Author Topic: Led master class; pt 2 of 3 (theory)  (Read 13391 times)

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commanda

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Led master class; pt 2 of 3 (theory)
« on: April 26, 2007, 08:29:29 AM »
This is a continuation of part 1 which can be found here:

http://www.fieldlines.com/story/2007/4/20/95351/1614


Let's look at some Vf figures for some real leds, and explore the consequences.


I bought a bag of 100 5mm White SuperBright leds off an australian seller on ebay.com.au.


The published specs are;

20,000 mcd at 20mA (which is also the max continuous forward current)

Vf = 3.0 typical

viewing angle 15 degrees

max power dissipation 100mW


I took a random sample of 20 units, and tested them for Vf at 2 different currents, 11.00mA and 18.00mA, at a room temperature of about 20 degrees C or just slightly above.


Without going into all the boring details, the summary is;

at 11.00mA

average Vf = 2.961

standard deviation(sd) = 0.049

min = 2.845

max = 3.024


at 18.00mA

average Vf = 3.068

standard deviation(sd) = 0.055

min = 2.939

max = 3.132


Average and sd calculated by a spreadsheet.

If you want to understand the significance of sd further, remember "Google is your friend".


This is actually really difficult to do, because as soon as you apply current, the leds start to warm up, and Vf starts to decrease (at 2mV per degC as I explained last week). So what I did was wait about a minute for each one to let the Vf stabilise at some voltage.


Now, whats the significance of all this?

Consider last weeks circuits, in particular 3 & 4, where we have multiple parallel strings powered by one constant current source.

Let's say in the first string of 3 leds, we just happen to randomly pick 3 leds that have a very low Vf, and conversely for the second string, pick 3 leds which have a very high Vf. If we set the value of Req to zero, the string with 3 very low Vf's are going to hog all the current. So, to make the 2(or more) strings of leds play nicely together, we need to set a value of Req greater than zero, but by how much?


At this point, I set up an experiment and did some testing. I picked 3 leds which had a Vf of Avg - 1sd (< 2.912 at 11.00mA) and 3 leds with a Vf of Avg + 1sd (> 3.010 at 11.00mA).

Req = zero, currents were 6.91 mA and 3.28 mA

Req = 22 ohms 6.66 and 3.68

Req = 47 ohms 6.46 and 3.98

Req = 100 ohms 6.20 and 4.37


I know the 2 currents in each case don't exactly add to 11.00 mA, this is due to the internal resistance of the meter and moving it from one string to the other. No prizes for guessing what would happen if we had assumed perfect leds, and set it up to run 20mA in each string of leds; after all, the spec is so many milliCandela at 20mA.


You can see that in the worst case scenario, it takes a large value of Req to swamp the differences in Vf of the 2 strings.

Knowing this, it should be obvious that a little effort in matching the 2 strings is probably worth it.


Now that I have these 2 strings, I want to go back to square one, and show what happens when you use the simple, single resistor per string, just for comparison. Using a certain online calculator, I input my specs. 12 volt supply (because it is a 12 volt battery, isn't it?), Vf = 3.0 volts, I = 20mA. It gave me a value of 150 ohms for the resistor.


at 11.0 volts, currents were 14.06 and 11.86

at 12.0 volts, 19.09 and 16.62

at 13.0 volts, 24.07 and 21.40

at 14.0 volts, 29.56 and 26.69

at 15.0 volts, 34.7 and 32.3


You can see that with a real lead-acid battery, during charge and discharge cycles, this simple approach leaves us pushing way too much current through the leds most of the time. This will result in very short lifetimes for the leds.


So, lets use the maximum battery voltage of 15 volts, and select a resistor based on that. The previously mentioned online calculator gives us a value of 330 ohms. Re-doing the previus test gives;


at 11.0 volts, currents were 7.38 and 6.45

at 12.0 volts, 9.99 and 8.94

at 13.0 volts, 12.63 and 11.50

at 14.0 volts, 15.31 and 14.12

at 15.0 volts, 17.96 and 16.71


Now we keep the current below the leds rated maximum continuous current; but the light output varies very noticeably as a function of the battery voltage.


This simple single resistor approach can be made better, as far as the led is concerned, by using less leds in the string and increasing the value of the resistor. Taking it to the extreme, we end up with one led per resistor, which is not very efficient. I suppose I could have run a test with 2 leds per string and one led per string, for the sake of completeness. But, even though the variability of current vs battery voltage would be decreased, it still would not be completely eliminated.


