Replacing light bulbs with LEDs - LED lightbulb information

Slalom

I would try a working and none working one on a multimeter to see if they are different resistance. The other possibility is that they are slightly different plugs, and don't make good contact..

grahamlthompson

Incorrect, the current will be higher but the
Incorrect the equation for ac power is Voltage x Current x Cosine of angle between voltage and current.  The current leads the voltage by 90 degrees through a pure capacitance. Cosine 90 degrees is zero so power is zero.  Your domestic meter measures power not VA and therefore energy used over time,   and that's what you pay for.   500VA at zero power factor is zero watts.

The vector current will be a tiny bit larger due to the quadrature component created by the capacitor, but the current in phase with the voltage will be unaltered.

Generating units are rated in Megawatts, as is the national demand. The rating of a generating unit is rated by the power of the prime mover (say a steam turbine).  The power factor at which the alternator works is controlled by the field excitation.  Provided the steam regulating valves aren't altered and the boiler firing is not adjusted then the power output of the alternator does not vary.

grahamlthompson

It isn't using more power.

The OP is trying to apply the DC power Voltage x Amps to a AC circuit.  If you stick a capacitor across a DC voltage after the capacitor has charged to the applied voltage the current is zero.  Not so with a ac applied voltage, the capacitor is charged and discharged on the positive and negative going voltage wavefom so a ac current is created.  The current is highest at the point where the voltage change over time is greatest (at voltage zero) and zero at the point where it changes least (at the peak of the voltage waveform), it is displaced by 90 degrees.   The means of measuring an ac ampere is determined by the same power being created when applied to a pure resistance.  Using average current does not work (average current of an ac waveform is zero).  Root Mean Square (RMS) is the answer.

For pure sine waves

A ac voltage of 1 volt applied to a pure resistance of 1 ohm will create a ac current of 1A and use 1 watt of power.

Assuming the frequency is 50Hz, a 1 Volt ac voltage applied to a capacitive reactance of 1 ohm will create 1 amp of current but zero power.  Capacitive reactance is inversely proportional to frequency so the same capacitor will have a falling capacitive reactance as frequency rises.  You can look up the formula.

Decoupled RF sockets use a series blocking capacitor, if the OP was correct then the resulting power output would be very much less.  As the frequency is high (capacitive reactance is low), the power output is not affected. An inadvertent imposition though of a 50Hz mains voltage will be massively reduced to a non lethal level. Ever wondered how neon mains detector screwdrivers don't give you a shock when you light the neon.

grahamlthompson

The key to choosing a led bulb is nothing to do with how the mains ac voltage is reduced to a stabilised DC current limited source to drive a led. Nor is it related to LED bulbs designed to work on low voltage 12V systems either.

The indicated wattage is the power that the bulb will use irrespective of the internal circuitry, the light output is measured in lumens.   The more lumens you get for each watt of input the more efficient the bulb is.  If a 5W led bulb outputs twice as many lumens as another also rated at 5W the former is twice as efficient as the latter.  If you can reduce the number of bulbs by 50% to get the same lumens output, then the running cost for the same on time will be 50% less.

It would be pointless specifying the VA used by a LED bulb or a low energy fluorescent lamp either,  in terms of energy used.

Here's a simple example this bulb generates 320 lumens for a power consumption of 4.5W and costs £4.99 .


*New* 4.5 Watt - B22 LED Golf Ball Shape Bulb - LED Bulbs

This one costs £5.76 but generates 440 lumens using only 4W.


*NEW* 4 Watt B22 Bayonet - Filament Standard Shape LED Bulb - LED Bulbs


Clearly the latter is significantly more efficient and will cost less to run, but costs more to buy.

How long would it take to recover the difference,  well assuming a energy cost of 15p/unit  (One unit is 1000watts/hour).

To recover 77p you need to save 15p/0.77 =  approx. 19 units.  19 units at o.5W will take 1000*0.5*19 = 9500 hrs.   At 6hrs/day approx. 4yrs.  However if you need more bulbs to get the right lighting level then of course at some point the extra cost will be justified.

