4 led's parallel resistor advice?

Started by davepedals, December 13, 2016, 07:29:01 AM

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davepedals


Using four 3mm super bright LED's parallel wiring, wanting them to remain bright as possible.  9V pedal adapter.  Would 2.2K be a good value? Suggestions?

INFO:

Usage voltage(V): 3.2-3.4
Current(MA): 20
View angle: 20 - 25
Luminous intensity(MCD): 18,000

Not sure on ac adapter current, perhaps 300 - 500 ma.

Many Thanks!
Dave
dave

bluebunny

The forward voltage and the maximum current values for the LEDs will tell you the minimum resistance value you can get away with for your given power supply.  Now whether that blinds everyone within a two-mile radius is another matter!  I'd say, try it out to find a value that suits (get your breadboard out!) - you'll have an answer is less time than it's taken the two of use to post this question and answer.  :)  (And give each LED its own resistor, or else they'll fight.)
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Ohm's Law - much like Coles Law, but with less cabbage...

davepedals

Good idea, I'll try different values, been up all night, not good for thinking!  So - no good if I put one resistor from positive then connect it to all positive leads? Multiple resistors will increase pcb size as they all need to lay down.  But I'll mess with it after some sleep.  Perhaps I can do a re-draw.  Thanks!
Dave
dave

EBK

Quote from: davepedals on December 13, 2016, 08:06:50 AM
Good idea, I'll try different values, been up all night, not good for thinking!  So - no good if I put one resistor from positive then connect it to all positive leads? Multiple resistors will increase pcb size as they all need to lay down.  But I'll mess with it after some sleep.  Perhaps I can do a re-draw.  Thanks!
Dave
If you put LEDs in parallel with only one current limiting resistor, generally only the LED with the lowest forward voltage will light up (for example, if you put a red and green led in parallel, only the red one will light up, which can be very useful in some contexts).  You may get mixed results when the forward voltages are similar. 
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Technical difficulties.  Please stand by.

davepedals

That makes sense, definitely have to do a redraw then.
Thanks!
Dave
dave

EBK

If your LEDs will be off-board, you could wire your current limiting resistors directly to the LEDs, saving board space.  I'd recommend covering the resistors and joints with heat shrink tubing if you go this route, to keep things neat.
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Technical difficulties.  Please stand by.

davepedals

So after a bit of research, apparently this is what happens:
dave

davepedals

Yes I'm building a separate PCB , that's really a great idea, I will try that for sure. Yes I always use the heat shrink tubing. It's a must!
dave

duck_arse

dave - put two leds in series, put that string in series with a current limit resistor. you want 4 leds? put two strings of *two-leds-and-one-CLR-in-series* in parallel.

or, put four leds in series with one CLR in series, job done (if you have enough forward voltage).
" I will say no more "

EBK

Quote from: duck_arse on December 13, 2016, 08:49:00 AM
or, put four leds in series with one CLR in series, job done (if you have enough forward voltage).
He doesn't have enough voltage for that.
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Technical difficulties.  Please stand by.

davepedals

Thanks, and yet another good idea! Seems I now have a few things to do today... and that's great since the humidity is 100%, not a very good painting day at all!  Note to Santa: please bring me a climate controlled painting booth this Christmas!
dave

davepedals

#11
Of course what I am doing is building some pedals for several other guitarists that I know , and adding my own light plate to them.  Some of them use pedalboard power supplies, others possibly use a single AC adapter.  Once again, I should've mentioned that the first place... Lack of sleep. Coffee! Coffee! Coffee! Sure I could just buy one already built, however it's cheaper for me actually , and I just enjoy building.
dave

Rob Strand

Duck_arse's suggestion also saves on battery power but if you put too many in series everything will go dim when the battery voltage drops.

