Electronic – Emitter follower for parallel LED switch

What you're talking about is not huge risk for a hobby project.. a few things to think about.

It's not really the \$V_F\$ at the operating current that matters, it's the \$V_F\$ at the maximum 'off' brightness that is acceptable. That voltage might be as low as 1.4-1.5V volt for a low-current red LED- in a dark room they can be quite visible with microamperes of current. Driving the output to the 3.3V (nominal level) gets us 3.3V on the output. A fresh alkaline cell with minimal load might be 1.63V/cell at room temperature (just measured one), so 3 would be 4.89V. That leaves you with 1.59V across the LED + resistor (nominal, not allowing that the 3.3V might be a few percent low).

That's way too much to ensure it's not emitting a whole bunch of light.

So, we tri-state it- that allows the output to go maybe a bit above Vcc without much current flowing. 300mV is safe, the datasheet says 500mV absolute maximum. At 500mV, we'd have 1.09V across the LED, probably enough to ensure it's off, at least under nominal conditions. The 'absolute maximum' figure is never a good thing to design to, but usually there's a caveat on this particular figure that allows that voltage or a bit more if the current is limited.

So, I do think this will work (with tri-state, not push-pull), and I also think it's acceptable enough for a hobby project assuming nobody is going to be using a battery eliminator on the circuit in the future*. Keep in mind the margin on the red LEDs is minimal and consider eschewing red in favor of yellow or orange. Or, simply add a silicon diode in series with the red LEDs (one diode can be used for several LEDs).

• If the ESD protection network in the ATMEGA328P does begin to conduct it will tend to raise up the 3.3V supply, out of control of the regulator. This is not a good thing for stability and could conceivably damage something, though the ATMEGA328P-M itself is rated for 5V operation.

I once did something like this in a commercial product (to drive a series string of LEDs with high Vf using a 5V constant-current output), but I designed a power supply with just the correct oddball voltage and appropriate temperature coefficient to match the LEDs and thus optimize the situation. I think the supply was around 8-9V. Worked a treat, easily from -20°C to 80°C (spec was 0-50°C).

I can only answer part of your questions.

1) I simply don't know how much current a Raspberry pin will put out, and Broadcom is very short on information. I recommend this: buy one 2803, hook it up with all the inputs tied together but no load, and measure the voltage at the inputs.

2) I can guarantee that with all the inputs tied together, you will draw less than 12 mA with an input of 3.3 volts. How can I say this? The spec you listed gives an input of 3.85 volts, not 3.3 volts. Lower voltage will give you lower current. But measure it. If the Raspberry cannot provide close to 3.3 volts when driving all 8 inputs, you may have extra problems with point 3.

3) Yes, you can pull 2.5 amps with one chip. I don't recommend it, but you can do it. If you do, you MUST provide a decent heat sink for the 2803. If you just use a bare IC you will burn it out. You should assume something like 1.3 volts across the 2803, and at 2.5 amps, your total power will be about 3.3 watts. The thermal resistance to ambient for this chip (section 7.4 in the data sheet) is 74 degrees per watt, so the heart of the chip will reach (3.3 x 74) + 20 degrees, or about 260 degrees C. Per the data sheet, the maximum is 125 C.

If your input voltage is low, the outputs will not turn on as hard, and the voltage drop across the outputs may be greater. In that case, the power dissipation of the chip will be higher, and you will need a bigger heat sink. But this is something that you will need to experiment with to find out. I suspect you'll be OK, but I'm not giving any guarantees.