Power factor

What is Power Factor

This is quite a "deep" subject when you get into all the detail, this article will attempt to cover some of the basics. Power factor (PF) is a concept that only applies to electrical loads being powered from an AC supply. It quantifies what proportion of the apparent power flowing into a load, is actually dissipated as real power.

With a DC supply, the power dissipated by a load is proportional to its resistance and the voltage applied to it. We have some simple relationships that we can apply:

Ohms law tells us that:

Voltage = Current x Resistance
or more commonly: V = IR
 
we also know 

Power = Current x Voltage
or W = IV

So combining these we can also get:

W = I x IR or 
W = I² R

So if we know something about our load - say that it is rated at 100W at 230V, we can deduce from it that it must draw 100/230 or about 0.4A, and its internal resistance must equal 230/0.4 or about 575 ohms.

Reactive power

In the simplest cases, the same calculations also apply to loads being powered from an AC source (like the mains). A resistive load being driven from the mains, will draw current that is in sympathy with the mains voltage; as it rises, the current will rise, and as it falls it will fall. At the zero crossing point it will be zero.

However things can get more complicated due to the effect of "reactive" elements in the load. These are typically components that have an inductance or capacitance. Inductors and capacitors actually store electrical energy (although in different ways). As you might imagine, having something that stores energy being fed from an AC supply, will cause it to charge and discharge in response to the ever changing applied voltage. So during one part of the mains cycle it can be "absorbing" energy, while in another it gives the stored energy back again. Reactive elements in a load can make the types of calculations that are easy to apply to simple resistive loads somewhat more difficult, because you can no longer assume that the current drawn will align with the alternating voltage.

By way of example, consider an extreme example: a load consisting of nothing but a capacitor being driven from an AC supply. Capacitors present an infinite resistance to a steady state applied voltage, and very low resistance to fast changing applied voltages. So with a sinusoidal input voltage, this will mean that the peak current will be drawn when the voltage is at the zero point in the cycle, and the zero current will be at the peak of the voltage cycle (i.e where the rate of change of voltage is zero). The effect is that the voltage and current waveforms still look the same, however they are no longer aligned (there is said to be a "phase shift" between them). Real measurable current is thus flowing into and out of the capacitor, but because it is being stored and returned rather than dissipated as heat, there is no actual power being dissipated in the load.

a capacitor is an energy store that consists of two "plates" separated by a gap. When you initially connect a capacitor to a voltage, electrons (i.e. a current) will flow into it - in effect charging it up. This energy is actually stored in the electrical field that is created between the two plates. As the capacitor charges, the rate of flow of electrons decreases, until the capacitor is fully charged and no further current flows (after all, the electrons can not actually flow over the insulting gap between the capacitor's plates to complete a circuit). If one were to reverse the polarity of the applied voltage however, the capacitor could discharge from its present state, and recharge with the opposing polarity.

Say we had a 5.5uF capacitor for a load, it reactance (i.e. AC resistance) is calculated using the formula:

            1
Xc = ----------------    
      2 x pi x f x c
 
With a frequency of 50Hz we would get: 1 / (2 x pi x 50 x 5.5x10^-6) = 578 ohms

So, from ohms law we can compute the current that will flow: 230/578 = 0.4A, but we also know that when the 
mains voltage is at its peak, the current is zero due to the phase shift, and when the voltage is zero, the
current is 0.4A. If we were to work out the overall sum of Voltage x Current for a complete cycle of the mains, 
we get a total of zero. 

Hence we have the odd situation where the VA of the load (i.e the magnitude of the product of the current and 
voltage but ignoring the phase relationship) looks like 100, but the real power dissipated is actually 0W 

How is a power factor expressed?

A power factor is expressed as a number between 0 and 1. A power factor of 1 (aka a "unity power factor") basically says the power in a load can be treated like a normal resistive load, and ohms law applies. A PF of 0 is a perfect reactive load, or one that actually dissipates no real power at all.

The power factor of our load can be expressed as:

                 Real Power
Power Factor = --------------
                    VA

So in our example above example the power factor would actually be zero. 

A load like this that is totally "reactive" would be unusual. In real world situations, loads can have both resistive and reactive components (and of those reactive components, the capacitive ones will have leading, and the inductive ones lagging phase shifts).

To compute the actual current flow into a load like this at any given time therefore requires "vector" arithmetic to take into account not only the reactance of the leading and lagging components, but also their phase shifts.

Other causes of low power factor

The above examples show the classical cause of non unity power factors. A typical real word example of a load that has a less than unity power factor as a result of these phase shift effects is an induction motor. Here a significant proportion of the current flow into the motor is actually reactive and does not get dissipated as work. So a power factor of 0.5 would not be uncommon. Standard linear strip lights are another common load with a poor PF.

There are other causes of low power factor. One such example is where current only flows in the load during some but not all of the mains cycle. giving rise to a non sinusoidal current waveform. Switched Mode Power Supplies (SMPS) typically rectify the mains and use this rectified AC to charge a capacitor. This has the effect of only drawing current near the peaks of the mains voltage, and none at all during the remainder. Although there is no phase shift between the voltage and current waveforms, the current waveform is distorted and not the same shape as the voltage waveform, hence the actual power dissipated by the load is not a simple product of Voltage and Current (since when dealing with AC, this calculation has the implicit assumption that the waveforms are both the same shape). Switched mode supplies are increasingly found in electronic appliances such as computer and electronic equipment, and also in energy saving light bulbs. The PF of a SMPS for a computer will often be in the region of 0.7. That found in an energy saving CFL may be as low as 0.1

Does a low power factor mean I am using more electricity?

Sort of. In a domestic situation a poor power factor will not result in you being charged for more electricity, although the load with a lower power factor will draw more current. Poor power factors are bad for distribution efficiency though, and can result in the mains supply waveform getting misshapen and noisy - so power supply companies tend to penalise big industrial users if they don't control their power factors.

If you want an analogy, imagine riding a bike up hill. You stick a certain about of push into the pedals to keep it moving overcoming resistance, and more to add the energy you are acquiring by climbing the hill. Imagine someone attaching a big spring to one pedal and the seat post, such that every time you push the right pedal down you also need to stretch the spring. As you can imagine this will take more "push" from you to keep riding. However that extra push is only required on the right pedal. When you push the left pedal you have the energy stored in the spring pulling up on the right pedal and hence working for you. So the result is the bike is harder to ride, but the total energy required to get up the hill is actually the same. This is similar to the effect of having a poor power factor as a result of large reactive elements in the load - the load still dissipates the same amount of energy, but it is harder to drive (i.e. needs more peak current flow).

Can you change or compensate for a poor power factor?

Yes, you can do what it called power factor correction (PFC). In the case of poor PFs caused by current phase shifts, you can add other reactive components to the load to try and offset the effects of the reactive components in the load. So if you had a reactive bank of fluorescent strip lights (which have an lagging power factor due to their inductive ballasts), adding a capacitor to create a leading reactive element can actually cancel out the effects of the inductors.

While this is worthwhile in an industrial setting where customers are usually charged based on their VA loading rather than their real power loading in watts, it is far less useful in a domestic one where the meter will give a reasonable indication of the actual power consumption regardless of the PF.

Note that recent EU legislation has stipulated that larger SMPS (i.e. over 25W) must now include PFC. However there will me lots of legacy equipment in use that does not include this for some considerable time.

I have some poor PF appliances like an old fridge, should I apply correction to them?

Generally no, because you are only charged for the real power used it will not make any difference to your bill. However, trading up to a modern low energy consumption one will help and may well pay for itself in energy saving alone.

See also

• Category: Electrical
• Wikipeadia article

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