0
\$\begingroup\$

enter image description here

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.

\$\endgroup\$
  • \$\begingroup\$ What are the failure modes for a capacitor? \$\endgroup\$ – Ignacio Vazquez-Abrams Mar 7 '17 at 20:15
  • \$\begingroup\$ It can short and destroy everything or open circuit and the circuit will stop working or anything in-between. It can also shock even if the circuit is disconnected from the supply. \$\endgroup\$ – skvery Mar 7 '17 at 20:30
  • \$\begingroup\$ How do you get this 1mA ? 120V/10k = 12mA. Also BJT and Zener diode is not needed and can be removed. And here you have an example circuit obrazki.elektroda.pl/7354559700_1358103251.png \$\endgroup\$ – G36 Mar 7 '17 at 20:40
  • \$\begingroup\$ "The 8050 NPN is used as a switch here. " Nope. It's used as a variable resistor, to drop the 24v to about 15 volts. (14.4 plus 0.6 volts Vbe). So this is a really horrible circuit, which attempts to drive the LEDs with a constant voltage. Bad circuit. Bad circuit. No biscuit for you. \$\endgroup\$ – WhatRoughBeast Mar 7 '17 at 20:47
0
\$\begingroup\$

Question 1, what is the purpose of R4 680? Since we already have c1 and it's 9.8k.

It is to prevent high inrush currents. Also to prevent high currents during switching transients on the supply.

Question 2, Since there is a 1mA current limit by the C1 and R4, what exactly is the current for LED1-LED4?

$$\lt 1~mA$$

Question 3, what is the current going through D5? less than 1mA I suppose.

Correct!

Question 4, if I want to increase the brightness of these LED, which are way too dim, which component do I have to change?

Depends...

If you remove the transistor and short 3 - 1 are the LEDs bright enough?

If not, increase the capacitor C1.

Otherwise adjust the variable resistor.

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.

10 - Still extremely dangerous and can cause death.

If you want a better answer... ask a better question.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.