Philips Electromagnetic Lamp manual

5

3.4 Power factor correction

Fig. 111. Lamp current (Il), lamp voltage (Vl) and mains voltage (Vm).

Vm

I

Vl

ϕ

This can be seen in Fig. 111, which is showing the lamp current Il, the lamp voltage Vl (both in phase with each other) and the sinus form of the mains voltage Vm.

The power factor of the circuit can be calculated by dividing the total wattage by the product of mains voltage and current. In formula:

P.F. = (Wl + Wb)/(Vm . Im) (1)

Without the parallel compensating capacitor the power factor of a gas-discharge circuit is approx. 0.5.

For the fundamentals of the voltages and current a so-called vector diagram can be made (see Fig. 112). Lamp voltage and lamp current are in phase and the voltage across the ballast is leading 90 electrical degrees to the current.The vectorial sum of lamp voltage and ballast voltage gives the mains voltage. Now we see that cos ϕ = Vl/Vm, which is less accurate than (1).

In any case the energy supply authority has to deliver an apparent power of Vm . Il to the system on which the distribution network must be based (cabling, transformers).

The energy meter only records the in-phase component Vm . Il cos ϕ, so the supply authority does not get paid for the so-called ‘blind’ part: Il sin ϕ .Vm (Fig. 113).

Fig. 112. Example of a vector diagram showing lamp voltage and lamp current in phase.

Fig. 113. Uncompensated circuit with lamp current and mains voltage out of phase.

				1.1 Vm	Vb
Vb				Vm
				0.9 Vm
ϕ						Il cos		ϕ
								ϕ
I	l	V	l			I
						l
						sin	j

Vm

Il Vl

For this reason, the supply authority demands compensation of the phase shift.

Where in general the ‘unadjusted’ power factor is about 0.50, it has to be compensated to a minimum of 0.85 or even 0.90.This is achieved by adding a capacitor across the mains. In contrast to an inductive

120