PRACTICAL TRANSFORMER CHARACTERIZATION

overview

Shown below is an equivalent circuit representation of a transformer. Note that this may represent either a single phase transformer or one phase of a three phase transformer - all methods shown may be used on either type.

This equivalent circuit is a common example of a what is known as a two-port network. For the sake of simplicity, analysis of the above circuit will not utilize techniques employed in solving two-port networks.

Obviously, the circuit shown above involves many variables, of which few are known in practical cases. The goal of transformer characterization is to accurately capture the values of these variables through practical testing. Namely, the transformer equivalent circuit can be fully characterized through two simple tests:

Open Circuit Test
Short Circuit Test

It is preferable to perform both tests at a point in time when the transformer its rated operating temperature; although sometimes this isn't possible (adds some error calculated values).

open-circuit test

This test involves only one winding of the transformer. Since no secondary current exists during this test the open circuit test, the secondary portion of the equivalent circuit can be eliminated. The new equivalent circuit for this test is then:

The corresponding test circuit is:

To perform this test one must obtain a variable AC power source and have a watt-meter suitable to measure the transformer during open-circuit conditions. Rules of thumb for equipment ratings required for this test are:

Voltage Source Range: 0 - Vp
Watt-Meter:
- True RMS
- Voltage >= Vp
- Current >= 0.1*Ip

Now we may begin to analyze our open circuit results which allows the construction of the excitation branch of our circuit. Firstly, we must consider a few important facts which simplify our analysis. Since the open circuit/excitation current is typically less than 10% of the rated current, only a small voltage drop occurs across Rp and jXp. We may therefore neglect this portion of the circuit entirely. Furthermore, since the copper losses in the winding are proportional to the square of the current, no losses are assumed in the winding.

Since Rc represents the real core losses and knowing:

We can then say:

The reactive power is calculated in a similar manner:

This gives:

The apparent power, which is the vector sum of the real and reactive power, is easily calculated from test values as:

The reactive power may then be calculated based on the real and apparent power:

short-circuit test

Now that the magnetizing branch of the transformer equivalent circuit has been developed, the remaining series-connected components must be determined.

The test circuit/setup is nearly the same as that of the open circuit test with one important exception; the secondary winding is shorted. It is noteworthy to mention that this test should have the excited winding be the higher voltage winding as it reduces the current requirements for the test equipment.

We can effectively ignore the relatively high impedance magnetizing branch constructed from our prior results/test since its associated losses rise with the square of the voltage. Since this test requires only a small fraction of the nameplate voltage to reach the nameplate current we can safely neglect this portion of the circuit.

Perhaps already noticed - this test involves both windings of the transformer meaning both the primary and secondary side parameters are inseparable. They are calculated instead as lumped parameters which accurately represent the total effect(s) from both windings:

Since this test is governed by current we have:

Similar to the open circuit test, the apparent power is calculated via:

The same goes for the reactive power:

The total reactance is then:

An important and widely used value, the percent impedance, can be obtained through this test and is calculated via the general form:

From this, the bolted fault current at rated primary voltage is:

real-world example

Apparent Power

Power Factor

Reactive Power

We may then calculate our equivalent circuit parameters as follows:

Core-Loss Resistance

Magnetizing Impedance

Apparent Power

Power Factor

Reactive Power

We may then calculate our equivalent circuit parameters as follows:

Combined Winding Resistance

Combined Leakage Reactance

The percent impedance and fault current can then be calculated as:

Percent Impedance

Fault Current

A bolted fault test was performed to highlight the significance and importance of transformer impedance and prove the accuracy of fault currents derived from it:

Since our terminal voltage during the fault wasn't exactly rated voltage, we can estimate the fault current simply by scaling our experimental results based on the terminal voltage difference. This is possible since the fault current equation is linear in nature:

Thus we can clearly see that:

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