LED lighting

[duncan webster 1Participant @duncanwebster1]

I've been swapping all domestic lighting from CFD to LED for a while now. The LED's don't last as long as they claim, but seem to manage a couple of years. I've now got 2 where the built in switch mode power supply has died, but the LEDs are still OK. Tonight I coupled one up to a transformer rectifier and fed it from a variac. With the DVM reading 40V the LEDs just light (can't go any higher). As there are 24 I guess they are in series. The master plan is to connect 2 (or more when more bulbs die) and a small value resistor in series, and then drive them off a LED dimmer switch. Is this daft? If not, can anyone suggest a suitable resistor. I imagine it's going to be a big wire wound job, so I don't want to 'try it and see'.

Once it's running I would measure volts and amps across the LED and adjust the dimmer to get 10W as the original bulbs.

[duncan webster 1Participant @duncanwebster1]

[clogsParticipant @clogs]

Duncan……

how do u know all this stuff…….?

have enough trouble trying to find LED strip lamps at a decent price…….hahaha

all credit to ya……..

[Robert Atkinson 2Participant @robertatkinson2]

Hi,

If there are 24 LEDs in the light then there must be 2 series strings of 12 wired in parallel. 1.7V (40/24) is not enough to light a white LED. For 10W wee need 0.25A at 40V So for resistor value (if we work on the peak voltage for a safety margin) the resistor needs to drop 320V – 40V at 0.25A so Ohms law gives us 320-40/0.25 = 1120 Ohms.
So say a 1000 ohm resistor. However the power rating of this resistor is 78 Watts (280×280 /1000) so it is not a practical solution unless you want a heater as well.

Robert G8RPI.

[AdrianRParticipant @adrianr18614]

Been a while since i fiddled with electronics, but if i remember right the forward voltage is about 2.2V so you will need about 63V for full output. The high output LEDs may have a higher forward voltage.

You are correct you will need a something to limit the current, but if you have too many LEDs in series the voltage across the resistor will be low. So its effect at limiting the current for voltage variation will be reduced. Plus of course you will be dissipating a fair amount of heat which defeats the purpose of energy efficiency.

What you can use for a current limiter is a filament bulb, find one of a suitable voltage and wattage and it will regulate the current for the LED. You would need to consider in rush, that could blow the LED. So a small inductor in series, and a capacitor in parallel to the LED will limit that.

[AJWParticipant @ajw]

I think what you need is a capacitive dropper, quite a common mod in vintage AC/DC radio sets that use a resistor as a dropper.

I think it's quite a science to work out the value but typically 3-5 uF is used. Info is available on the web but I can't remember where!

Alan

[John HaineParticipant @johnhaine32865]

A capacitive dropper would need to be 3.3 microfarads. Value is 1/(2 x pi x F x Z) for impedance of Z – insert Z=1000 and F = 50, assume 2pi=6. Remember you would need a bridge rectifier. But what a cludge.

Over on eBAy you can probably buy a separate replacement constant current LED driver for not a lot of money.

[HowiParticipant @howi]

What are you going to use them for?

LED bulbs are so cheap now why bother.

A lot of time/effort for very little gain IMHO

[Geoff TheasbyParticipant @geofftheasby]

LED lights are great, and very versatile. For domestic situations straight replacement is good, but the dearer ones £3-ish are better than the 'bargain' types. Philips Hue are expensive but can change colour and intensity. For the workshop, I buy reels of LEDs at £8 for 500, on a self adhesive strip, with suitable resistors, often run on 12 volts from a 'wall wart'. These strips can be stuck up anywhere, or fed through clear plastic tubes to make a strip light.

[Frances IoMParticipant @francesiom58905]

The common cause of failure in LED lamps is heat – either the actual LED (usually only for high power devices mounted on poor heat sink) or more usually in the mains powered LED, in the circuit hidden in the base to convert from mains to the low voltage required. Cheap LED lamps use cheap components that soon fail under high temperature as the LEDs used in cheap + higher priced lamps are usually manufactured in the same line

[BazyleParticipant @bazyle]

As Frances mentioned heat is a problem so if using strips like Geoff mentions stick them to a bit of aluminium angle which works as a heatsink and a mounting and as a reflector and as a glare shield.

I wonder if LEDs really save. I find myself more likely to leave my hall and porch light on now they are LED as I can more easily justify leaving them on if I go out and have the convenience of seeing the keyhole when I get home.

[AdrianRParticipant @adrianr18614]

I was going to suggest using a capacitor to provide the reactance, but there was mention of using a dimmer. A dimmer works by turning on part way through the voltage cycle. This results in a very rapid rise in voltage. This rapid increase creates a a LOT of high frequency harmonics. Well into the 10's KHz

As a capacitors reactance is 1/(2 * pi * f * C) where f is the frequency and C is the capacitance

The reactance of the 3.3 microfarad capacitor at just 1KHz would be 50 ohms. This would result in a huge spike in voltage and current in the LEDs. The LEDs would do a very good magic trick and vanish in a flash of light and smoke.

