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Offline Caplin  
#1 Posted : 23 April 2007 21:25:13(UTC)
Caplin


Joined: 23/03/2005(UTC)
Posts: 2,497
Location: Denmark

Hi,

I bought a Viesmann #3556 Leds kit (three leds for 14-16 volts + three resistors each 1.8 kOhm).
If I am to connect the three leds in series, what is the formular for calculating the single resistor to be used in the cathode end for the same amount of lux per led confused
Regards,
Benny - Outsider and MFDWPL

UserPostedImage
Offline pa-pauls  
#2 Posted : 23 April 2007 22:24:30(UTC)
pa-pauls


Joined: 08/06/2002(UTC)
Posts: 1,812
Location: Norway
Hi Benny

U (Volt) = I (amp) * R (resistor)

So in your case R (resistor) = U (Volt) / I (amp)

Pål Paulsen
Märklin Spur 1 Digital, epoche 3
Offline alonso231gery  
#3 Posted : 23 April 2007 22:34:56(UTC)
alonso231gery

Greece   
Joined: 24/08/2002(UTC)
Posts: 3,945
Location: Hellas (Athens)
Ohm's law.
An outsider.
I'm looking for the owner of that horse. He's tall, blonde, he smokes a cigar, and he's a pig!
Offline Mikael  
#4 Posted : 23 April 2007 22:44:09(UTC)
Mikael

Denmark   
Joined: 10/09/2004(UTC)
Posts: 953
Location: Copenhagen, Denmark
It depends on the leds. It is white leds, which actually are blue ones. There is some variation among white and blue leds, but I will assume they are 3.5V each. So one led with 1.8kOhm and 16V supply will give (16V - 3.5V) / 1800 Ohm = 6.94 mA.
So the resistor for three leds in series is: (16V - (3 * 3.5V)) / 6.94mA = 792 Ohm. The nearest standard resistor value is 820 Ohm.
Offline Caplin  
#5 Posted : 23 April 2007 22:51:31(UTC)
Caplin


Joined: 23/03/2005(UTC)
Posts: 2,497
Location: Denmark

Hi again,

Thank you for your promp replies. Smile
Regards,
Benny - Outsider and MFDWPL

UserPostedImage
Offline pa-pauls  
#6 Posted : 23 April 2007 22:58:49(UTC)
pa-pauls


Joined: 08/06/2002(UTC)
Posts: 1,812
Location: Norway
Yes Nikos !
Pål Paulsen
Märklin Spur 1 Digital, epoche 3
Offline TTRExpress  
#7 Posted : 24 April 2007 04:28:23(UTC)
TTRExpress

United States   
Joined: 06/04/2006(UTC)
Posts: 655
Hi Benny,

Since you will have all 3 LEDs in series the light intensity (lx, lux, or illuminance) will <u>not</u> be the same for each LED. If you want the intensity of the LEDs to be the same you will want them to be connected in parallel. That way you will not need to worry about a single resistor to reduce the power to the LEDs to give the same intensity.

The voltage and current will be the same for each LED (resistor). The equivalent resistance will be 1/R(eq) = 1/R1 + 1/R2 + 1/R3 = 1/1800 + 1/1800 +1/1800 = 1.66 mOhm

You may want to refer to the following:
http://www.physics.uogue....ohm.intro.parallel.html

Series circuits are also described.

Hope this helps.

Regards (a Scot in Wisconsin)
Regards (a Scot in Wisconsin),

Maurice [ETE, TTRCA, IG-TRIX Express, Maerklin-Insider & TRIX Profi-Club]
Offline Lars Westerlind  
#8 Posted : 24 April 2007 09:25:08(UTC)
Lars Westerlind


Joined: 19/10/2001(UTC)
Posts: 2,379
Location: Lindome, Sweden
What do you mean Maurice? Those three diodes are equal, meaning that the same current through them will give the same light intensity, if connected in series. If you connect them in parallell it would work, but you would have a higher risk of variations, and the way you suggest very large ones. I guess it's because of contact resistances, but nevertheless, the way you point out you'll probably have almost all current through one or two of the LEDs, and nothing to the third. From experience, trying to be smart :-(

You could connect them in parallell, if each diode also keeps it's orignal resistor in series with it's diode. In this case variations in resistance will also influence the brightness, but far less. The pure series schema Benny suggested seems by far the best to me.

