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Denny
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This is more of a theoretical question than something I want/need to do, but would you get less/no flicker from strings of 1/2 wave LEDs if you removed the "rectifier" from the string and connected the string to the DC board -- CTB16D?

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Denny wrote:

This is more of a theoretical question than something I want/need to do, but would you get less/no flicker from strings of 1/2 wave LEDs if you removed the "rectifier" from the string and connected the string to the DC board -- CTB16D?

Most likely a set of Line Voltage stringer LEDs will not work with the CMB16D.. They will require high voltage.
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Richard Hamilton

Denny wrote:

This is more of a theoretical question than something I want/need to do, but would you get less/no flicker from strings of 1/2 wave LEDs if you removed the "rectifier" from the string and connected the string to the DC board -- CTB16D?


Even if LOR would work, you aren't going to lessen the flicker. As you may know, a diode by design is effectively a half-wave rectifier. That's where the origin comes from in electronic equipment.

Some strings only have LEDs as the only component and don't have rectifiers. The only way you are going to get rid of peripheral vision flicker is to have a "FULL wave" rectifier in the string. Just buy strings that have them built in (a little more expensive). They look better and you lessen the chance that you will need a "snubber" inline to have them fade properly.

I have seen some strings with capacitors in them which appear that they attempt to hold a charge for the 33 milliseconds that AC current is not flowing in the direction that lights the LEDSs. Of the 3 brands I've seen so far, they don't work that well.
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Denny wrote:

This is more of a theoretical question than something I want/need to do, but would you get less/no flicker from strings of 1/2 wave LEDs if you removed the "rectifier" from the string and connected the string to the DC board -- CTB16D?

The string of LED's wired for 110V will not work on the DC controller unless you reduce the number of LED's per string.

Each LED's (except some RED's) will be like adding a 3.5 volt mini. Also the current has to be controlled as LED's are current devices. Normal is around 20ma but can go up to 40 ma per string.

Now I mentioned (Except RED) some brands use white LED's with a Red Lens others use RED LED's, if they use RED LED's then the voltage drop will be 2.2 volts @ 20ma.

How can you tell? One way is look at all your different colored strings of 50, 70, 100.

If the strings are wired as 25 or 35 LED's in series then the LED's are 3.5 volts, but if the RED's or other colors (maybe orange, yellow, amber) are wired as 50 or 70 in series then they are 2.2 volts. Look for the tell tail three wire to two wire spots in the wiring as most LED's today are sealed and you cannot remove a LED to check how long the string will be.

This goes for both Half wave and full wave LED's strings

If you want to use these on the DC board, then you need to know this as you need to add a resistor (ugh) in series with the LED string to make sure you do not let to much current destroy the LED. Same thing using Mini's except the life of the bulbs will be shortened and the intensity will be brighter. LEDs will not work long if you over drive the LED above 40 ma normally. The intensity will not be noticeable if over driven, it just will stop working.

Now one thing about an LED when it goes bad is is does not completely short out, it will have a voltage drop somewhere around the 1.5 volt level but no light, this helps if one or several might go bad as the increase in voltage and current accross the other LEDs will not be as great.

To calculate the resistor you need to know the voltage you are powering the set of 8 channels on the DC board. The CB16D can handle up to 65 volts DC, that is higher than what I would recommend, 24 volts DC is better and easier to find power supplies at that voltage.

So you take your voltage minus how many LED's and this is the voltage you need to drop across a resistor.

24 - 21 (6 LED's @ 3.5V) = 3 Volts (That can be a mix of 3.5 and 2.2 LED's also)

Ohms law says I=E/R or R= E/I (in Amps)

3 volts/.020ma = 150 ohms.

Watts = volts * amps

3 volts * .02ma = .06 watts so a 1/4 watt resistor would work fine, always use try to double or more your watt size.

You can parallel strings on each channel if six (per my example) is not enough.
Just treat each string on that channel as a separate series string and calculate the resistor for each string.

Examples 1:

15 LED's on one channel.

15 / 3 = 5 LED's per string.

5 * 3.5 Volts= 17.5 Volts.

24V - 17.5V = 6.5V

6.5V / .02ma = 325 (use 330 ohm)

6.5V * .02ma = .13 watts (use 1/4 watt or 1/2 watt)

Example 2:

14 LED's on one channel:

14/3 = 4.66666 ( try to keep the strings as close to equal number as possible)

2 strings w/ 5 LEDs = 10 LEDs
1 string w/ 4 LEDs = 4 LED's

So we already calculated the first two series strings in example 1, we only need to figure out the third string. Lets add a 2.2V LED to this string.

3 @ 3.5V = 10.5V
1 @ 2.2V - 2.2V

10.5V + 2.2V = 12.7V

24V - 12.7V = 11.3V

11.3V / .02ma = 565 ohms (nearest resistor value is 560 ohms)

11.3V * .02ma = .226 watts (use 1/2 watt resistor)

If you want to learn more here is two good websites.

http://www.the12volt.com/ohm/ohmslaw.asp

http://www.logwell.com/tech/components/resistor_values.html

There is another way of doing this without calculating resistor values but that is for amother post.

