simulate this circuit – Schematic created using CircuitLab
I was watching a video about the capacitive dropper sources and saw the schematic but the guy forgot to tell what value capacitor C2 is, so what value(s) could it be in order to get 3.3 V at the output?
C2 is in parallel with C3, 470 µF which is there to filter the ripple voltage, so any value for C2 much lower than that will probably not be significant. It could be that it's something like a low value film capacitor meant to filter high frequencies better than the electrolytic type likely used for C3 would.
It is also in parallel with a 3.3 V Zener diode and a 470 Ω resistor. The Zener is meant to regulate the output voltage to 3.3 V, so the capacitor value wouldn't affect the voltage much if at all anyway. There is some worry about the Zener having no current limiting resistor outside of R2 and to some degree R1 although that one is bypassed by C1 for AC current. That may be enough, but the circuit should be simulated and things like current and power dissipation for the various components checked to see if they are reasonable.
Update: Now that I can see the video it appears that C2 is a small electrolytic cap, the video author calls it a 'filter cap'. You can't see the markings but judging by the size compared to the 470 µF cap I'd guess something between 10 µF and 100 µF, maybe go with 47 µF. I'm not sure why they included this capacitor, it's pretty much redundant.
Also he doesn't explain how he came up with the value for C1, but the value used would have an impedance at 60 Hz of $$ \tag{1} Z_{C1} = \frac{1}{2\times\pi\times 60 Hz\times 3.3\mu F} = 803.8\Omega $$ so there is more current going through R1 than through C1, in simulation I get 70 mA for the cap and 120 mA for the resistor. That's $$ \tag{2} P_{R1} = 120mA^2\times 470\Omega = 6.768W $$ In simulation I get 6.9 W,and R2 is worse, averaging over 9 W. So if you use 1 W resistors like the video recommends they're going to get quite hot.
Conclusion: It's probably not a very good circuit and the guy in the video doesn't explain much of anything, he just says 'do this'. Safety is also lacking, he says something along the lines of 'this should be an X2 cap but I don't have one so oh well', and he doesn't appear to use an isolation transformer (a $40 Variac with banana jacks hacked onto it's side won't substitute for one). Showing people how to do something dangerous and then just adding a 'if you try this you're on your own' disclaimer is not very responsible. You might want to avoid the videos from that channel.
That's a nuts-o circuit with 470Ω in parallel with the capacitor. Xc at 60Hz for 3.3uF is about 800Ω so the parallel combination will have an impedance of more than 400Ω, pretty much dominated by the resistor.
With 120VAC in the power dissipation in the 470Ω resistor will be of the order of 7-8W each. Much more than the resistors he shows are safely capable of- they would burn through the breadboard. The capacitor is doing nothing good- it's a resistive not capacitive dropper.
It's normal to have a resistor (such as 470kΩ) in parallel with the capacitor to discharge it so little Johnny doesn't get shocked if he pulls the plug at an unfortunate point in the AC cycle and proceeds to touch or stick the prongs into his mouth. However I've reviewed part of the video and the creator clearly says and writes 470Ω. At least he warns the viewers that the circuit can kill.
Please do not muck around with this kind of circuit, particularly with this kind of source for information. You can, for a song, buy a really nice safety-agency approved USB-C supply that will efficiently and safely supply low voltage that is isolated from the mains. If you need higher voltages, get a trigger board and use PD USB-C or an open frame power supply with appropriate precautions. If you need 3.3V rather than 5V, a linear regulator can be used.
Ceramic or film capacitor, 100 nF or 1 uF. Any voltage will work. It's there to absorb high-frequency spikes. It has a lower impedance than the 470 uF electrolytic capacitor across it.
As shown, the primary resistors = 470 Ω are poorly chosen.
Removing the both 470 Ω Rs, reveals even the primary cap is conducting too much current. To make an effective C dropper, one needs to make major changes.
There are many solutions.
This one is more efficient.
Recommendations. Define maximum power dissipation in each component before selecting and during verification. Define max. acceptable ripple and verify. Choose high quality components. Using a primary resistor as a fuse is OK.
