BadNewsBears
Diamond Member
Ever touched metal in a car, while holding a hot lead . 12v at say oh 120 amps.
Sure the metal in a car is grounded, but do you think that ground suplies power?
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How to: Wire 2-6 red LEDs to a home outlet or batteries
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Ever touched metal in a car, while holding a hot lead . 12v at say oh 120 amps.
Sure the metal in a car is grounded, but do you think that ground suplies power?
You would want the 3V batteries in series to get 6V from them. If you use the red LED's in your eBay link, the maximum current you want to put through them is 30mA. The 150 ohm resistors I used in my diagram will limit the current to each LED to about 27mA.
How it works is each LED expects 1.8-2.2 VDC (according to your link), at a maximum of 30mA. So with a 6 volt source, you want to pick a resistor value that will limit the current to something under 30mA while dropping ~4 volts across it. To keep from sending too much current through the LED, I like to pick a resistor value that will allow for slightly less current to flow.
So assume the maximum current you want going through each LED is 25mA. You want 4 volts across the resistor, so using ohm's law where resistance = volts/amps, you get R = 4V/0.025A = 160 ohms. Since 160 ohms isn't a very common resistor value, I picked the closest common value - 150 ohms.
If you wire the LEDs in parallel, then you just need to add another resistor/LED pair in parallel with everything else that's there. In a circuit, everything that's in parallel with each other is at the same voltage potential. So each resistor/LED pair wired in parallel will be using the same 6VDC provided by the source. Using the resistor values I provided, you want to add 27mA to the amount of current being drawn for each resistor/LED pair.
The amount of time that those 3V batteries will last depends on how many LED/resistor pairs you have hooked up to it. The power rating on those batteries is 540mAh, which I believe means that the battery is capable of supplying 540mA for one hour before it's no good (someone correct me if I'm wrong here). So that means that with one resistor/LED pair, drawing 27mA, the batteries will last ~20 hours. If you use two resistor/LED pairs, that will draw 54mA, meaning the batteries should last ~10 hours. If you want the batteries to last a little longer, you'll need to use bigger batteries such as AAA or AA batteries. Though, of course, that means needing to find a place to put them that won't detract from the flower.
As for using the flower itself as one of the conductors, that should work as long as you check to make sure it will conduct electricity well. However, you may have some problems trying to solder to it - some metals just don't take solder very well. Regarding the concerns of touching it, with only having 6V going through it, and especially if you're powering it with batteries, you shouldn't have any worries about touching it.
JW
CaptainKahuna
Platinum Member
Hey now, don't hijack my thread!
yes, I have actually.. Nothing will happen. Ever.
Car batteries are harmless. 12 volts will almost never shock you, regardless of amps. Voltage is like electrical "pressure". The resistance of your skin is too high. At 12v, the pressure isn't great enough to flow any current.
Also, most car batteries are capable of outputting considerably more than 120A.. more like 300.
😛
Spencer278
Diamond Member
To increase the life time of the batterys you could wire it in series and not use a bunch of resistors but then you will run into problems adding more LEDs and if one LED dies you take out the entire strand. I would recommend putting the LEDs in series and parral and experiment with finding the number of LEDs you can have in series before the start to dim alot I would guess at least 5 or 6.
FrustratedUser
Lifer
This is easier than you think. You can run it from 120VAC directly.
Estimate each LED has a drop of ~1.5V
6 x 1.5V = 9V
Estimate ~10 - 15mA of current throough the LED.
120V - 9V = 111V. This is the voltage drop you need over the resistor before the LEDs.
Ohms law:
111V / 15mA = 7400ohms
Ohms law again: (Power dissipation in resistors)
P = R x I^2 ---> P = 7400 x 15mA^2 = 1.67W
So you need to buy a 2W power resistor (with the value closest to 7400 ohms) + 6 LEDs to make this work, which should be way less than $10.
Estimate each LED has a drop of ~1.5V
6 x 1.5V = 9V
Estimate ~10 - 15mA of current throough the LED.
120V - 9V = 111V. This is the voltage drop you need over the resistor before the LEDs.
Ohms law:
111V / 15mA = 7400ohms
Ohms law again: (Power dissipation in resistors)
P = R x I^2 ---> P = 7400 x 15mA^2 = 1.67W
So you need to buy a 2W power resistor (with the value closest to 7400 ohms) + 6 LEDs to make this work, which should be way less than $10.
FrustratedUser
Lifer
Real men use 120VAC.
FrustratedUser
Lifer
... and he could power around 70+ LEDs at once. :thumbsup:
imported_FishTaco
Golden Member
Try candle power forums
They got a bunch of people there who know LEDs. I would start in the homemade and modified light section.
They got a bunch of people there who know LEDs. I would start in the homemade and modified light section.
AlienCraft
Lifer
While your skin may have enough galvanic resistance to insulate you from the shock, it certainly will burn if held in contact with a direct short between two poles of a battery say a wrench or a ring< Common disfigurement of auto mechanics.
AlienCraft
Lifer
We need to know the voltage of your LEDs. That P.S. may be all you need, Unmodified even, depending on the voltage of the LEDs in question.
The voltage of the LEDs is mentioned in the eBay listing. Unless he wants to burn out the LEDs, then he'll definitely need some resistors to limit the current going to the LEDs. Otherwise, the LEDs will draw as much current as is available to them. With a 3V 1000mA power supply, the LEDs would attempt to draw all 1000mA and burn out as a result. Even if the LEDs were 3 volts, you would still need some sort of current limiting resistor in series with the LED to prevent it from blowing out (unless you had enough LEDs hooked up so that when combined, they drew 1000mA).
JW
TitanDiddly
Guest
I'm no EE or anything, but I don't think that's right.
OP:
Most LEDs draw about 20ma. 5V is the most common voltage; they're actually just 3v ones with built-in resistors. You could probably run it off the 5V line in your computer if you have it near the case. If you are using 3v LEDs, 2 AA batteries will do the trick. Otherwise, find a 5v wall wart and just wire everything up in parellel(assuming they're 5V LEDs)
Spencer278
Diamond Member
Your correct as long as the voltage across the LED is with in the given spec you will not harm it. You don't need a resistor unless you need to reduce the voltage, and a better way to reduce the voltage is to added LED in series.
KingNothing
Diamond Member
Ok I finished my attempt to add a red LED to the rose. It looked pretty cool, especially with the lights off. No pics, sorry. I'm not going to do any more like that though for two reasons:
1) Too big a pain to get the wires glued down to the inside of the spiral. This could possibly be addressed with a better, faster-drying adhesive.
2) In my opinion, wires snaking down the spiral detracts from the looks too much. The only way I would consider adding an LED would be if I could use a watch battery or something and hide the entire assembly inside the rose.
1) Too big a pain to get the wires glued down to the inside of the spiral. This could possibly be addressed with a better, faster-drying adhesive.
2) In my opinion, wires snaking down the spiral detracts from the looks too much. The only way I would consider adding an LED would be if I could use a watch battery or something and hide the entire assembly inside the rose.
BadNewsBears
Diamond Member
Delcotron cs144 150 amps. at 2100 rpm's it puts out 150 amps never more. Now it can go up to 14 vdc for charging, and when it starts to fail it may even reach 21vdc. When I went outside grabbed the bare brake light lead I have hanging out tapped the breaks while holding onto the steel body I got a good zap.
BTW My battery is a dual optima red top with an isolator. Like 800 CCA apiece. But they are isolated from my alternator 80% of the time.
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