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Step 8Wiring up multiple LEDs in parallel
Unlike LEDs that are wired in series, LEDs wired in parallel use one wire to connect all the positive electrodes of the LEDs your using to the positive wire of the power supply and use another wire to connect all the negative electrodes of the LEDs your using to the negative wire of the power supply. Wiring things in parallel has some distinct advantages over wiring things in series.
If you wire a whole bunch of LEDs in parallel rather than dividing the power supplied to them between them, they all share it. So, a 12V battery wired to four 3V LEDs in series would distribute 3V to each of the LEDs. But that same 12V battery wired to four 3V LEDs in parallel would deliver the full 12V to each LED - enough to burn out the LEDs for sure!
Wiring LEDs in parallel allows many LEDs to share just one low voltage power supply. We could take those same four 3V LEDs and wire them in parallel to a smaller power supply, say two AA batteries putting out a total of 3V and each of the LEDs would get the 3V they need.
In short, wiring in series divides the total power supply between the LEDs. Wiring them in parallel means that each LED will receive the total voltage that the power supply is outputting.
And finally, just some warnings...wiring in parallel drains your power supply faster than wiring things in series because they end up drawing more current from the power supply. It also only works if all the LEDs you are using have exactly the same power specifications. Do NOT mix and match different types/colors of LEDs when wiring in parallel.
OK, now onto to actually doing the thing.
I decided to do two different parallel setups.
The first one I tried was as simple as it could be - just two 1.7V LEDs wired in parallel to a single 1.5V AA battery. I connected the two positive electrodes on the LEDs to the positive wire coming from the battery and connected the two negative electrodes on the LEDs to the negative wire coming from the battery. The 1.7V LEDs didn't require a resistor because the 1.5V coming from the battery was enough to light the LED, but not more than the LEDs voltage - so there was no risk of burning it out. (This set up is not pictured)
Both of the 1.7V LEDs were lit by the 1.5V power supply, but remember, the were drawing more current from the battery and would thus make the battery drain faster. If there were more LEDs connected to the battery, they would draw even more current from the battery and drain it even faster.
For the second setup, I decided to put everything I had learned together and wire the two LEDs in parallel to my 9V power supply - certainly too much juice for the LEDs alone so I would have to use a resistor for sure.
To figure out what value I should use I went back to the trusty formula - but since they were wired in parallel there is a slight change to the formula when it comes to the current - I.
R = (V1 - V2) / I
where:
V1 = supply voltage
V2 = LED voltage
I = LED current (we had been using 20 mA in our other calculations but since wiring LEDs in parallel draws more current I had to multiply the current that one LED draws by the total number of LEDs I was using. 20 mA x 2 = 40 mA, or .04A.
And my values for the formula this time were:
R = (9V - 1.7V) / .04A
R = 182.5 Ohms
Again, since the variety pack didn't come with that exact value resistor I attempted to use the two 100 Ohm resistors bundled together in series to make 200 Ohms of resistance. I ended up just repeating the mistake that I made in the last step again though, and wired them together in parallel by mistake and so the two 100 Ohm resistors only ended up providing 50 Ohms of resistance. Again, these LEDs were particularly forgiving of my mistake - and now I have learned a valuable lesson about wiring resistors in series and in parallel.
One last note about wiring LEDs in parallel - while I put my resistor in front of both LEDs it is recommended that you put a resistor in front of each LED. This is the safer better way to wire LEDs in parallel with resistors - and also ensures that you don't make the mistake that I did accidentally.
The 1.7V LEDs connected to the 9V battery lit up - and my small adventure into LED land was completed.
If you wire a whole bunch of LEDs in parallel rather than dividing the power supplied to them between them, they all share it. So, a 12V battery wired to four 3V LEDs in series would distribute 3V to each of the LEDs. But that same 12V battery wired to four 3V LEDs in parallel would deliver the full 12V to each LED - enough to burn out the LEDs for sure!
Wiring LEDs in parallel allows many LEDs to share just one low voltage power supply. We could take those same four 3V LEDs and wire them in parallel to a smaller power supply, say two AA batteries putting out a total of 3V and each of the LEDs would get the 3V they need.
In short, wiring in series divides the total power supply between the LEDs. Wiring them in parallel means that each LED will receive the total voltage that the power supply is outputting.
And finally, just some warnings...wiring in parallel drains your power supply faster than wiring things in series because they end up drawing more current from the power supply. It also only works if all the LEDs you are using have exactly the same power specifications. Do NOT mix and match different types/colors of LEDs when wiring in parallel.
OK, now onto to actually doing the thing.
I decided to do two different parallel setups.
The first one I tried was as simple as it could be - just two 1.7V LEDs wired in parallel to a single 1.5V AA battery. I connected the two positive electrodes on the LEDs to the positive wire coming from the battery and connected the two negative electrodes on the LEDs to the negative wire coming from the battery. The 1.7V LEDs didn't require a resistor because the 1.5V coming from the battery was enough to light the LED, but not more than the LEDs voltage - so there was no risk of burning it out. (This set up is not pictured)
Both of the 1.7V LEDs were lit by the 1.5V power supply, but remember, the were drawing more current from the battery and would thus make the battery drain faster. If there were more LEDs connected to the battery, they would draw even more current from the battery and drain it even faster.
