Embarrassing question. What resistance to make this lamp/led brighter?

MisterBill2

Joined Jan 23, 2018
11,640
The TS wants to connect it to 12 volts. A.K.A. possibly an automobile. In which case, running a 6V LED on 12V will require additional resistance. However, one must take into account the Vf and internal resistance when adding an external resistor. Or am I the only one who took it that way?
Follow the link to the data spec in post #1 and it states rather clearly No external resistor required, 6.0 volts @ 15 mA. So it has either an internal resistor or an internal current regulator. ALL the colors take 6.0 volts at 15mA. So the resistor for 12.0 volt use is 400 ohms, or use a 390 ohm (Orange white Brown) resistor and push the current a small bit.
 

ThePanMan

Joined Mar 13, 2020
359
I have already seen the data sheet. And I already know that there is no need for an external resistance. Therefore, as you said, there must be some form of current limitation internally. IF it is a resistor then it's safe to assume it's possibly 267Ω. The data sheet (poor excuse for a data sheet at that) shows different colors and different - I'm assuming - intensity, where Green is the brightest and Yellow is the dimmest. The DS shows 15mA across the board. So that would suggest that whatever is inside there, the LED's are different because they apparently produce different light levels. With such a poor DS I'm guessing it's not a current limiting circuit but rather a simple resistor specific to each lamp assembly.

Since you seem to be so big on telling others to "READ THE POSTS", perhaps this will refresh your memory:
i am using these dialight 610-2221-120f 6v lamps from a carboot sale I picked up and have already burned through a pair in getting them to be brighter with what I thought was the proper resistance.
So just a guess here, but he hooked up a 6V indicator to a higher voltage; probably 12 volts, because he said it's working on 9V.

Here's the numbers on a 9V source on an unknown LED with an unknown resistor - remember, we have no choice but to assume at least one more value. I'll stick with the 2Vf number. (9V-2Vf)/267Ω=25.9mA. Assumably, at 26mA the LED is surviving. But if he hooked them up to 12V then (12V-2Vf)/267Ω=37mA. IF SO - POOF goes the LED. If not immediately, in rather short order. Thus, the hypothesis supports the reported results. "have already burned through a pair". And he never said what resistance he was using. Admittedly, there can be other reasons why adding the wrong resistance would still burn things out - but again, the TS has not said what exactly what he did.

This is what the TS is doing right now - laughing at us and watching us argue points that we disagree upon.

[[ personally removed by self - My apologies for going on the attack]]

Yes, the data sheet DOES say "No external resistance is needed" WHEN OPERATED ON SIX VOLTS! But the TS ISN'T operating them on 6V. Last report it was 9V successfully, but it also seems he wants to supply 12V, as his blown lamps indicate. So I asked him (to no avail) if this is an automotive application. IF SO - then the automotive voltage can vary all the way up to 14.5V. In some cases even higher.

Who's not reading?
 
Last edited:

Tonyr1084

Joined Sep 24, 2015
6,864
I'm guessing it's not a current limiting circuit but rather a simple resistor specific to each lamp assembly.
With 15mA on all the units, it's possible there's a constant current circuit other than a resistor. But since the TS is running this on 9V and would seem to have greater light output - it's probably a resistor. Still, if it IS a constant current circuit, 1) the brightness would not change, 2) running it on 12V would probably blow out the CC circuit. I think Pan is right - it's probably a resistor.
 

MisterBill2

Joined Jan 23, 2018
11,640
I have already seen the data sheet. And I already know that there is no need for an external resistance. Therefore, as you said, there must be some form of current limitation internally. IF it is a resistor then it's safe to assume it's possibly 267Ω. The data sheet (poor excuse for a data sheet at that) shows different colors and different - I'm assuming - intensity, where Green is the brightest and Yellow is the dimmest. The DS shows 15mA across the board. So that would suggest that whatever is inside there, the LED's are different because they apparently produce different light levels. With such a poor DS I'm guessing it's not a current limiting circuit but rather a simple resistor specific to each lamp assembly.

Since you seem to be so big on telling others to "READ THE POSTS", perhaps this will refresh your memory:

So just a guess here, but he hooked up a 6V indicator to a higher voltage; probably 12 volts, because he said it's working on 9V.

Here's the numbers on a 9V source on an unknown LED with an unknown resistor - remember, we have no choice but to assume at least one more value. I'll stick with the 2Vf number. (9V-2Vf)/267Ω=25.9mA. Assumably, at 26mA the LED is surviving. But if he hooked them up to 12V then (12V-2Vf)/267Ω=37mA. IF SO - POOF goes the LED. If not immediately, in rather short order. Thus, the hypothesis supports the reported results. "have already burned through a pair". And he never said what resistance he was using. Admittedly, there can be other reasons why adding the wrong resistance would still burn things out - but again, the TS has not said what exactly what he did.

This is what the TS is doing right now - laughing at us and watching us argue points that we disagree upon. I must say: You claim to be an engineer. Well, I claim to be a doctor, a lawyer, a movie star. I'm an astronaut and I own (a) bar." Lyrics Here. Truth is - on the internet I can be anything I want to be. Or claim to be. But for an engineer, you sure do think in "Grid Lines" rather than seem to be able to grasp thinking in "Abstract". With you it's either black or white (grid line thinking) whereas others here clearly show the ability to think in abstract - the ability to see shades of grey as well as myriads of colors.

Yes, the data sheet DOES say "No external resistance is needed" WHEN OPERATED ON SIX VOLTS! But the TS ISN'T operating them on 6V. Last report it was 9V successfully, but it also seems he wants to supply 12V, as his blown lamps indicate. So I asked him (to no avail) if this is an automotive application. IF SO - then the automotive voltage can vary all the way up to 14.5V. In some cases even higher.

Who's not reading?
Given that the external resistor needs to drop 12 volts down to six volts at 15mA the series resistor for that calculates to 400 ohms, unless my midnight math is off. V/I =R=6/0.015=400.So that is how to run it from a 12 v source.
 

ThePanMan

Joined Mar 13, 2020
359
So the resistor for 12.0 volt use is 400 ohms, or use a 390 ohm (Orange white Brown) resistor and push the current a small bit.
OK, Sir, my humble apologies. Yes 400Ω ADDITIONAL to the internal resistance. OK, I thought you were saying (12V - 2Vf)/0.015=400Ω. Case of misunderstanding.
 

bassbindevil

Joined Jan 23, 2014
384
Carboot sales in the UK (England) are swap meets in parking lots where people sell stuff out of (or around) the trunk ("boot") of their car.
Anyway, Dialight must have used the old type of LEDs where 15 mA doesn't make them painfully bright. If you've already burned one out, might as well pry open the case and stuff a more efficient LED inside.
 

MisterBill2

Joined Jan 23, 2018
11,640
OK, Sir, my humble apologies. Yes 400Ω ADDITIONAL to the internal resistance. OK, I thought you were saying (12V - 2Vf)/0.015=400Ω. Case of misunderstanding.
Sometimes I neglect to make myself clear. That happens more when I am answering very late or very early at night. It slips away that others may not be on the same thought train. Sorry about that.If the TS can open some of the fried ones it should be possible to install a much brighter LED.
 
Top