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Topic: LED Letters Christmas Ornament (Read 204 times) previous topic - next topic

larrywal32

Hi all - am planning to make a PCB christmas ornament that is 3 letters (my wife's initials) with LEDs all along the letters.  Wanted to double check my design planning and calculations.

I'm thinking ideally 8-10 LEDs / letter.  So that would be 25-30 LEDs total.  I'd like to power it from 2 AAs.

So I'm assuming I need to power all LEDs in parallel from the battery source (which will have a basic switch on it).

If I use this 3MM LED from Digikey (Datasheet: https://www.lumex.com/spec/SSL-LX3054SRD.pdf), then the steady current would be 30mA * 30 LEDs = 900 mA of steady current. 

In theory, with V=IR, that would mean I have 4=0.9*R, so I'd need a resistor of 4.5 ohms.  Rather than have 30 resistors, I'd plan to have one single resistor feeding the parallel branches of LEDs.  That LED would have about a 2V drop on it (because each LED has a forward voltage of around 2).  That means that via W=AV, I'd need to handle 0.9A * 2V = about 2 Watts.  So I'd probably bump that up for some buffer and need a 4.5 ohm, ~5 Watt resistor.

First question: Is my math and plan accurate above?

Second question -- clearly 900mAs can't be sourced realistically from 2 AA batteries.  So I think I can jack up the resistance on the resistor to reduce the current.  The brightness would be less, but in theory maybe it's an ok balance for a christmas ornament.  I can probably test a single LED at lower currents to see what would be acceptable.

So if I plan for 10mA per LED, then I'd only use 300mA of steady current.  That means 4=0.3*R, so R=13.3 Ohms, and wattage could drop to about 2 with padding (e.g. 0.3A*2V = 0.6W).

But since AAs have mAH of around 2.5, then I could run for about 8 hours before running dry.  That's not enough.  I figure the batteries need to last for 3-4 weeks realistically, of about 5 hours per day.  So if I could get to ~100 hours I'd be good.

But that would mean I'd have to roughly take the current down to 1mA per LED, which I can't imagine would be enough.

So SECOND QUESTION - am I hosed?  Is what I'm doing impossible without lowering the # of LEDs?  Are there different "low current" LEDs? 

Open to thoughts - I'm a beginning but would love if my thought process above is right.


TheMemberFormerlyKnownAsAWOL

You posted in "programming".
I can't see any code.
Please don't PM technical questions - post them on the forum, then everyone benefits/suffers equally

PerryBebbington

#2
Oct 26, 2019, 09:26 pm Last Edit: Oct 26, 2019, 09:28 pm by PerryBebbington
Quote
I'd plan to have one single resistor feeding the parallel branches of LEDs.
I'm afraid you need 1 resistor per LED. The Vf of an LED varies slightly from LED to LED, the one with the lowest Vf will take most of the current leaving the others dim if you only use 1 resistor for a group of LEDs.


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In theory, with V=IR, that would mean I have 4=0.9*R, so I'd need a resistor of 4.5 ohms.
I don't know where you get 4 from.
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That means that via W=AV, I'd need to handle 0.9A * 2V = about 2 Watts.  So I'd probably bump that up for some buffer and need a 4.5 ohm, ~5 Watt resistor.
The 2W you have calculated is dissipated in the LEDs, not the resistors. For the resistor power rating you need 0.9^2 * R.


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So if I plan for 10mA per LED, then I'd only use 300mA of steady current.  That means 4=0.3*R, so R=13.3 Ohms, and wattage could drop to about 2 with padding (e.g. 0.3A*2V = 0.6W).
There's that mystery 4 again.
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Since AAs have mAh of around 2.5
I think you need to check that....


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That would mean I'd have to roughly take the current down to 1mA per LED, which I can't imagine would be enough.
Stop imagining, get some LEDs and try it. The only person who can tell you if they are bright enough is you (or maybe your wife).