Summary of parts one and two.


The stated forward voltage of an led is not a cast in stone constant. It will vary from led to led, even from the same batch. It will vary with the amount of current you pass through it. It will vary as an inverse function of temperature; approx 2mV per degree C. Remember that when you plan on stuffing a whole heap of leds in a tiny, un-vented space.


The light output is a function of current.

Take careful heed of the rated maximum continuous current. Exceed this and your leds will die an early death.


There is no perfect led driver circuit. Cheap, efficient, simple. Pick one.


The true art of design is balancing all the compromises to achieve the best result in a particular situation. If you have grasped the concepts presented, you will be in a better situation to maximise the compromises for your next led lighting project.


In part 3, I will build a prototype of circuit 4, show how we work out a few values, and at least one method of matching the leds for the parallel strings.


Amanda

« Last Edit: April 26, 2007, 08:29:29 AM by (unknown) »

SamoaPower

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Re: Led master class; pt 2 of 3 (theory)
« Reply #1 on: April 26, 2007, 06:52:42 AM »
Amanda,

I commend your efforts on this post series to help straighten out some of the misconceptions of LED application, particularly the use of a current regulator. I have yet to understand why some want to use a voltage regulator for current devices.


I did a similar study on the issues about seven years ago and came to similar conclusions. This resulted in a number of arrays that are still in use today, some of which run 24/7.


One of my objectives was to eliminate power-robbing resistors. I also purchased large batches (100) of LEDs and graded them for Vf. This usually resulted in a statistical bell curve. If one is willing to do this and the string matching, resistors can be eliminated.


It should probably be pointed out that, in your typical series/parallel array, it's only important that the sums of the individual series strings be matched. This allows some latitude in choosing the individual LEDs in each series string so that all LEDs in the total array do not have to be matched to each other.


There is another way to configure an array. Using a 9-LED array as an example:





The parallel/series offers some advantages. I have had three individual LED failures out of over 100 in use in the past seven years. All of these have failed open circuit. One usually thinks of the majority of diode failures being short circuit. I guess this is true for power diodes that have been overstressed, but I haven't seen it for LEDs that are well controlled.


In the more usual series/parallel hook-up, a single LED failure results in three LEDs out of action. In the parallel/series, a single-failure (open circuit) results in an increase in current in the remaining LEDs of the parallel group (two, in this example) but they are still functional. If one is conservative in setting the array current, this is often of no consequence. Also, the more LEDs in the parallel group, the less the current increase in each.


Matching in the parallel/series is different in that the LEDs in a parallel group need to be matched to each other (I think 10% is adequate), but the parallel groups do not have to be matched to each other.


I think elimination of resistors is worthwhile.


Looking forward to Part Three.

« Last Edit: April 26, 2007, 06:52:42 AM by SamoaPower »

ghurd

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Re: Led master class; pt 2 of 3 (theory)
« Reply #2 on: April 26, 2007, 07:30:18 AM »
Nice explanation.


I am glad to see someone else point out reasonable quality, modern white LEDs have a Vf well below the "3.6Vf" usually used on the internet.

I believe the so commonly quoted 3.6Vf is a major cause of failure in DIY LED projects.

G-

« Last Edit: April 26, 2007, 07:30:18 AM by ghurd »
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SamoaPower

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Re: Led master class; pt 2 of 3 (theory)
« Reply #3 on: April 26, 2007, 07:55:06 AM »
I failed to point out that for the single failure case in the series/parallel configuration, besides losing light from three, the current in all of the remaining six increases.

« Last Edit: April 26, 2007, 07:55:06 AM by SamoaPower »

finnsawyer

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Re: Led master class; pt 2 of 3 (theory)
« Reply #4 on: April 26, 2007, 08:25:13 AM »
The solution for that would be to move the current regulator to the source side (where I would put it anyway), and then place small resistors at the ground side of each string.  Add circuity to monitor the resistor voltages.  When the voltage of a string drops to zero due to a diode failure in that string, the regulator then drops the current to the proper value for the remaining strings.  It seems to me that this would be the proper reason and use for putting in a current regulator.  What is the point of having it if it doesn't adjust for diode failures?  
« Last Edit: April 26, 2007, 08:25:13 AM by finnsawyer »

Gtrax

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Re: Led master class; pt 2 of 3 (theory)
« Reply #5 on: April 26, 2007, 03:52:18 PM »
Though you seem to have successfully used parallel connected diodes in practice, it is unlikely any in parallel were able to share the current as resistors would. What really happened was the diode with the lowest Vf would start conducting first, and hog the current until the voltage across its own bulk resistance, added to its Vf, went above the Vf of the next diode, and so on.