FWW: I pretty much never put semiconductors in parallel it generally means trouble.  However I noticed those 9 LED torches wire the White LEDs in parallel.  I measured the LED VI curves and they have an effective internal series resistance of 30 ohms or so - which agreed with the white LED datasheets.   The torches rely on this resistance and add a single small resistor to back the current off (like 2R2 feeding the 9 LED's) .
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

GGBB

Quote from: davepedals on December 13, 2016, 07:29:01 AM
Usage voltage(V): 3.2-3.4
Current(MA): 20

With 9V supply, those LEDs have too high a voltage drop to put all 4 in series. But 2 will work fine. So use two resistors, each in series with two LEDs. By my estimation, the resistors should be about 150R for max brightness but that's cutting it close to frying an LED especially if your supply is over 9V. 180R is a safer bet.

http://www.digikey.ca/en/resources/conversion-calculators/conversion-calculator-led-series-resistor
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amz-fx

What color are the LEDs? (It makes a difference since the Usage voltage specified may not be the same as Vf.)

regards, Jack

davepedals

The superbright blue from Tayda.   I have about 300, time to use a few!
Emitting color: Blue
Diameter: 3mm
Lens color: Water clear
Usage voltage(V): 3.2-3.4
Current(MA): 20
View angle: 20 - 25
Luminous intensity(MCD): 18,000.


Already finished redrawing my PCB, I'm gonna go with EBK's idea and just solder LEDs directly to the Resistors.  Parallel wiring.  Board is only about .080" square, plenty of room ::) in the 1590A.
dave

amz-fx

In this case (since they are blue) the usage voltage is close to the same as Vf.

I've seen datasheets for red LEDs that state that the operating voltage is 3v, but the Vf will be less than 2v.

Parallel wiring is a good way to go.  Two LEDs in series with a separate dropping resistor for each pair will use less current, but if you have the available power then not much difference overall.

regards, Jack

GGBB

Quote from: amz-fx on December 13, 2016, 09:32:16 PM
I've seen datasheets for red LEDs that state that the operating voltage is 3v, but the Vf will be less than 2v.

OT, but this is interesting to me. Is there a technical explanation for what's going on with that (I'm sure there is)? But not too technical?  :icon_redface:
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PRR

> those 9 LED torches wire the White LEDs in parallel.

They buy many thousand LEDs and sort them for forward voltage.

Or as you note, they buy marginal rejects with high parasitic resistance and hope this semi-equalizes the currents.

Also I have seen these not-light all the LEDs, new or after a few uses. They just do not care.

As we do not have the economy-of-scale (or anonymity) to use these techniques, we can mainly stack to 50%-70% of battery voltage (maybe two blue) and run multiple stacks to fill our area. 3.3V twice is 6.6V. 9V minus 6.6V is 2.7V. 2.7V/0.020A is 135 Ohms. I would grab 120 Ohms. We know there is some parasitic resistance, also that a "20mA" LED won't blow-up at 21.6mA.

So that is two strings of 2 with a 120r for each. Total demand of 40-43mA is not a strain for a wall-wart (would be hard on a battery).
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Rob Strand

QuoteThey buy many thousand LEDs and sort them for forward voltage..
I was thinking they buy a batch and hope for the best.

QuoteOr as you note, they buy marginal rejects with high parasitic resistance and hope this semi-equalizes the currents.
I extracted the series resistance from the datasheet of a few white LEDs and it actually has the 30 ohms or so in the typical data.

QuoteAlso I have seen these not-light all the LEDs, new or after a few uses. They just do not care
I'm not surprised.    The soldering is pretty bad too.
I've seen no LEDs light (LOL!).   The PCB at the front has an aluminum ring which screws down on the copper and the case connects to the battery.   Aluminum is a recipe for pretty bad connection but aluminum onto copper is a disaster.

QuoteWe know there is some parasitic resistance, also that a "20mA" LED won't blow-up at 21.6mA.
To test the water I calculated out the worst case LED current for the LED torch, and not surprising,  with new batteries it was over the general maximum for the white LEDs.  And that was assuming equal current sharing.

QuoteThey just do not care.
Very true.   Not at all.

To be honest I've got to the point where I'm sick of the made in china junk.   I changed the battery in mother's watch recently and the whole face caved in because it was made from ultra-thin aluminium.    The hour ticks were copper pressed into slots in the aluminum, no bonding , so they of course popped out.

I look back at the HP test equipment of the 70's and 80's and they are true examples of technical know and quality.    These days the only "know how" is knowing how to deceive!
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.