For the second setup, I decided to put everything I had learned together and wire the two LEDs in parallel to my 9V power supply - certainly too much juice for the LEDs alone so I would have to use a resistor for sure.
To figure out what value I should use I went back to the trusty formula - but since they were wired in parallel there is a slight change to the formula when it comes to the current - I.
R = (V1 - V2) / I
where:
V1 = supply voltage
V2 = LED voltage
I = LED current (we had been using 20 mA in our other calculations but since wiring LEDs in parallel draws more current I had to multiply the current that one LED draws by the total number of LEDs I was using. 20 mA x 2 = 40 mA, or .04A.
And my values for the formula this time were:
R = (9V - 1.7V) / .04A
R = 182.5 Ohms
Again, since the variety pack didn't come with that exact value resistor I attempted to use the two 100 Ohm resistors bundled together in series to make 200 Ohms of resistance. I ended up just repeating the mistake that I made in the last step again though, and wired them together in parallel by mistake and so the two 100 Ohm resistors only ended up providing 50 Ohms of resistance. Again, these LEDs were particularly forgiving of my mistake - and now I have learned a valuable lesson about wiring resistors in series and in parallel.
One last note about wiring LEDs in parallel - while I put my resistor in front of both LEDs it is recommended that you put a resistor in front of each LED. This is the safer better way to wire LEDs in parallel with resistors - and also ensures that you don't make the mistake that I did accidentally.
The 1.7V LEDs connected to the 9V battery lit up - and my small adventure into LED land was completed.
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For my project I am going to us a 9 volt battery and have 20 LEDs and each LED has a voltage of 3.2-3.4V at 20 mA each. I want to run every LED in a parallel circuit.
How would I go about doing this? What size resistors would i need? I bought 1/4 watt resistors could i use those? or would i need to get a 1/2, 1/4, 1/8, ect.?
Also do i need to have 20 different resistors for each LED or could i just use one at the bottom of the parallel circuit?
Do I have to keep in mine about anything else? I would be using a 9 volt 1000+ mAh battery
Thank you in advance!!!
Thanks in anticipation
Prewired 12V LEDs
High quality LED, soldered to a resistor for 12V (including 12V automotive) use. Strong connections are neatly wrapped with heat shrink and attached to colour coded wire terminating in stripped tinned connections.
Technical Spec
High Power 3mm LEDs - 7000mcd
12V DC
Presoldered Resistor
20 mA
20 cm Wire Length
High Intensity - Low Power 0.36W
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You can do it like this:
+12 ------------------------------------------------------------------
| |
R R
| | o o o as many as you like
led led
| |
gnd ------------------------------------------------------------------
Concerns:
- multiply the current per LED (shown on the specs for each LED, above you say it is .020A) times the number you want (100?) = 2A. Therefore get a power supply that can provide 12V for at least 2A (more is ok). Repeat this formula if the current is different to determine the max ps current needed.
- I wouldn't group a big pile of these RIGHT TOGETHER of in a confined space. Each dissipates .36W (almost all in the resistor), so three or four tightly together (or out of air flow) will be a watt of power, and will start getting a little hot. For this issue, a much better solution would be to create your own LED 'cluster' of series LEDs and a single resistor setting the voltage. If you created 33 sets of three LEDs on the same line with a single custom resistor value to limit to 20mA, then your power dissipation would go way down. This is because the power would go toward light instead of being boiled away in the resistor - resistors just waste!
So this:
12V+ ------- led ------- led ------- led ------- resistor ------- gnd
3.2V 3.2V 3.2V 12v-9.6=2.4V
2.4V /.020A = 120ohms
2.4V x .020A = .048W
Note this dissipates lots less in the resistor. and there are only 33 resistors instead of 100. it uses the voltage much better.
But if you can easily get air to these guys, you'll be fine.
So: in your example above you have a 5V supply, and an LED that uses 20mA at 3V, then the voltage 'left over' across the resistor on any one leg is 2V. 2V/100ohms = .020A (20mA). Thus each leg would correctly function with a 100ohm resistor.
Each leg draws the 20mA as a separate subsystem, as it's own circuit. You should see it as if the + point in the drawing above were two independent batteries. The only limitation would be the maximum current available at the + point... remember that mA are thousandths of Amps, so a power supply rated 1A @ 5V would provide enough power for 50 LEDs... 50 x .020A (20mA) = 1A... batteries can do much better than 1A without much voltage reduction.
I hope this helps you think about this in a way that allows you work with LEDs and resistors in the future. Please reply with any questions or comments.
Sorry I wasn't clear. I was referring to step 8 where he used the 9V and wired the two LEDs in parallel.
R = (9V - 1.7V) / .04A
R = 182.5 Ohms
So in that case it would be a 200ohm resistor for each correct?
As soon as you supply a voltage higher than the CHARACTERISTIC VOLTAGE of the LED (about 1.7v for a red LED) the LED will want to take a very high current and it is this CURRENT that will over-heat the LED.
See 30 LED Projects on Talking Electronics website:
http://www.talkingelectronics.com/
http://www.talkingelectronics.com/projects/30%20LED%20Projects/30%20LED%20Projects.html
Colin Mitchell
However just for two resistors you can do (R1 x R2)/(R1 + R2) which is why your 100/2 = 50 worked in the previous step.