Here is a little background on LED's...
LEDs are semiconductors, diodes in particular. The current flowing in an LED is an exponential function of voltage across the LED. The important part about that for you is that a small change in voltage can produce a huge change in current. The current and voltage in a resistor are linearly related. That means that a change in voltage will produce a proportional change in current. Current versus voltage is a straight line for a resistor, but not at all for an LED.
Never connect an LED directly to a battery or power supply! It will be destroyed almost instantly because too much current will pass through and burn it out.
LEDs must have a resistor in series to limit the current to a safe value, for quick testing purposes a 1k resistor is suitable for most LEDs if your supply voltage is 12V or less. Remember to connect the LED the correct way round! LEDs must be connected the correct way round, the diagram may be labelled a or + for anode and k or - for cathode
To choose the correct resistor for a particular LED use the following:
An LED must have a resistor connected in series to limit the current through the LED, otherwise it will burn out almost instantly.
The resistor value, R is given by:
VS = supply voltage VL = LED voltage (usually 2V, but up to 4V for blue and white LEDs) I = LED current (e.g. 20mA), this must be less than the maximum permitted; 20mA is typical.
If the calculated value is not available choose the nearest standard resistor value which is greater, so that the current will be a little less than you chose. In fact you may wish to choose a greater resistor value to reduce the current (to increase battery life for example) but this will make the LED less bright.
If the supply voltage VS = 9V, and you have a red LED (VL = 2V), requiring a current I = 20mA = 0.020A, R = (9V - 2V) / 0.02A = 350Ω, so choose 390Ω (the nearest standard value which is greater).
Ohm's law says that the resistance of the resistor, R = V/I, where: V = voltage across the resistor (= VS - VL in this case) I = the current through the resistor
So R = (VS - VL) / I
Hope that helps,
-Robert
I have no idea what source that the info Robert gave you above comes from. A 2 volt LED can easily be tested with a AA battery, they put out 1.5 volts under load, but an LED draws little current. I know, I've done it.
I built a simple rig to test LED's out of two AA battery holders, three alligator clips, and some wire. Tie the battery holders in series, connect an alligator clip lead to each end, and one to the point where the holders are tied in series. Insert a battery in each holder. Use the center clip and one end clip to connect to the LED being tested. If it doesn't light, reverse polarity. If it still doesn't light, use the clips connected to the outside of the battery holders, reversing polarity if necessary. It's a simple way to identify the anode and cathode, and the operating voltage of an LED.
I've never heard Robert's statement about NEVER connect a LED directly to a battery before. Given Ohm's law, as stated by Robert, using a 1.5v battery will give a value of R = 0, i.e. R=(1.5 - 2.0)/I. Since the value of (Vs - VL) cannot be less than zero, any number for I, no matter how large, divided into zero will result in a value of zero for R. Using a 1k resistor will resulting in no current flowing to the LED.
I use a small rechargeable 12 volt battery with a 1k resistor and clip leads. The resistor allows 10ma current. Most LED's for modeling have a 20ma max current. LED's are current devices. I never worry about voltage.
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.
TomDiehl I have no idea what source that the info Robert gave you above comes from. A 2 volt LED can easily be tested with a AA battery, they put out 1.5 volts under load, but an LED draws little current. I know, I've done it. I built a simple rig to test LED's out of two AA battery holders, three alligator clips, and some wire. Tie the battery holders in series, connect an alligator clip lead to each end, and one to the point where the holders are tied in series. Insert a battery in each holder. Use the center clip and one end clip to connect to the LED being tested. If it doesn't light, reverse polarity. If it still doesn't light, use the clips connected to the outside of the battery holders, reversing polarity if necessary. It's a simple way to identify the anode and cathode, and the operating voltage of an LED.
