I am thinking of using 2 LEDs in Series to create the lightning bolt for my project.
My question is this? If I use a 5mm bulb, does the greater the MA increase the brightness of the bulb? What happens if I use a smaller resistor?
Harold
The diameter of the device (is is no longer called a bulb) is irrelevent. Some LEDs (new Christmas lights, for rxample) are very bright, Others much less so.
What do you bean by "in series"? Do you think to put them on once circuit, or do you want them to flash one after the other (a good Idea, it would seem to me for lightening). LEDs wired in series is not a good practice, they should be across the circuit each in series with its own resistor.
LION "Rule of Claw" is 12v dc power supply and 1000 ohm resistor. Him worts not about calculations the numberants of which him knows knot. Reduce this number to make it brighter, increas this number to make it less bright. Titrate to taste.
ROAR
The Route of the Broadway Lion The Largest Subway Layout in North Dakota.
Here there be cats. LIONS with CAMERAS
Running LEDs in series reduces bulb brightness. Better to have each fed with the rated voltage. It does vary, so best to consult the docs with the LEDs. Running them at the stated max for brief should be fine.
On the other hand, running LEDs somewhat below the rated max extends their life, If they're bright enough, toning them down can help if the installation is like some I do where it will be a pain to take it apart to ever replace the LEDs.
Mike Lehman
Urbana, IL
There are LED's and light bulbs. I have been seeing occasional references to LED light bulbs which have built in resistors.
LED Resistor Values
Measured with 12.2 VDC supply using 20 ma max current LED's. Values will vary a ittle depending on actual DC supply value.
Resistor Current
1000 ohm 9.0ma
750 ohm 12.45ma
680 ohm 13.12ma
510 ohm 16.25ma
You could use 470 ohms but I like to stay at 75 percent of componenets.
There are 25 ma max LED's I believe.
You need some more info on LED's. Look at the below link.
http://www.members.optusnet.com.au/nswmn1/Lights_in_DCC.htm
Rich
If you ever fall over in public, pick yourself up and say “sorry it’s been a while since I inhabited a body.” And just walk away.
hwolfIf I use a 5mm bulb, does the greater the MA increase the brightness of the bulb? What happens if I use a smaller resistor?
in general, an LED gets brighter as the current (ma) increases. But like many devices there is a limit above which, the LED is damaged. I think the limit is ~20 ma for many LEDs, but check the specs for the ones you have.
With a constant voltage across the circuit, current will increase as resistance is reduce. See LED circuit.
greg - Philadelphia & Reading / Reading
Sorry for not expressing myself a bit better about the LEDs may not be brightness balanced in series.
RR_MelUsing almost any electronic component in series is not the thing to do...
Where did you get that idea? There are many cases where series is better.
RR_Mel...mixing manufactures will cause different current to flow through the LEDs in series...
No, you're mistaken. Current is constant throughout a series circuit, so LEDs in series will have the same current flowing through them.
RR_Mel..Different runs or batches from the same manufacturer can have differences too. Between batches the efficiency may also change effecting brightness...
There will be slight differences, but no more than you would get from wiring them in parallel with their own resistors.
Many LED strip lights are wired in groups of three LEDs in series. LED Christmas lights are wired in groups of 35 - 50 LEDs in series. Wiring LEDs in series is perfectly acceptable and a great way to reduce the current draw (a 3.5 watt 50 LED string of Christmas lights would draw 175 watts if the LEDs were all wired in parallel).
You can put LED's in series. I have many times but I have been using LED's since 1972 so I know how to experiment if concerned with same brightness. Usually LED's in the same batch are close but I still check them.
All things equal, the current will be less. I=V/R, where I=current, V=voltage, R= resistance. The voltage is the source voltage minus the voltage drop and the voltage drop will double having two LEDs in series. You'll want to use a smaller value resistor to get the same current.
Just remember, in series all elements get the same current, which is why when you calculate the resistor size for a specific current, it also limits the LED to the same current. Or when you put an LED in series with a Tortoise motor you don't need a resistor because the Tortoise motor is already only allowing 15ma or so. And voltage in series add - which is why when you calculate the resistor, you take the power supply voltage minus the LED voltage drop.
