最近在想，要不要把这篇文章翻译出来，不过，深知自己的水平还是有限的，怕到时失去了原味。等下星期比较有空了，再来试试看吧。原文也开始真正的进入主题了。好了，不说废话，进入原文。

Step 7Cube size and IO port requirements

To drive a LED cube, you need two sets of IO ports. One to source all the LED anode columns, and one to sink all the cathode layers.

For the anode side of the cube, you'll need x^2 IO ports, where x^3 is the size of your LED cube. For an 8x8x8 (x=8), you need 64 IO ports to drive the LED anodes. (8x8). You also need 8 IO ports to drive the cathodes.

Keep in mind that the number of IO ports will increase exponentially. So will the number of LEDs. You can see a list of IO pin requirement for different cube sizes in table 1.

For a small LED cube, 3x3x3 or 4x4x4, you might get away with connecting the cathode layers directly to a micro controller IO pin. For a larger cube however, the current going through this pin will be too high. For an 8x8x8 LED cube with only 10mA per LED, you need to switch 0.64 Ampere. See table 2 for an overview of power requirements for a LED layer of different sizes. This table shows the current draw with all LEDs on.

If you are planning to build a larger cube than 8x8x8 or running each LED at more than 10-ish mA, remember to take into consideration that your layer transistors must be able to handle that load.

Step 8IO port expansion, more multiplexing

We gathered from the last step that an 8x8x8 LED cube requires 64+8 IO lines to operate. No AVR micro controller with a DIP package (the kind of through hole chip you can easily solder or use in a breadboard, Dual Inline Package) have that many IO lines available.

To get get the required 64 output lines needed for the LED anodes, we will create a simple multiplexer circuit. This circuit will multiplex 11 IO lines into 64 output lines.

The multiplexer is built by using a component called a latch or a flip-flop. We will call them latches from here on.

This multiplexer uses an 8 bit latch IC called 74HC574. This chip has the following pins:

• 8 inputs (D0-7)
• 8 outputs (Q0-7)
• 1 "latch" pin (CP)
• 1 output enable pin (OE)

The job of the latch is to serve as a kind of simple memory. The latch can hold 8 bits of information, and these 8 bits are represented on the output pins. Consider a latch with an LED connected to output Q0. To turn this LED on, apply V+ (1) to input D0, then pull the CP pin low (GND), then high (V+).

When the CP pin changes from low to high, the state of the input D0 is "latched" onto the output Q0, and this output stays in that state regardless of future changes in the status of input D0, until new data is loaded by pulling the CP pin low and high again.
To make a latch array that can remember the on/off state of 64 LEDs we need 8 of these latches. The inputs D0-7 of all the latches are connected together in an 8 bit bus.

To load the on/off states of all the 64 LEDs we simply do this: Load the data of the first latch onto the bus. pull the CP pin of the first latch low then high. Load the data of the second latch onto the bus. pull the CP pin of the second latch low then high. Load the data of the third latch onto the bus. pull the CP pin of the third latch low then high. Rinse and repeat.

The only problem with this setup is that we need 8 IO lines to control the CP line for each latch. The solution is to use a 74HC138. This IC has 3 input lines and 8 outputs. The input lines are used to control which of the 8 output lines that will be pulled low at any time. The rest will be high. Each out the outputs on the 74HC138 is connected to the CP pin on one of the latches.

The following pseudo-code will load the contents of a buffer array onto the latch array:

// PORT A = data bus
// PORT B = address bus (74HC138)
// char buffer[8] holds 64 bits of data for the latch array

PORTB = 0x00; // This pulls CP on latch 1 low.
for (i=0; i < 8; i++)
{

PORTA = buffer[i];
PORTB = i+1;

}

The outputs of the 74HC138 are active LOW. That means that the output that is active is pulled LOW. The latch pin (CP) on the latch is a rising edge trigger, meaning that the data is latched when it changes from LOW to HIGH. To trigger the right latch, the 74HC138 needs to stay one step ahead of the counter i. If it had been an active HIGH chip, we could write PORTB = i; You are probably thinking, what happens when the counter reaches 7, that would mean that the output on PORTB is 8 (1000 binary)on the last iteration of the for() loop. Only the first 8 bits of PORT B are connected to the 74HC138. So when port B outputs 8 or 1000 in binary, the 74HC138 reads 000 in binary, thus completing its cycle. (it started at 0). The 74HC138 now outputs the following sequence: 1 2 3 4 5 6 7 0, thus giving a change from LOW to HIGH for the current latch according to counter i.