Simple LED Circuit
| Author: Wayne Eggert Date: 02/27/06 Difficulty: Basic |
|
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:
Comments:
| help me |
| Posted 09/27/11 5:53AM by vasim786 |
|
hello friend..i have 12 volt battery..then how can i calculate current(i) and r ohm..i know v=ir but i have only value of voltage..i m nt student of electronic.pls guide me |
| Wiring LED |
| Posted 08/17/11 1:25AM by Sylvi418 |
|
I use plenty of LED Bulbs for my car inside and parking lights 4 LED in series. I am from Srilanka Sylvi418 |
| help. |
| Posted 07/31/11 4:20AM by venemvette |
| what if i want to use a 1.8v led what resistor do i need |
| voltage drop |
| Posted 04/29/11 4:41PM by Anonymous Techdoser |
| currently I have a circuit that drops from 11.8vdc to 8.6vdc when 26 standard bulbs are turned on. Will replacing these standard bulbs with 12v leds bulbs reduce the voltage drop |
| simple throw-away led lights.... |
| Posted 03/10/11 1:12AM by Anonymous Techdoser |
| What happens if the led diode is 3.2V and the battery is only 3V? Obviously it does not draw the extra .2V, but what effect does that have on the circuit? I am just using 10mm leds with a 3V lithium battery. I tested on out and it stayed lit for 2 weeks. If I only need them for 1 day, can I use more then one light per battery? If so, do I need to be concerned with resistors, capacitors etc? Or can i simply attach them all directly to the one battery? My physics books are beginning to look like ancient languages and do not typically explain such easy things. :) Please help! |
| Solar Blinking LED |
| Posted 12/24/10 9:39AM by pulsar2121 |
|
Hi, I liked your article here. I am trying to get an LED to blink 60 times a minute. The LED is from a garden light. It uses a 1.2V rechargeable AA battery. I also want to connect it to the garden light's solar panel and daylight sensor. What do you think would be a good circuit for this application. |
| cant undersant how it's (6v-3.2v) |
| Posted 09/15/10 2:50AM by Amartya_karan |
|
sir i cant understand why it will be (6v-3.2v) i know the ohm's law. but i cant understand why you doing this (6v- 3.2v) please give me the full method. |
| info on LED |
| Posted 05/12/10 5:24AM by mmkkhh |
|
i feel much happy to read the comment as above which have give me much info. thanks more info , , i will welcome ,, ,khurshid |
| Converting av to dc to be used to power LED's |
| Posted 03/15/10 9:00PM by Anonymous Techdoser |
| Is there a 'simple' way to convert 24vac to dc voltage to drive a LED? I fully understand the need to use the correct resistor value to reduce 24 volts to the required 2.2 volts. But how to change the darn voltage to dc???The whole package must fit inside a door bell button. |
| Re: without a resistor |
| Posted 11/20/09 5:53AM by AceBHound |
| Yes, that's correct -- you can hook 2 leds up without a resistor since the voltage drop will be greater than the power source voltage. You could even try 3 leds in series and as long as the LEDs can still operate around 2v they'll light (though dimmer than if they were getting the full 3-3.2v each). If you add too many LEDs though, you won't meet minimum voltage requirements and they won't light. Have fun! |
| without a resistor |
| Posted 11/18/09 1:31PM by spaceshipman |
| so am I right that if I hook up 2X 3V LEDs in series to a 6V battery I wont need a resistor? I'm also a beginner! |
| To Math Confusion |
| Posted 09/01/09 1:48PM by Anonymous Techdoser |
|
The 18A that the PSU puts out is not the component in the calculation that you should be using. What you are looking for is the consumption of the LED which is typically 20mA or .02A. Thus your calculation should be (12V-3.2V)/.02A = 440ohms. The resistor that you would need is a 470 ohms resistor. They are color coded as : Yellow, Violet, Brown, Gold. Note you should verify the voltage of the LED as every color has its' own voltage. Also verify the mA used by the LED. |
| Try this One |
| Posted 08/30/09 7:45AM by Anonymous Techdoser |
| use the same circuit with a lemon or two. Stick a piece of copper wire in one end and a galvanized nail in the other. You should have enough current to light the led. Trick add more lemons(batteries)to make it brighter. Then try other foods around the house. Why do some work and others don't. What even works better as a load is one of those little digital clocks from the dollar store. It takes very little current to make one of those start working and it is useful also. McGuyver will be proud of you. |
| Math Confusion |
| Posted 07/27/09 10:39AM by Anonymous Techdoser |
| Okay, going back to the simple circuit above... 6V - 3.2V / .02A = 140ohms. So, if, for some reason, I wanted to power an LED off of a PSU at 12V/18A... 12V-3.2V / 18A = 0.5ohms. That can't possibly be right! What have I done wrong, please? |
| 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 |
| Power rating |
| Posted 05/25/09 7:02PM by Anonymous Techdoser |
|
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? |
| To extend to 10LED with 9V |
| Posted 05/25/09 6:30PM by Anonymous Techdoser |
|
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. |
| Ok... help me extend this. |
| Posted 03/13/09 6:15PM by Anonymous Techdoser |
|
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. |
| 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. |
| I can't get it! |
| Posted 01/05/09 9:36AM by Anonymous Techdoser |
| The title says it all. |
| 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. |
| Simple LED Circuit |
| Posted 04/15/07 4:34AM by deltaboy12000 |
|
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... |
| Just a small note. |
| Posted 10/21/06 2:42PM by TruthByDenial |
|
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. |
| I figured it out, before I found this site. |
| Posted 10/21/06 2:39PM by TruthByDenial |
|
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 |
| 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 |
| 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! |
| URGENT |
| Posted 05/01/06 5:01PM by Anonymous Techdoser |
|
YOU MUST HAVE INSERTEED SUMTHING ELSE INSTEAD OF BATTERY CHEQUE YOUR PANTS |
| Diodes. |
| Posted 04/24/06 8:49AM by Anonymous Techdoser |
|
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). |
| help. |
| Posted 04/16/06 3:13PM by Anonymous Techdoser |
| maybe you've got your resistor in backwards? |
| 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. |