Getting semiconductors to share current equally when put in parallel can be done by using series resistors to mostly determine the share, or in the case of power transistors, using low value emitter resistors to adjust the bias by a d.c. feedback mechanism. This is the common way of parallel connecting power transistors in amplifiers. With diode rectifiers, nobody would try to increase the current capability by sticking another diode in parallel, because you cannot tell if one diode is taking most of the current. I think the same might be true of LEDs.


Even in the series connected sets, if there is any excess in the supply beyond the added  up Vf voltages, it needs to be lost across a regulator (say a current regulator) of some kind. In the case of diodes, a resistor will do nicely, because the regulation is automatically provided by the Vf of the diodes, and the resistor dissipates the same as a regulator would do. Making the LED's themselves get hot seems unnecessary and LED life-shortening.

« Last Edit: April 26, 2007, 03:52:18 PM by Gtrax »

ghurd

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Re: Led master class; pt 2 of 3 (theory)
« Reply #6 on: April 26, 2007, 04:39:04 PM »
I'm not sure you aimed this one at me.


I believe the intent of Amanda's original post was to waste less power in resistors, meaning resistors will be replaced with something else.

And I expect she will do a great job with it.  She always does.


Amanda's Part #3 I expect will contain a method to avoid all issues related to imbalance between paralleled series strings of LEDs, and I look forward to seeing it because I usually use a simple resistor for these situations.

Resistors work perfect for the situation I am usually in, but it is certainly not always the best solution.  More so if the system battery is large.


If the intent was more toward Samoa, 3 failures of 100 in 7 years is not too bad either.

LED failure is directly related to ma, in my experience. Meaning I would not mind an LED failure if it saved me battery power for 'X' years.


Again, my standard disclaimer...

G-

« Last Edit: April 26, 2007, 04:39:04 PM by ghurd »
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commanda

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Re: Led master class; pt 2 of 3 (theory)
« Reply #7 on: April 26, 2007, 06:05:36 PM »
Gtrax,


If this is aimed at me, I can only assume you haven't actually read part 1, or perused the circuits presented there.


Amanda

« Last Edit: April 26, 2007, 06:05:36 PM by commanda »

commanda

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Re: Led master class; pt 2 of 3 (theory)
« Reply #8 on: April 26, 2007, 06:17:49 PM »
First, as SamoaPower pointed out, running the leds correctly, you get very few failures.

Second, the simpler approach to make it more failsafe, would be to use one current source per string. I don't see why it would be advantageous to "add circuitry to monitor the resistor voltages". A similar or even less component count could be achieved using seperate current sources.


Running multiple strings in parallel off one current source is simply the least component count solution consistent with generally very good results. Certainly far better than the single resistor approach, especially when powered from a typical lead-acid battery.


Amanda

« Last Edit: April 26, 2007, 06:17:49 PM by commanda »

SamoaPower

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Re: Led master class; pt 2 of 3 (theory)
« Reply #9 on: April 26, 2007, 06:21:09 PM »
Gtrax,

What you say about diodes is true if you assume the "ideal" diode with a sharp knee in the curve. LEDs however, typically have a much "softer" knee with Vf much more dependent on current. By actual measurement, Vf in a parallel group, that are matched to within 10% exhibit current matching at least that good or better. It can be better because the curves themselves are not identical.


A 10% delta in current isn't perceptable to the eye as a light intensity difference, at least, not to mine. If the basic array design is conservative, the current deltas are not significant to heat or lifetimes.


Why add extra parts if they're not necessary in a practical sense?


Two of my arrays of 30 LEDs each, 10 Parallel, 3 series, 12V and no resistors are still going strong after 6 and 7 years, running 24/7.


 

« Last Edit: April 26, 2007, 06:21:09 PM by SamoaPower »

Flux

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Re: Led master class; pt 2 of 3 (theory)
« Reply #10 on: April 27, 2007, 01:15:50 AM »
I can't add much to this, if driven properly the failure of leds is low and not a real issue.


