Are these LEDs suitable for the sammichSID?

MIDIbox SID
1

Having read that ultrabright LEDs are not suitable for the sammichSID and Wilba recommending LEDs in the 20mcd range, I came across these:

http://www.soselectronic.de/a_info/resource/pdf/king/L-934MBDK.pdf

Can I use them with the sammichSID without any (resistor) modification?

Sorry for the noob question, electronics is all new to me, but I’d like to figure it out.

Thanks in advance!

2

alright, let’s have a look at the specs:

Forward Voltage : 4.0V

Forward Current : 20mA

the sammichSID design feeds the LEDs with 5V. That is one Volt too much.

Kirchhoff’s circuit laws (google/wikipedia that) say that the current flowing through the resistor is as high as the current flowing through the LED (IF=20mA)

R=U/I

R=1V/20mA

R= 1V/0.02A

R=(1/0.02)Ω

R=50Ω

So 50Ω should be the minimum resistance you should put in series with the LED. You can experiment with some different values to reach the light intensity you want.

You can always go above 50Ω, but you should never go below. Decreasing the resistor value will increase the current which will in turn heat up the LED and kill it.

regards

matthias

edit by nILS: Nope, that’s wrong. Try again. Also note the 220Ohm resistors on the CS PCB.

3

lol

Blue is evil, looks cool, bad for eyes.

FWIW you can use ultrabright LEDs but the problem is if you put too high a resistor to lower the brightness, then you get this strange artifact where the previous column of LEDs (in the matrix) will show a “ghost” glow.

What you linked to looks good…

The resistor in series with a LED is to limit current, voltage drop across the resistor will be total voltage (in this case 5V) minus forward voltage drop of the LED (in this case 4V). I=V/R means current equals voltage drop divided by resistance… if you use the stock 220 Ohm resistors, that’s 1/200 = 0.005 amp = 5mA. That may be bright enough… you probably will need to see it in a circuit to decide… but treat that as your lower limit… if you want max brightness for this LED (20mA) then you do some algebra and turn I=V/R into R=V/I where V=1 and I=0.02 amp (20mA)… 1/0.02=50 Ohms.

Now you could stick 50 Ohm resistors in there, but the problem is, the shift register IC won’t deliver 20mA to all pins, i.e. if a whole row of 8 LEDs are lit, so sticking 50 Ohm resistors in there will probably not give you the maximum brightness you could get with this LED. Therefore stick with 220 Ohm, it will be slightly less than max brightness for this LED, but this will offset the evil eye-burning effect of blue LEDs as well as make them all be consistently bright, instead of (possibly) different brighness depending on how many in a row are lit.

edit by nILS: You suck at the explaining.

edit (this time _really_ by nILS): You suck at making jokes others will understand :wink:

4

WTF IS WRONG WITH YOU, Nils?

http://www.elektronik-kompendium.de/sites/grd/1006011.htm ?

I must admit that I forgot to mention that the 220Ω Resistor should be *replaced* in my case.

Now I’m very excited to read your explanation :slight_smile:

5

I’m mightily busy studying so I couldn’t be fukt explaining the problem, and was hoping for someone like you or Wilba to find it and fix it, which has happened now :slight_smile: Just quickly wanted to make sure the wrong/incomplete info wasn’t gonna be used.

6

Okay the crossed lines confused me a little bit:

So basically, with the stock 220Ω resistor everything will work fine, except I will not reach max brightness. But that’s okay, I’ll guess.

Thanks so far for your help!