identificationled
This LED (and 8 others like it) form the lighting component of a ceiling fan assembly. Unfortunately, they put out a very dim and cold light, and I'm interested in replacing them, but I figure I need to determine what I'm starting from before looking for replacements (although any suggestions on what to look/search for are welcome).
A potentiometer is also called a variable resistor and rheostat. Those 100Ω ballast resistors and the potentiometer are the same as a "Resistive dimmer". They will not damage an led. They are the simplest way to control the current to a led segment.
Each of your 100Ω resistors are limiting current to ((12V - 1.35V - 1.35V) / 100Ω)) 93 mA and dissipating ~0.86W each (I hope they are 1W resistors, or they will heat up and fry quickly). Combined, that is 6.88W in heat for the resistors, and 744 mA. That's essentially a 12.5Ω load. Since the leds are pretty low forward voltage, and the resistors are burning a very large amount in comparison (2.7V vs 9.3V), you have plenty of headroom to change the resistance. Increasing the resistance lowers the current, which in turn dims the leds.
A single 10Ω 5W resistor will (roughly) halve the current (51mA) and brightness to each led, and since it is also taking up ~1/2 of the power, will help dissipate the heat nicely, as each of the 100Ω resistors will only be dissipating 0.44W instead of 0.86W. I saw 10Ω instead of 12.5Ω, as a 10Ω will be easier to find. You could use a 20Ω for 35mA to each led string. Or a 5Ω for 66mA each. Since leds are logrithmic, it's not an exact scaling of brightness.
Add a multi-throw switch, and you can chose between them.
simulate this circuit – Schematic created using CircuitLab
The reason you fried your pot was that it couldn't handle your poorly designed panel. Frankly, you should have just stuck them all in series with a single resistor. 12V Source Voltage - 10.8V Forward Voltage (1.35 * 8) = 1.2V to drop. 1.2V / .093A = 12.9Ω. 1.2V * 0.093A = 0.111W. A single ~13Ω 1/4W resistor would have been perfect to set the upper limit and Most pots could handle a measly 1/4th~1/8 Watt.
You need 3 batteries minimum, and 3 resistors. With 3 batteries, try
simulate this circuit – Schematic created using CircuitLab You have 2 problems. The really fundamental one is tieing the LEDs in parallel. They operate at different voltages. The red has the lowest voltage, and it's hogging the current from the others. You need a separate resistor for each LED. Red has the lowest voltage, green next, and blue the highest. The current (and brightness) is determined by the resistor, so for the same supply voltage you really ought to have different resistor values. This is not entirely true, since different LEDs can have different efficiencies, but it's a good starting point.
The other problem you have is using a 3 volt battery. This is just about the blue LED voltage. When you first connect the blue LED, the battery has just enough voltage to light the LED, but this quickly gets exhausted. When you disconnect the battery, it recovers and will briefly work again.
Best Answer
What Oli says, plus:
**BE AWARE THAT THE WHOLE LED CIRCUITRY MAY BE AT MAINS VOLTAGE.
EVEN IF IT MEASURES AT LOW VOLTAGE RELATIVE TO GROUND IT MAY BE NON ISOLATED AND MAY EFFECTIVELY BE AT MAINS POTENTIAL HAZARD WISE.
ACT ACCORDINGLY AT ALL TIMES IF YOU HAVE NOT PROVED WITH CERTAINTY THAT THE ABOVE IS NOT TRUE.
See "ANSWERING THE ACTUAL QUESTION" below - but do read the rest as well.
If there are markings on the LEDs it would help greatly to know them.
Summary:
Determine LED forward voltage by measurement.
Determine LED string current by measurement
Choose appropriate LED.
Some such LEDs have 3 dies in the pkg which are brought out separately on eg pins 1-6, 2-5, 3-4 as viewed here and the user can join them in series. This does not seem to be the case here. The pins MAY be connected underneath but it seems unlikely.
Note that the top left and bottom pins (1 & 4) connect to the adjacent contacts by wider copper tracks while the rest have apparently unconnected tracks.
POSSIBLE COMPLICATING FACTORS:
Be aware that the LEDS may be being fed with DC with a strong 2 x mains frequency component if an attempt is being made to power factor control the load (unlikely but ...).
Also, as part if a ceiling fan, they may have a low voltage winding acting as a transformer winding from the motor. This may lead to unusual waveforms.
If you can determine how steady the DC feed to the LEDs are it would be good. As they MAY be floating wrt ground this may involve a floating differential measurement involving mains. Not hard but you do want to know what you are doing and there are safety aspects to consider.
The following is written on the assumption that the LED supply is reasonably steady conventional DC. Worst case you would get voltage and current readings that are wrong. If LED Vf varies by much from 3.3V say then you may suspect interesting happenings. Lowest Vf you are liable to see with white LEDs is about 2.7V and highest is about 3.7V. More like 3.3 +/- someis usual. (The Raijins I mention below run at about 2.95V at rated current as a part of their overall awesomeness. This low is rare at full current. )
ANSWERING THE ACTUAL QUESTION
Measure voltage across contacts (same as pins 1 & 4).
This will be 3.xx Volt OR a multiple of this.
Record and report.
Multiple occurs if there are 2 or more dies in series in package. Happens.
Name your 9 LEDS LA LB LC LD LE LF LG LH LI say.