They are current driven devices so to me the cost of a transistor for each string makes more sense than paralleling strings on one current source. If you do parallel them then the linked scheme gives you a little better chance if one fails. To keep failure to low levels you need to work below the limiting currents anyway .


For a resistor to act as a current source it needs to drop significant volts in relation to the diode drop so incorporating resistors in a current loop to save a transistor seems pointless.


If you want to go to more complicated circuits with current sense resistors, yes that works fine, but then they are individual current loops and need their own control components.


I am pleased to see someone trying to lay to rest this crazy idea of using a voltage regulator to drive a current operated device. I doubt that we shall get universal agreement about the merits of any one circuit, but I look forward to the final part, which I am sure will be a good practical solution.


Flux

« Last Edit: April 27, 2007, 01:15:50 AM by Flux »

finnsawyer

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Re: Led master class; pt 2 of 3 (theory)
« Reply #11 on: April 27, 2007, 07:50:05 AM »
I agree one should keep it simple.  I didn't check to see if anyone suggested this.  If they did I apologize for the duplication.  In part one I suggested using a Zener diode.  I'd like to update that by adding a junction transistor in the common emitter configuration.  The led strings (with limiting resistors) are put in parallel in place of or in addition to the normal emitter resistor to ground.  The collector of the transistor is connected to the 12 volt supply as is one side of the resistor that supplies current to the Zener.  The base current for the transistor is supplied from the junction between the Zener diode and its supply resistor.  The base current is set to about 1/10 the normal Zener current.  The transistor should have a dc current gain of at least 100.  Most common Zeners are .6 watt devices.  We could probably bias an 8.2 volt zener at .05 amps for a battery voltage of 12.6 volts.  The maximum emitter current would then be .55 amps, which would allow a large number of strings to be paralleled.  The voltage across the strings will be the Zener voltage minus the base to emitter voltage of the transistor, which is small will change very little even if some strings burn out as long as the base to emitter voltage is biased properly.  That is, the base current must be kept high enough so that a slight change in base current due to a slight change in emitter current does not change the base to emitter voltage significantly.  Then all strings will continue to be biased properly if one burns out.  Since the base to emitter voltage of a transistor has a diode dependence with base current this is achievable by adding an emitter resistor in parallel with the strings.    
« Last Edit: April 27, 2007, 07:50:05 AM by finnsawyer »

commanda

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Re: Led master class; pt 2 of 3 (theory)
« Reply #12 on: April 27, 2007, 04:46:36 PM »
finsawyer,


You're still describing a voltage regulator.


We could probably bias an 8.2 volt zener at .05 amps


That 50 mA (wasted) could be powering 3 strings of leds. With an 8.2 volt zener, emitter voltage of 7.6 volts, you're only going to have 2 leds in series in each parallel string, for a total Vf of 6 volts. With a 12 volt battery, thats 6 volts (times the total led current) consumed as heat by the transistor and series resistors.


You will also find that your 8.2 volt zener changes its voltage as the nominal 12 volt lead acid battery goes from flat (11 volts) to equalisation charge (15 volts). Plus the transistor Vbe which will decrease with temperature, leading to more led current.


Please have another look at circuit 4 in part 1. There is no wasted current (except for about 12 microamps through Rg). The circuit overhead in terms of voltage is about 1 volt, so a string of 3 leds (9 volts) will work perfectly down to a battery voltage of 10 volts. And the negative temperature coefficient. If it gets hot, the current decreases. And the current is constant over a wide range of input voltage.


Amanda

« Last Edit: April 27, 2007, 04:46:36 PM by commanda »

finnsawyer

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Re: Led master class; pt 2 of 3 (theory)
« Reply #13 on: April 28, 2007, 08:53:19 AM »
First off, I need a correction.  The circuit is an emitter follower.  You say it is a voltage regulator to which I say, "So what".  If it does the job, who cares.  And it does the job.  To begin with, the numbers I gave would allow some 40 strings in parallel for a total of 80 Leds.  Based on that, the overhead of the presence of the Zener diode doesn't look that severe.  As you reduce the number of strings you can reduce the current through the zener diode.  There is always going to be some overhead.  It is up to the user to decide what is acceptable.  Vbe for the transistor will be around .3 volts.  That makes it about 4% of the Zener voltage or of the voltage across the strings.  To consider temperature effects due to Vbe you need to divide by 25 to get the effect on the voltage across the strings.  Not really a concern.  If it is you can put a germanium diode in series with the Zener diode.  That should help.  If one string becomes an open, the total string current will drop.  This will result in a very slight drop in Vbe, which again should not have any appreciable effect on the voltages across the strings.


