Simple LED Circuit
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Author: Wayne Eggert Date: 02/27/06 Difficulty: Basic |
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Description
This is the "Hello World" of Electronics -- a simple circuit that's sole function is to light up an LED light. I just got my first breadboard a few months ago and have been meaning to put it to use, so I figured what better way than to light a uber-l337 blue LED up.
Materials
(1) Solderless Breadboard (optional)
(1) 5mm LED (any color)
(1) 330ohm resistor
(1) 4 AA Battery holder
(4) AA 1.5v batteries
A Quick Word About Breadboards
Solderless breadboards are a great invention for electronic hobbyists since it allows components to be placed and moved without the need to solder and desolder if something in your circuit changes. They are fairly inexpensive ($10-$30 depending on the size you need) and considering the amount you will use them if you're into tinkering they're worth the investment. For this simple of a project a breadboard is kinda overkill, but if we were to create a variable-blinking led (with a 555 timer, potentiometer, capacitors, resistors) it would start to get messy having components directly soldered to eachother or having jumper wires holding them together. Another cheaper option with breadboards are the solder-variety, which aren't meant for moving components around much (if at all) after they have been soldered to the board.
Building The Circuit
I have included the schematic for the circuit below. I used a 6v power supply consisting of (4) 1.5v AA batteries since I had a battery holder already. The only thing missing was the value of the resistor required to limit the current. Since LEDs have very low resistance, without an external resistor to limit the current they will burn up. To find out the value of the resistor needed, we use Ohm's Law.
V=IR (voltage = current * resistance)
(Vsource - Vled) = IR
The specs on my 5mm blue LED say it can operate at 3.2-3.6v at 20mA current. Since we know V & I, we can solve for R.
V/I = R
(6V - 3.2V) / 0.02A = R
140ohm = R
So at a minimum we need a resistor of at least 140ohms to protect the LED from burning up. A 330ohm resistor should be more than adequate for most LEDs.
Here's the schematic we'll be implementing:
Updated 5/27/2009: The schematic will look very complicated, but once you get used to some of the electronic component notations it will get easier. For those that are new to schematics, I have created a pictural representations of what the schematic is saying. I hope this helps everyone that is new to electronic diagrams.
When building the circuit just make certain that you have the LED oriented properly (anode to cathode) as in the schematic since LEDs are polarity-sensitve. You can use a different voltage power source, just substitute the voltage in the equation above to find the resistor value you'll need.
Summary
As I mentioned previously, this was just a very basic circuit so that I could begin putting my breadboard to good use. It doesn't get much simpler than this =) Still, it was nice to see the LED light up in all its glory and I now know I didn't buy dud LEDs a few months ago. Good luck!
Comments:
| I can't figure it out!! |
| Posted 04/16/06 3:12PM by Anonymous Techdoser |
| I've read the instructions like a millions times and I can't get it to work!!! LOL!! This schematic is too complicated. |
| help. |
| Posted 04/16/06 3:13PM by Anonymous Techdoser |
| maybe you've got your resistor in backwards? |
| Diodes. |
| Posted 04/24/06 8:49AM by Anonymous Techdoser |
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Two probably errors: -Battery polarity is backwards (fix: switch battery leads to circuit) -Diode is placed in reverse bias, backwards (fix: take out diode, flip it or turn 180 degrees, replace diode.) One possible error: -Something (breadboard, resistor, diode, battery) is broken or out of power. Note: Diodes require 0.7V to bias properly (light up). |
| URGENT |
| Posted 05/01/06 5:01PM by Anonymous Techdoser |
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YOU MUST HAVE INSERTEED SUMTHING ELSE INSTEAD OF BATTERY CHEQUE YOUR PANTS |
| lame |
| Posted 05/03/06 12:03AM by Anonymous Techdoser |
| I can't figure it out so I decided to maek a necklace out of all my resistors!!! It is so kewl LOL!!11! |
| Resistor Direction |
| Posted 10/12/06 8:28PM by PUREMUNY |
| Please help. I am trying to hook up a 2.4v LED light to a 2 AA battery pack. Which direction does the resistor go? Which end of the LED? Positive or negative on the battery pack. Thanks |
| I figured it out, before I found this site. |
| Posted 10/21/06 2:39PM by TruthByDenial |
