This is a very interesting and simple classical Non-isolated power supply. Before asking the my question, let me try to list the things that I think I know about this circuit. correct me if I am wrong.
C1 is a 274 (270nf) capacitor, with a 120vac/60hz input, it's equalivent to 9.824k ohm resistor. (XC = 1 / (2pfC))
D1-D4 is a AC rectifier to convert AC into DC.
D5 (ZR) is a 24v Zener diode to stabilize the voltage at that point to 24v. Because the current at that point is about 1mA (120v-24v/9.8k+.68k)=.9mA, the zener need to be at least 24v*1ma= 0.025 watt.
The 8050 NPN is used as a switch here. Ib is about 0.75mA and with a DC gain of 120, Ie should be about 9mA
The 4 white led have a 14.4v voltage drop.
Question 1, what is the purpose of R4 680? Since we already have c1 and it's 9.8k.
Question 2, Since there is a 1mA current limit by the C1 and R4, what exactly is the current for LED1-LED4?
Question 3, what is the current going through D5? less than 1mA I suppose.
Question 4, if I want to increase the brightness of these LED, which are way too dim, which component do I have to change?
Question 5, I know this is very dangerous circuit because it's not isolated, but since there is a 24 zenzer diode, so the voltage at the led leads are probably less than 24v. How dangerous is it to touch those led leads? 1-10, 10 being the most dangerous.