How to build a 8x8x8 led cube (English version)

Step 3 Build the control circuit (first part)

Short introduction
The control circuit is basically composed of:
- 8 shift registers of 8 bits (total 64 bits) that enable / disable each column of the cube
- 1 shift register connected to 8 NPN transistor to enable / disable each layer.
- 2 connectors with 40 and 34 pins for connection to the cube
- 1 connector with 2 pins for connection to the parallel port

Why do we connect the 8 NPN transistor to the shift register of layers? As we know, the leds of each layer have a common cathode (-), then we must enable them to connect the signal mass. The problem is that the output of the layer shift register have a high level signal (logic level 1 is equivalent to Vcc) in any pin enabled. We need a component that, given a signal at logic level 1, enables the passage of the current. This component is the transistor. It works as a switch: if the pin of the base is set to logic level 1, the pin collectors and emitter are connected together.

Tools needed:
- Soldering iron
- Soldering tin alloy
- Pincers

Materials needed:
- 9 shift register of 8 bit (I have used 74HC164B)
- 9 slots for shift register
- 8 NPN transistor (I have used 2N3904)
- 70 resistors 100 ohm
- 8 resistors 1500 ohm
- 1 prototype board
- 2 connectors with 40 and 34 pins for connecting to the cube
- 1 connector with 12 pins for connection to the parallel port
- 3 connectors with 2 pins for connecting the power switch, the led power and power supply
- 1 red / green led (power led)
- 1 button (power switch)
- 1 male connector for power supply
- 3 little cables with 2 wires, ended with a connector (like those used in PCs to connect the leds of the front panel to the mother board)
- Electric wire suitable for soldering (thin, malleable, and resistant at the same time)

Once you have got all the components, you can begin the construction of the control circuit. Use the scheme that you can found in the pictures of this step.

SEE PICTURES FROM 20.00 TO 20.14

First of all you must begin by placing the components above the base (or use the same disposition that I have chosen), to realize their actual size. Once you have positioned all the components, you should draw rectangles around them with a pencil to outline their actual space. So, when you re-position them back, you will be sure of their place. This is useful as you have to flip the base for soldering.

I would suggest you to solder the components in this sequence:
1 - Slots for shift register
2 - Resistances of 100 ohms and 1.5 Kohm
3 - Transistors
4 - Connectors

Now, with mooooooooooooooooooore patience, make the connections between the main components. In a second stage, connect each resistor/transistor to the connectors for connecting the cube. Take note of all matches OUTPUT SHIFT REGISTER -> CONNECTORS PIN. This will be essential when we will solder the various wires to each terminal of the cube.

There are several tips that you can follow:
- Cover 3/4 of the perimeter of the base with a track which you connect to the signal mass. It is always useful to have a mass nearby.
- For connections between 2 or more adjacent points use soldering iron to connect directly these points each other
- Do not insist too much with the iron on a single hole of the base, otherwise the pitch copper could fall off.

Comments

[kieth]

wow. thanks guys!!!!

[huoyunjie]

Its cool!
But Im thinking how let it make easily!

[MrGentlemen]

I Have a Question about the vcc connection:

Did this Circuit doesnt need a VCC connection to work?

The only positive volt connection is just on the led, and nowhere else.

the shift register needs vcc too, but Pin 14 is nowhere connected

[guusberens]

I have a question about the wiring diagram.
The single circuit left I do not understand.
Could someone maybe able to explain what that meant?

Thank you

[dexterac]

hey, i was planning to use 7401/7400 instead of BJT switches and supplying logic 1(voltage 5) to each layer (Z coordinate) by a port. So selecting a layer can be done by 8 bit directly and selecting a led from 64 leds of a layer by NAND gates. This can also allow selecting multiple LED's in one go. But i am confused with what resistors to use for biasing 7400/7401 with avr and if should i directly connect the 8 layers to a port or use some driver ic or something.

[mmirek88]

Nice work. But I found some irregularities. The transistor 2N3904 has maximum current 200 mA, but the one layer of 64 LEDs in your design (switched by 2N3904) needs cca 20 mA * 64 = 1280 mA. And the next: maximum current through VCC/GND of 74HC164 is 50 mA, but 8 LEDs powered by 74HC164 need 20 mA * 8 = 160 mA. Didn't I overlook some trick? :-)

[twofouroh]

He's only enabling one column at a time. That's 8 LEDs, so the max current would be 160mA. At ~96 mW, it's OK on power rating also, especially since it only has a 12.5% duty cycle.

[mmirek88]

That explains everything. But in that case each LED has only 1,56 % duty cycle. That causes very low brightness of LEDs, I think.

[twofouroh]

Found his printed circuit diagram, much more clear than the handwritten one.  He is activating an entire layer through each 3904, but his duty cycle is still <=12.5% on each 3904, that would still be a max average power dissipation of 96 mW, but instantaneous max of 768mW.  Since it is clocking at MHz speeds, it should be fine.

[padrini]

esto es muy cool this is an amazin place to learn thank averu body