An autotransformer is a transformer having part of its winding included in both the input and output circuit. A simple autotransformer connection is shown in Fig. 1. Note that because there is a common portion of the winding (Section 1-2), there is no isolation between the input and output circuits. Also note that there is only one winding.

As shown in the diagram, the line current is 10A for a 2000VA output (10A x 200V). The load current is 20A at 100V or 2,000VA output.

The winding is tapped at 100V so that the autotransformer functions as a 200V-to-100V stepdown transformer. While this is an acceptable transformer connection, its application is rather limited because of the absence of isolation between input and output circuits.

Note that the load current flows in the opposite direction as the line current, which is typical in all transformers. The current in Section 1-2 is, therefore, the difference between these two currents, or 10A (20A load minus 10A line).

Calculating equivalent size

The equivalent size of the autotransformer in the diagram below can be found as follows.

Section 2-3 VA equals 100V times 10A, or 1,000VA. Section 1-2 VA equals the quantity of the load current minus the line current times 100V[(20A-10A) x 100V], or 1,000VA.

Therefore, the equivalent physical size is equal to Section 1-2 VA plus Section 2-3 VA, divided by 2[(1,000VA + 1,000VA)/2], or 1,000VA. Thus, we have a transformer whose equivalent size is 1,000VA but supplies a load of 2,000VA. Are we getting something for nothing here? Not really. The transformation noted here is, in reality, only half of the load kVA rather than all of it, as it would be with an isolated transformer.

The following equation can be used for calculating the equivalent size of any autotransformer.

Equivalent physical size = [([V.sub.H] - [V.sub.L])/[V.sub.H]] x [kVA.sub.load] (equation 1)

where [V.sub.H] = input voltage

[V.sub.L] = output voltage

Referring back to the diagram and inserting the known values into Equation 1, we have the following.

Equivalent physical size = [([V.sub.H] - [V.sub.L])/[V.sub.H]] x [kVA.sub.load]

= [(200V - 100V)/200V] x 2kVA

= 1kVA

Note that the larger the transformation ratio, the larger the transformer's equivalent physical size and the smaller the transformation ratio, the smaller the equivalent physical size. For example, let's suppose we have an autotransformer with a 200V-to-50V ratio feeding the same size load kVA (2kVA). The equivalent physical size is as follows.

Equivalent physical size = [([V.sub.H] - [V.sub.L])/[V.sub.H]] x [kVA.sub.load]

= [(200V - 50V)/200V] X 2kVA

= 1.5kVA

Now let's suppose we have an autotransformer having a smaller transformation ratio, 200V-to-190V, but feeding the same size load kVA. Its equivalent physical size is as follows.

Equivalent physical size = [([V.sub.H] - [V.sub.L])/[V.sub.H]] x [kVA.sub.load]

= [(200V - 190V)/200V]/200kVA

= 0.1kVA

As you can see, the larger the transformation ratio, the less economical an autotransformer becomes. As a result, autotransformers with transformation ratios over 2 are seldom used.

A small quiz

To help you understand autotransformer applications, let's take a small quiz. Please study the transformer connections shown in Figs. 2, 3, and 4. Fig. 2 is a wiring diagram of an isolated transformer, including the input and output voltage at each winding. Fig. 3 shows this same isolated transformer connected as a stepdown autotransformer, while Fig. 4 shows it connected as a bucking autotransformer.

Suppose you want to step a 132V line down to 120V. Which autotransformer connection should you use and what physical size transformer is required to handle a 10kVA load?

Answer. If you chose Fig. 4, you're wrong because this connection calls for 132V on a 120V winding. The voltage on the 12V winding would be 13.2V. Thus, you would end up with 118.8V (132V - 13.2V) on the load side because this is a bucking connection. As a result, the transformer would overheat because the input voltage is 10% high.

The correct answer is Fig. 3 because this connection uses both windings at their rated voltage, and the exact voltage ratio required is provided.

Now for the equivalent physical size. Maybe we should rephrase this question. We actually want to know the kVA rating needed for our isolated transformer so that, when connected as an autotransformer, it will handle the 10kVA load. Referring back to Equation 1 again and using the Fig. 3 connection, we have the following.

Equivalent physical size = [([V.sub.H] - [V.sub.L])/[V.sub.H]] x [kVA.sub.load]

= [(132V - 120V)/132V] x 10kVA

= .91kVA

Thus we need an isolated transformer rated at .91kVA and connected as shown in Fig. 3 to handle our 10kVA load. In practice, we would choose a standard rating that is close to but slightly higher than that calculated. In this instance, we would choose a 1kVA transformer.