduction is

${{S}_{cond}}={{V}_{1}}{{I}_{2}}=\left( 440 \right)\left( 90.9 \right)VA=40\text{ }kVA$

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