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chuckd

Why not do full wave in the box?

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I'm sure everybody out there would like to use LED lighting, but it's quite expensive as we all know. Further, the full-wave LED's (with/without) resistor seem to be much better to control with LOR and other boxes. However, the full wave LED's are EXTREMELY expensive!!!

So, it got me thinking... why not do the full wave circuitry before the LED strand, and therefore convert the entire strand to full wave? This would be quite easy to do in any of the LOR boxes (new circuit cards, but still easy to do). Then we can all buy half wave LED's and run them with the benefits of full wave, at 1/5 the cost!

OR, simply add a simple bridge rectifier between the output and the strings. Wouldn't this convert a half wave strand to full wave, at a very, very small cost per strand (less than $1).

Moronic, wishful thinking, or a good idea?

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I asked about adding a bridge rectifier to a half-wave string on PlanetChristmas. Of course PC's search never works properly and I can't find it now.

About half the people said it would work, and half said it wouldn't, as I recall...

-Tim

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Well, I'm not sure, but there isn't any extra circuitry in a normal half-wave LED string, is there? Simply, the LED's are turning on for the positive half of the cycle, and not the negative half. This of course is why we can actually see the blinking on and off of a half wave string.

I think I'm going to have to try it and find out. Even being lazy and buying a whole bridge rectifier (instead of discrete diodes) I'll only spend less than 50 cents trying it out.

I'm real curious about why some of the people responded that it wouldn't work. Obviously, if you plugged a full wave string into this circuit with the wrong polarity, it wouldn't work at all, but this obviously isn't for that purpose.

It would probably be best to implement it externally, with an included bleed resistor as well.

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I wish I could find the thread (I guess the only answer is to keep bugging Chuck about getting (and keeping) search working).

Some people thought that the different halves of the string had the diodes facing opposite 'directions'. Others thought the rectified voltage would be off, blowing up the diodes. I'm a software guy, not hardware so I have no idea really...

-Tim

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Well, if half the string is in fact wired in the opposite direction, then it wouldn't work. All it would take is connecting up a DC source and seeing if only half the string lit up. A bridge is not going to blow up the LED's, because they're faced with a reverse polarity anyway each half cycle already. They're made to withstand that much reverse voltage.

Even if it's all wired the same direction (hopefully), you'd have to be careful which way the plug was plugged in with a full wave rectifier. Otherwise, everything will stay off completely.

I'll probably connect up a DC source first and check it out before I build anything.

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I like the idea and I can't think of any reason why it would not work.

I don't buy the comment from whoever told you that it would blow out the diodes if connected wrong. I agree with you that the stran simply won't work if connected wrong. They are already being subjected to a back biased 120 VAC during part of the AC waveform, so blowing them out isn't something that should happen.

My only concern is that if you are experimenting, leave the LOR out of the experiment for risk of blowing out a channel if you build the full wave rectifier wrongly.

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I don't have any 200V diodes on hand, so I'm going to make it really easy and simply order a 200V bridge rectifier from Digikey and give it a shot. Today I hooked up my DC power supply to see if half the strand would light up, but my power supply only goes up to 30 VDC, so nothing lit at all no matter which way I connected it up. Probably not enough juice, I'd guess.

I'll let you guys know what happens.

By the way, I'm using the Diogen brand of LED lights that have normal plug connectors. They are definitely half wave LED's.

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Any LED strings that use capacitive current limiting won't work after a full wave rectifier, as you then no longer have the alternating voltage that pushes the charge in and out of the cap, and through the string each cycle..

For half wave resistive limited strings, the string will light, but you will be running the string at about twice the power it was designed to handle.. After all, the string was designed to be on half the time, and off half the time. Adding a full wave rectifier up front causes it to be on all the time, and thus have twice as much apparent brightness, and twice as much heat to dissipate from the tiny LED junction... Some strings might have the design reserve to survive this, others may not... Even if they survive, you have cut into the designed reserve tolerance of the string, and substantially increased the odds of failure.

