light bar easy question

Thread Starter

frozenlead

Joined Mar 30, 2008
19
I'm building a few light bars for my dorm.

Right now, it'll be two light bars put together, one white, one blue, each with 7 led's.

Now, if i put all 7 LEDs in parallel, and the LEDs are rated at 5v 30mA,
then if i give the circuit 5v and 210mA, none of them will burn out will they?

It seems too simple. A lot of people say you put resistors in series with LEDs, but i'm not using series to avoid a huge voltage need.

thanks
 

Externet

Joined Nov 29, 2005
1,410
Each led should use a resistor in series, but if you really want to use them without resistors, you must supply the leds with not more than the Vf forward voltage from their datasheet. Which is usually lower than 5V.
Some may be brighter than others that way, but works.

If you do not know the forward voltage but you know the current you want them to operate, connect the correct resistor to one led to make it wwork at that current, measure the voltage across the led; and that is the voltage you have to supply directly to the led after removing the resistor.
Every color has a different value.
Miguel
 

Thread Starter

frozenlead

Joined Mar 30, 2008
19
I mistaked - the reverse voltage was 5V, not forward, it doesn't list the forward.
Well, my power supply is 9v and 210mA, so if I throw in a good 25Ohm resistor before the 7 LEDs in parallel, this would reduce the voltage a little greater than 5V, thus giving the parallel LEDs less than 5v to work off of, right? Perhaps 30 would go farther?

Or should I go ahead and use a bit more?

I don't actually have any of the parts yet, I want to get the design down first, so I can't test. Or should I order, then put it together?
 

blocco a spirale

Joined Jun 18, 2008
1,535
With a 9V supply and white or blue LEDs, put a 180 - 220 Ohm resistor in series with each LED and connect them in parallel.
 

Wendy

Joined Mar 24, 2008
21,847
LEDs are current devices, so they should always have something to limit the current flowing through them, it can be a resistor or a constant current device. They also drop a relatively fixed voltage, if you have several from the same family next to each other they should drop the same amount of voltage each, the variations in forward voltage drop will translate into variations in current between different LEDs if they are wired in parallel with a common resistor, which is pretty common in things like LED flashlights and whatnot.
 

SgtWookie

Joined Jul 17, 2007
22,201
Red and blue LEDs usually have a Vf around 3.6v to 4v. Let's just say 3.8v for the moment.

If I read your statement correctly, each light bar is going to have 7 LEDs, two light bars per assembly, for a total of 14 LEDs, each rated for 30mA. 30mA x 14 = 420mA. Oops, that's twice as much current as you have available.

But, since you're using a 9v supply, you can run two LEDs in series, 7 strings in parallel.
You're still going to need current limiting resistors, one per string.
Rlimit = (VoltageSupply - VfTotal) / LEDcurrent
Rlimit = (9v - (2 x 3.8v)) / 30mA
Rlimit = (9 - 7.6) / 0.03A
Rlimit = 1.4 / 0.03
Rlimit = 46.666... Ohms
The nearest standard value is 47 Ohms. This page has a table of standard resistance values.
Now you need to calculate the wattage of the resistor.
P = EI (or, Power in Watts = Voltage x Current)
P = 1.4 x 0.03
P = 0.042 Watts
For reliability, you double the result.
P = 0.084 Watts
You could use a 1/10 Watt or larger resistor on each string.

The blue LEDs will likely have a Vf somewhat lower than the white LEDs. In order to be able to use all the same resistors, you can run one white and one blue LED in series.

See the attached for an example schematic.
 

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Thread Starter

frozenlead

Joined Mar 30, 2008
19
i think there's been a misunderstanding -
i'm building two different bars. I want to be able to use white when it's not so late, and blue when it's really late, as to not disturb my roommate as much, so the total current is 210mAh per circuit, not 420.

Okay, so I got that i have to use a resistor in each chain of the parallel circuit, which is 7 chains. Since what we've resulted is wit 47 Ohms per branch, for two LEDs, I think i'm correct in saying that using 23 or so ohms for one in a branch, which makes me feel good, because it's what i said before - except i need 7 of them, not one.

I guess i don't understand that part. Voltage across series is added..so, considering my original circuit, one resistor in series with the 7 parallel leds, the resistor should lower the voltage for the leds, which would each receive the same voltage after reduction (and, consequently, the same current because they're proportional).
Why doesn't this work out? Is there something about an led i'm missing?
 

SgtWookie

Joined Jul 17, 2007
22,201
OK, let's re-visit the formula I put in my prior reply:
Rlimit = (SupplyVoltage - VfLEDtotal) / LEDcurrent

If you used 23 Ohms as a current limiting resistor in each parallel string, you would wind up with:
I = E/R
I = (9V - 3.8V) / 23 Ohms
I = 5.2/23
I = 226 mA :eek: Bye bye, LED's.


