 Joined: 25/08/2008(UTC) Posts: 633 Location: Montlouis sur Loire, France
|
Hi all,
I've just converted to digital a 3021 and will replace the original bulbs by LEDs.
Will carry out a test with 3mm yellow LEDs. I wonder what is the best way to connect the leds: in series (resistor-led-led) or in parallel (each led with its resistor).
For connecting them in series, had thought about using a 800 ohm resistor, while the value for parallel connection should be 900 ohm.
Is this OK? |
|
|
|
|
|
|
 Joined: 08/01/2006(UTC) Posts: 2,218 Location: Shady Shores, TX - USA
|
marklin has there surface mount boards in series 3 yellow/white led's circuit board should have about a 1k ohm resistor for the red lights and a 780ohm for the 3 front lights. see if this link helps https://www.marklin-user...98&SearchTerms=37090 |
DT
Now powered by ECoS II unit#2, RocRail
era - some time in the future when the space time continuum is disrupted and ICE 3 Trains run on the same rails as the Adler and BR18's. |
|
|
|
|
|
Joined: 10/02/2006(UTC) Posts: 3,997
|
resistor - led - led will consume less current, but the voltage will need to be sufficient to drive both - about 6V or more for two white LEDs. You will have plenty of voltage so I would go serial. for calculating the resistor we need to know the forward voltage of the LEDs and the supply voltage and how much current you want to run them at. I suggest 16mA or less. I have a table at http://layout.mixmox.com/1/Resistors for 12V DC supply and white LEDs that use 3V each. |
|
|
|
|
|
|
 Joined: 16/08/2006(UTC) Posts: 5,382 Location: Akershus, Norway
|
I normally use a somewhat bigger resistor than actually required, as it doesn't have a great influence one the brigthness, and reduces the current draw.
For normal LEDs (2 Volt) I use a 1,2 KOhm resistor. That should give 600 ohm if two LEDs are connected in series, 400 ohm with three LEDs and 300 ohm with four.
For the bright white LEDs in my camera car I used 1000 ohm per LED, LEDs connected in parallell. Two LEDs in series gives 500 ohm, three 333 ohm, four 250 etc.
The only disadvantage I can see with LEDs connected in series is that if one LED is broken (very rare) all the LEDs go dark. Trouble shooting is required to find the defect one.
When LEDs are connected in series, make sure that all the LEDs are identical, otherwise their brighness may be unequal.
Edit: added LEDs to avoid misunderstanding. |
Best regards Svein, Norway
grumpy old sod
|
|
|
|
|
|
 Joined: 12/01/2002(UTC) Posts: 2,578 Location: Sweden
|
Quote:[size=1" face="Verdana" id="quote]quote:Originally posted by intruder
<br />
For normal LEDs (2 Volt) I use a 1,2 KOhm resistor. That should give 600 ohm if two are connected in series, 400 ohm with three and 300 ohm with four.
If you mean 600 Ohm for two LEDs etc. it is not correct. But maybe you mean that you can connect two 600 Ohm resistors in series to get 1200 Ohm ?
|
|
|
|
|
|
Joined: 16/08/2006(UTC) Posts: 5,382 Location: Akershus, Norway
|
Sorry, Per.
I did not explain myself good enough.
You are correct. I meant one 600 Ohm resistor if two LEDs are connected in series and so on. |
Best regards Svein, Norway
grumpy old sod
|
|
|
|
|
|
Joined: 12/01/2002(UTC) Posts: 2,578 Location: Sweden
|
Quote:[size=1" face="Verdana" id="quote]quote:Originally posted by intruder
<br />Sorry, Per.
I did not explain myself good enough.
You are correct. I meant one 600 Ohm resistor if two LEDs are connected in series and so on.
Your clarification avoids misunderstanding. But the calculations you propose are incorrect. For one normal LED you can use a 1.2 KOhm resistor, if it is used in a digital lok where the total voltage is 20 V. 2 V is over the LED and 18 V over the resistor. This gives you 18/1.2 mA = 15 mA of current, which is reasonable. If you have two LEDs in series, you get 2 x 2 V = 4 V over the LEDs and 20 V - 4 V = 16 V over the resistor. To get the same current and thus the same brightness you should then have 1.2 x 16/18 kOhm = 1.067 kOhm. With three LEDs you get 2 x 3 V = 6 V over the LEDs and 20 V - 6 V = 14 V over the resistor. The resistor should then be 1.2 x 14/18 kOhm = 0.933 kOhm And so on. Note: 2 V over the LED is only an example. It depends on what type of LED it is.
|
|
|
|
|
|
Forum Jump
You cannot post new topics in this forum.
You cannot reply to topics in this forum.
You cannot delete your posts in this forum.
You cannot edit your posts in this forum.
You cannot create polls in this forum.
You cannot vote in polls in this forum.