I hope this helps.
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Richard Hamilton

Nice detail Dennis. There was some info in there that I was not aware. I've never taken the time to analyize LED voltage and current draw individually.

Just a note for other folks. A thought came to mind as I read the reply... Although we usually refer to line voltage as "110", in reality it a stanadarized "rms" voltage of 120 voltage in America and might vary from 117-120 in some areas. so your line voltage should be closer to that value. Thus, when calculating total current draw for a system, 120 vac is the value to use.

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Thanks for all the valuable input. Like I mentioned, I have no intention of doing this, but the responses give a lot of useful information for future reference. I have been concentrating on lights too much, I think. Woke up this morning and for some reason, I had been doing this in a dream. :D Just had to find out if it could be done! Man, I need to get a life or take a vacation!

Something Dennis said, about the two sets of 35 LEDs in a string. The other day, I bsically cut a set of 70 clear LEDs from Home Depot in half, giving me 35 LEDs per section. Each section has one of those "blister rectifiers" on it. One is at the male plug end and the other at the female plug end. I had never noticed this before. Would I be correct in assuming that this string is actually two independent strings that were sharing the same neutral return? After putting a male plug on the one string and attaching an independent neutral to each string, I plugged them in and they both lit up as individual 35 LED strings. Based on what Dennis said above, should I put a resistor in series on each string?

Why did I do this? Well, a couple years ago, I bought a set of lawn lights that never worked. I decided to experiment and cut all the sockets out of the lawn lights and replaced them with the LEDs mentioned above. This gave me two 35 LED sets of lawn lights vs. one 50 mini light set. I soldered each LED into the string. I was amazed when they both actually lit up. Incidentally, I also found out why the lawn lights did not work -- the wire was broken in three places. I think the two wires were twisted so tightly, it actually broke the wires. You couldn't see the breaks unless the wires were "untwisted" from each other as the breaks were on the inside.

Man, I need to get a life or take a vacation!!!!!:)

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Denny wrote:

Thanks for all the valuable input. Like I mentioned, I have no intention of doing this, but the responses give a lot of useful information for future reference. I have been concentrating on lights too much, I think. Woke up this morning and for some reason, I had been doing this in a dream. :D Just had to find out if it could be done! Man, I need to get a life or take a vacation!

Something Dennis said, about the two sets of 35 LEDs in a string. The other day, I bsically cut a set of 70 clear LEDs from Home Depot in half, giving me 35 LEDs per section. Each section has one of those "blister rectifiers" on it. One is at the male plug end and the other at the female plug end. I had never noticed this before. Would I be correct in assuming that this string is actually two independent strings that were sharing the same neutral return? After putting a male plug on the one string and attaching an independent neutral to each string, I plugged them in and they both lit up as individual 35 LED strings. Based on what Dennis said above, should I put a resistor in series on each string?

Why did I do this? Well, a couple years ago, I bought a set of lawn lights that never worked. I decided to experiment and cut all the sockets out of the lawn lights and replaced them with the LEDs mentioned above. This gave me two 35 LED sets of lawn lights vs. one 50 mini light set. I soldered each LED into the string. I was amazed when they both actually lit up. Incidentally, I also found out why the lawn lights did not work -- the wire was broken in three places. I think the two wires were twisted so tightly, it actually broke the wires. You couldn't see the breaks unless the wires were "untwisted" from each other as the breaks were on the inside.

Man, I need to get a life or take a vacation!!!!!:)

Yes, the 70 light strings are actually two 35 light strings. (35 Lights * 3.5V = 122.5Volts

By cutting the string in the middle (the 2 wire twisted section) you have two strings and only need to put a male plug on the two cut wires of the second string. If you do not need to have a female plug on the first string just cut the two wires at the base of the last socket on the first string and put some shrink tubing over the loose cut wire, this will give you 4 inches of wire to connect your plug to the second string. I hope this was explained correctly for you.

No additional wires or resistors are needed to make both string light, just add the plug to the second string.
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Thanks Dennis, that is the way I did it. By the way, are you going to set that terrific G-Gauge train up again this year?

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For anyone having issues with LED type lights, I have found by putting a regular night light in the circuit seems to solve a lot of issues. My LEDS dim and brighten just as smooth as regular lights with the night light in the circuit. No need to solder or modify anything! It seems the extra load by the one light seems to have resolved all of the issues I have had with any LEDS so far.

Hope this helps!

Sorry I just noticed you were asking about DC connections, this is for AC connections only!

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Richard Hamilton

hi Dinosang,

Yup we are talking about two different things, but your remedy does work for getting leds to ramp smoothly. There are other details threads about this topic elsewhare on the board and I have a detail solution on my own board as well.

We don't like to use night lights because it spoils the effect. We use "snubbers" instead and they are very easy to make. The LEDs we bought this year won't need any of that stuff.

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Denny wrote:

Thanks Dennis, that is the way I did it. By the way, are you going to set that terrific G-Gauge train up again this year?

They want me back again this year.

Thinking about it. Its a lot of work.

For some of you here is a short video of one train display.

http:// http://www.vimeo.com/user343675/videos/tag:gscale

Dennis
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