"You will also find that your 8.2 volt zener changes its voltage as the nominal 12 volt lead acid battery goes from flat (11 volts) to equalisation charge (15 volts)."


You do not provide any numbers to justify this statement.  My experience with Zener diodes has been just the opposite.  The current versus voltage curve for a typical Zener diode is as close to a right angle as you could hope to show on an oscilloscope.  The entire purpose of using the Zener diode is to provide a constant voltage over a wide range of currents and they are very good at that.  My numbers were intended to be ballpark numbers, and as such will be affected by assumptions made about the range of operation.  It is my opinion that this simple but robust circuit would satisy the needs of many home brew projects.  And it's cheap.  A single Zener diode, several resistors, and one transistor could safely power several strings.    

« Last Edit: April 28, 2007, 08:53:19 AM by finnsawyer »

commanda

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Re: Led master class; pt 2 of 3 (theory)
« Reply #14 on: April 28, 2007, 05:04:03 PM »
I always knew there would be some people who just wouldn't get it.


And I wasn't seeking suggestions for circuits on how to drive leds.


Amanda

« Last Edit: April 28, 2007, 05:04:03 PM by commanda »

finnsawyer

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Re: Led master class; pt 2 of 3 (theory)
« Reply #15 on: April 29, 2007, 07:18:35 AM »
This is why I don't get into discussions involving design of circuits.  There are numerous ways of doing these things, and each has its advantages and disadvantages.  I simply wanted to flag the use of Zener diodes to establish a stable voltage reference.  And yes, they can be cascaded to improve the net result.  
« Last Edit: April 29, 2007, 07:18:35 AM by finnsawyer »

Rembrant

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Re: Led master class; pt 2 of 3 (theory)
« Reply #16 on: April 30, 2007, 04:19:20 PM »
 Hello.. my name is Luke. I have been lurking here for quite a while reading up. I normally wouldn't jump right into a conversation like this but it seems to me that some of you are confused about the nature of Light emitting diodes.(LEDs)

   Firstly, contrary to popular belief, LEDs are !!NOT!! current devices. As Amanda has already shown an LED will not draw more current without a subsequent increase in voltage. The use of any LED at more than it's intended voltage will reduce the MTTF(mean time to failure)regardless of the current it draws! This is called electron migration and it is true of all silicon devices. A proper LED driver <IS> voltage regulated with all LEDs arranged in a parallel configuration. It is common to see LEDs arranged in series Parallel configuration because most manufactures are looking to reduce operational costs. If your concern is with efficiency then current regulation will do no better and resistors are worse. With efficiency in mind you should match all the LEDs and string them up with no regulation and no resistors.

  There is more than one way to skin a cat I suppose. My intentions here are not to offend anyone, but, perhaps, to shed some light(pun intended :0) on the devices we are working with, and my opinion on the subject as I understand it. If indeed regulation is required the most efficient option is to use A PWM controller. This is a two fold attack that will reduce the duty cycle of the LED(increase MTTF) and increase the working voltage of the Device requiring no kind of regulation.
« Last Edit: April 30, 2007, 04:19:20 PM by Rembrant »

commanda

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Re: Led master class; pt 2 of 3 (theory)
« Reply #17 on: April 30, 2007, 05:42:00 PM »
Luke,


You're the one that is confused.

So confused that I really don't know where to begin to further respond to this.


A proper led driver is current regulated with all leds in series.


Your complete mis-understanding of the operation of Led's is precisely why I wrote this series of articles.


To repeat what I wrote back at the very start of part one.



Led's are current operated devices. The voltage drop across them is nothing more than a practical nuisance.


Amanda

« Last Edit: April 30, 2007, 05:42:00 PM by commanda »

Rembrant

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Re: Led master class; pt 2 of 3 (theory)
« Reply #18 on: April 30, 2007, 07:24:48 PM »
Again, my intention is not to offend. If your logic was correct, you could power an LED from a supply of 0 volts at .6 amps and it would function without failure. This is not the case. It will emit no light with no pressure(voltage). Conversly, an LED is closer to a pure "watt" device with limits. In your case a limit of about 3volts at 20ma. exceed either of these parameters and you will reduce the MTTF. This is no secret. There is no mystery behind it. The silicon in your LEDs are rated for 3volts at .2amps, Period. Exceed this and you will reduce MTTF. Alternatively, If one LED is rated for 3volts at .2amps and powered by a supply of 3volts and .2amps and fails the part is defective and should be replaced.

