PF for LEDs

[charlie b]

When calculating the total load on a lighting panel that will only feed LED fixtures, what power factor value should I use? I should like to hope that all LEDs are 1.0 PF, but that might be asking too much.

 

[don_resqcapt19]

I see numbers from .4 to .95

 

[Rock86]

As per Energy Start Requirments... " For lamp power >5W, power factor must be ≥ 0.70 ". Page 4 of https://www.energystar.gov/ia/partn...s/Integral_LED_Lamps_Program_Requirements.pdf

 

[Besoeker3]

When calculating the total load on a lighting panel that will only feed LED fixtures, what power factor value should I use? I should like to hope that all LEDs are 1.0 PF, but that might be asking too much.

I don't know if this helps:
"Energy Star requires LED lamps of greater-than 5-Watts to have a minimum power factor of 0.7 "

My field was/is in the industrial field so lighting loads were generally not significant.

 

[charlie b]

So if the contractor tells me they bought a lighting inverter rated at 1000 "watts," and if the load is "750 VA," I might conjecture that I don't need to know the inverter's VA rating or the LED's PF rating to know that we are not overloaded. But I am not sure that is correct. Any comments?

 

[Rock86]

I don't know if this helps:
"Energy Star requires LED lamps of greater-than 5-Watts to have a minimum power factor of 0.7 "

My field was/is in the industrial field so lighting loads were generally not significant.

I figured I would give him a base. The original poster could research about the number's they are working with. Generally we don't calculate pf with regards to lighting loads.

 

[WasGSOHM]

I see numbers from .4 to .95

With the LED itself being purely resistive and the LED driver being a 95% efficient switch-mode supply, 0.4 sounds way low.

 

[Rock86]

So if the contractor tells me they bought a lighting inverter rated at 1000 "watts," and if the load is "750 VA," I might conjecture that I don't need to know the inverter's VA rating or the LED's PF rating to know that we are not overloaded. But I am not sure that is correct. Any comments?

I'd agree that you wouldn't need to know the pf in that case. your True Power is greater than your Apparent Power... you'd need a -0.01 PF if and a reactive power of greater than 661VARs to cause an over load... if my math is correct.

 

[Besoeker3]

With the LED itself being purely resistive and the LED driver being a 95% efficient switch-mode supply, 0.4 sounds way low.

A LED, or light emitting diode, is non-linear rather than restive.

 

[GoldDigger]

So if the contractor tells me they bought a lighting inverter rated at 1000 "watts," and if the load is "750 VA," I might conjecture that I don't need to know the inverter's VA rating or the LED's PF rating to know that we are not overloaded. But I am not sure that is correct. Any comments?

IMO it all depends on the circuit design. There are siuations where an inductive or capacitive or distortion component of the net load could cause problems for the inverter even when the VA is less than the nominal W limit.
This is known to happen with the voltage regulation of some generators.

Sent from my Pixel 4a using Tapatalk

 

[WasGSOHM]

A LED, or light emitting diode, is non-linear rather than restive.

It's not inductive or capacitive so I'd call it a non-linear current-dependent resistor.
The current through it should be in phase with the voltage across it.

It might have some current-dependent capacitance if reverse-biased, but as an LED it's forward biased.

 

[gar]

210111-1720 EST

I measured several:

Eight foot tube with internal ballast was 0.98 .
A dual 4 ft from Costco was 0.97 .
A CREE 9 W was 0.94 .
An unknown screw base 9 W was 0.82 .
A Feit 9 W was about 0.72 .

To make the measurements I used a Kill-A-Watt EZ, a rather good inexpensive instrument.

.

 

[WasGSOHM]

And it now seems to me that a load that draws current pulses that are "in phase" with the voltage waveform still does not give a PF of 1.0.
So the LED dimmer may push the PF off of 1.0.
If you know of a graphical way to calculate PF based on V and I waveforms, I'd like to see it.
I searched on
power factor FWB
but it didn't help much. Maybe I should have said
power factor SCR

 

[gar]

210111-2023 EST

WasGSOHM calculate a lot short time power values over a full cycle, add up these values, then divide by the number of samples to provide the average value of real power. In other words do an integral calculation.

The rest of the calculation should be obvious.

.

 

[WasGSOHM]

For the current pulses, squaring is not involved?

Thanks, I'll find an example, do the procedure, and check my work.

 

[WasGSOHM]

Ignore the above.

