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LED Matrix Questions

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krich

New Member
I'm trying to put together an LED matrix. I understand the math around picking a resistor for use with an LED, but this matrix project is pushing the envelope of my knowledge on the subject.

I found this LED wizard online that does all the math for me. I know the equations to do this myself, but it's nice to see the same results from another source. (My inputs for the wizard are 12, 1.5, 100, 56 if you want to see the schematic I'm talking about)

For reference, I plan on putting together a 7x8 LED matrix. The DC source is 12V and the LED parameters are: Vf = 1.5, If = 100mA (infrared LEDs).

I have a couple concerns. I've learned the hard way that nothing in the electronics world is exact. My DC source will probably not be exactly 12V, my LEDs Vf will probably not be exactly 1.5V and will vary from LED to LED, etc.

So, my first concern is whether I should try to put 8 LEDs in series as the wizard suggests since I may have variations in Vf, or have lower than 12V source due to ripple, current demands, variations in different wall warts (I have one that's rated at 12V, but only gives me 10.5V), etc. Would it be smarter to do 7 LEDs in series and use a bit larger resistor so that I have plenty of Voltage to drive the LEDs?

Next, the wizard put in a 1ohm resistor. I don't have any of these. Is such a small resistor strictly necessary? This LED matrix will be on all the time (pulsed), so I don't want to overheat/short it just to save the hassle of finding a few low value resistors.
 

colin55

Well-Known Member
Have a look at this project:

5x7 Display

<mod edit: self promotion deleted. If you have content to share, please post directly to ETO. Do not link to your own personal website>

and see if this is what you are doing.
If you want to scan more LEDs, see 15x7 Display

The choice of "dropper resistor value" is totally different for scanning operations. You did not say if the array is going to be scanned.
 
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kchriste

New Member
Forum Supporter
I'm trying to put together an LED matrix. <snip!> The DC source is 12V and the LED parameters are: Vf = 1.5, If = 100mA (infrared LEDs).
What are you attempting to do with an infrared LED matrix?
 

3v0

Coop Build Coordinator
Forum Supporter
The 5 x 7 is a matrix and what you are working with is several strings of LEDs connected in parallel. Not the same thing.

LEDs are strange things. They drop the voltage but do not limit the current like a resistor will. So you need the resistors to limit the current to safe levels for the LEDs.

Try searching the web and forum for previous posts on the subject. If you do not find anything post again and I will give you a hand.

Hint: If you can afford to waste a bit more current you can use fewer LEDs per string and larger resistors. Ohms law.

3v0
 

Ubergeek63

Well-Known Member
I'm trying to put together an LED matrix. I understand the math around picking a resistor for use with an LED, but this matrix project is pushing the envelope of my knowledge on the subject.

I found this LED wizard online that does all the math for me.
Why didn't you ask on the linear1 forums?
 

krich

New Member
What are you attempting to do with an infrared LED matrix?
Just putting together an IR illuminator for day/night surveillance camera that I have (with a rather puny IR output). I happened across a number of IR LEDs at a good price, so it seemed like a fairly doable project for my current skill level.
 

krich

New Member
Try searching the web and forum for previous posts on the subject. If you do not find anything post again and I will give you a hand.
Man, I've tried. The search terms are so common that it's really difficult to pull up something relevant to my specific questions. My google-fu is apparently not up to the challenge.

Hint: If you can afford to waste a bit more current you can use fewer LEDs per string and larger resistors. Ohms law.

3v0
As you probably saw in my original post, this is what I was toying around with. I'm not in a power save situation, so I can afford to waste a bit of current for a simpler, more reliable solution. Thanks for the confirmation. I think I'll go down this route and see how it turns out.

Thanks all!

Ken.
 
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