Author Topic: Led master class; pt 3  (Read 6115 times)

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commanda

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Led master class; pt 3
« on: May 01, 2007, 07:48:56 AM »
This is a continuation of part 2 which can be found here:

http://www.fieldlines.com/story/2007/4/26/82929/6080


Which was itself a continuation of part 1 which can be found here:

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


In part 3, I will build a prototype of circuit 4, throw in a few component values, and at least one method of matching the leds for the parallel strings.


In part 2, we saw the necessity to match the combined Vf in each string if we want to use one current source to power multiple strings of leds. With the best of intentions, I figured that for a string of 3 leds, I could divide my batch of 100 leds into 3 piles; those where Vf was less than avg - 1sd, those where Vf was greater than avg + 1sd, and all those inbetween. According to textbook stats, I should have about two thirds of my leds in the middle pile, with the rest split evenly between the other 2 piles. My initial plan was to use one led from each pile to make each string of 3 leds. But Murphy was going to have none of that, and a change in room temperature from one day to the next meant I had 4 leds in the lowest pile, 39 leds in the highest pile, and 56 leds in the inbetween pile.


So I decided to build my sample using only leds from the inbetween pile. What I wanted was 2 strings of 3 leds. Now this is not the only way to achieve this end result; I don't even claim that this is the best way; it was simply how I chose to do it. I set up my breadboard with circuit 4,





set Rs at 47 ohms to give me a total current of 11.00 mA, set Req at 47 ohms for no particular reason, and picked 6 leds at random from the middle pile to populate it. I connected a voltmeter to the lower leg of each of these equalisation resistors. If the current through each string of leds was exactly the same, the meter would show a voltage differential of zero. By swapping pairs of leds between each string, I managed to get the differential voltage down to several millivolts. I then changed Rs to 15 ohms to increase the current source to 35mA, and used 2 meters to measure the current in each led string, and was satisfied to find that the current in each string was 17.x milliamps, with a difference of approx 0.3mA. I also varied the supply voltage from 11.0 to 30.0 volts, and was satisfied to see the current change from 34 mA to 36 mA, with absolutely no discernable change in light output.


This technique of swapping leds can be expanded beyond 2 strings, with a little imagination.


Just for interests sake, I changed Req to zero, and found that the difference in the 2 currents was less than 1 mA.


I then did a pencil layout of the components to fit it on veroboard, and built it, being careful to keep each of the 2 strings of leds together. I mounted the two equalisation resistors (Req) vertically, and in such a manner that I could use the lower legs as connection points for my voltmeter, as a final check, and confirmed the voltage differential of less than 10mV.

















As a final check, I waved the hot-air gun at it whilst monitoring the overall current drain, to confirm the negative temperature coefficient characteristics. This is the same test I use for the 24 bit ADC's on our commercial beltweighers. Whilst not as fine grained a test as firing up the environmental chamber, it is at least a useful test, and confirmed the negative temperature coefficient, with the current dropping by several milliamps.


It would be very interesting to repeat this test on the simple single resistor setup.


In part one, I presented a number of circuits. The first 3 are interesting as an aid to understanding how we finish up at circuit 4. But circuit 4 is far superior in terms of stability, particularly thermal, with its small but very important negative thermal coefficient.


Component values for circuit 4.


from Ohm's law, R = E/I

for Rs, the voltage will be the Vbe of the transistor. At the extremely low current levels we're working at, will be a little less than the typically quoted 0.6 volt. I measured 0.525 on my prototype.

So, for 35mA, Rs = 0.525 / 0.035  = 15 ohms


Rg, is not critical. As a rule of thumb, I use 1M.


Req, again not critical. Depends on how well you can match the strings of leds, and also partially on how much headroom you give yourself between design current and maximum continuous current. If using one current source per led string, can be omitted completely.


Number of parallel strings and failure mode.

It has been my experience that;



  1. : when properly fed, leds don't normally fail.
  2. : on those rare occassions when leds do fail, they fail to open circuit.


The upshot of this is that running only 2 strings in parallel off one current source is not that good an idea. If one led fails to open circuit, that string will draw no current; and the other string will draw all the current. In the presented prototype, failure of one led would probably result in destruction of the 3 leds in the other string. Good thing leds don't normally fail very often. Once one got past 3 or 4 parallel strings, it would be possible to leave enough headroom between design current and max continuous current, that failure of one string wouldn't push the other strings beyond their maximum current. But that's one of the things that has to be weighed up on a case by case basis;  whether the small savings in component count is worth the risk versus using a seperate current source for each string. There is no one perfect solution for all applications.


I know this series started out as a two part series, and this is already part 3; but in part 4 (which will probably be a diary entry) I will hopefully present a slightly more ambitious led light. Running off 24 volts, multiple strings each of 6 leds with individual current sources, built on veroboard, stacked, with the leds mounted edge on, stuffed inside a small jiffy box. Just need a gooseneck or articulated arm and I'll call it a desklamp.