I remember now that I knew this stuff when I was 19 but that book is long gone. I shoulda' kept it along with the thermodynamics book (because we are having household humidity problems).

I'll dig up this RMS stuff online.
I now believe the 0.4 PF, I was thinking linear circuits which we hardly have anymore.

 

[Flicker Index]

And it now seems to me that a load that draws current pulses that are "in phase" with the voltage waveform still does not give a PF of 1.0.
So the LED dimmer may push the PF off of 1.0.
If you know of a graphical way to calculate PF based on V and I waveforms, I'd like to see it.
I searched on
power factor FWB
but it didn't help much. Maybe I should have said
power factor SCR

Watt is still volt times amps, but these are not Vrms x Irms. Draw the current and voltage waveform on the same graph so they overlap together. Start the computation where voltage is at 0° and continue for several complete cycles. If you were to break 360 degrees in to 36 sections (whole cycle spans across 36 columns on graph paper), then you do V x I column 1, V x I column 2 . . . V x I column 36, add them all up and divide the sum by number of columns and that will get you the average power for one cycle. Such a tedious process is done by computer these days or was done in an analog multiplier.

Power factor is the ratio of the above (real power) divided into (Vrms x Irms) and it is always between 0 and 1.

 

[Besoeker3]

It's not inductive or capacitive so I'd call it a non-linear current-dependent resistor.
The current through it should be in phase with the voltage across it.

It might have some current-dependent capacitance if reverse-biased, but as an LED it's forward biased.

Yes, non linear Power factor is not really applicable for this.

 

[Flicker Index]

So if the contractor tells me they bought a lighting inverter rated at 1000 "watts," and if the load is "750 VA," I might conjecture that I don't need to know the inverter's VA rating or the LED's PF rating to know that we are not overloaded. But I am not sure that is correct. Any comments?

https://hubbellcdn.com/ohwassets/HLI/dual-lite/resources/literature/0603484_dl_cli_brochure.pdf

I don't really understand the way lighting inverters are rated either. Look at page 4. These small lighting inverters are rated the same VA and W, yet they specify a PF rating of 0.44. These two conditions can not be met simultaneously. The LPS26, for example is 20W/20VA with PF rating 0.44. This power factor rating is consistent with straight forward switch mode power supply input that uses a bridge + capacitor. A simple old school mass market corkscrew 18W CFL is probably 18W/36VA with ~0.5PF.

My guess would be that real power is limited to 20W, but the back-end of inverter is built with enough margins to handle the poor power factor reasonably anticipated small load, for example, using the lighting inverter to fire up one off the shelf 18W corkscrew CFL as emergency fixture. By not putting 20W/45VA, they avoid unnecessary support calls about this.

Their larger units are rated at PF=0.8.

https://hubbellcdn.com/catalogpage/dlt_syn_psg.pdf

1kW/2kVA is practically the same thing as 1kW with PF rating of 0.5. If we're talking about a genset, the VA determines the alternator sizing, but it's the real power that determines the load on the prime mover, thus the fuel consumption. Similarly, it's the real power that matters when it concerns the runtime.

 

[texie]

https://hubbellcdn.com/ohwassets/HLI/dual-lite/resources/literature/0603484_dl_cli_brochure.pdf

I don't really understand the way lighting inverters are rated either. Look at page 4. These small lighting inverters are rated the same VA and W, yet they specify a PF rating of 0.44. These two conditions can not be met simultaneously. The LPS26, for example is 20W/20VA with PF rating 0.44. This power factor rating is consistent with straight forward switch mode power supply input that uses a bridge + capacitor. A simple old school mass market corkscrew 18W CFL is probably 18W/36VA with ~0.5PF.

My guess would be that real power is limited to 20W, but the back-end of inverter is built with enough margins to handle the poor power factor reasonably anticipated small load, for example, using the lighting inverter to fire up one off the shelf 18W corkscrew CFL as emergency fixture. By not putting 20W/45VA, they avoid unnecessary support calls about this.

Their larger units are rated at PF=0.8.

https://hubbellcdn.com/catalogpage/dlt_syn_psg.pdf

1kW/2kVA is practically the same thing as 1kW with PF rating of 0.5. If we're talking about a genset, the VA determines the alternator sizing, but it's the real power that determines the load on the prime mover, thus the fuel consumption. Similarly, it's the real power that matters when it concerns the runtime.

I looked at that link for the small units and, yes, it says the OUTPUT W and the VA are the same. It seems to me that that PF of .44 is the INPUT PF. That seems explain the discrepancy.