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One of the few things that I found Interesting is that, from negative to positive, You would have this {Wire (-) Connects to LED (-)} {The other leg of LED, LED (+) connects to one leg of resistor. (There is no way to put a resistor in backwards)} {The other leg of the resistor plugs into wire (+). So it goes. - wire, LED, Resistor, + wire. Hope this helps as I have just figured it out, and I'm just starting to learn about electricity and circuitry |
| Just a small note. |
| Posted 10/21/06 2:42PM by TruthByDenial |
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If your led doesn't light up. or you are confused. ∙Make sure your battery isn't dead ∙Flip the LED ( It only conducts current one way.) ∙You're resistor is fine, as long as its directly connected to positive wire. |
| Simple LED Circuit |
| Posted 04/15/07 4:34AM by deltaboy12000 |
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LED's cathode is identified by: 1. the short lead of the LED 2. the flat side of the case of the LED 3. the large internal metal side of the LED. If further understanding of LEDs is wanted look up N-type and P-type material. This is very advanced reading but it the science as to why LEDs only "work" when forward biased. Also in the digital world keep in mind its sometimes required to reverse bias a diode... |
| Doesn't matter. |
| Posted 05/23/08 2:02PM by Anonymous Techdoser |
| The resistor can be on either end... The only thing that matters is that the polarity is right on the LED. Put the resistor anywhere you want. If the LED lights up voilla, if it doesn't reverse the pins and it will. |
| I can't get it! |
| Posted 01/05/09 9:36AM by Anonymous Techdoser |
| The title says it all. |
| question about power rating on resistor? |
| Posted 02/25/09 10:12AM by Anonymous Techdoser |
| How do you know what power resistor to use? I mean 1/4 watt, 1/8 watt, 1/2 watt, etc.? I just bought a breadboard myself and this question has come up several times already. Any help would be greatly appreciated. |
| Ok... help me extend this. |
| Posted 03/13/09 6:15PM by Anonymous Techdoser |
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I have a 9v battery and want to light a series of LEDs (up to 10). 9v-3.2v = 5.8 5.8/20mA = 5.8/.02A = 290 ohms? But how does a series of LEDs affect it? If I hooked up 3 LEDs... you can't just keep subtracting. (9 - 3.2 - 3.2 - 3.2). Yes, I'm an idiot. I got a D in physics electromagnetism 20 years ago. |
| To extend to 10LED with 9V |
| Posted 05/25/09 6:30PM by Anonymous Techdoser |
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I assume you've identified that your led's forward voltage is 3.2V (they are not all the same, i'm working with IR LEDs right now that are 1.2V). If you have 3.2V LEDs then you can put two in series with the resister (for 1.2V LEDs, you could use 5-6 in series). However, for more you must make a parallel path with another resister and two LEDs. But you are right, for the resister look at the total series chain for voltage drop. (9 - 3.2 - 3.2) / 20mA --> 2.6V / 0.02A = 130 ohms (plus some margin for life time if needed). You may not use a common resister for two parallel pairs of LEDs. If one pair has a slightly smaller forward voltage total, then one side or the other will take more amperage and could keep one side off, and possibly blow out the other side. |
| Power rating |
| Posted 05/25/09 7:02PM by Anonymous Techdoser |
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These are the two most basic equations to electricity (they are useful to anyone and everyone, not just engineers and electricians): Power/watts(P) = Volts(V) * Current/Amps(I) Volts(V) = Amps(I) * Ohms(R) Typically written as: P=V*I V=I*R In the example a 6V bat is being used with a 3.2V LED, and a 330 ohm resister. This is all you need to know. To get the resister power, we need the resister voltage and current. ResisterV = BatV - LoadV = 6 - 3.2 = 2.8V ResisterI = (I=V/R) = 2.8V / 330ohms = 0.0085 or 8.5mA Now Resister Power = 0.0085A * 2.8V = 0.0238W Less than a tenth of a Watt is required. What if you choose the max rated power of 20mA with the 140ohm resister what is the power then? |
| larger batterys and several LEDs |
| Posted 06/12/09 9:08AM by Anonymous Techdoser |
| i would like to set up a string of some where around 200 leds and have them chase 4 or 5 light up at a time down the string i'm assuming ill have to use a car battery to get the voltage and ampreage that i need to do this successfully but am un sure of how to sequence them to work properly. i would like to see 5 on 5 off 5 on 5 off and so forth down the string |