I have two LEDs in the circuit green LED turn on when the switch is on also capacitor is charged, Now I want to turn on the red LED by using capacitor when the switch goes off. Please help me, how I can do that. Thanks
Maybe something along these lines. When SW1 is closed, C1 charges through D4 + D2 + R3 and D3 (green) is illuminated. R3 limits the current when SW1 is initially closed.
When SW1 is opened, D3 goes dark and D1 (red) is illuminated, and the current through it decreases (more or less) exponentially towards zero.
In this simulation, the switch is opened at T = 1 second.
Answered by Spehro Pefhany on November 21, 2021
4 Asked on July 23, 2020 by nachonachoman
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