Is this the way they rate CFLs?

[bhvm]

I just got hold on One of the Phillips 11W CFL.
We've been using CFLs as long as I remember.

Lets Glance at its specs-

Phillips GENIE (india)
11W 220~240V If=80mA
Cool Daylight, 570LM, 52L/W

This is where the similarity ends...

You divide 570 by 11W and you get 51.8 LM/W... Nice

But peek closer at Specs-
It mentions 80mA at 220v... and Thats 17.6 Watts and not 11W !!


I we take into account 570Lm at 17.6W, the efficency is measley 32.38W!!


What the hell is going on?

[bhvm]

Here is the proof-

[lctorana]

I've wondered about that. I think it's a power factor thing.

I think what the label is saying is that it's 11W, but 17.6VA

CFLs present a highly capacitive load, skewing the power factor by that much. Some, I was informed by our Data Centre electrical bloke earlier this week, have a PF even less than 0.5!

For most users, and probably ALL domestic comsumers, I expect this is practically irrelevant, as thay are billed by the kW-hour, not the kVA-hour. But some industrial users are billed by the KVA-hour, and are thus responsible for their own power factor correction.

So you would be only billed at the 11W rate, but 17.6VA of electricity must be generated to supply it. Is that bad design? Of course it is. But it's cheaper to make that way. So far, only Europe has laws on acceptable power factor limits for CFLs.

[bhvm]

Dear sir,
I appreciate the input.

being unknown to all these terms, would you please explain me in detail about the PF etc. we are talking about here?

It about just zoomed above my head.

[Light Sabre]

This might be over your head to, but take a look anyway: Power factor - Wikipedia, the free encyclopedia. This one is a little simpler: Power Factor - facts vs fiction. Someone here may be able to explain it in even simpler terms.

[jtr1962]

This is about as simple an explanation of power factor as I've seen

Granted, power factor is not an easy concept for those not familiar with electronics, but it is quite relevant these days with all sorts of non-resistive loads on the grid.

[bhvm]

Much appreciated.

BTW, what would be PF of an LED (say, Luxeon Star) Connected to battery. Lets assume absense of any driver/circuitry in between

[jtr1962]

Originally Posted by bhvm

Much appreciated.

BTW, what would be PF of an LED (say, Luxeon Star) Connected to battery. Lets assume absense of any driver/circuitry in between

Power factor is only applicable for alternating current power sources. A battery is a direct current source, so the power always equals the current times the voltage. The current through an LED connected to a battery is steady (once thermal equilibrium is reached), so for all intents and purposes you can use the usual P=V*I equation. An LED driver is a different, somewhat more complicated story. Input current may oscillate at the driver's oscillation frequency, and you may need an RMS meter to correctly measure current in order to calculate input power.

[2xTrinity]

Originally Posted by bhvm

Much appreciated.

BTW, what would be PF of an LED (say, Luxeon Star) Connected to battery. Lets assume absense of any driver/circuitry in between

As JTR said, the power factor only applies for AC. The reason is the following:

Power = Current * Voltage

this is true when talking about instantaneous current and voltage. In the case of DC, the instantaneous current or voltage is the same at all times. For AC, both voltage and current oscillate as a wave, and there may be a phase difference between these waves.

Power = I * V * cosine ( phase between I and V)

The cosine term is the "power factor"

If the current and voltage waves are synchronized with each other, phase difference is zero, and cos(0) = 1. This Lets say a device has a power factor of zero. This means current and voltage are 90 degrees out of phase. So at times when current is flowing, there is no voltage across the load. This means no power is consumed inside the device. I*V at particular instances is zero. However, if you were to look at the "average" current and "average" voltage over time, and not in an instant, and multiply them together -- you would still have a non-zero value of VA.

Even if a device is drawing no power however, the higher VA means that the current-carrying capacity of the power grid will still have to be higher to handle running your device, even if it isn't consuming any power inside your home.

[bhvm]

Now I see....
Leds are superbly efficent at DC sources.

But to drive them at AC, we need transformers,Constant current sources and what not. These things can distort AC and also bring up the PF issue.

Hmmm Hmmm....

Good explanation,Thanks a lot all of you!

[ponygt65]

Just an FYI OP...(since the topic of verbiage is at hand), lm/w = efficacy, not efficiency.