Electrical – Using 1 IR receiver for 4 LED controllers

The forward voltage for an IR LED is much lower than for a visible light LED, typically around 1.3 V, but rising if you push real high currents through them, like > 100 mA. There seems to be no reason why you couldn't place two of them in series, especially if your Vcc would be 5 V. If your Vcc comes from a pair of AA batteries though, the voltage drop of two LEDs + the transistor's saturation voltage may come close to Vcc and that could limit the output current.

The two outputs to drive the four LEDs are to avoid overloading the microcontroller's output. Or better, should avoid overloading. A 120 Ω resistor means 35 mA base current per transistor, and that's too much already for the AVR, let alone the 70 mA which it will draw now.

The 2N3904 is not a good transistor for this either: it's only rated at 100 mA and the low hFE necessitates the high base current. A BC337-40 has an hFE of minimum 250 at 100 mA collector current, then 5 mA base current should be enough to drive it. A base resistor of 820 Ω will allow you to drive all four resistors from 1 pin. The BC817 is also rated at 500 mA.

Alternatively you could use a FET to drive the LEDs. A PMV20XN can handle several amperes and has an on-resistance of only 25 mΩ so it will dissipate hardly any power. 1.5 V gate voltage is sufficient for 2.5 A.

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A note about current limiting. Usually we'll have a resistor in series with the LED for that, but if you look at the schematic of a commercial remote control that resistor is often missing, because they count on the batteries' internal resistance for that, and then they save another 0.001 dollar per remote controller.

This is not a good idea if you power from a mains powered voltage regulator. That will limit the current, but at a too high level, and if it doesn't destroy the LED immediately it will severely limit its lifetime. So a small series resistor is recommended. At 5 V supply and 2 LEDs in series you'll have a voltage drop around 2.9 - 3.0 V, so for 100 mA you need a 30 Ω resistor. Peak power will be 300 mW, but at a 50 % duty cycle average power is only 150 mW, then a 1/4 W resistor will do.

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.