So for two LEDs in series, you subtract both forward voltage ratings. Say they are 2.5V, and you are using a 12V power supply. For one LED, it's 12-2.5, or 9.5V that the resistor has to handle. With two LEDs, it's 12-2.5-2.5, or 7 volts. Current is the same, so if you want 15ma with one led, the resistor is 9.5/.015 or 633 ohms (have to use the closest actual value), with two LEDs, it's 7/.015 or 467 ohms (again, have to use the closest standard value). If you had 3 LEDs in series, it would be 4.5V/.015A or 300 ohms. If ther LEDs are 3.5V, you can go one more on a 12V supply, adding a 5th one would drop more than 12V and it wouldn't work very well.
Each of the 3 LEDs with the 300 ohm resistor in this example would be just as bright as 1 LED with the 633 ohm resistor because the same current is flowing through each LED.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
Alot of great info to absorb. Using ohms law when I put 2 LEDs in series it cut the size of the resistor in half.
hwolf Alot of great info to absorb. Using ohms law when I put 2 LEDs in series it cut the size of the resistor in half. Harold
Do some more reading. With two LED's in series, the resistor value does not change.
As an example, one LED, one 1k resistor.
Two LED's in series, one 1k resistor.
Three LED's in series, one 1k resistor.
For any other voltage circuits, I just used a 1k pot with an ma meter. Used the closest standard value resistor.
Bottom line, one resistor with two LED's in series.
I have done as many as three in 12 volt circuits with one, 1k resistor.. I don't have to bother doing the math.
You could use a 1k pot set at max resistance and slowly lower the pot to a brightness you like.
When you work at the componentt level you usually have to experiment.
richg1998 hwolf Alot of great info to absorb. Using ohms law when I put 2 LEDs in series it cut the size of the resistor in half. Harold Do some more reading. With two LED's in series, the resistor value does not change. As an example, one LED, one 1k resistor. Two LED's in series, one 1k resistor. Three LED's in series, one 1k resistor. For any other voltage circuits, I just used a 1k pot with an ma meter. Used the closest standard value resistor. Bottom line, one resistor with two LED's in series. I have done as many as three in 12 volt circuits with one, 1k resistor.. I don't have to bother doing the math. You could use a 1k pot set at max resistance and slowly lower the pot to a brightness you like. When you work at the componentt level you usually have to experiment. Rich
Not true at all. Use this calculator to see ....
http://ledcalculator.net/
Using a 12 volt supply and running a white LED rated at 3.4 volts / 20ma, a single LED would use a 430 ohm resistor to operate at its max potential. Two in series would use a 270 ohm resistor and three inseries would use a 91 ohm resistor.
Mark.
¡ uʍop ǝpısdn sı ǝɹnʇɐuƃıs ʎɯ 'dlǝɥ
Now I am getting confussed. Here is what I did. Let me know if this is correct?
Vs = 12 Vd = 3.3 Current 20Ma
12-3 = 8.7 divided by .020 = 435 Ohms
If I use a second LED the 3.3 changes to 6.6
12 - 6.6=5.4 Divided by .020 = 270 Ohms
hwolf...Using ohms law when I put 2 LEDs in series it cut the size of the resistor in half...
richg1998Do some more reading. With two LED's in series, the resistor value does not change. As an example, one LED, one 1k resistor. Two LED's in series, one 1k resistor. Three LED's in series, one 1k resistor.
If you want each individual LED to be as bright as the single LED would be, then the resistor value most certainly does change. You are increasing the voltage drop, which decreases the voltage across the resistor, which requires a different value for the same current.
Going from one LED to two will not neccesarily cause the required resistance to be reduced by half, it depends on the supply voltage and the voltage drop of the LEDs.
hwolf Now I am getting confussed. Here is what I did. Let me know if this is correct? Vs = 12 Vd = 3.3 Current 20Ma 12-3 = 8.7 divided by .020 = 435 Ohms If I use a second LED the 3.3 changes to 6.6 12 - 6.6=5.4 Divided by .020 = 270 Ohms Harold
You are correct EXCEPT, if 20ma is the MAX current, you need to drive the LED at something less than that.