If you run seven LED's in parallel, you're going to need much larger resistors to drop the current down. Since your supply is 9v, you would really be much better off running two LEDs in series. That way, you will dissipate much less power in the current limiting resistors. You could run 8 LEDs in four parallel strings with 4 current limiting resistors, and have power left over.

Let's try it with one LED per string:
Rlimit = (9v - 3.8v) / 30mA
Rlimit = 5.2 / 0.03
Rlimit = 173.333 Ohms
180 is the closest standard value. Now calculate the resulting current:
I = E/R
I = 5.2/180
I = 28.9 mA (rounded up)
Now the power consumed:
P = EI
P = 5.2*.0289
P = 0.15028 Watts, or 150.28 mW. If you refer to the prior post, you'll see that the limit resistors in the series circuit were only consuming 0.042 Watts. If you run them in 7 parallel strings, you will be wasting nearly 3.6 times as much power in the resistors.
Going on with the power calculations, let's see what wattage resistors you'd need. Doubling 0.15028 = 0.30056; you would need to use 1/2 Watt resistors.

When you are running LEDs in parallel, you must have a current limiter in each string; otherwise you will very likely wind up with a "domino effect" of one string burning out, the remaining LEDs getting more current would rapidly burn out. Sort of a Chernobyl effect. :eek:

Have a look at the attached schematic. Each string of LEDs would use 30mA, or 120mA total; well within the capability of your 9v 210mA supply. S1 switches back and fourth between white and blue. You could use a SPDT center off switch (ON-OFF-ON).
 

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Thread Starter

frozenlead

Joined Mar 30, 2008
19
Okay, i've mulled it over a few times and i think i'm getting it.

I've never encountered the equation you used for Rlimit - i was just working with the resistance equation.

Those resistors are 47 ohms? And since in each string there are two leds, wouldn't one be brighter (even if slightly) than the other? Or would it not be discernible?

This is all assuming those Vf are correct...so if I go through the procedure to find it when i get the parts, i suppose i might be changing a few things.
 

SgtWookie

Joined Jul 17, 2007
22,201
Okay, i've mulled it over a few times and i think i'm getting it.
Take your time. :) Better to be sure about things beforehand than to go off willy-nilly and let the magic smoke out of things. :eek:

I've never encountered the equation you used for Rlimit - i was just working with the resistance equation.
It's pretty standard. You usually want to have at least a volt "headroom" above the total Vf of the LEDs wired in series. If the supply is not regulated, you really should have more headroom - but your power consumed in the limiting resistors will rapidly increase.

Now, you didn't say if your supply was regulated or not. If it is a "wall wart", then it most likely is not regulated; it will output 210mA @ 9v. If the load is less than 210mA, then it's output may be much higher. You will need to verify that.

Those resistors are 47 ohms?
That is what I calculated based on two LEDs with a Vf of 3.8 at 30mA.
And since in each string there are two leds, wouldn't one be brighter (even if slightly) than the other? Or would it not be discernible?
Not at all. In a series circuit, if 30mA goes in one end, it comes out the other. You don't "lose" current in a circuit.
This is all assuming those Vf are correct...so if I go through the procedure to find it when i get the parts, i suppose i might be changing a few things.
Exactly.

You say the LEDs you ordered only have a Vr specification of 5v.
Normally, the Vr is given along with a MAX Vf @ current, and "typical Vf", which is always lower than the MAX Vf.
It is the typical Vf that you want to use in the calculation. Most of the LEDs you receive will be within a few % of the "typical Vf", but your mileage may vary.

I purchased 240 blue LEDs that were on tape (for automatic insertion). I assume they came from the same batch. I measured each LED's Vf with a constant current of 22mA. The minimum Vf was 3.73, max was 4.12, median was 3.86. There were just a few that were as much as 5% above or 5% below the median; those would be noticeably brighter or dimmer unless they were paired with LEDs that had Vf's with a correspondingly higher or lower Vf.

You can make a constant current source with your 9v supply, a fixed resistance, and an LM317 or LM317L voltage regulator.
You connect your +V to the Vin terminal, connect 40 Ohms of resistance between the Vout terminal and the Vadj terminal, and your constant 30mA will come out of the Vadj terminal. This is documented in National Semiconductor's datasheet for the LM117/LM317.
Since 40 Ohms is not a standard resistance, you could use two 20 Ohm resistors in series, or two 82 Ohm resistors in parallel, or three 120 Ohm resistors in parallel.
 
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