Cliff

Wow..just got notified of my old post.

Most of your post I agree and I don't see any contradiction apart from the bottom line.

But let me be clear with a quick summary.
Firstly using a capacitor and its reactance to match a low voltage LED is not a green way to go. As you point out the consumer, still pays the right bill as the meter measures watts.

The problem is passed on to the generating plant.
Most of the generating sets I have dealt with are rated in KVA and not watts for the very reason that a bad power factor will load the generator. You cannot rate a generator in watts because you don't know the type of load and its PF. If the voltage and current waveforms are in phase then it works efficiently. If they are not the effective power in watts of the generating set is reduced.
So,for example you could have a 1000kw engine. 1000kva electricity output (100% efficiency)
Lets assume it is supplying a small village using LEDs with a reactive capacitor dropper. Each house would pay the right bill in terms of watts, but the power plant would see a difference in kilowatts and KVA. The difference, is due to the the power factor.

The generator engine could still be producing 1000kw of power but the output from the alternator might only be 900kw. That was my point about keeping the power factor close to 1 , i.e. keeping the volts and amps in phase.  That is the most green solution. So I still maintain that a small switch mode psu is the way to go.

I had a quick Google and I think this better explains the implications of LED lighting on the National Grid.
Power Factor and Solid State Lighting – Implications, Complications and Resolutions

Cliff

No, all my post was about AC power.

Just a quote from the article I linked to...and explains the problems with LED reactive droppers.

The ratio of the reactive power to the real power is called power factor (PF). This basically means that for an equivalent real power consumed by a highly reactive load, for example 5W, the actual current that the grid needs to supply to the load in order to provide the real power has to be higher than the real power by the power factor ratio. For the previous 5W example, for a load with a PF of 0.5, the grid needs to provide 2x the current actually required by the load at any given time. This adverse impact on the power grid does not apply to incandescent lighting, which is purely resistive and has a unity power factor.

grahamlthompson

There is several  massive flaws in your argument.

1 The efficiency of an alternator is well into the 90%'s. I hope your 1000% is a typo   The alternator is connected to the system via a transformer with on load tap changing. As I already said changing the excitation of a alternator does not alter the power output in any measurable way, but does alter the output current and voltage.

Such a device as you envisage would have a massive copper loss , and be grossly inefficient in the first place.

2 The system has static voltage compensation equipment that can generate or absorb mvars as required.

3 The high voltage system itself in terms of mvar requirements is by far the greatest factor in the requirement to operate alternators over excited (generate mvars) or under excited (absorb mvars). Under excitation has implications for synchronous stability (a wholly different and complex topic).  A overhead line has series reactance and shunt capacitance.  The lagging mvars required due to the line inductance is proportional to the square of the transmitted load. The leading mvars generated by the shunt capacitance is proportional to the square of the system voltage at that point.  A typical 400Kv line is mvar neutral at approx. 60% of the full load capability (known as the line natural load).  A  cable has much higher shunt capacitance than series inductance, the natural load of a cable is way higher than it's thermal rating.  There becomes a length of cable that the capacitive mvar requirement current exceeds the cable thermal capacity. (One of the main reasons for using DC for high voltage interconnections like the cross channel links to France). At light load conditions the system generates more leading vars than can easily be coped with.  One technique is to switch out lines not required for security thus increasing the load on others.

4 By far the highest reactive load is lagging generated by devices like electrical motors etc. Capacitive load is rare.  A fluorescent light has a large capacitor to power factor correct the inductor used to generate the tube strike voltage. Industrial consumers who do pay for VARs install large capacitor banks not inductors.

5 Any appliances you have in your home with power factors less than 1 will be reactive.  Adding capacitance will improve the power factor presented at your meter not reduce it.

