The snubber on a triac helps the triac turn off.
Whether you need a snubber has nothing to do with the how much power your load consumes. If you have an inductive load, you need a snubber to get that triac to turn off and stay off -- to avoid unwanted turn-on -- even when your load is very low power.
Typically the control circuit tries to turn off a triac by pulling the other end of a resistor tied to the triac gate "up" to VCC, the same voltage as the cathode A1.
The triac remains "on" until the current through the triac reaches zero, which may be as much as 10 ms later.
At that later time, there is zero current through an inductive load, and therefore zero energy stored in the magnetic field.
(When we use a NPN transistor or MOSFET transistor or a relay contact to turn off an inductive load, we have to somehow deal with the "flyback voltage" produced when the energy stored in the magnetic field in the load is dumped.
We don't have to deal with this energy dump when we use a triac, and so the complete system using a triac+snubber typically ends up simpler and cheaper than these other ways of switching AC mains power to a load).
When the triac finally turns off, the voltage across the load rapidly changes to near zero, and the voltage across the triac rapidly changes to nearly the instantaneous mains voltage.
(At the instant the triac turns off, the instantaneous current through the load is near zero, but with an inductive load the instantaneous absolute voltage across the load is close to the maximum peak instantaneous mains voltage).
The voltage itself is not a problem -- before the triac turned on, and after the triac has been turned off for a while, the full mains voltage is applied across the triac A1 and A2 pins indefinitely, without any problems.
The rapid change in voltage causes problems -- the rapid change in voltage at the anode A2 is coupled through unwanted parasitic capacitance inside the triac to the gate of the triac, turning the triac back on.
To avoid this unwanted turn-on, we add a snubber to reduce the rate of the change in voltage at A2.
Lowering the change in voltage reduces the current through that parasitic internal capacitance.
We can't reduce that current to zero, but we can keep it low enough that the resistor connected to the gate terminal keeps the gate voltage close enough to A1 -- keeping the triac turned off when it is supposed to be off.
Another way to avoid this unwanted turn-on is to choose one of the newer "SNUBBERLESS" triacs that have much smaller parasitic capacitance inside the triac.
The entire point of the a optocoupler is to allow electrical separation of a circuit. It's literally a led, a little tube, and a phototransistor. You could make one yourself. You don't have to share a common ground, but you can if you want.
As for a short on the phototransistor size, for most circuits, especially low voltage low current ones you have nothing to worry about. Unless you connect a high voltage high current line that way exceeds the optocoupler's absolute maximum specs, enough that they cause the optocoupler to melt/explode/catch on fire.
Best Answer
Do the math.
The output is driving 16 V across 4.7 kΩ, so is passing 3.4 mA.
On the input there is 5 V across 150 Ω and the LED. According to the 4N25 datasheet, the forward drop of the LED can be 1.5 V. The current is therefore ((5 V) - (1.5 V)) / (150 Ω) = 23 mA.
You are getting a current transfer ratio of (3.4 mA)/(23 mA) = 15%. Now again look in the datasheet and see that the minimum guaranteed CTR is 20%. Something is therefore not as expected.
The most likely culprit is that the digital output is not really 5 V when you try to draw 23 mA from it. That's a lot to expect a digital output to do. Check its datasheet.
Let's say for example (your job to look up the actual value), that the digital output can source 10 mA and drop to no less than 4.5 V in the process. After the 1.5 V drop across the LED, the resistor must be at least (3.0 V)/(10 mA) = 300 Ω.
Now let's look at what the output should be for this example. With 10 mA in and 20% CTR, you can expect 2 mA output current. To drive a resistor all the way to 18 V, the resistor must be at least (18 V)/(2 mA) = 9 kΩ.
So change R1 to 300 Ω and R2 to 10 kΩ, and everything should work for this made-up example. Look in the datasheet to see what the actual digital output source current and voltage at that current are, and plug in the numbers for your case.