/Lars

PS. Your calculations seems rather strange to me too. 1/R measured in mOhm? of course it should be 1/kOhm. And R is easily calculated to 1/3 of 1k8 =&gt; 600 ohm without any formula.
/Lars
Offline DaleSchultz  
#9 Posted : 24 April 2007 17:05:24(UTC)
DaleSchultz

United States   
Joined: 10/02/2006(UTC)
Posts: 3,470
Rules of thumb.
Never connect more than one LED in parallel without each one having its own resistor. (slight variations will cause the neighbour to fail)
Connecting them in series is doable only if the supply voltage is sufficient to drive them all.
Dale
Intellibox + own software, K-Track
My current layout: https://cabin-layout.mixmox.com
Arrival and Departure signs: https://remotesign.mixmox.com
Offline Caplin  
#10 Posted : 24 April 2007 18:24:02(UTC)
Caplin


Joined: 23/03/2005(UTC)
Posts: 2,497
Location: Denmark
Thank you all for replies.

Actually I copied Sveins (intruder's reply 2007/04/10) circuit board for the triangel setup of three leds except my leds are in series and 2 x 1k8ohm in parallel (only R's on hand) and it works with a mains plug-in power pack with 16 vac 2 amps out. Some flickering occurs as expected.

Unfortunately, when I connected it to the yellow and orange wires of the decoder (rear lights) it did not illuminate! I had tried before with just one led and resistor connected to the decoder and it was ok then.

UserPostedImage UserPostedImage

Back on the mains plug-in power pack it still worked. It appears that I have blown the decoder rear light circiut! [xx(] The directions and sound circuits are still working ok.

Please note that the cold white glow is dampened by the use of two layers cut from an ordinary plastic chateque for paper.

UserPostedImage UserPostedImage

I still have AUX2 circiut from the decoder unused (AUX1 is reserved for the smoke gen.).
I wonder if it can be programmed to handle the rear lights!



Regards,
Benny - Outsider and MFDWPL

UserPostedImage
Offline Lars Westerlind  
#11 Posted : 24 April 2007 21:30:54(UTC)
Lars Westerlind


Joined: 19/10/2001(UTC)
Posts: 2,379
Location: Lindome, Sweden
have to ask,
did you try the LEDs in correct direction? The function output of the decoder is always of LOWEST potential; Common return wire is positive!
As it works in analog, you seem not to have turned any of the LEDs at least.

/Lars
Offline Caplin  
#12 Posted : 25 April 2007 00:16:34(UTC)
Caplin


Joined: 23/03/2005(UTC)
Posts: 2,497
Location: Denmark
Quote:
[size=1" face="Verdana" id="quote]quote:Originally posted by Lars Westerlind
<br />have to ask,
did you try the LEDs in correct direction? The function output of the decoder is always of LOWEST potential; Common return wire is positive!
As it works in analog, you seem not to have turned any of the LEDs at least.

/Lars
Hi Lars,

I am HAPPY that you asked. When I swapped the yellow and orange wires it works! [:I][:I]

Apparently I took orange (function ground) for being lowest potential by heart. Shows the importance of reading the instructions very careful the first time when one is trying new stuff!! That was an eye opener! Happy that no damage was done.

Regards,
Benny - Outsider and MFDWPL

UserPostedImage
Offline intruder  
#13 Posted : 25 April 2007 00:37:01(UTC)
intruder

Norway   
Joined: 16/08/2006(UTC)
Posts: 5,532
Location: Akershus, Norway
You are not the first (and definately not the last) to forget about the common plus, Benny.
I'm glad you made it work.

We have seen examples on calculation of resistors in parallell. This is "my" easy way, if only two resistors (several calculations are required if more than two):

(R1*R2)/(R1+R2)

As you know, two identical resistors gives half the resistans, 3 gives 1/3, 4 gives 1/4 and so on. If two or more different resistors, your total resistans is allways lower than the lowest resistor.
Best regards Svein, Norway
grumpy old sod
Offline Lars Westerlind  
#14 Posted : 25 April 2007 00:50:19(UTC)
Lars Westerlind


Joined: 19/10/2001(UTC)
Posts: 2,379
Location: Lindome, Sweden
:-)
Good to hear Benny.
/Lars
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