 You can check these parameters by powering all of your LEDs with a regulated 3volt supply and if it draws more or less than .2 amps the part is defective and not to be used. Thus, irrelevant in this discussion.

  Of course I understand that no LED is perfect because of imperfect manufacturing processes. However, it is up to each individual to preform their own quality control measurements and decide whether the part falls within the intended parameters for the application. Another problem here is that everyone is concerned with mean voltage of the power source at 12 volts. When you should be concerned with the peak voltage of said power supply witch is indeed around 15volts not 12. NOTE!!! I am not saying that your method won't work, only fundamentally incorrect.
« Last Edit: April 30, 2007, 07:24:48 PM by Rembrant »

Rembrant

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Re: Led master class; pt 2 of 3 (theory)
« Reply #19 on: April 30, 2007, 07:52:22 PM »
since I can"t edit my reply, I will say this. I was wrong the LEDs you posses are .11 amps still the same theory applies. With no exceptions. Sorry.....
« Last Edit: April 30, 2007, 07:52:22 PM by Rembrant »

commanda

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Re: Led master class; pt 2 of 3 (theory)
« Reply #20 on: April 30, 2007, 07:54:28 PM »
It's MTBF, Mean Time Between Failure.


20mA is 0.02 amps.


.2 amps is 200mA.


Amanda

« Last Edit: April 30, 2007, 07:54:28 PM by commanda »

Rembrant

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Re: Led master class; pt 2 of 3 (theory)
« Reply #21 on: April 30, 2007, 08:08:17 PM »
^Ok my mathematics are incorrect. I'm sorry, I misplaced a decimal and transposed a digit. Again I apologize. I guess it's just not a good idea around here to contradict a long time poster even if members of the community are being misled. I won't make any further comments in this discussion, not to fear. Please continue. I would love to learn something I wasn't taught in the university(UWF) where I obtained my EE degree.

Lets just say the world of the internet is a strange place everyone is an expert.
« Last Edit: April 30, 2007, 08:08:17 PM by Rembrant »

SamoaPower

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Re: Led master class; pt 2 of 3 (theory)
« Reply #22 on: April 30, 2007, 11:20:32 PM »
It seems that the conventions taught at your university (UWF) are significantly different than those taught at mine (Stanford).


Certainly, "it takes two to tango". Since Mr. Ohm's law has yet to be repealed, to my knowledge, both voltage and current are factors in LED operation. However, the convention, as I understand it, is to identify the dominant parameter that controls the primary function of the device to be able to say whether it's a "current device" or a "voltage device".


A LED's primary function is to produce light. It's quite clear from the following representative graphs that current is the dominant parameter that determines light output, hence a "current device".





Voltage is only spec'd over a range of 1.5:1, where current covers two decades. It's essentially a near constant voltage device. It's obvious from this that current is the parameter to control.


"...even if members of the community are being misled."


I think you owe Amanda an apology. If anyone is leading the uninitiated astray, it's coming from comments like yours.


No response is required.

« Last Edit: April 30, 2007, 11:20:32 PM by SamoaPower »

TomW

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Re: Led master class; pt 2 of 3 (theory)
« Reply #23 on: May 01, 2007, 05:00:06 AM »
Samoa;


Plus, I always feel I am being "shined on" when someone insists on mentioning their supposed education and degrees. My experience is if you have to brag it up its probably because you are a poser.


I happen to have a couple degrees, too. USC and Cal Tech but I don't go around waving the sheepskin around to "prove" my point. Besides, that was long long ago and experience has taught me a lot more than the classes.


I was interested in this thread because I feel there is a lot of "unfact" around about using LEDs and I had hoped this might clear things up some.


As far as long term posters having more credibility than a new face: Duh, of course contributors get more credibility over time than a newbie waving its sheepskin and claiming expertise. Commanda has contributed a LOT to this forum over time. Sensible advice and shared circuits, to boot.


I really think we need to raise the bar to entry here. We are attracting entirely too many know it alls and experts with no social skills and often less correct knowledge.


Cheers.


TomW

« Last Edit: May 01, 2007, 05:00:06 AM by TomW »