In theory, if you doubled the resistance of the current limiting resistor, you should be fine running full wave.. However, you can't measure the current limiting resistor without isolating it from the string of LED's. You could use a true RMS hard wired meter to measure the current on half wave, then insert different sized resistors in series with the full wave configuration until you achieved the same RMS current...

The reason that you will always have a current limiting device in a LED string has to do with how LED's behave as compared to incandescent lights. Once an incandescent light is over 10% brightness, it is reasonably close to a linear load. The deviation from linear is that the hotter it gets the higher the resistance.. For a true linear load, 100% of rated voltage would be 100% rated current, 90 would map to 90, etc.. With the thermal coefficient, 90% power is probably 90.5% of rated current, 110% voltage is probably around 109.5% brightness.. In any case, small variations in voltage applied result in small differences in brightness, and power to dissipate...

With LED's, (without considering current limiting) the stated/rated voltage is the voltage across the LED at the rated current. The voltage to current curve is extremely non linear. From zero volts to about 3/4 of the rated voltage current climbs with voltage, but it will be a tiny percentage of the rated current. At some point below normal operating current, the current starts growing exponentially for the voltage applied, to the point where a 10% variation in voltage (which is totally in spec for US line power) may result in a 10 fold increase in current through the LED, and a LED failure... This non linear voltage to current relation on LED's is why there will always be some sort of current limiting device in a LED string. Inserting a linear load in series, like a resistor is one way to do it. Using a capacitor to limit how much charge flows on each cycle is another way to do it.. In theory, an active current regulating circuit could be used, but I have not heard of any being identified in commercial strings yet, and I don't expect it to ever be cost effective...

- Kevin

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First off, we're talking only the simple, cheap half wave LED strings here. The whole point is to get full wave without the full wave cost.

There's two points of power capacity in a LED string, the first being the wire, and the second being the LED's themselves. IF the designer of the string is running the LED at such a high current that it has to be turned off half the time just to be within tolerance, then they have a very poor design indeed. Of course, I'm just guessing at this point, but I can't imagine a designer implementing it this way. If the full wave current is still in tolerance, then the MTBF of the bulb will not be affected significantly at all.

Also, the current versus brightness curve is definitely not linear, so it seems silly to run the LED 'half wave hot' just to squeak out a little extra brightness. For example, running a standard LED at 10ma or 20ma; I'd challenge anybody to tell much of a difference in brightness.

Second, the wire is definitely rated to string quite a few strands together. So to be safe, I guess you could divide the number of suggested end-to-end strands in half.

Just a side not here: I just tested the theory, and there is preliminary good news and bad news here. The good news is that for my 35 light strand of Diogen LED's, it worked perfectly. The jitter is gone, and I have a full wave set that does appear brighter. It also dims quite nicely with LOR too! Unfortunately, in my 70 light strand, half the LED's are in fact wired in reverse, so only half the set lights up. I have no idea if other brands of LED's are wired in this fashion or not.

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chuckd wrote:

Unfortunately, in my 70 light strand, half the LED's are in fact wired in reverse, so only half the set lights up. I have no idea if other brands of LED's are wired in this fashion or not.


Shouldn't that be a simple "snip 2 wires in the middle and reverse" fix?

-Tim

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I like to think of it as a "snip it and flip it" fix! :)

Now, I can buy 70 light C6 strings for $23.50 (blue), and I can buy 35 light C6 strings for $14.25 each (two strands would make 70 lights for $28.25). So basically, is it worth a savings of $4.75 per strand to get out the soldering iron and go to town? Hmmm.

What I do know is that I've seen "full wave" lights, 70 lights, for $99!!!! Yikes!

Even if you're lazy and use a bridge rectifier instead of diodes, the rectifier costs about $0.50 apiece.

Now a question.... where in the heck is the resistor(s) located in an LED strand. I thought it might be in the base of each bulb, but it doesn't seem to be in my Diogen light strands. Hmmmm. I'm wanting to investigate the individual bulb current draw a bit further before I feel totally warm and fuzzy about this.

