led
Is it possible to control a bicolor LED with just one pin of a microcontroller?
Instinct says NO, because you can have one end grounded and the other going to the micro's pin, allowing you to turn it on or off but not change its color.
But maybe someone has a better idea?
You can easily do this in a small FPGA, provided that you use a shift registers like a 74xx595 to reduce the number of FPGA I/O Pins. Even the smallest Xilinx Spartan-3 could do it. Now, what you want to do with a 70x70 LED display might place additional demands on the FPGA size.
If you wanted to not use an external shift register(s), then you need a lot of pins on the FPGA. A minimum of 140 I/O pins, but possibly 210 or more. That still is not terrible, but will push you into a BGA which you might not want to deal with at this point.
With external shift registers, you might be able to use a single MCU. The trick here is that you want to connect the shift registers to some sort of SPI serial port, and then use DMA to feed the SPI interface. You don't want to bit-bang the serial interface for this. A typical 8-Bit Micro is probably not going to be enough for this. There are lots of ARM Cortex M0, M3, and M4's that are super cheap and would be a good option.
You could also use external latches (74xx374) instead of shift registers with an MCU. You could probably bit-bang an interface to latches, but using shift registers and SPI with DMA would be superior.
I have done a lot of multi-CPU systems and I can safely say that you do NOT want to do this if you can avoid it. There are a lot of synchronization issues that creep up when using several CPU's, and if you don't know what to pay attention to then you will likely just end up with a sub-par result and a lot of frustration.
If you examine the table you can see a reasonably linear relationship between resistor current and LED current. This is implied in the data sheet where it says that LED current is 400 times Rset current. So, given that Vset is constant at 1.2 volts, for an Rset of 4990 ohms, Iset will be 0.24mA and Iled will be 400 x higher at 96mA (not far off 100mA). At mid scale (say) 1050 ohms, Iset will be 1.143mA and Iled should be 457mA - it says 500mA in the table so it's not far off. At max LED current (1A) Iset is 2.185 mA and the LED current would probably be 400x higher at 874mA.
What I'm saying is that you could use an op-amp and a transistor to extract a controlled current from the Rset pin and you would find this to be significantly more linear than trying to delinearize a resistor: -
Iload in the diagram would connect to the Rset pin of the chip. Rset in the diagram above would ideally be slightly lower than 549 ohms - maybe a 470R would be fine. Vset in the diagram would force a current to flow into the collector that is Vset/470R and, for 400mA LED current would be approximately 0.47 volts.
Remember that the CAT4101 uses current mirrors to produce the overall target LED current and therefore feeding in a variable constant current is going to be more accurate that choosing a value for Rset no matter how delinearized.
Best Answer
You can do this with a bi-color LED that has the two LEDs back-to-back if you connect one LED terminal to an intermediate voltage eg 2.5V on a 5V design and connect the other side to the MCU via a suitable resistor (I used 560R).
Then a low output gives one colour, high gives the other and tri-state leaves the LED off. Pulse width modulation will allow you to control brightness (switching output between active and tri-state) or mix the colours (switching output low to high).
You can adjust the intermediate voltage to compensate for different LED forward voltages too.
I used an LM2904 op-amp to provide the intermediate voltage - it works with supply voltage down to 5V. There are plenty of other devices that can operate at lower voltages and still sink and source enough current to drive the LEDs.