Another new incandescent lamp coming out?!?!

Anders Hoveland

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CFL pollutes the inside of your home with “dirty electricity” just by turning it on.

Umm. What?
Yes, I know. :ironic:
They are actually referring to electromagnetic field fluctuations emanating from the ballast. There is all sorts of controversial research and speculation that exposure to this may not be good for you. I am not going to further discuss it here, it's a topic for another thread.


I also emailed them this:
I do not believe they are underdriving their bulbs. I think that "power disc" just helps to prevent a sudden surge of current when the bulb is turned on, leading to increased lifespan.

The main thing that leads to increased efficiency is the fact they are using IR halogen technology in the bulb.
 
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CoveAxe

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I really wonder what exactly that "power disc" does.
Maybe it has a diode and little capacitor to turn the AC into a steady DC current?

A capacitor would have a negligible effect. The power level here is just too high. I doubt it's anything more than a diode.

It's also not going to have an effect on how the bulb turns on or off. All the diode is going to do is chop one half of the voltage wave, so current is only flowing during half the cycle, so the filament is going to be running cooler than usual. That's where the increase in lifetime comes from.

I had a stupid idea to increase efficiency. Just run 3 bulbs in a series circuit, with each bulb designed for a third of the outlet voltage

I'm not sure how this ends up being more efficient. It looks like it would be a wash.

I really do not think the bulb being discussed in this thread is very underdriven. That would have a huge negative effect on efficiency.

That's why the the whole idea sounds suspect. Because they advertise this great efficiency and this insanely long life. You can't have both.
 

Anders Hoveland

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A capacitor would have a negligible effect. The power level here is just too high. I doubt it's anything more than a diode.
Such a capacitor would not need to be very big. Outlet AC is 60 cycles per second. So for a 50 Watt bulb, this capacitor would only need to store less than a joule of power. (and this is assuming full power of the bulb is going through the capacitor)

Probably just some sort of simple circuitry to gradually even out the increase in current when the bulb is first turned on.
 
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CoveAxe

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Such a capacitor would not need to be very big. Outlet AC is 60 cycles per second. So for a 50 Watt bulb, this capacitor would only need to store less than a joule of power. (and this is assuming full power of the bulb is going through the capacitor)

What you're asking for here is not possible in anything close to resembling the powerdisc package shown. You will also need a bit of circuitry to extract every last drop of power out of that efficiently, and it's also going to have to work at >100C temperatures since it's in direct contact with the bulb. Oh, and it's going to be expensive (a few dollars at least).

This is neglecting the fact that a capacitor is completely unnecessary: The filament has enough thermal inertia that skipping a half cycle is not going to be noticeable (hence why traditional dimmers work so well on incandescents).
 

Anders Hoveland

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So you are saying maybe the diode just cuts down the current by half for the first second or two after the bulb is turned on?
 

mvyrmnd

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Yes, I know. :ironic:
They are actually referring to electromagnetic field fluctuations emanating from the ballast. There is all sorts of controversial research and speculation that exposure to this may not be good for you. I am not going to further discuss it here, it's a topic for another thread.

I know what they're referring to, and there's not much to discuss. In over 50 years of research there is zero evidence to show that non-ionising EM radiation (at household or even industrial levels) has any negative effect on people, plants or animals. There's a lot of pseudoscience and anecdotes out there, but we must remember that the plural of anecdote is "anecdotes" not "data".

Non-Ionising EM can hurt you - if you decide to hug a Kilowatt-power FM transmitter - but a lightbulb ballast is not going to affect you at all.
 
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idleprocess

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I know what they're referring to, and there's not much to discuss. In over 50 years of research there is zero evidence to show that non-ionising EM radiation (at household or even industrial levels) has any negative effect on people, plants or animals.
There's an old management-consulting scam template subject to creative variation that goes like this:
  • Claim that some metric effects sales/profits/costs
  • Teach a business to measure the metric
  • Subsequently tech the business to control that metric
  • Pocket fat consulting fees

This kickstarter campaign seems to change it up a little by using it to pitch a consumer product:
  • Convince consumers that EMF is bad
  • Sell consumers an EMF meter (bet the needle tries to break off the post at the far end of the scale when it's pointed at a CFL!)
  • Sell solution that reduces EMF, sprinkled with some feel-good boilerplate about mercury being a super-toxin that maybe-ish poisons thousands-ish of children each year
  • Profit as the rubes send in their greenbacks

There's a lot of pseudoscience and anecdotes out there, but we must remember that the plural of anecdote is "anecdotes" not "data".
I'm going to speculate that a large slice of their target market gets their "news" from sources that amount to editorial that just happens to be flattering to their worldview, thus will take anything that supports their preconceived notions as gospel while ironclad evidence against it is cast as mere opinion.

We still have segments out there insisting that wi-fi and cell-tower radiation causes maladies like cancer and have for decades, so why would they let inconvenient science get in the way of core identity-defining beliefs?
 

idleprocess

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The fact that you the user must peel-and-stick the "power disc" yourself suggests that they're white-labelling commodity halogen A19 bulbs from some contract manufacturer with all the snake oil packed into said "power disc".
 

The_Driver

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Concerning the cost of IRC-bulbs: to the average household the cost of halogen bulbs is not very important. The bulbs last quite a while with their 3000h rating and only cost around 2,50€ per piece from German online shops (3$). The electricity costs over the lifetime of the bulb are much higher. 35W * 3000h = 105kWh => 105kWh * 0,25€/kWh = 26,25€ (32$).

Replacing a 35W standard halogen bulb with a 25W IRC reduces the electricity costs by about 8€ (11%) for 3000h. So in that sense they are worth it if the bulbs are used a lot.
 