Mark R. richg1998 hwolf Alot of great info to absorb. Using ohms law when I put 2 LEDs in series it cut the size of the resistor in half. Harold Do some more reading. With two LED's in series, the resistor value does not change. As an example, one LED, one 1k resistor. Two LED's in series, one 1k resistor. Three LED's in series, one 1k resistor. For any other voltage circuits, I just used a 1k pot with an ma meter. Used the closest standard value resistor. Bottom line, one resistor with two LED's in series. I have done as many as three in 12 volt circuits with one, 1k resistor.. I don't have to bother doing the math. You could use a 1k pot set at max resistance and slowly lower the pot to a brightness you like. When you work at the componentt level you usually have to experiment. Rich Not true at all. Use this calculator to see .... http://ledcalculator.net/ Using a 12 volt supply and running a white LED rated at 3.4 volts / 20ma, a single LED would use a 430 ohm resistor to operate at its max potential. Two in series would use a 270 ohm resistor and three inseries would use a 91 ohm resistor. Mark.
Don't need the calculator.
Very true for me and others. A 1k resistor allows about 9 ma current through three LED's in series with 12vdc.
I have gone as low as 510 ohms for about 16ma through all three LED's in series for more brightness.
270 ohms would be pushing it for a 20 ma LED.
I rarely use less than 510 ohms.
Do the measurements, not the calculations.
My NCE Power Cab patch panel has a single red LED with 1k resitor across the DCC voltage.
Rich - Your post seemed to indicate that the resistor stays the same regardless of one, two or three LEDs .... but what you didn't state was that the more LEDs you use, the dimmer they will be at the same resistor value.
If you want three LEDs to each have the same brightness as a single one (regardless of whatever that level of brightnes IS), the resistor value HAS to change.
This is what I have done with one, two or three LED's. Adjusted the resistor value to suit me. Been a few years and don't remember all the values.
I used an approximate LED voltage for simplicity. Very easy to do with a pot or junk box of resistors.
I do know the value I put in a diesel with three LED's was around 870 ohms that suited me.
Couple times I used a tiny 1k trim pot.
If there are 3 resistors in series, and each is 3.4V, there will be a 10.2V drop. If there is only 1 LED, there is a 3.4V drop.
If the power supply is 12V:
12V - 3.4V = 8.6V. 8.6V/1000 (1K) = 8.6ma for the LED.
12V - 10.2V = 1.8V. If you use the same 1K resistor, 1.8V/1000 = 1.8ma, which should result in a noticeably dimmer set of LEDs.
If you want the same 8.6ma, then the resistor is 1.8V/.086a (8.6ma) = 21 ohms. If you reverse the work to check it, 21 ohms * .086a = 1.8V
1----A-----B----C----2
If the power is applied at 1 and 2:
Current through A = current through B = current through C
Voltage across A + voltage across B + voltage across C = total voltage across 1 and 2
In a parallel circuit, which is not easily drawn with regular ASCII characters, it is exactly the opposite. For 3 LEDs in parallel, the voltage across each one is equal. The current drawn by all three is the sum of the current drawn by each one. It is with parallel LEDs that you want to use a resistor on each one, in which case each resistor would be the same 1K resistor, resulting in 8.6ma of current through each resistor/LED pair. You don't want to use 1 resistor for a set of parallel LEDs because if they are different, one will take the lion's share (not Br. Elias) of current and they may not all be the same brightness.
We've all been using Ohm's Law to calculate the resistor values. This series and parallel stuff is the other guy - Kirchoff's Laws.
(who may some day in the next 10 years stop posting this every few weeks and just put a section on his web site with the explanations and just link it each time)
The problem is the direction this thread has taken ....
The OP wants to wire TWO Leds in series to illuminate a lightning bolt. I would tend to think he would want them to be as bright as possible.
All these different resistor values "to taste" are irrelevant here. Use the online calculator to run those LEDs as bright as they can .... safely.
Thanks Mark
As I stated in the original post. I need to get the brightest light. As a point of info , there are two bolts of lightning but not to worry they will be on seperate curcuits
Thanks for all your help. I think I have enough info to make it right.
For a lightning simulator, you'll probbaly want LEDs that are much brighter than the ones typically used for LED headlights. Such as this one:
http://www.radioshack.com/10mm-ultra-high-brightness-white-led/2760005.html
And run it right up near the 20ma limit.
