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Discussion Starter #1
I'm putting together some new LED lights for a mod I'll be posting later but I've run into a snag with the resistors needed. The circuit needs 39ohm 1W resistors but as usual Radio Shack doesn't carry any 1W resistors under 100ohms.

So here's the question: should I be concerned about using a resistor with a lower wattage rating than the circuit needs? I realize it comes down to the heat generated by the circuit but should I be concerned?

For those like me, here's the math:
3x LED 3.2v 100mA
13.5v source

(13.5v - (3 x 3.2v)) / (100mA / 1000) = 39ohm
13.5v x .1A = 1.35W

One option I have is to combine some 20ohm 1/2W resistors in series. I know this will combine the resistance but I wasn't sure if it would combine the wattage as well.

I've already sourced the necessary 1W resistors from the guy I purchased the LED's from but before i buy them I'd prefer to use what I have on hand.
 

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so are you hooking the leds in series with one another or are you hooking all their anodes and all their cathodes together (parallel)?

i think that the math may be slightly off but i want to make sure that i am on the same page as far as how the led's are wired in respect to one another.
 

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Discussion Starter #3
I'm wiring three series of three LED's each. Each series needs 39ohms 1W resistor. I've never wired a parallel LED circuit as I understand it's not quite as efficient.

The LED's themselves are higher power than I'm used to dealing with. Previously I've used just high output 5mm LED's but these are multi-chip flux LED's, 50,000mcd, which is why they're drawing 100mA each. That's what's putting me over the wattage on the resistors :-(
 

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if each led is 100mA and there are 3 in each circuit then the total current draw of each circuit is 300mA not the 100mA that you used to calculate power of resistor needed. so you are actually going to need a resistor 3x that of your calculation below. P=IxE where I=300mA and E=12v (or 13.5 or whatever you want to use as it will vary) so P=.3Ax12v or P=3.6w

so you will need 4 160ohm 1 watt resistors wired in parallel. that will bring the resistance to 40 ohms and the power capability to 4 watts
 

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Discussion Starter #5
This is why I always ask before I solder :)

Thanks for the refresher on my math. As luck would have it, the Shack does have 160ohm 1W resistors in stock. I'm not thrilled about having to use 4 per series but I'll manage to squeeze them onto the board somehow.

Now it's off to inhale some flux!
 

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its nice to see someone using ohms law on here. :) the MECP basic study guide has a good section on it as well as a bunch of car audio information as well.
 

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This reminds me of my first car, a well used Dodge Dart. It started fine, but died when I released the key to the run position. From the maintenance book I realized that the start circuit bypassed a large, low resistance resistor. I walked to the nearest Radio Shack and bought a couple of their high power resistors, which are packaged in 4" ceramic rod (something like 10 ohms, 10 W). I wired those in parallel and drove home. Latter I bought the proper replacement from a dealer.

So, yes, if you are pushing the power ratings of the available resistors, think parallel.
 

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Sorry Yoda, I have to disagree with your logic here. If you have 3 LED's in series, and each will have a voltage drop across it of 3.2V with 100 ma. going thru it, then the voltage drop across 3 in series will be 9.6 volts with 100 ma. going thru the whole string. Since the series string will be dropping 9.6 volts, and the source voltage is 13.5 volts, that leaves 3.9 volts that must be dropped across the resistor. R=E/I 3.9V/.1A=39 ohms. You can't draw 300 ma. thru a diode that is only good for 100 ma. (it will burn up), and if they are in series they will all have the same current going thru the string.
Wattage wise, P=ExI 3.9V x .1A = .39 wats to be actually disapated by the resistor. A good rule of thumb for resistor wattage is twice the power that is being disapated, so a 1 watt 39 ohm resistor is a good choice for each series string of 3 LED's.
Coffee was right!
Joe
 

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Coffee was just a little off in his wattage calculation. The resistor only has to disapate the power of 3.9v x .1A = .39Watts. All the rest of the power is disapated by the LED's.
I still think it should be a 1 watt 39 ohm resistor for each string of 3 LED's in series.
Joe
 

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the resistor is used to drop the voltage down to the appropriate level as well as limit the current on the circuit. and the voltage drop needed in the circuit will not need to be as great with there being 3 LED's in series (each acts as it's own "resistor" as it disapates both heat and power/light) but the current draw is added up for each in the circuit. 100mA leds are pretty heavy as most are in the 20-30 mA range, and then add the fact that you have 3 in series to that equation... try a 1 watt 40 ohm resistor on there, and let me know how that works out for you ;-)
 

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If I am understanding what you trying to do then there is 2 ways to wire the resistor/s you need.


