Electronic – Help identifying this circuit and understanding how it works

The way it's hooked up now it looks like at best the gate of the FET will be at half the drain voltage. If there's enough voltage on the drain it might get into a linear region where there's some drain current flowing, but it's not going to fully enhance the FET. You really want to hook the side of R2 that's attached to the drain to an independent supply voltage that can give you enough Vgs to turn the FET on fully. (For example to the + side of the 9V battery, not the drain of the FET.)

Yes, the "diode" is the LED.

There is no such thing as "wide open"- the current at the LED is reflected (within limits) at the transistor by the ratio "CTR" = Current Transfer Ratio.

If you put 5 mA through the LED you get somewhere between 2.5mA and 20mA through the transistor (until it saturates)- that's what the minimum/maximum CTR figures of 50% to 400% at If = 5mA and Vce = 5V mean. Vce =5V means that it's far from saturation. So if you use a resistor in series that limits the current to (say) 1mA (eg. 5K on a 5V supply) you'll have it saturate with 5mA into the LED. Note the since they vary over an 8:1 range, the manufacturer has ranked some of them and marked them in different "bins"

N : 50 to 400 (%)

H : 80 to 160 (%)

W : 130 to 260 (%)

Q : 100 to 200 (%)

L : 200 to 400 (%)

Naturally, the N version will tend to be the cheapest since the CTR can vary over the widest range, and includes the worst-performing units (50-80% CTR).

The transistor in the NEC part is rated at 70V and you should not put more than that across the Emitter-collector. There are higher voltage rated solutions.. for example the Sharp PC851XNNIP0F is rated at 350V.

You should make sure you have plenty of CTR- it degrades with temperature and with time (as the internal LED fades). Putting extremely high currents (like 30mA) through the LED will hasten the deterioration.

A proposed model of optocoupler aging is life \$\propto \frac{1}{Ie^{\frac {-E}{kTj}}}\$

Where k is Boltzman's contant 8.62\$\times 10^{-5} eV/K\$

Tj is junction temperature

E is the activation energy of approximately 0.15eV

so if you increase the current you not only get a decrease due to the current itself but an exponential decrease due to self heating. If I plug some plausible numbers into that equation, I get more than a 1000:1 reduction in life at 20mA vs. 5mA, with the same ambient temperature.

Note that if a relatively high current is only seen with a very short duty cycle (perhaps fitting your application) then the life is hardly impacted. It's the current and temperature integrated over time that causes the deterioration.

Bottom line is that you want to keep the current as low as is reasonable to make the thing work (and since operation is guaranteed at 5mA, that's not a bad place to start). You should also make sure it will work at (say) 3mA so even if it ages a bit it will still continue to work. In some cases you may have to buy more expensive optos with higher CTR than the cheapest ones if you need long life. Anecdotally, there is significant difference between different manufacturers' products. Personally, I tend to stick with the best-known Japanese makers.