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. Can somebody suggest your feedback on this circuit.
- 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