LED Matrix Question

[Ioannis]

No, you cannot parallel LEDs Diodes or Bipolar transistors (only FET). Since each of them is not a resistor the current cannot be shared among them. You sure need a resistor in series.

But you already have post this as in the link http://www.injektilo.com/LEDMATRIX/8...%20Example.JPG

Ioannis

[Dave]

mekohler, Why not series connect the LED's and use only 1 resistor. 1.8 volts x 5 = 9 volts. The driver you have is open collector so you can use 12 volts for the LED source and use say a 150 Ohm resistor in series and that will give you 20Ma. for the string.

Dave Purola
N8NTA

[Ioannis]

Thats what I meant but was not very clear. Anyway the link http://www.injektilo.com/LEDMATRIX/8...%20Example.JPG is just that in picture! Seems to be the best choice right now.

Ioannis

[mister_e]

seems to have a problem with the link

i think it could be http://www.injektilo.com/LEDMATRIX/8...%20Example.JPG

now let's show the schematic <IMG SRC="http://www.injektilo.com/LEDMATRIX/8%20Led%20Matrix%20Example.JPG">

[T.Jackson]

No, you cannot parallel LEDs Diodes or Bipolar transistors (only FET). Since each of them is not a resistor the current cannot be shared among them. You sure need a resistor in series.

Depends on the junction drop. If you have two LEDs perfectly matched, say 1.75V ea (or thereabouts) - both will share equal current. Tolerance levels are considerably quite remarkable nowadays. No need to spend hours with your meter looking for matched BJT pairs for that destined to be perfect amplifier.

[mekohler]

Thanks for the help guys.

I think I settled on constructing the matrix like this:


And the schematic is:



Let me know if you see any problems. Also, the book I am learning from recommended using the74LS174 decoder. and has provided example code.

[sayzer]

You seem to have good progress.

Here is a similar approach.

<img src="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=2045&d=1190816000">

[mekohler]

Thanks!

If I'm only using RC0 - RC4 as my outputs on the PIC (ommitting RC5)...the declarations of PORTC = 0b111111 and PORTC = 0b011111 do the same thing, bcs the low 0 or 1 bit isn't connected to anything, correct? I can leave it like that, or should I omit that bit and just use 0b11111?

[sayzer]

In PBP, PORTC = %11111, for instance, will set RC0, RC1, RC2, RC3 and RC4.

This should be the same in your "0b11111" example. Since you do not care about RC5, and possibly RC6 and RC7, then you can use one of your three "PORTC = 0b111111" or "PORTC = 0b011111" or "0b11111" examples.

They will all set RC0, RC1, RC2, RC3 and RC4 anyway.

[Ioannis]

Depends on the junction drop. If you have two LEDs perfectly matched, say 1.75V ea (or thereabouts) - both will share equal current. Tolerance levels are considerably quite remarkable nowadays. No need to spend hours with your meter looking for matched BJT pairs for that destined to be perfect amplifier.

Sure technology advances forward but just to be sure, i'd put a small resistor there. You never know...

Ioannis

[T.Jackson]

<img src="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=2046&stc=1&d=119088114 8" align="center" border="1">

Diagram above depicts two 1A power diodes paralleled together in attempts of obtaining 2A. As you can see, only a minor difference in drop (300mV in this case) - will mean that one of them is going to be working harder. In the case of what's illustrated above, D2 will be working around 5% more than D1.

If anyone doesn't agree with this, by all means speak up.
<br/>

[Dave]

mekohler, I don't think your design using an 74LS174 will be able to drive the current you are anticipating using. Even with only 20 Ma. per led string in parallel, thats 40Ma. An normal 74LS series part can at best case source 5Ma. You are going to need a high power sourcing device such as a PNP transistor to source the required 40+ Ma. I beleive after your design has been prototyped you will find that if you are multiplexing the displays at any kind of rate you will lose almost all of the intensity. Therefore you will have to increase the current to the display system. Most wallboard displays you see in stores for advertising use anywhere from 100 to 500 Ma. as the display is multiplexed and switching at high speeds. I suggest looking at Allegro semiconductor as they have a variety of sourcing drivers available...

Dave Purola,
N8NTA

[mekohler]

mekohler, I don't think your design using an 74LS174 will be able to drive the current you are anticipating using. Even with only 20 Ma. per led string in parallel, thats 40Ma. An normal 74LS series part can at best case source 5Ma. You are going to need a high power sourcing device such as a PNP transistor to source the required 40+ Ma. I beleive after your design has been prototyped you will find that if you are multiplexing the displays at any kind of rate you will lose almost all of the intensity. Therefore you will have to increase the current to the display system. Most wallboard displays you see in stores for advertising use anywhere from 100 to 500 Ma. as the display is multiplexed and switching at high speeds. I suggest looking at Allegro semiconductor as they have a variety of sourcing drivers available...

Dave Purola,
N8NTA

Actually, it is labeled incorrectly. My book says to use 74S138:

"In this application I used a 74S138 decoder to provide the column negative power supply select. The selection of the S technology decoder was selected over the more common LS technology to sink the most current (LED's fully lit) as possible"

Does that make any difference? The low level output current for the 74s138 is 20mA...still, according to you that is way too low. The ULN2803A is the transistor array for the "rows", which http://www.electronelec.co.uk/uln2803a.htm says outputs current to 500mA

Should I just place a second ULN2803A after the decoder?

[T.Jackson]

Those LEDs in the matrix arrays are very efficient. You'll get "satisfactory" light output from only 2-3mA for ea LED. The 74LS138, if you sink current from it, it might be adequate enough.

[Wilson]

Diagram above depicts two 1A power diodes paralleled together in attempts of obtaining 2A. As you can see, only a minor difference in drop (300mV in this case) - will mean that one of them is going to be working harder. In the case of what's illustrated above, D2 will be working around 5% more than D1.

If anyone doesn't agree with this, by all means speak up.
<br/>

I cannot see where the 300mv difference in drop is. Looks a bit like 30mv

[T.Jackson]

You're right, one too many 0's - thanks for pointing it out. I'll see If I can mod the post.