I have used a arc welding simulator circuit to simulate arc welding flashes. Bought it on line about ten years ago. Could work for lightning simulation.
Google it.
There are super bright LED's that require much more resistance. Some here say as much as 3k in their locos.
The SMD type in white might look nice but there are different colors in lightning strokes having done severe storm research number of years ago.
No idea what your display will look like though. There is cloud to ground, between clouds, in a cloud, etc.
Running LEDs in series reduces bulb brightness.
Ummmm NO.
Running leds cumulatively drops the voltage, NOT current. The resistor you put in the circuit will determine the current. So it you have two 3.2 Volt LEDs running in series, you will need at least 6.4V to get them to light. (Personally I like a 10% tolerance so 7.4V would be my reccomendation)When they rate LED lifetimes, they are typically 2,000 to 80,000 hours at their rated maximum. (Typically 20ma, but less for nano or smaller leds...READ THE TECH SPECS) And even then LED's don't burn out like incandecents. They fade. So after their rated life is over (say 20,000 hours typical) they will be at half brightness or their new life.So I wouldn't worry about replacing them unles you intend to run your LEDs 40 hrs/week for a year solid.
Don - Specializing in layout DC->DCC conversions
Modeling C&O transition era and steel industries There's Nothing Like Big Steam!
DigitalGriffin Running LEDs in series reduces bulb brightness. Ummmm NO. Running leds cumulatively drops the voltage, NOT current. The resistor you put in the circuit will determine the current. So it you have two 3.2 Volt LEDs running in series, you will need at least 6.4V to get them to light. (Personally I like a 10% tolerance so 7.4V would be my reccomendation)When they rate LED lifetimes, they are typically 2,000 to 80,000 hours at their rated maximum. (Typically 20ma, but less for nano or smaller leds...READ THE TECH SPECS) And even then LED's don't burn out like incandecents. They fade. So after their rated life is over (say 20,000 hours typical) they will be at half brightness or their new life.So I wouldn't worry about replacing them unles you intend to run your LEDs 40 hrs/week for a year solid.
But the voltage factors into the equasion. One LED using a 1000 ohm resistor will be brighter than two LEDs inseries with a 1000 ohm resistor.
So, the statement running them in series reduces brightness is correct assuming you are using the same resistor.
That is why I am saying, get a pot and experiment. Pick the nearest standard value for permanent installation. I don't remember all the details as it has been some years. No doubt off on some things.
It can be tough for those without electronics background but with Google, a lot of details can be found on the Internet. Cobble up a variable voltage power supply, some clip leads, couple different pots, 1k, 5k, various resistor between 270 ohms and maybe 3k.
Be adventurous. That is part of model railroading.
Many of us did that before the Internet when all we had were model railroading magazines. Our Google was the local library.
Added question
I just bought bright LED with resistors and bridge rectifiers 7-19 v output.
I also bought a Miniatronics Arc welder LED
Hear is my question. I have two wart power supplies. One is 9v output and the other is 12v Output. In looking at the wire one has dashes and the other wire has writing. How do I know which side is positive and which side is neg?
If I touch the wrong way will it burn out the LED or just not light?
Be careful with wall wart supplies. There are two types - regulated and un-regulated. Regulated means the voltage rating listed on the pack will be the voltage supplied regardless of the load on it (up to its maximum rating of course).
An un-regulated supply may say 12 volts, but it needs a specific load on it to provide that twelve volts. If it doesn't, the actual voltage on the wires will be considerably higher than what it says.
For a quick test, if the LED is wired backwards, it won't hurt it - it won't light either.
richg1998 That is why I am saying, get a pot and experiment. Pick the nearest standard value for permanent installation. I don't remember all the details as it has been some years. No doubt off on some things. It can be tough for those without electronics background but with Google, a lot of details can be found on the Internet. Cobble up a variable voltage power supply, some clip leads, couple different pots, 1k, 5k, various resistor between 270 ohms and maybe 3k. Be adventurous. That is part of model railroading. Many of us did that before the Internet when all we had were model railroading magazines. Our Google was the local library.
Not sure why you keep going off on tangents. The OP wants the LED to be as bright as possible. The online calculator does just that. No need for experimenting.