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LED Cubes and Current Management

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m1tch37

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Hi,

I have mastered a 8x8 LED matrix using two shift registers, and now i want to add another dimension to it. I have thought long and hard about it and have done a fair amount of research. I plan to make a 64x8 array. The illustration below I made a while ago, but helps understand what im thinking.

**broken link removed**

This design will use 9 shift registers, 8 to make the 64 columns, and 1 to make the 8 rows. This is based off the advice given to me from someone else who had successfully used this design.

However, it's not the design that puzzles me, but how i mange the current to stay within the limits of the ICs and the LEDs. I don't understand what i have to put between the shift registers and LEDs. I understand the need for 220-ohm resistors, but i have read in some places i need transistors to take the large load of the ICs when dealing with matrices the size of mine. I have seen a few different ways, from simply having lots of transistors to a 'Darlington Transistor Arrays' or LED Drivers. Futhermore, **broken link removed**, a commercial 4x4x4 tri-colour cube, clearly doesnt have a resistor on every LED, which i thought was a requirement.

I have found no definitive answer for what i place between the Shift Registers and the matrix, and with my lack of experience, i have just got confused. Does anyone know how i should manage the currents as to not blow the ICs or LEDs? (And idealy, save me the need to solder 64 transitors and/or resistors, as done in the Hypnocube)

Kind Regards,
Mitch
 
Another way is to use drivers which limit the current to each LED internally. A MAX7219, for instance, does this. It uses a single resistor to set the output drive current for all the outputs. Coincidently, this chip will control one 8x8 array per chip, so you'd need eight of them. At DigiKey they average about $5 US each in single-piece quantities. Here's a link to datasheets: **broken link removed**.

Jeff
 
each of the 64 columns need to be able to source enough current to light one led ... this is easy for the 74hc595 which can source something like 50ma, so you are back at your 220 ohm resistor.

each of the eight rows need to be able to sink the current of (worse case) 64 leds. figure on 20ma per led, thats roughly 1.3 amps. a row of eight transistors controlled by your ninth shift register should handle this task just fine... you could run the eight row transistors with a 4017 - so you're assured that only one row can be on at a given time, then you'll need a clock and rst line from the uC.
 
You might consider using serial-to-parallel constant current sinking drivers to eliminate the LED current limiting resistors.

The design concept below drives the Strobe and Output Enable lines on the driver ICs with a PWM signal which allows me to multiplex the 8 bit bus for use as row drivers during display on time and as data lines for loading the driver ICs during display off time (blanking interval). The PWM signal also provides full fade-to-black brightness control.

Take a look at the Allegro A6275 (8 bit) or A6279 (16 bit) constant current driver ICs.

Mike

cube-drawing-png.16472
 

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where is the thing if you are sequencing one of the time then you only need one resistor per leg, On the other end if more maybe be on at the same time randomly the forget about voltage and concentrate on a current source for the maximun at one time ON.
 
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