The Information comes from a DC Electronics Book! I'm an Electronics Engineer and design LED Signs; like at Football Stadiums. Yes, two AA batteries (3.0 Vdc) in series will light an LED. The data, I supplied allows one to calculate the proper resistor that will limit the current. Without the datasheet on the LED, one assumes a max current of 20mA for full brightness. Using two AA batteries, you should still be using a 56 Ω resistor to limit the current. Although the LED will not blow using 3.0 vdc as quickly as with say a 12 volt supply, it will still cause latent damage to the LED. Don't believe me, as a rule of thumb, different color LEDs require different forward voltages to operate - red LEDs take the least, and as the color moves up the color spectrum toward blue, the voltage requirement increases. Typically, a red LED requires about 2 volts, while blue LEDs require around 4 volts. Typical LEDs, however, require 20 to 30 mA of current, regardless of their voltage requirements. The table below shows how much current a typical red LED will draw at various voltages.
Notice that this LED draws no current under 1.7 volts; the LED is "off". Between 1.7 volts and about 1.95 volts, the "dynamic resistance", the ratio of voltage to current, decreases to 4 ohms. Above 1.95 volts, the LED is fully "on", and dynamic resistance remains constant. Dynamic resistance differs from resistance in that the curve isn't linear. Just remember that this non-linear relationship between voltage and current means that Ohm's Law doesn't work all that well for LEDs.
Notice how steep the slope is - almost vertical. LEDs have a much more vertical slope than do normal diodes (but not as bad as laser diodes).This means that a tiny increase in voltage can produce a large increase current, and lots of smoke. In the above-mentioned LED, 2 volts is required to drive the LED properly, but as little as 2.04 volts could destroy it. To keep the current down to a reasonable level, a series resistor must be included in the circuit.
Google this subject, you'll see I'm not making this up,
Silver Pilot I've never heard Robert's statement about NEVER connect a LED directly to a battery before. Given Ohm's law, as stated by Robert, using a 1.5v battery will give a value of R = 0, i.e. R=(1.5 - 2.0)/I. Since the value of (Vs - VL) cannot be less than zero, any number for I, no matter how large, divided into zero will result in a value of zero for R. Using a 1k resistor will resulting in no current flowing to the LED.
Basic physics says that an LED that needs 2.0 Vf. is not going to light up using a 1.5 Vdc battery. The 1KΩ resistor is a safe choice for testing, when using higher power supplies like; 9-12 Vdc.
An LED has very low internal resistance. This means that left to itself, an LED will pass so much current that it will burn up. They require an external resistor to limit the current. Is a Series Resistor Really Necessary? In a word, no. However, neither is a seat belt. Both are "cheap insurance" against disaster. The formula I supplied was to allow someone to choose the proper current limiting resistor. Since John stated the LED's were for a Searchlight, I figured that they would operate off of a higher power supply than a couple of batteries.
Don't believe me, Google is your friend, or any basic electronics book.
My head hurts after reading all that! Felt like I was back i High School in Mr. Wizards Science class
"Robertzep sez:
Never connect an LED directly to a battery or power supply! It will be destroyed almost instantly because too much current will pass through and burn it out"
Funny, I've had LEDs connected directly to a 3v supply now for 5 or 6 years with no difficulty and have not lost an LED yet.
Oh, well.
Roger Hensley= ECI Railroad - http://madisonrails.railfan.net/eci/eci_new.html == Railroads of Madison County - http://madisonrails.railfan.net/ =
Depends on what kind of batteries you are using. With enough internal resistance, the batteries won't allow the current to get too high, especially when the voltage is right at the LED's threshold (see the graph)
The chart does approach vertical as the voltage grows, so Ohm's Law works well enough when the voltage is well above the LED's forward voltage, like when using a 12 volt power supply.
One correction - the current rating for the LED is a MAX, not a 'required'. You want to keep it at or below that value. Typical white LEDs used for loco headlights, connected to 12 volts via a 1K resistor results in 9-10ma in the LED, most of those LEDs are rated at 20-25ma max, the 9-10ma results in plenty of brightness and long long life for the LED being operated at half its maximum rating.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
Since it appears that everyone thinks I made this stuff up, I am directing the subject to the web.
http://led.linear1.org/why-do-i-need-a-resistor-with-an-led/
http://answers.yahoo.com/question/index?qid=20090102154845AAyjb7U
I'm not affilated with either of these sites!