6 Switch Mode Power supplies themselves require power factor correction.


http://www.digikey.co.uk/Web Export/Supplier Content/Lambda_285/PDF/TDKLambda_pfc_switchmode_powersupplies.pdf?redirected=1




I ought to know what I am talking about, 40 yrs in Electricity Supply, most of it in National Grid operation and planning.  I worked at several National Grid control centres including National Control.

grahamlthompson

The small capacitor will not make anything like the difference in PF you state,  the leading vars generated will by tiny and in any case will help to correct the largely lagging (and much larger) effect of your other electrical kit.  A capacitor is the exact opposite of a reactive load.  Inductance is created by a back emf generated by the current in an adjacent conductor.  Basically a coil of wire. In a pure inductor  the current lags the voltage by 90 degrees.  Adding a capacitor with the same var generation will cancel out the effect of the inductor.  Indeed switch mode power supplies require capacitors to correct he PF anyway.



You can't make such massive and incorrect assumptions as to the effect of LED lighting on the supply to your house without looking at what else you have connected.

The reduction in system losses by the reduced loading of LED bulbs will more than compensate for any small losses arising from the PF not being 1.  This is completely ignored in the opening paragraph which states 17.5% of global power is for lighting.   If it was all replaced with LED the figure could fall to around 1.45%.

Did you actually measure the PF of the different bubs, I suspect not. Had you done so I reckon the bulbs you reckon had a superior PF would in fact make the overall position worse as they would be inductive rather than capacitive. Prepared to bet the current drawn by the two types of bulb of the same wattage would by minutely different. Not the rather ridiculous suggestion that one would draw twice the current of the other.

As to your post being about AC power, it clearly was not.  Voltage x Current where the voltage is not DC does not give you power.

Cliff

I am not sure why you can't just agree?  

What I have said, and the link, which I only found today by co-incidence, stills stands. Although, as you have pointed out the capacitive load might be corrected by other inductive loads in your house which drag the phase the other way- good point- but has been mentioned earlier.

As  for saying.. - well it does if the PF is 1  for instance if the load is resistive.
A kettle with a PF of 1 and running 10 amps at 240v RMS will give you 2.4kW.
W = PF ×A × V
Agreed?
I get the impression you are being argumentative for the sake of it?


A proper switch mode supply converts to DC first through rectifier- so in theory the PF should be approaching 1.

In summary, what are you actually saying?

Are you saying we can ignore the PF? Or just for small loads?
What if everyone has a small LED with a capacitor reducer- which is likely in the future.
Are you poo pooing my premise that the smps is a better solution? More friendly to the grid?
You disagree with the linked article?

grahamlthompson

I can't agree because everything you say is incorrect.  Of course if the power factor is 1 then Voltage x Current = power. It not a trivial point, a load does not have to a pure resistance to have a pf of 1, all it needs is to have equal and opposite lagging and leading var loadings. It's the only case where this is true and totally ignored in your original post.

There is no coincidence, all the kit in your house that uses ac induction motors (fridges, freezers, washing machines etc), will have a considerably larger lagging var load than any number of LED bulbs, which if they had a leading power factor of less than 1 would actually be an advantage.

A switch mode power supply has required filters which affect the pf and also introduce unwanted harmonics into the system.

switch mode power supply basics - Google Search

As I already said you posted a lot of information, that was inaccurate and basically just guessed.

You said the smps bulbs were much better because they had a pf closer to 1, without a shred of evidence.  Had you actually measured the current taken by each design of bulb and found a difference that at least would have had some sort of believability.

There may well be other reasons why using a smps may be a superior option, one that comes to mind is bulb longevity.  That though then becomes an economic argument as to the extra cost being worthwhile.  If the cheaper design option persuades more people to buy led bulbs and if as I suspect they have a leading pf compared to a smps one of 1 or slightly lagging then yes they would be a lot more friendly to the grid.

Without actually dismantling one how would you know which was which ?  Never seen this information specified for any bulb.  All you really need to know is bulb power rating and lumen output.