Power factor of a retrofitted LED unit in a fluorescent fitting

[Guineafowl]

TL;DR Summary

How does leaving the ballast and capacitor in circuit affect power factor of LED lights?

A poster on another forum has fitted a large number of 230V 50Hz 24W nominal LED tubes, apparently designed to be plugged straight into the old fluorescent fittings.

He has found, by clamp meter measurement, that the new LED lights appear to be consuming 80W, not the claimed 24W.

A number of us pointed out the difference between apparent power as measured by the clamp meter, and real power as charged by the power company, and the possible effects of leaving the ballast and capacitor in circuit.

An engineer from the supplier has visited, and merely measured the consumption by clamp meter again. I would be very interested in the input from members on here.

May I link the thread here?: https://www.mig-welding.co.uk/forum...florescent-led-replacement-efficiency.128634/

edit: I should add that this class of LED tube must have a minimum power factor of 0.5. In my scribblings, I have (for now) modelled them as 24W resistive loads. It might be more meaningful to assume a 0.5 (leading?) PF and calculate the complex impedance of the tubes as such.

 

[Baluncore]

The inductors were there to keep current flowing through the tubes for a greater angle, not just at the peak of the voltage cycle.
The capacitor was there to neutralise the inductance, to partly restore the PF.

The clamp current being measured now is the reactive capacitor current. That will be of the same magnitude as the original tube current through the inductor. That explains why the clamp current shows the VA of the original tubes. Depending on the type of metering used and tariff, they may actually be paying for the reactive current.

Removing both the inductor and the capacitor will reduce the reactive current, and the circulating energy. It will minimise the bill for lighting, by an unknown amount, depending on the metering.

 

[Guineafowl]

Thanks. The OP is, I believe, on a domestic three phase tariff, meaning only real power is charged. This may change in the future.

The height of the ceiling means that removing the inductors/caps would be a significant upheaval. Given that the ‘extra’ current is not being charged for, it looks like the job would not be worth it.

May I quote, or at least link to your post?

 

[Baluncore]

May I quote, or at least link to your post?

Feel free.

 

[Tom.G]

I have a magnifier/desk lamp at my work station with an 8 inch Circ-line LED lamp in it. The LED lamp is a plug-in replacement for the original fluorescent tube, the ballast is still in the circuit.

I plugged it in to a KIL A WATT meter and here are the numbers with the LED:
V = 118.5
A = 0.42
W = 32.1
VA= 49.8
PF= 0.64

If I did the math right, that shows a phase angle of 50°.

Cheers,
Tom

 

[Guineafowl]

I have a magnifier/desk lamp at my work station with an 8 inch Circ-line LED lamp in it. The LED lamp is a plug-in replacement for the original fluorescent tube, the ballast is still in the circuit.

I plugged it in to a KIL A WATT meter and here are the numbers with the LED:
V = 118.5
A = 0.42
W = 32.1
VA= 49.8
PF= 0.64

If I did the math right, that shows a phase angle of 50°.

Cheers,
Tom

Would that be an electronic ballast?

 

[Tom.G]

Would that be an electronic ballast?

I haven't taken it apart, but:
It used a conventional starter for the original lamp.
The housing size where the ballast seems to be is 3 x 3.5 x 2.5 inches in size.
It is about 12 to 15 years old.

I vote for a magnetic (transformer) ballast.

 

[Guineafowl]

I haven't taken it apart, but:
It used a conventional starter for the original lamp.
The housing size where the ballast seems to be is 3 x 3.5 x 2.5 inches in size.
It is about 12 to 15 years old.

I vote for a magnetic (transformer) ballast.

…and when can we expect the stripdown and analysis?

 

[Tom.G]

'Fraid not, the steel case, presumably with the ballast, is crimped together.