[VicParticipant @vic]

I don’t know how they get away with the false longevity claims. Every LED lamp I’ve bought so far has failed long before it should have.

[duncan webster 1Participant @duncanwebster1]

Having read the above I think that is going in the 'not quite such a good idea' box. However, to clear up some issues;

the 40V was rectified but not smoothed, so the peak would be 56V or so. This must have been enough to light them up dimly. If it were 2 lots of 12 I think I'd have let the magic smoke out

I wasn't thinking of using a capacitor, but a trailing edge dimmer with a say 100 ohm resistor. If 10W at 240 V is 40 mA, the power in the resistor is way less than 1W. When I come across a decent size 100 ohm resistor I might resurrect the idea

Why bother? I just can't bear seeing things go to waste if they can be used for something else, and it keeps the grey matter churning

Thanks for the replies it clarifies one's thinking.

[Harry WilkesParticipant @harrywilkes58467]

Swopped my R80 spots in the kitchen middle of last year had them for quite a while the real bulbs that were in from day 1 lasted very well years in fact but I found the modern halogem ones didnt last long. I took the plunge and purchased some new fitting with LED lamps from Ikea, to date sy 6 monhts ish later I have had 3 of the new LED lamps fail so I take them back and get a free replacment how long they will keep doing this remains to be seen !

Not a fan of LED lighting ok when they are being used you maybe saving money but then with the cost of replacing the sort lived lamps rekon you have lost any saving.

H

[John RuddParticipant @johnrudd16576]

My new house was fitted with the halogen capsule type lamps, as they were quite power hungry at 50w a piece, I bought non-dimmable and dimmable replacement bc lamp bulbs from Screwfix. Around £10 for a box of 5. Got the 4000k versions, so plenty of light….as for longevity, Inguess we will see, but they a guaranteed for a reasonable number of hours…..

[Frances IoMParticipant @francesiom58905]

I use the LED 600×600 panels as used in many offices – cool white for workshop + kitchen, warm white for living areas – some have been in use for well over 3 years – I suspect the failing LEDs are those used to replace spotlights – there were many cheap (too cheap) versions on sale as the prices dropped – I suspect the cost savings were made in the per lamp power supply and it is these rather than the actual LEDs that failed.
I also replaced several of the CFC 2D + 4D fittings with LED replacements about 5 years ago – again all are still working – I replaced these early on as even tho the CFL were low power the fittings were prone to be left on whilst I was away from the premises for a couple of weeks.

The other cause of failure as for many electronics, is poor supply especially too high a voltage or spikes – this I suspect is more likely to be found in rural areas at a significant distance from the transformer sub station or near to intermittent high power devices.

Edited By Frances IoM on 08/03/2019 15:12:14

[AdrianRParticipant @adrianr18614]

Duncan,

The 10W is at the voltage of the LEDs, If say you need 60V, 10W is 167mA. Using a resistor you still need to have 167mA. The resistor will be dropping 180V @ 167mA, so would be need to be 30W.

A transformer converts the power, ignoring losses in the transformer, 240V @ 42mA in, 60V @ 167mA out .

[Alan VosParticipant @alanvos39612]

Posted by AdrianR on 08/03/2019 17:16:48:

The 10W is at the voltage of the LEDs, If say you need 60V, 10W is 167mA. Using a resistor you still need to have 167mA. The resistor will be dropping 180V @ 167mA, so would be need to be 30W.

Which, for anybody who has not spotted, would mean 40W consumed, not 10W. You have to consider the efficiency of the control gear. 25% is not good!

[duncan webster 1Participant @duncanwebster1]

Posted by AdrianR on 08/03/2019 17:16:48:

Duncan,

The 10W is at the voltage of the LEDs, If say you need 60V, 10W is 167mA. Using a resistor you still need to have 167mA. The resistor will be dropping 180V @ 167mA, so would be need to be 30W.

A transformer converts the power, ignoring losses in the transformer, 240V @ 42mA in, 60V @ 167mA out .

Ah yes, forgotten that. The (burnt out) switch mode power supply increases the current from the mains, so it's definitely not a good idea. I wasn't going to drop all the voltage in the resistor, the dimmer would have reduced the effective voltage.

[Geoff TheasbyParticipant @geofftheasby]

AdrianR, on a point of information, yes, SCR dimmers did that, if suppression circuitry wasn't designed in. However, using a zero-crossing circuit, which switches the current on or off at the very point at which the mains sine wave (Ha!) goes from +ve to -ve, results in little interference, and it is easily suppressed.

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