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Topic: RGB SMT LED Cube, resistors, drivers, and shift registers. (Read 24 times) previous topic - next topic

Hippynerd


Am I right in thinking that for each plane you have a sepreate cathode connection and that all the cathodes are not connected together on each plane? Otherwise it would never work would it.

Have you done any tests to see how these LEDs look when they are on, if you need diffusers or what current you need to run them at.
Once you know the current for each LED you can then work out how much current to sink with the cathode sinks, you might need a FET here.

You then can look into how to provide that current at the anodes.

You are correct about the each plane having a separate cathode connection (4 cathodes, 48 anodes), but all the cathodes per plane are connected (thats 48 cathodes tied together per plane as the LEDs have 6 leads)

I did many tests on the LEDs at 3.3vs to figure out the forward voltage on the LEDs. I figure at 5vs I would need about 100 ohm on the red, 150 on the blue/green. At 3.3v, I need around 50-75 ohms on the red, and maybe something on the blue, im not too sure. I had a an LED hooked up to the 3.3v with a couple resistors, with the 3 leds lit, it seemed a bit blue, and Im guessing that it should be white.

Nearest I can tell, the forward voltage on the red is 2.15, and the blue and green are about 3.3v (That is to say, if I want 20 ma per LED, then those are the approximate voltages)

I have been experimenting with making some voltage dividers to try to hook up some shift registers, to get them to output 3.3v my lack of parts is making it hard, but I have found  bunch of tiny 4.7k resistors, and I'll be messing around with them this morning.

Grumpy_Mike

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but all the cathodes per plane are connected (thats 48 cathodes tied together per plane as the LEDs have 6 leads)

You mean anodes here don't you. It is the cathodes that are negative that are connected together.
The LEDs are capable of 60mA so you intend to run them at 20mA, is that right.

Do you know when you start multiplexing them they will look dimmer because they will only be on for one quarter of the time. One way round that is to boost the current.

If you are using resistors then there is no point in running the shift registers at a reduced voltage, you just alter the value of the resistor.
So given a 5V feed what values do you come up with for a 20mA current?

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my lack of parts is making it hard

Resistors are less than a penny each unless you pay vastly inflated prices.

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I've been reading about MAX7219/7221 chips, they are common cathode drivers, Is there a good reason why I shouldnt try to find some of those?

The reason is that you cannot control the brightness of each individual LEDs. Also they will not work with the way you have wired it up.


Hippynerd


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but all the cathodes per plane are connected (thats 48 cathodes tied together per plane as the LEDs have 6 leads)

You mean anodes here don't you. It is the cathodes that are negative that are connected together.
The LEDs are capable of 60mA so you intend to run them at 20mA, is that right.

Do you know when you start multiplexing them they will look dimmer because they will only be on for one quarter of the time. One way round that is to boost the current.

If you are using resistors then there is no point in running the shift registers at a reduced voltage, you just alter the value of the resistor.
So given a 5V feed what values do you come up with for a 20mA current?

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my lack of parts is making it hard

Resistors are less than a penny each unless you pay vastly inflated prices.

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I've been reading about MAX7219/7221 chips, they are common cathode drivers, Is there a good reason why I shouldnt try to find some of those?

The reason is that you cannot control the brightness of each individual LEDs. Also they will not work with the way you have wired it up.




I do mean I have 48 cathodes tied together per plane (aka common cathode) There are also 48 anodes per plane that need to be individually controlled, they are connected to the copper wires in my photo, the silver colored wires are the cathodes. Basically, the cathodes connections hold each plane together, and the 48 copper wires hold the planes up. Using copper for the anodes (+), and zinc coated wire for the cathodes (-), was not an accident, I did that intentionally to make it obvious which is + and which is -.

Resistors around here are about 25 cents each, so 48 resistors will cost $12. assuming I only buy the right ones, and dont have any problems. I ordered shift registers online a month ago, and they cost $5/20. A shift register, and a resistor are basically the same cost to me. 48 resistors will make things quite messy too, I would probably want to hide that mess under the cube, but keep the shift registers, and nano, neatly next to the cube, for display, but im still not sure how I will mount/display the cube.

Since there is no way to run the shift registers at the low voltage that the red LEDs need, I will need a minimum of 16 resistors, or a constant current driver. I havnt messed around with PWM yet, but I expect to do PWM, shiftPWM or otherwise...

I havnt done much testing at 5v, my guess is to get 20ma, I will need 100ohm for red, and 150 for blue and green. Im still not too sure about how much current each LED needs for even brightness/color.

The whole point of running the shift registers at lower voltage is to reduce the amount of parts. If I only needed 4 resistors, I could live with that, but 48 is going to expensive, ugly, and more effort.

I have a few of these ULN 2003a chips, they are transistor or diode arrays? Could those be helpful? 

Hippynerd

making voltage dividers with 0402 resistors is very difficult, but I managed to make 3 of them with 4.7k resistors, and soldered them up to my old cubes shift registers, and it lights up, but its not working right, its acting crazy. It could be my dividers are not working right, I dunno.

Grumpy_Mike

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Resistors around here are about 25 cents each,

WHAT!!!

You need to order them on line, it is sewing your view of electronics and forcing you into making silly design decisions.

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It could be my dividers are not working right,

I doubt it, that sounds more like faulty wiring or bad software.

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I will need 100ohm for red, and 150 for blue and green.

That sounds wrong.
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the forward voltage on the red is 2.15

Again that sounds too high, but assuming it is right:-
So the voltage across the resistor is 5 - 2.15 = 2.85V so to have that voltage with 20mA needs a 2.85 / 0.02 = 142.5 Ohms
Similarly for a voltage drop of 3.3V you need 5 - 3.3 = 1.7 then 1.7 / 0.02 = 85 Ohms

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