Good Luck with your projects...I'll be keeping my knowledge to myself! It seems to keep me employed and sought out by the Electronics Industry.
RobertzepSince it appears that everyone thinks I made this stuff up,
Robert, one or two people is not everybody. I suspect that many like me, read your postings and appreciated them.
Simon Modelling CB&Q and Wabash See my slowly evolving layout on my picturetrail site http://www.picturetrail.com/simontrains and our videos at http://www.youtube.com/user/MrCrispybake?feature=mhum
Robert:
A agree with Simon. There will always be "spirited discourse" on this forum and opinions will be fiercely defended. However, the are a large number who "lurk and learn" and appreciate the discussion.
Joe
rogerhensley Funny, I've had LEDs connected directly to a 3v supply now for 5 or 6 years with no difficulty and have not lost an LED yet.
Exactly, your 3 V supply is current limited or regulated, other wise the LED's would have burned out. If your supply exceeded the current limit of the LED's then you would have had to put in a resistor to preserve the LED's. Either you were aware of this, or were lucky.
As Robert has eloquently pointed out, the resistor ensures that you don't exceed the current rating of the device.
There are many on-line resistor calculators http://ledz.com/?p=zz.led.resistor.calculator or
http://led.linear1.org/1led.wiz plus dozens more. You will be very hard pressed to find one that says you don't need a resistor!
..... Bob
Beam me up, Scotty, there's no intelligent life down here. (Captain Kirk)
I reject your reality and substitute my own. (Adam Savage)
Resistance is not futile--it is voltage divided by current.
Robert, Its nice when others step forward with more technical data. Many articles and websites simplify the info for the masses. I, Randy Rinker, and others have also contributed, with similar KISS comments as the comeback. However, that thinking will sometimes lead to the "head scratching" puzzlement that occurs when much is assumed and a problem arises.
For those who don't believe in Robert's statements have been successful not by knowledge but by the situation: the type of batteries, the LED itself, etc. Sooner or later you run the risk of failure that wont be understood given limited knowledge.
For eight years, I taught electronics and ran into "doubting Thomases" with interesting, if not untrue theories on how electronic devices actually work. I say this to all, don't discard the rest of the story or data just because you don't think it applies. I wish I had a dime for everytime I was approached to repair or troubleshoot something by someone with the "it cant be that hard" or "I don't need to know that or care" mentality.
Now, don't get mad, try to understand from someone who knows the difference. It's like watching a car run full speed into a brick wall, knowing the outcome, wanting to stop the driver, but your attempts to flag them down are futile.
We are not bragging, trying to show what we know, professing superiority, or any of that nonsense that shows up in these threads. We are trying to help; to prevent a situation where someone is left unhappy or frustrated.
Many people say "be careful or you'll let the smoke out of that _________", referring to the damage that can occur if something is improperly installed or configured. In electronics there are almost no second chances to human error. We experienced folks (that do it for a living every day) can tell you stories that'll curl you hair. Most of you only deal with this stuff when you have to, in small quantities. In many cases your experiences are good. In this hobby, that's the ideal, wanted situation. But, with changes in technology and people breaking new ground with "creative engineering" things can happen. It's not fun to "hit the wall". We geeks that care don't want to see it happen. IMO, the detailed explanations can lead to more understanding and a wider knowledge base.
Carl the techo-geek with a heart and always a lending hand.
Carl in Florida - - - - - - - - - - We need an HO Amtrak SDP40F and GE U36B oh wait- We GOT THEM!
Robert, I'm an Electronics Technician and work with this type thing on a more practical level. When you post an absolute and put it in red lettering, expect that to be disputed, especially if others have done the opposite of your absolute. My post says that I have connected LED's directly to a battery with no adverse effects, I also tie three of the 3 volt ones in series, then parallel these and tie the whole thing to a 9 volt wall wart. The only resistance in the circuit is the LED's themselves. And I have found that Ohm's law does work with LED's if you consider their forward biased resistance the same as any other resistance.