Amanda

« Last Edit: May 01, 2007, 07:48:56 AM by (unknown) »

BigBreaker

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Re: Led master class; pt 3
« Reply #1 on: May 01, 2007, 10:19:53 AM »
Sorry, posted this on part 1 by accident:


Amanda - any thought to adapting this circuit for a wind turbine driven battery charging?


We can think about the battery as a current sink.  While the voltage across a battery only changes a little, the output voltage of a wind turbine changes a lot.  The driver circuit raises "ground" for the battery so that the difference between the turbine voltage and the "top" of the driver is 13ish volts - always - because the current is limited to acceptable level.


The "direct connect" method of charging a battery from the rectifiers of a wind turbine tends to stall the machine as the current rises too fast.  The voltage difference between the turbine to ground and the battery to ground is too high.  Current is determined by the resistance in the line and the change in voltage from the turbine to the battery.


Putting your circuit between the battery and the turbine would hold down the current at lower RPM and allow the rotor to speed up.  The next step would be to use the dimmer version to modulate the current according to turbine RPM, MPPT-lite.


A "cut off" is needed for when the battery voltage rises above the ref level and push the current into some heaters.  You wouldn't want the mill to freespin in open circuit.

« Last Edit: May 01, 2007, 10:19:53 AM by BigBreaker »

jimjjnn

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Re: Led master class; pt 3
« Reply #2 on: May 01, 2007, 11:07:37 AM »
I think Amanda is just providing theory in his LED postings. It may not apply to any particular power source, wether wind,solar or any other power source.

« Last Edit: May 01, 2007, 11:07:37 AM by jimjjnn »

Flux

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Re: Led master class; pt 3
« Reply #3 on: May 01, 2007, 11:13:34 AM »
I agree with Jim, this is too far off subject to discuss here. Your final bit makes sense, there is no future in a linear constant current regulator for a wind generator but the basic idea is there.


Flux

« Last Edit: May 01, 2007, 11:13:34 AM by Flux »

BigBreaker

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Re: Led master class; pt 3
« Reply #4 on: May 01, 2007, 12:30:22 PM »
A bit off topic, appologies.


I studied circuits in uni and think about them more abstractly.  Perhaps I'll just say that current sources are useful and leave it at that.

« Last Edit: May 01, 2007, 12:30:22 PM by BigBreaker »

ghurd

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Re: Led master class; pt 3
« Reply #5 on: May 01, 2007, 03:14:46 PM »
Again I am happy.  

This time I am happy to read 17 to 18ma has "absolutely no discernable change in light output".

There is no reason to use LEDs past 17ma.


I hope this will keep DIY people from making a 20 or 25ma per LED light, because that's what the data supplied says.

I doubt it, though I do have hope.

G-

« Last Edit: May 01, 2007, 03:14:46 PM by ghurd »
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mtbandy

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Re: Led master class; pt 3
« Reply #6 on: May 02, 2007, 11:02:40 AM »
That's interesting, there must be a 'break even' point then with current draw vs light output, where it is more efficient in terms of lumens/watt to use more LEDs together rather than supply more current to just a few (ignoring LED lifetime here). Has anyone come across an efficiency vs current graph or similar, by any chance, for white LEDs? Maybe it could be measured by the heating of the semiconductor junction or something...
« Last Edit: May 02, 2007, 11:02:40 AM by mtbandy »

ghurd

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Re: Led master class; pt 3
« Reply #7 on: May 02, 2007, 01:55:27 PM »
Last I checked, Max L/W was about 1 or 1.5ma for standard 5mm white LEDs.

Not quite cost effective at the time.  "Cost Effective" changes quick.

The less the ma, the longer the life.
« Last Edit: May 02, 2007, 01:55:27 PM by ghurd »
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s4w2099

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Re: Led master class; pt 3
« Reply #8 on: May 05, 2007, 09:39:19 PM »
Will it not work simply using an LM317 to regulate the voltage and use a standard series resistor with the LEDs?
« Last Edit: May 05, 2007, 09:39:19 PM by s4w2099 »

commanda

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Re: Led master class; pt 3
« Reply #9 on: May 05, 2007, 11:14:59 PM »
Yes it will work.


There were numerous comments about this back in part 1.


Amanda

« Last Edit: May 05, 2007, 11:14:59 PM by commanda »

s4w2099

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Re: Led master class; pt 3
« Reply #10 on: May 06, 2007, 04:17:25 PM »
Ohh ok, I did not see that. Thanks a lot. Still using the resistor is a waste of energy.
« Last Edit: May 06, 2007, 04:17:25 PM by s4w2099 »

commanda

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Re: Led master class; pt 3
« Reply #11 on: May 06, 2007, 06:14:14 PM »
If you're concerned about wasting energy, you need to go straight to a dc-dc converter.

And you still won't get 100% efficiency.


Circuit 4 with the fet will give the best results consistent with simplicity and minimal parts count. It will also "waste" less energy than using a voltage regulator followed by resistors.


Amanda

« Last Edit: May 06, 2007, 06:14:14 PM by commanda »