Have name increasing in order of connection if in series.
Probably in series but not certain.
Measure voltage from LA in to LI out - ie voltage across string. If this is 10 x the voltage across one LED all are in series. eg if Vf La or Lb or Lf ~= 3.3V then you'd expect 10 in series to be ABOUT 33 volt. There will be some variations. If Vmax across string = ABOUT 5 x Vf = say about 17 Volts then you have series / parallel arrangement. You'd expect them to be all in series.
Hopefully they are not all in parallel - unlikely but happens.
IF all in series and Vsring ~= 10 x Vf of one LED then
Set a meter to measure milliamps.
Connect meter probes across a single ON LED with all working. Do this briefly, measure current note effect. Remove. It is extraordinarily unlikely that this will do any damage. If it did then your system is prone to such damage at any moment spontaneously as LEDs can and do go short-circuit in use.
That LED should go out and meter will measure string current.
Note if remaining 9 LEDs get slightly brighter.
Remove meter
If LEDs are all in series you now know LED forward voltage and string current. If LEDs are not all in series you can do a version of this as long as not all in parallel.
You now know the voltage and current of LEDs that can be used as replacements.
Tell us what you measure and we can advise best superior replacement.
UNTIL you do the above you could find that a replacement was "just wrong".
It may not light at all or may die instantly.
Both are genuine real-world possibilities.
Look for pin compatible parts, same number of die per package (1 ~= 3V, 2 ~+6.xV, 3 ~= 10V) and. most important, maximum possible number of lumens at available current. If you want a yellower light you'll want a loer "cplour temperature" - say more like 3000 degree K than the 6000+ that bluish LEDs have.
Be prepared to be disappointed IF current is low. IF these are 20 mA LEDs as they might be then may power may be 20 ma x 3.3V say x 10 LEDs = 0.66 Watt. The best LEDs will have outputs combined of the equivalent of 5 to 10 W of incandescent lighting. Higher currents = higher power = more light.
The VERY best LEDs you can buy will give you about 100 lumen at 0.66W.
The best you can get at low currents:
If Vf for one LED is 2.5V to 4V say (either extreme unlikely) and if string current is under 50 mA (say 15 mA - 30 mA most likely) then you will get more light per available energy than from almost anything else on earth in the size range by using Nichia NSPWR70CSS-K1 LEDs - Nichia codename = Raijin.
These are NOT surface mount but may be able to be adapted - the pin positiin match and can be surface soldered - the height may be excessive in your application.
Photos of illumination need great care to be objective and this is rushed, but this is what you can achieve with 1 x Raijin at about 130 milliWatts. More comment on this LED if of interest.
If you have more than 50 mA available then the Raijin is not the best choice.
Luminous efficiency = l/W = lumenr pr Watt.
A Cree CLP6S-WKW or CLP6S-MKW has been suggested as an example of an LED that youd fit in this locaion. It would, but see my comments else where re Vf and current.
However
Cree make some utterly superb LEDs. This is not one of them, alas.
Vf per LED is strangely high at 4V typical and 5V max !!! - they may even include an internal resistance!. This is too low to be 3 x die in series so each ie must be 4V+!.
To improve brightness you want more lumen so need more lumenr per Watt in a given situation.
Reasonably hidden in the data sheet is the lumen output = 9000 mlm = 9 lumen at 3 x 50 mA on page 2.
This translates into a very low efficiency, alas.
Power = say 4V x 50 mA x 3 LEDs = 600 mW.
Light = 9 lumen so l/W = 9/.6 ~= 15 l/W (lumen per Watt).
This is "not good".
Triple this on the assumption that data sheet is misread and that 9000 mlm is the spec for one LED (which I think is unlikely) and you still only get 45 l/W = not good.
The Raijin I mentioned above gives 160 l/W at 20 mA and 120 l/W at 50 mA.
so at 50 mA a single Raijin outputs just less light than all 3 LEDs in the CREE and at 25% the power (as Vf = 2.95V as opposed to 4V or 5V.)
So
Please measure vf and Istring as above so we can know where to go next.
A measurement "trick"
If you do not have an oscilloscope or cannot safely measure a floating point then the following MAY help to test whether LEDs are fed with steady DC. They probably are, but ... .
Cheaper multimeters, when measuring DC volts on an AC range, will often display double the actual voltage. eg on a 10 VAC range 3 VDC may display as 6 VAC. We can use this "feature" to advantage.
See if you have a meter that does his. Measure eg a battery on a DCV and ACV range. If ACV = 2 x DCV we have a "magic meter" that can be used as below.
Using a magic meter as above, Measure voltage across ONE LED using DCV and ACV ranges. If ACV = 2 x DCV then voltage is essentially pure DC. If ACV strays from 2 x DCV then AC is probably present. AC readings higher or lower than 2 x DCV may occur depending on AC + DC components. Not a certain test, but a potentially useful one.
NSPWR70CS-K1 / Raijin sources:
LEDRISE - prives in Euros but c... dig ... dig ... aha, Hong Kong based.
Available in 1's here
Nichia Raijin Superflux Warm White LED 25.5lm 70° 50mA NSPLR70CS-K1
Brief review and comments here - the real lifetimes are better than their tests - they are abusing them somewhat thermally.
Anyone know a US or UK source?