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Old 12-22-2002, 06:28 PM   #1
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Do you double current or volts in ohms law in parallel wireing?

Im using parallel wireing for a pair of 3.6volt leds with a current of 20mA off a 9volt supply. Using R=V/i, is i suppost to be double meaning R=(9-3.6)/.04 ? Or should it be R=(9-7.2)/.02 ?
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Old 12-22-2002, 07:06 PM   #2
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If you wire them in parallel then your curent would be double as your resistance will be half.
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Old 12-23-2002, 12:42 AM   #3
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Actually, Ohm's law doesn't come into this.

It depends on what happens when you parallel something - if you parallel resistors, then they divide, if you parallel capacitors, then they add.

What is the question you are really asking?

What current limiting resistor will you need? The same as with one since the current draw from both LED's is going to be negligible compared to the percentage of variance in the resistor you're going to get.

How much power will you need? the 9v supply should be fine.
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Old 12-23-2002, 03:11 PM   #4
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I think I see what you are asking, and iff you run two LED's in parrellel of a 9 volt power suppl\y, you will probably fry the LED's. You would be better off running them in series, that way they can handle 7+ volts without a problem. In parrallel, they would still only need 3.6 volts (which is real high for 1 led???) however, your total current draw is going to be 40 milliamps. Negligable as boomer said. Do yourself a favor and by a 5K pot and figure out where the LED's get bright, back off just a scoch, measure with voltmeter(with the ohm check) then get a resistor with that value.
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Old 12-23-2002, 04:25 PM   #5
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You will always need a resistor for current limiting. As you can see the value you need is fairly low so 5k is fairly excessive, If I read your message right then I would suggest the attached diagram... not the bottom illustartion.

Oh BTW you can parallel more as you want with the circuit safely with this method...

Question I have is you said the LED are 3.6V, are they LED clusters or are they single LED with such volt drops? Cos most are between 0.7V - 1.0V, please check this as you may have to adjust the resistor value.

remember also if you are looking for brightness you cant really drive themwith D.C. You will need to pulse drive the LED so you can incease the input power while not damaging the LEDīs ... okay that is a few more components but it really depends on what you are after...
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Old 12-24-2002, 08:57 PM   #6
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racinZrooky,

I guess from your equation that you want to calculate the resistance of the resister required for your LED. Am I right?

If you want to connect LED in parallel then you better have seperate resistor for each LED. However the current consumtion will be doubled because you have 2 loop of circuit. Each R value will be R=(9-3.6)/0.02 = 270 ohms. The total current consumtion is 40 mA.

If you want to save some 20mA for extra runtime (half a second or so from my calcuation) and save some weight then you can connect them in series also. You last equation is correct for series connection and you need only 1 resistor. R=(9-7.2)/0.02=90 ohms.

good luck,tifosi
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Old 12-26-2002, 04:34 AM   #7
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Tifosi,

Sorry but the bit you said about more run time by wiring them in series is total rubbish!

If 2 LED are working at their desired parameters how can one way save power by wiring them differently??? You mention current but forgetting power!

1 LED consumes (P=VxI) then P = 3.6x0.02 = 0.072W
then 2 in parallel then would mean 0.072x2 = 0.144W

Yes there is more current drawn but the voltage drop is at 3.6V.

Now when connected is series power consumption for both LED =

P=(3.6+3.6)x0.02 = 0.144W so it is exactly the same right?

The only reason I suggested in prallel is becasue the fellow can add more LED if he wishes without having to re-calculate different resistor values everytime and he cant do that for 3 or more LEDīs to have the same brightness!Also in parallel means should one LED circuit gets damage then the other LEDs will still work but in series one goes then all goes, which is important in certain circumstances if driving in the dark!!! at least you can still see your car with one headlight!!! Think in real cars each headlight are in parallel and independent of each other for safety reasons...

BTW weight saving of one resistor? For parallel connection the resistor power rating is (IxIxR) 0.02x0.02x270 = 0.108W so you can do down to 1/8W resistor pacakages and that really wont weight you down cos they are smaller than a grain of rice! (actually about half a grain in size!) and anyone who wants to use 2 sets of LED are not going to worry about the having a grain of rice on their car!!!
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Old 12-26-2002, 05:45 AM   #8
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Talking I was kidding

double posted

Last edited by tifosi; 12-26-2002 at 06:15 AM.
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Old 12-26-2002, 06:14 AM   #9
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Talking I was kidding

hi sonneteer,

I was kidding about the weight and run time. Who will care about 0.5 second and 0.05 g ? Not me. It's my false not to make it more obvious (about being jokeing) and forgot to put some smilely faces

However if you connect 2 LED in parallel they will surely consume more enegy because you also lost the power in resistor too. You forgot about it in your first paragraph but mentioned about it in your second paragraph.

The correct calculation are:

parallel:
1 LED+R consume power (P=VI) = (3.6+5.4)*0.02=0.18W
2 LED+R consume 2 times power of 1 LED+R = 0.36W
* 5.4V is voltage drop in resistor

series:
P=VI= (3.6+3.6+1.8)*0.2 = 0.18W
* 1.8V is voltage drop in resistor

It's obvious that connecting them in parallel will consume more energy because of the extra energy wasted in the current limiting resister. That right?

About the runtime. I assume that mod motor will consume 30A = 9*30=270W so the extra energy 0.18W energy wasted is 100*0.18/270 = 0.067% and 0.067% of 300 seconds is a whooping 0.2 seconds!! That's just a roughtly estimate runtime to be used in my failed joke. Please don't be serious about that.

You're correct about the LED damage issue but I never seen an LED that damage in correctly designed circuit before.

Next time please being more careful about the word "rubbish"

regards,tifosi

Last edited by tifosi; 12-26-2002 at 06:17 AM.
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Old 12-26-2002, 11:50 AM   #10
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Tifosi,

Sorry I thought you were referring to circuit 3 in my diagram when I made that comment. I generated that diagram becasue that is what I thought PitCrew was suggesting to do with one resistor and 2 LEDs in parallel which would mean the current would double to 40mA. As for a complete in series circuit which in that case you are quirte right.

Drawing on several interpretations on what has been said to racinZrooky, the examples I was using should help him to calcualte the values and not having to guess, and should he want to add more lights he can expand on the simple circuit before embarking in a combined series/parallel circuit.

I thought you were joking cos there was a but really in a TC or truck any LED installation ( eg headlights) can be quite easily damaged in a crash that is why I suggested parallel wiring especially if car is to be driven in the night! I didnt mean the LED will blow by themselves...
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Old 12-26-2002, 09:04 PM   #11
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hi sonneteer,

I wasn't mean to posted anything againt your comment at all. I just want to explain the misunderstanding in racinZrooky calculation. It might be useful for him in the future and could help some high-school student out there that may confuse with the same thing.

BTW I though that your diagram is very helpful and should help anyone that want to install LEDs in their car .
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Old 12-27-2002, 04:15 AM   #12
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hey no probs Tifosi.... there is always more than one way to skin a cat!!! (or always more than one way to wire up some LEDīs!!!) Every applications will have its merits and faults, I think we both have shown all of them!!!

Still trying to figure what kind of LED that has such a large volt drop of 3.6V.... hey racinZrooky! can you shine a ligt (ahem) on the type of LED you are using!!!
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Old 12-27-2002, 07:15 AM   #13
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sonneteer,

try this link.

http://www.lumex.com/

Not exactly 3.6V but close enough.

tifosi
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