Oh, and ++Karma; for such a lovely idea :)








Deva_Rishi

You should get you post moved, Drop the plan of using batteries, use an old phone charger instead, depending on the voltage put as many leds in series so they still light up sufficiently.
To 'Correct' you have to be Correct. (and not be condescending..)

larrywal32

#4
Oct 26, 2019, 09:53 pm Last Edit: Oct 26, 2019, 10:15 pm by larrywal32
Sorry folks - I meant to post this in general electronics.  Both Perry and I have reported to moderator and asked to move it.  Once that's done I will respond to above.

larrywal32

Ok - now that it's in the right forum - thanks so much Perry.  Answers:

  • I'm a doofus - everywhere I used 4, I meant 3.  Because 2 AA batteries in series should be 3 Volts, I think - 1.5 + 1.5
  • Agree with you about trying it :).  Was just trying to figure out a plan of attack in theory to help me have a model for what I expected to see on my breadboard.  It's how my brain works...


To Deva - WRT old phone charger - I'm trying to avoid plugging into AC power.  I thought about putting as many LEDs in series as possible so they could light up sufficiently, but with 2 AA in series and a ~2V forward voltage drop, at most I could do 2 LEDs in series, if that.  I guess having groups of two LEDs in series still might not be bad as it would reduce the overall current by a half if I could do it.  Another thing to test on the breadboard.

...For two resistors in series, that only needs one resistor, right?

Thanks again folks.  Hopefully Karma pays off and my wife thinks a geeky circuit board ornament is cool :).  Odds are on my side, she married me knowing I like tinkering....

PerryBebbington

#6
Oct 27, 2019, 09:15 am Last Edit: Oct 27, 2019, 10:19 am by PerryBebbington
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I'm a doofus - everywhere I used 4, I meant 3.  Because 2 AA batteries in series should be 3 Volts, I think - 1.5 + 1.5
:)
Yes, 3V.

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In theory, with V=IR, that would mean I have 4=0.9*R, so I'd need a resistor of 4.5 ohms.
Swapping 3 for 4; 3V = 0.9A * R is wrong because you have ignored the voltage across the LED, which you should subtract from the battery voltage. According to the data sheet you linked to they have a Vf of 1.85V, so the calculation becomes:
(3 - 1.85) = 0.9 * R, which is (3 - 1.85) / 0.9 = 1.27 Ohms.

Note that in the data sheet it says Vf is typically 1.85V with a maximum of 2.5V. This is why you can't put them in parallel with one resistor feeding all of them, the ones will the lowest Vf will take all the current.

You need a resistor for each LED, use the equation above with the current for 1 LED instead of for all the LEDs

For the resistor power rating you need the voltage across the resistor * the current, which is (3 - 1.85) * 0.9 = 1.035W

There are some basic electronics principals that I suspect you are not aware of or if you are you have not fully understood. These are:
Ohms Law
Kirchoff's circuit laws
Once you understand these and how they relate to each other the above becomes simple.

Please share photos of your finished project :)



Deva_Rishi

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To Deva - WRT old phone charger - I'm trying to avoid plugging into AC power.  I thought about putting as many LEDs in series as possible so they could light up sufficiently, but with 2 AA in series and a ~2V forward voltage drop, at most I could do 2 LEDs in series
Yes i suspected as much, but 2 AA's are not going to get you 100hrs, so then i would go for more and bigger batteries.
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Note that in the data sheet it says Vf is typically 1.85V with a maximum of 2.5V. This is why you can't put them in parallel with one resistor feeding all of them, the ones will the lowest Vf will take all the current
another good argument for putting many in series, and trimming the channels individually, LEDS can usually be produced with a decent tolerance though, finding one with a vF above 2.0v within a set of 30 is not all that likely.
To 'Correct' you have to be Correct. (and not be condescending..)

wvmarle

If I use this 3MM LED from Digikey (Datasheet: https://www.lumex.com/spec/SSL-LX3054SRD.pdf), then the steady current would be 30mA * 30 LEDs = 900 mA of steady current. 
I'm normally powering my LEDs with 3-10 mA. Actual brightness varies a bit per LED type.
Note that with 2xAA you can only use red and green, maybe yellow, definitely not blue or white LEDs. Those require ~3.5V Vf.

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Second question -- clearly 900mAs can't be sourced realistically from 2 AA batteries.
They can happily supply 2-4A of current. Just not for a very long time. This current is no problem, and you can probably reduce it to some 100-300 mA and have still sufficient brightness.

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But since AAs have mAH of around 2.5, then I could run for about 8 hours before running dry.  That's not enough.  I figure the batteries need to last for 3-4 weeks realistically, of about 5 hours per day.  So if I could get to ~100 hours I'd be good.
Then you'll have to look at using D cell alkaline batteries... or replace batteries every few days.
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

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