The circuit in the image is to detect AC mains is present when the wall switch is ON. When wall switch is in closed position the DC current will pass through it and return to the circuitry.Th e transistor circuitry will help to switch ON the LED bulb when mains voltage (230v) is not available. Each LED will require 3.2V and consume 25mA. I am planning to use two 2N3906(PNP) transistors and one 2N3904 (NPN)transistor. enter image description hereCan somebody suggest your feedback on this circuit.

  • \$\begingroup\$ You give a lot of details, but I can't for the life of me understand what goal you're trying to reach ultimately. What is the aim? "detect AC mains is present when the wall switch is ON"... Fine, does it mean "light up a few LEDs when there is 230VAC between L and N"? If yes, it can be achieved with a much simpler circuit. \$\endgroup\$
    – dim
    Commented Jun 13, 2018 at 14:38
  • \$\begingroup\$ The 100k resistor in parallel with the diode looks strange to me, why is it there? You might want a 1M ohm discharge resistor across the 1uF capacitor. The left PNP might break due to high voltage across the left 1uF cap. There's no proper current limit set for the base current of the NPN and for the right PNP, for both add a base resistor. Why not solve this in a much simpler way by using a relay. Another solution can be to use an opto-isolator, then you only need one NPN or PNP and a simple capacitive dropper for the 230 V side. \$\endgroup\$ Commented Jun 13, 2018 at 14:40
  • \$\begingroup\$ Suppose if AC mains are ON & wall switch is OFF(OPEN) LED should not glow. Then only circuit will detect actual mains failure which I believe cannot be accomplished with opto-isolator or relay. \$\endgroup\$
    – GA A
    Commented Jun 13, 2018 at 15:17

1 Answer 1


Design Review

  • 100k is redundant with 1M both in series
  • 1M part must be rated for 3kV impulse and better with equivalent Z(f) cap
  • LEDS are too low impedance for // operation and must be in series
  • power transfer is thus inefficient and won’t work as expected due to Impedance divider
  • 1 uF is far too small to sustain power to LED ESR=1/Pd per device yet too big as voltage divider
    • eg 3.3V @25mA =85mW , ESR= 12 ohms or less , thus <<12us decay for 1 LED only

busted. Topology NG


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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