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Julian Holtz

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I would guess that the "power disc" is an NTC.

http://en.wikipedia.org/wiki/Thermistor#NTC

These things have high resistance when cold and little resistance when hot.

Incandescent bulbs behave the other way round. When the filament is cold, it has little resistance, therefore you have a current surge when turning it on. This current surge is damaging to the filament. Maybe you noticed that most of the times when an incandescent bulb burned out, it was right at the moment when it was switched on, and almost never while being in the midst of continous operation.
An NTC can eliminate the current spike at the beginning. By heating up the filament in a longer time than an instant, it gets damaged less, and we get an increase in lifespan. The current flowing through the NTC heats it up, the bulb does the same, so that during usual operation the resistance is very low.

Some wall adapters and power supplies have NTCs as well to limit the current surge when the input capaitors are initially charged.

But when reading their description again, it could also just be a diode. Or Both.

Some other claims of them are BS, of course.
 
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Anders Hoveland

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Maybe I am just showing my ignorance here, but how does half-wave rectification reduce the voltage from 120V to 85V with minimal loss?

(I can see the equations, I am just hoping someone can explain why it works to me)

e186337a3603ccd921b4a862f15cf3a9.png


Vdc, the DC or average output voltage
Vpeak, the peak value of the phase input voltages
Vrms, the root-mean-square value of output voltage
 
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Steve K

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120vrms has a peak voltage of 169.7v (just multiple by the square root of 2).

half of 169.7v is 85v.

That's the math to get you from 120vrms to 85v... but... there's a difference between 85v peak and 85vdc. If you filtered the rectified voltage, you would get 85vdc, but that is only if you don't draw any power from it. Once you draw power, you'll discharge the filter cap and get a lower average voltage.

Most calculations deserve to be checked by actually building the circuit and confirming the assumptions behind the calculations.

edit.. I'm not sure how half wave rectifying the AC produces 85v. It just produces half of a sine wave. The peak is still 169.7v, but the power is less. Mostly, it'll just produce more ripple than full wave rectification. </edit>
 
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Steve K

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Maybe I am just showing my ignorance here, but how does half-wave rectification reduce the voltage from 120V to 85V with minimal loss?

(I can see the equations, I am just hoping someone can explain why it works to me)

e186337a3603ccd921b4a862f15cf3a9.png


Vdc, the DC or average output voltage
Vpeak, the peak value of the phase input voltages
Vrms, the root-mean-square value of output voltage

where did you get those equations? DC voltage has nothing to do with AC voltage. The purpose of rms voltage is that 120vrms will deliver the same power to a load as 120vdc.... this is the method of comparing AC voltages to DC voltages when your concern is delivering power to a load.
 

Steve K

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the tutorvista site is just wrong. I'd recommend picking up a proper EE text.

Not to be pedantic, but your equations say that the rms voltage is the peak voltage divided by 2, not the square root of two.

also.... I don't know where "Vdc = Vpeak/pi" comes from. It may be true for some specific circumstance, but it's certainly not generally applicable.

I'd recommend visiting the Linear Technology site, www.linear.com , and download a copy of their LT Spice program. That'll at least let you see what the waveforms are, and get instantaneous values for voltage and current.

In the half wave rectification situation, the peak voltage is still 169.7v. There is no such thing as half wave voltage... are you trying to find the rms voltage? That's easier to calculate. The power delivered to a resistive load is cut in half due to half of the voltage waveform being cut to zero. As a result, the rms voltage will be reduced by 0.707 when compared to the full wave rectified rms voltage. (recall that power = V^2/R, so dividing power by 2 will divide V^2 by 2, which is the same as dividing V by 1.414. In these equations, V is the rms voltage).
If the incoming voltage is 120Vrms, then the half wave rectified rms voltage will be .... 84.8Vrms. That must be what the 85v is referring to... the rms voltage and not the DC voltage.
 
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CoveAxe

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120 multiplied by square root of 2, then divided by 2 gives 84.85

So I am assuming that is the equation to go from 120v to 85v

since 169.7v is the peak voltage of average 120v AC power, and the half wave voltage would be half of that... ???

I am still confused :help:

This wiki about rms explains how that number is calculated.

The big takeaway from using a half-wave rectifier is that you reduce the amount of power by half because you eliminate one of the half cycles. This makes the filament run cooler and last longer, at the expense of efficiency and brightness.
 

Anders Hoveland

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The big takeaway from using a half-wave rectifier is that you reduce the amount of power by half because you eliminate one of the half cycles. This makes the filament run cooler and last longer, at the expense of efficiency and brightness.
I think you misunderstand. Using the same filament and just reducing the power would reduce efficiency.
Being able to use a lower voltage and designing the filament to use that lower voltage allows an increase in efficiency.
I have already explained it, but to state it again, a lower voltage allows the filament to be shorter and thicker for any given wattage. Shorter=higher temperature (because wattage is distributed over a smaller length), Thicker=longer filament lifespan
 

mvyrmnd

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I think you misunderstand. Using the same filament and just reducing the power would reduce efficiency.
Being able to use a lower voltage and designing the filament to use that lower voltage allows an increase in efficiency.
I have already explained it, but to state it again, a lower voltage allows the filament to be shorter and thicker for any given wattage. Shorter=higher temperature (because wattage is distributed over a smaller length), Thicker=longer filament lifespan

Your point is correct, from a technical standpoint, but you fail to note that they're selling the sticky-on bits for people to use with regular bulbs as well, which are not optimised for lower voltages and therefore will run at lower efficiency.

They're still selling snake oil, even if it's in a fancy jar.
 

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