Can anyone suggest the correct way to wire a LED light circuit for a aquarium (i.e nightlight)
I need to run 4 Blue LEDs which need to be mounted in a plastic panel (I can use led mounting bezels). I have 240v avaliable in the hood, or could plug in a "main adaptor" into a nearby socket.
What is the easiest way to light up 4 leds, what size resistors would be required etc?
It doesn't really address your query ... but yesterday while mooching about Homebase I came across a set of four 15mm LED 'lights' in a box for £24? Blue or White lamps were on sale. They also had 30mm lamps (set of four) for £34? ...
I looked into the box to see what was in it -memory says it was four lamp assemblies which plugged into a 'black-box' 13A plug. I couldn't quite deduce which dimension the 15mm/30mm referred to - it may have been the size of hole the assembly required. IIRC there was 10m of cable (?). I didn't really pay much attention ... just thinking 'Hmm' wonder where/what/why I could use those?'
So the answer to you question; "What is the easiest way to light up
4 leds," might be get down to Homebase! [Other sheds may be available].
If you wish to wire them direct to the mains, then you'll need a diode - say an 1N4148 - wired in parallel but opposite polarity to the LED, and then a capacitor in series with this combination. 100nF 250 v working.
But I'd personally use a low voltage arrangement this close to water. The value of the current limiting resistor is calculated as follows:-
Vs-Vf R = ----- I
Where Vs is the DC supply voltage, Vf is the forward voltage drop, and I is the required forward current.
(You'll need the spec for the LED to find out the Vf and I if you intend driving them as hard as possible)
However, in general, 2 will do for Vf and 0.01A for I
So for a 12 volt supply, 1k ohm should be fine.
i used a old car battery and some 17 amp resistors to light the leds on my fish tank.
shokk
Since when was resistance measured in amps?
Mains adaptor would be the simplest I guess (you can run a led from the mains directly if you really want to, but I think given the amount of water involved I would be less inclined to!).
The specific answer to your question is "it depends". You need to know two things, the nominal forward current of the LED in question (Maplin catalogue will tell you in most cases), and the voltage of your supply.
If the supply is unregulated (i.e. most plug in "wall warts"), then its output voltage will tend to be higher than that indicated on the box when you are only drawing a small current. Hence it is probably a good to measure it.
So say for example you had a 12V supply, and an LED that wants 5mA. The voltage across the resitor will be 12V - 0.6V (approx) i.e. the supply voltage less the forward bias voltage of the LED. The current through it needs to be 5mA. So given V = IR, you get 11.4 / 0.005 = 2280 ohms. Hence a 2k2 miniture res will do fine.
Note you can buy LED with built in current limiting resistors designed to be attached directly to a specified voltage supply.
Hi
Put them in series with a resistor. Feed off a bridge rectifier. Put a small electrolytic (100uF will prob do) in paralled with the LEDs if you want to minimise the 100Hz flicker.
Budget to feed the LEDs 20mA at the RMS voltage of 240V. LEDs have a forward voltage drop of a couple of volts so ignore here.
V=IR => R=240/0.02 = 12k ohms. Resistor power dissipation is: I^2 * R = 4.8W so use a 5W or bigger.
Rate the resistor and bridge (and capacitor if used) at at least 400V to give some margin for spikes (remember, 240V is RMS, peak at the top of the sine is about 340V).
Treat the whole circuit as a lump of humming mains. You did mention "aquarium"? My inclination would be to pot the whole lot in epoxy with just the wires coming out and the tips of the LEDs showing. Or use a waterproof box.
If potting in epoxy, you could use a metal cased resistor, bolted to a flat bit of metal and set near the surface of the expoxy - should get rid of the heat - 5W isn't much unless it's well insulated, in which case it may melt things.
HTH
Timbo
For an aquarium, I'd counsel strongly against running direct off the mains - safety being the main reason, but efficiency also. The 4 blue LEDS in series want no more than 15V across them, leaving 225V to be dropped in a bleedin' great series resistor. Nah.