1. You can put 1 resistor with each set of 3 leds. (3 resistors)

or

2. You can put 1 resistor for all 3 sets of 3 leds. (1 resistor)


Your wiring could look like this.....

1.

V-------D----D----D-----R-------G (D=LED, R=Resistor, G=Ground, V=12v/13.5v source)

You would make 3 of these and connect the Vs together and the Gs together and then connect the V to 12v source and the G to Ground.

In this circuit the resistors would need to be 39ohm, 1/2 watt resistors.

or

2.

. . . . . . . . . . . . . .. .. /--V3------D----D----D-----G
. . .V1---R------V2---l---V3------D----D----D-----G (ignore the periods they are for spacing.)
. . . . . . . . . . . . . .. .. \--V3------D----D----D-----G

This way you would only use 1 resistor in series with the 3 parallel LED strands.

V1 would get connected to 12v source.
You would connect the three V3s with the V2 and the G to Ground.

In this circuit your resistor would need to be a 13ohm, 1.17watt or 2 watt.
or
Any combination of resistors wired in series to give you 13 ohms, like a 10 ohm and three 1 ohm all 2 watts.

Of course....

E=I/R where E=voltage, I=current, R=resistance

and

P=I/E where P=watts, I=current, R=resistance


 

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Yoda E SC said:
the resistor is used to drop the voltage down to the appropriate level as well as limit the current on the circuit. and the voltage drop needed in the circuit will not need to be as great with there being 3 in series (each acts as it's own "resistor" as it disapates both heat and power/light) but the current draw is added up for each in the circuit.


In a series circuit the voltage value of each load (three 10 volt loads equals 30 volts needed) and resistances are added, current stays the same as there is only one path for the current to take.

In a parallel circuit the voltages are all the same and the current of each leg is added to get the total current.

The resistances are a little more difficult to calculate.

Resistance of a parallel circuit uses this formula:

[SIZE=+1]1/Rt = 1/R1 + 1/R2 + 1/R3 +...[/SIZE] [SIZE=+1]Rt = R (t)otal

[/SIZE]
[SIZE=+1]


[/SIZE]
 

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you could make that 13 ohm, 1.17w resistor by putting 3 39 ohm resistors in parallel.

In effect this is making the 3 separate circuits, with an optional cross link between the resistors and the LEDs.
 

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:roll:

guys, go back and review your posting before these are forever locked, there's a a good bit of incorrect information in some of these posts. But, there is also some good information too, I'll leave it for you to decide...

Honestly, guys it's just OHMS law - the very basics of electronics we're talking about here.
 

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Resistance of a parallel circuit uses this formula:

[SIZE=+1]1/Rt = 1/R1 + 1/R2 + 1/R3 +...[/SIZE] [SIZE=+1]Rt = R (t)otal

[/SIZE]
[SIZE=+1]


[/SIZE]
also if the resistors are the same value it is R/N where R=resistance and N=number of resistors in the circuit. but again, this only works when the resistors are the same value.
 

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Discussion Starter #16 (Edited)
Ok, I've settled on two series of 3 LEDs connected in parallel to a single resistor as suggested by BigTzElement, mostly because it works with the resistors I have on hand. Here's the circuit and math for everyone to double check. I upped the voltage to 14v to make sure it's designed to handle the max output of the E's electrical system.

. . . . . /-> D--D--D--G
14v---R1-|-->
D--D--D--G

D = 3.2v 100mA
R1 =
22ohm 1W or 2x 47ohm .5W parallel

R1 = 14v-(3*3.2v) / ((2*100)/1000)
R1 = 4.4 / .2
R1 = 22ohm

RW =
14v-(3*3.2v) * ((2*100)/1000)
RW = 4.4 * .2
RW = .88W

22ohm 1W or 2x 47ohm .5W

Does that all look right to everyone?
 

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Yes, that would be correct.

22ohm 1 watt should be fine.

I would however make sure it is dissipating the heat load like you want. Run the circuit and make sure it is not getting too hot. If it is use a 2 watt resistor.

:)

 

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And one other tidbit. You have to be careful to match the forward resistance of high power leds if you run them in parallel at capacity. The diode with the lower actual resistance will end up taking more load than the higher resistance one and end up being over spec. Either allow a little fudge factor or measure your diodes to make sure they match. That's why cheap arrays are usually in series.

Sometimes, tolerance is intolerable.
 
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