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Now I want to investigate whether or not I'm risking burning something out. Maybe klb can help out here (or anybody else who happens to be a electrical geek like me).

My 70 count Diogen light strings draw 3.6 watts in their 'out of the box' form, meaning only half the LED's are on at any given time. Each LED unit then has to dissipate 3.6/70, or about .05 watts of power (50 mw).

My new rectified strand should draw about 7.2 watts of power, since all the bulbs are now on during each half cycle. (Remember, the RMS voltage/current of a fully rectified sine wave is the same as a regular sine wave).

This means that each LED unit of my fully rectified strand should now dissipate about 1/10 of a watt of power. This is 100 mw.

A lot depends on the location of the current limiting resistor here. IF it's in the bulb unit, then it will share the load with the LED, and a high output LED can handle somewhat less than 100 mw just fine.

IF there's one per string (or one per half of each string), then I'm still dissipating power in the resistors, and the LED is handling less than 100 mw. 'Should' be fine, but without knowing specifics of the string, I'm not sure.

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The resistor will be in series with some number of LEDs. From the sounds of it, probably in groups of 35. Different brands and different colors will have different junction voltages.. If it is a 2 volts, a 35 LED string is 70 volts, leaving about 45 volts dissipated across the limiting resistor.. Using rough numbers, about 1/3rd of the power is going into the resistor... I suppose you could measure the voltage across 10 of the LEDs to give a good idea of the actual junction voltage and work it out from there..

Another way to look at it is the current instead of the power.. From what I recall, most LEDs are rated for maximum operating current, not maximum power draw... 3.6W would be about 31mA for the string of seventy, but each half string is only seeing half of that, or about 15mA... Unfortunately the number that comes to mind is for LEDs used for indicator lamps, not lighting, and that is typically 20mA... Not sure what the design max current will be for lighting...

Also, before I forget, there is a reason that the two halves of the string are reversed to each other.. If you run a load that only draws power on one half cycle, it will to some extent impact the transformer feeding your house. As the current goes back and forth, the magnetic flux cycles back and forth.. When the current only goes one way, the magnetic flux keeps building up one way, until it saturates the armature of the transformer. At this point the transformer starts to rebalance the flux by converting the imbalanced magnetic flux to heat... For small imbalances like short strings of LED's, it probably doesn't matter much... If you had a bunch of half wave LEDs on the same circuit, and they did not alternate the halves of the strings, it could be an issue...

BTW, when looking at the pricing be sure you know the units.. One place I have seen full wave LED's in strings of 50, the pricing is depending on color is from about $100 to $150, but that is per 12 strings of 50...

- Kevin

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I'm the same way.... used to small circuit board LED's that are designed to draw 20 ma. With a 5V DC supply and a 2.7 or 3.3K resistor, you're pretty much in business.

Checking on some of the high output LED's, I see that they often draw 60 ma on upwards. 100 ma on up is not uncommon. I went with power numbers because from what I know, it's power dissipation that ultimately 'melts' an LED. Of course, this is a function of the junction voltage and current drawn.

I guess if you modify a 70 count string to work with full wave rectification, then you probably don't want to use it without the rectifier. I find it a bit odd that Diogen's 35 count strings are all polarized the same way. Probably a unit assembly that is simply tied together for a 70 count string, I guess.

The place I saw full wave strings for $99 (blue, I think), seemed pretty clear it was for a unit of 1. I sure hope that it was for 12 strings. If that's the case, then I'm probably wasting my time on this little project! :)

Thanks for all the input.

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Why not use 5 watt power resistors across each channel and wire them internally via the terminal strip in the controller?

I am currently using a 4 watt night light plugged into a 3 way tap at each of the channels that have "non dimming" LED strings attached and it works fine.
I'm not really happy taking the chance that if a bulb burns out and goes unnoticed that I'll toast my LED's but it will get me by untill I either get LED's with resistors or full wave circuits or an alternative such as yours proves to be effective.

The controller needs to have a resistive load in order to reduce the voltage output.
Adding the night light achieves this simply and effectively.

The trade off of the power reisitor would be 5 watts and heat?

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