Electronic – How to decipher zener diode encoding

Those are not Zeners. They are Schottky diodes.

Shottkies have a lower voltage drop than regular Si junction diodes, about 0.3V instead of 0.7V. They are also generally faster.

Their purpose is to keep the inputs within the supply rails.

The BJT power dissipation is minimal - any current it takes comes through R2 which is 15kohms. Supply is 3.3V and if half appeared across the BJT (max power transfer) it would dissipate 182uw!

The easiest way to picture the action of the BJT and D1 is by considering that the transistor cannot turn on too much because if it did, D1 won't have enough voltage across it to conduct current to the BJT's base and turn the transistor on hence, assume that the base is only very slightly forward biased at about 0.4V and therefore D1 must have 2.4V on its cathode.

And here's where you are going to hit problems with the low supply voltage....

On the original example it ran from 9V and you could assume the collector voltage would be at 2.4V + (say) 2V higher (due to R1 volt-drop). With 4.4V on the collector you can calculate collector current (\$\frac{9v - 4.4v}{15k\Omega}\$ = 307uA). And from this you could make a reasonable assumption about base current being 30 times lower\$^1\$ at 10uA. 10uA through a 220kohm resistor produces 2.2V across it and this isn't far away from the 2V I assumed by sticking my finger in the air.

But, on the reduced supply voltage design there isn't the headroom to assume the collector is 2V higher - you can barely assume 0.5V and this is making me think that the base current will be more like 2uA with a collector current of more like 60uA. This 60uA will drop 0.9V across the 15k resistor and clearly this doesn't figure because it would only leave 2.4V on the collector and not enough.

Therefore base current must be even smaller and I have doubts about the noise it will produce if any BUT good luck.

\$^1\$ Quite a few BJTs have poorer current gain at quite small collector currents so assuming 30 for \$H_{FE}\$ isn't unreasonable.