7-segment blue LED display at 3.3V

Functional Reasons

Placing LED's in a parallel-series combination like this is a technique to average out the brightness given constant voltage. It helps present a more even looking illumination cheaply (as opposed to active constant-current circuitry).

The Q-point of the two LED's in parallel will be determined by the voltage assigned to the pair by the current allowed through the 4 series steps.

Diodes in Parallel

1. All diodes have a forward voltage drop that is non-linearly related to current.
2. This curve is not identical even for diodes made in the same batch

The final Q-point will be V1 = V2 (diode voltages equal), I1 ≠ I2 (currents unequal).

However, in your case the parallel combination is itself in series.

That means that I1 + I2 = Ip is the same for each parallel pair of diodes in the same string.

Light Emitting Diodes

In the case of LED's in the forward region, brightness is almost linearly related to current. So holding Ip constant gives you approximately the same amount of light from the pair of diodes even though it is difficult to know how the light output will be distributed between them.

Extending this to the string of pairs, you cannot find a closed form solution (I might be wrong, but I don't see one) so you would need to solve it iteratively (I like Newton-Raphson) like SPICE does.

In general, however, the more diodes in each parallel group and the more parallel groups in each string, the closer the light output of the parallel groups approach each other given a batch of LEDs, normally distributed, by light output.

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).