Much better to use one of those many spare wallwarts (see previous thread), do the calculation Dave Plowman quotes so that you put 0.01 or
0.015 A across the LEDs. Only part of Dave's figures I umbly suggest he's got wrong is the forward voltage drop: while "about 2V" is the right figure for red, orange, and green 'uns, the fancy-pantsy blue ones have a drop of 3.5-4V.Cheers, Stefekski
You're not wrong. ;-)
I simply pasted most of the article from a previous one I wrote, since it comes up often enough. I'll update that part.
Not disputing your logic in striving for efficiency, but I'd wager a walwart will waste a few watts just by being plugged in (hystereses loss assuming it's a linear PSU).
Timbo
I would counsel against an 1N4148 - it's a signal class diode and has a fairly low reverse voltage capability (75V I think). ie if your LEDs go open circuit and the diode see 340V peak for 1/2 cycle it will fail permanantly.
1N4004 would be better (400V reverse max).Mind you, if your diode goes open circuit, your leds will blow anyway...
Avoiding mains will solve many issues.
Timbo
In message , Stefek Zaba writes
Why does the OP want blue LEDs?
Is it for him or the fish?
Few fish can see red (according to David Attenborough)
Whatever you use will waste a few watts. Switched-mode power supplies will waste less than pure resistance methods. If you want efficiency then you probably want to use a capacitor instead of a resistive dropper. Blue and white LEDS need a forward voltage of about 4v. You should probably aim to provide 20 to 30 mA
That's not a fish tank - it's a fish fryer.
You're right about the wallwart loss: but using one makes it Someone Else's Problem to design the wart so it dissipates the small amount of heat safely, rather'n worrying about dumping the heat in your own components - if wirewound dropper resistor, balancing metal-cased-better-heat-losing-but-case-probably-needs-earthing versus ceramic-cased runs-hotter-for-same-nominal-rating, and looking up the mfr's data to discover that the nominal 5W rated wirewound will indeed happily dissipate 5W, and is 400VDC/240VAC rated, but will rise to a temperature of 120 degrees C when so dissipating, and finding you need to go up to a 25W rated component to get a reasonable merely-quite-warm-to-the-touch setup. (Can you guess who went through this sort of exercise in looking for a suitable dummy load for a mains PIR? ;-)
Stefek
I must admit to lifting the whole thing about using them direct from the mains from a book. I've never needed to try it. I wondered if it was a closer 'match' to the LED in forward voltage drop.
And 1N4148 are so cheap anyway. I bought a thousand - slightly tangled up
- for a quid.
I'd also council for at least a 1000V rating on the 0.1cap if expecting it survive the mains for long. Mains filter caps have that rating, and still get punctured all the time.
As previously stated, an led is not a creature of the mains. A cheapo DC-out black box would be favourite
At a given voltage and current, the same power will be dissipated, regardless of the power rating of the resistor. What you wanted was a resistor with a lower thermal resistance from case to ambient to limit the temperature rise. The 25W one probably provided that purely by being bigger with a larger surface area, but it has nothing to do with the power rating (other than the side effect that higher power resistors are often bigger). Alternatively just heatsink to the 5W resistor.
MBQ
MBQ wrote: >
Absolutely right - my point was that going by the "headline" rating alone doesn't tell you all you should know; specifically, what temperature the resistor will rise to when dissipating the power you require. The 25W (say) component has a lower temperature rise for a given power dissipation because of the larger surface area, as you say - which it has precisely to dissipate the higher powers it's "rated" to, but at temperatures which may be unacceptable in the particular physical design one is considering.
Cheers, Stefek
No way in this case. You'd also need a reverse diode across the LED because it won't take the reverse voltage.
You're not basically wrong but LEDs tend to have higher forward voltages eg a yellow one will be about 2.1V, blues tend to be about
3.5V (from the RS catalogue)
Indeed but blue ones are probably hard to come by.
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