Difference between revisions of "Power factor"

Revision as of 22:58, 5 March 2010

What is Power Factor

This is quite a "deep" subject when you get into all the detail, so this article will attempt to just 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.

What is the relevance of this to DIY?

Although power factor is a fairly complex subject to do with technical detail, there are occasions where it may crop up in a DIY setting. The most common cases when designing circuits, and when taking electrical measurements. In circuit design, a low power factor may mean that higher than expected currents will flow, and hence larger cable conductors sizes will be needed. When attempting to measure current flow or voltage drop, a low power factor may result in you getting erroneous readings, as it may fool many test meters.

Reactive 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.

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.

Resistive load with AC supply

Here the image shows a short section of an AC supply plotted against time. The voltage goes through one complete cycle (of which there are typically 50 per second in the UK).

It can be seen that with a resistive load, the current flows back and fourth through the load in exact synchronisation with the voltage.

Many domestic appliances will present this sort of load, including many heaters, filament lamps, and the type of electric motor commonly used in vacuum cleaners or similar appliances.

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.

The red sections of the bottom graph, show when power is being dissipated in the load, or in other words the power is actually doing "work". If you imagine averaging the power transfer graph over a full second, you would end up with a positive number that would be the actual power consumption of the appliance

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.

Capacitive load

Capacitive load with AC supply

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 applied voltage, and very low resistance to quickly changing voltage. 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.

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.

As you can see from the power graph, power appears to both flow into the load (red section) and then half a mains cycle later, it flows out of the load. So on average, there is no power dissipated in the load at all.

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.

Inductive load, click to enlarge

Inductive load

A similar situation exists with an "Inductive" load. Although inductors store energy like capacitors, they do so in a magnetic field around a coil of wire. Since the wire is coiled, the magnetic field produced by the current flowing in the wire will also interact with adjacent coils. As the current flowing through the coil changes, the associated magnetic field changes. This changing field will induce a current flow in the wire. However the induced current flow caused by the changing field, is in the opposite direction to the flow in the circuit. So the inductor tries to oppose changes in current flowing through it, by counteracting them using its stored energy. This means inductors will offer a low resistance path to stable voltages, but increase their resistance to changing ones - the opposite of a capacitor.

The main difference resulting from this is that the current waveform lags the voltage waveform rather than leads it. (see diagram to the right)

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 capacitive and inductive examples above, 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.

Composite load, Click to enlarge

Composite loads

In the real world, electrical loads are rarely just reactive, they are usually combined with a resistive element as well. This can give rise to the situation where there is a misalignment between the alternating voltage and the current drawn as before, but not as pronounced. If you look at the resulting power graph, you can see that in this example, most of the power flowing into the load is power transfer that is doing some useful work, but a small proportion is being returned as reactive power.

An appliance like an electric fan heater may present a load like this. The bulk of the current drawn will be supplying the resistive heater element. However a small amount will be driving the fan typically driven by an induction motor.

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.

A Switched Mode Power supply

Switched mode supplies are increasingly found in electronic appliances such as computer and electronic equipment, and also in energy saving light bulbs.

Switched Mode Power Supplies (SMPS) typically rectify the mains and use this rectified AC to charge a capacitor. The capacitor is then in turn used to power the later parts of the power supply circuit. Each mains cycle the capacitor it "topped up", but this can only happen when the mains potential is above that of the capacitor, which means this has the effect of only drawing current near the peaks of the mains voltage, and none at all during the remainder.

Unlike in the examples shown above, there is no phase shift visible between the voltage and current waveforms, the current waveform is distorted into a square shape and is no longer the same shape as the voltage waveform. This change in shape also means that the current waveform is no longer made up from just a single frequency, but is in now made from a single fundamental frequency with a whole bunch of other frequencies mixed in with it.

The following diagram shows how a single pure frequency will be distorted as you add in harmonic components:

The blue line represents the main fundamental frequency. All the other smaller waveforms are harmonics (i.e. multiples of the fundamental frequency). If these are all added together, you get the resulting square looking purple wave. The more harmonic frequencies we add in, the cleaner looking the square wave will become.

The actual power dissipated by the load is no longer 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).

As you can see from the power section of the graph, power is only dissipated in the PSU in short regular bursts.

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

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.

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

Sort of. In the sense you will be drawing more current that you would with a high PF. However in a domestic situation a poor power factor will not result in you being charged for more electricity.

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).

What about this device that claims to save money?

For industrial users of electricity, PFC can and does save them money. This has lead to some makers of gadgets and gizmos to promote them to domestic customers for the same reason. Alas, PFC alone is unlikely to save you much if anything in electricity, since the standard UK electric meters are actually very good at only measuring the real power used, and not being fooled by reactive currents that may also be drawn from (and returned to!) the supply.