# Power factor LED getting improved through Ecodesign.

Power factor (λ) a subject that comes up in many discussions. Lets have a look at power, power factor and the resulting issues. First we have three kinds of power, apparent power, reactive power and true power.

Apparent power is the power calculated by multiplying current with voltage. This gives the power in VA. This power is mot registered by the standard power meters. A standard power meter works on true power and registers the true power you take from the net. This is what you pay for. The apparent power is in general higher that the true power.

Measuring the power is not always easy. For a simple resistive load like a incandescent lamp standard multimeters can be used. They will register correctly the current and voltage and in this case you can calculate the true power by multiplication of the two. An incandescent bulb is resistive but LED products or fluorescent lighting are not and there you need more sophisticated measurement instruments to measure things like true power.

Reactive power is basically the power that the product gives back into the power grid. So we have three kinds of power and these can be used to determine the power factor.

The definition of the power factor is the factor resulting from the division of the true power by the apparent power.

What is the difference between power factor and cosines ρ? The latter describes phase shift and is only to be used for sinusoidal currents. If the current has a different shape as is normal for electronics this is no longer usable for calculation of the true power.

When the you calculate the true power the formulae would be P=IxUxλ. The power factor λ plays a significant role in determining the true power. Lets look at an example;

A light source of 60 W, incandescent bulb, will be at a power factor of 1. The current will be I=P/(Uxλ). For a this bulb the current would than be 60/(230×1)= 0,26 A.
The replacement LED bulb could be around 10 W. If the power factor is around 0,5 the calculation would be as follows;
10/(230×0,5)=0,087 A
When the power factor would have been 1 the current would be 0,043 A and this is the hurt point.

The problem with low power factor is the current. It goes up and this current has to be generated by the power company and causes additional losses in the grid. Cables need to be thicker and transformer bigger etc etc simply to be able to provide the extra current needed caused by low power factors.

This should be cause for concern? Yes and no. If you replace a existing bulb as we described above the actual current goes significantly down. Even if the power factor is only 0,5 it is still going down significantly and thus it does not contribute in new problems. If you replace good performing fluorescent systems (f.e. T5) it may be different. These systems have already a good performance and the energy reduction is not as high as compared with incandescent bulbs. The power factor of these systems is generally quiet good so using low power factors here can significantly increase the current in the installation. Result could be blowing the fuse or even that the cables are no longer able to support the currents.

The positive view; if you have a high power factor and replace significant amounts of lighting in a building with these high power factor products it could reduce you current to such an extend that it is possible to lower your current connection to the grid and significantly reduce you fixed fees to the energy supplier.

One note needs to be made and that is that these larger systems generally already have a better power factor because they need to comply with the EMC directive and are above 25 W.

The ErP directive however has set new rules;
P<2 W; no requirement
P above 2 till 5 W; λ > 0,4
P above 5 till 25 W; λ > 0,5
P above 25 W; λ > 0,9

These rules apply to LED lamps. Clear is that LED light bulbs must comply with this requirement. But how about luminaires? More in part 2.

Jacob.

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