The do’s and don’ts of insulation power factor testing – part 1

Electrical Tester – 16 February 2018

Author: Jill Duplessis, Global technical marketing manager and Editor

Power factor (PF) (or dissipation factor (DF)) and capacitance tests are widely used to assess the condition of the insulation in transformers and other electrical assets. This short series of articles provides a handy guide to the why and how of PF/ DF testing. Apart from this introductory article, the material will be presented in the form of Do’s and Don’ts, which hopefully will make it easy to access and to understand.


In simple terms, PF/ DF and capacitance tests are health assessment tools for electrical insulation, but it’s only fair to point out that they are far from the only tools that can be used for this purpose. Other options include, for example, insulation resistance measurements, dissolved gas analysis, and dielectric frequency response (DFR) tests. Performed in combination with each other, these tests give a very accurate picture of insulation health and performance but, for reasons that may be technical or financial, such comprehensive testing may not always be possible. In such cases, PF/ DF tests on their own can still yield invaluable information.

PF/DF and capacitance is a ‘2 in 1’ test – the two test parameters, that is, (1) power factor  (or dissipation factor) and (2) capacitance, relate to each other and are conveniently measured at the same time but there is some degree of separation that must be applied both to understand and analyze them.

Capacitance testing

To provide a sound foundation for our exploration of PF/DF testing, we need first consider capacitance testing. One of the principal functions of electrical insulation is to store electrical energy when subjected to an electrostatic field, which is exactly what a capacitor does. It is possible, therefore, to model an asset as a capacitor. Taking a transformer as an example, the windings and the core correspond to the two plates of the capacitor and the insulation between the uniquely energised parts of the transformer corresponds to the capacitor’s dielectric.

With this in mind, one way to assess how well the insulation of an asset is behaving is to trend how the capacitance of the “asset as a capacitor” varies over time. Let’s look at some basic theory to see why this is useful.

For a parallel plate capacitor with two similar plates, the capacity is proportional to the area of the plates and inversely proportional to the distance between them. In other words, if the geometry of the capacitor changes, so will its capacitance. In our transformer-as-a-capacitor example, it’s very clear that we’re not talking about a parallel plate capacitor. The geometry is much more complex but, for our purposes, that doesn’t matter. The same rule applies – if the geometry of the transformer changes so will the measured capacitance.

The capacitance of a capacitor depends, however, not only on its geometry but also on the permittivity of the dielectric. If the dielectric is a vacuum, as it usually is taken to be in textbook examples, the permittivity will be Σ0, the permittivity of free space. We don’t usually need to know the absolute value of Σ0 when testing electric assets – but just in case you’re curious – it’s around 8.854 x 10-12 farads per meter.

What we do need to know is that the dielectric in the assets under test is rarely a vacuum. In a transformer, for example, it’s often paper that is impregnated by an insulating fluid. This will have a permittivity different from that of free space. Transformer oil impregnated paper typically has a permittivity between 3.2 and 3.5 times that of free space, and this is usually expressed by saying it has a relative permittivity, Σr, between 3.2 and 3.5.

Summing all of this up, we can say the capacitance of a capacitor depends on its geometry and the relative permittivity of its dielectric. If these factors don’t change, the capacitance will stay the same. So, when we trend capacitance measurements made on a transformer over a period of time, we would hope to see no significant change.

If there is a change, however, we now know that it may mean one of two things: either the transformer geometry has changed, possibly as the result of a heavy fault current leading to severe winding deformation, or the permittivity of the dielectric has notably changed. In every case, however, a significant capacitance change is almost certain to be indicative of a problem that’s in urgent need of further investigation.

Capacitors, insulation and losses

Although simply measuring the capacitance of a transformer or other asset can, as we have seen, provide useful information, assessing the characteristics of that capacitance often provides information that’s even more useful. Let’s explain. Capacitors store energy and, when subjected to an AC voltage, a perfect capacitor would charge up during, say, the positive half cycle of the supply and would then, during the negative half cycle, return to the supply system all of the energy it had stored while charging. With a perfect capacitor, no energy would be lost.

In the real world, of course, there are always losses. The main reasons for these are polarisation processes, which relate to the movement of atoms and molecules in the dielectric when it is subjected to a changing electric field, and leakage current through the dielectric. In both cases these losses mean that some of the energy stored by the capacitor is lost in the form of heat. If the “capacitor” in question is actually a transformer or other asset, the losses relate to the asset’s insulation and examining the losses can reveal a lot about the insulation condition.

The behaviour of the asset’s insulation (the dielectric in the “capacitor”) is most easily understood with the aid of a simple vector diagram.

When an AC voltage is applied to an insulation system, a current will flow. In the diagram, this is represented by the vector IT, total current. The total current is the vector sum of two components, IC, the capacitive current and IR, the resistive current. The resistive current represents the losses. For insulation in good condition, the capacitive current will be much greater than the resistive current. Of interest are the two angles marked δ and θ in the diagram, and particular, the tangent of the angle δ, which is the dissipation factor (DF) of the insulation (although it is often known simply as tan δ); and the cosine of the angle θ, which is the insulation power factor (PF).

It can be seen that as the resistive current becomes smaller, the angle δ approaches 0 so DF also approaches 0. With decreasing resistive current, the angle θ approaches 90º, which means that PF also approaches 0. In fact, for perfect insulation, DF = PF = 0. As the insulation ages, the resistive current (IR in the vector diagram) increases, so PF and DF will also increase. As an aside, it is worth noting that provided IR is small compared with IC, PF and DF will be almost equal numerically.

Why PF and DF?

Hopefully it’s clear that both DF and PF give information about the condition of the insulation in an asset, since they both increase when the losses in the insulation increase. If, however, DF and PF are measures of losses in the insulation, why not simply measure those losses directly?

The answer is that the results obtained by direct measurements of losses depend on the size of the asset. If the insulation system of asset A is twice as large as that of asset B, but is in exactly the same condition, the insulation in asset A will have twice the losses of the insulation in asset B. In other words, there is no easy way of comparing directly measured losses between systems of different sizes because, in reality, there is no way of accurately quantifying the size of the insulation systems within assets.

PF and DF, however, are relative measurements that are unaffected by the size of the insulation systems, which means that the PF and DF results for different assets can be meaningfully compared. But what exactly does relative mean? For PF, relative losses mean the amount of energy lost to heat relative to the total amount of energy present in the system – both stored energy and lost energy. In other words, referring back to the vector diagram, PF compares IR with IT. For DF, relative losses mean the amount of energy lost relative to the amount of energy stored by the insulation. In other words, DF compares IR with IC. PF and DF measurements are equally useful although some organisations may have a preference for one or the other.

In the next article of the series we will start to explore the practicalities of PF and DF testing, using a Do’s and Don’ts format, dealing first with safety issues and general insulation testing knowhow.