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Topic: Basic maths: 10k? resistor, 5V voltage, 100µF cap=0.76 second LED fadeout (Read 3777 times) previous topic - next topic


How come you're not using a transistor ?  You can make it fade out nicely then!

Did you refer to me? Because i want to learn this stuff :) Transistors will come later  8)


Yes, the missing 0.25j goes into RF radiation from the magnetic field changes as the charge bounces between the caps, into resistive losses in the wires, losses in the dielectric, and losses in the ESR of the capacitors.

If you had two perfect capacitors with superconducting plates and superconducting wire and no losses in the dielectric, it would take a long time to stop ringing, but eventually it would radiate away as RF.
Steve Greenfield AE7HD
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Jack Christensen

It might be interesting to connect the circuit as below, use the digital pin to charge and discharge the capacitor, and the analog pin to measure the voltage across the cap. It probably won't be perfect, but I'd bet it would come pretty close to producing the classic charge/discharge curve (example here).


Yes mace, the led discharging through the cap gives a short pulse not so much fade. .

Using a transistor you fill a capacitor with the button press and as that slowly discharges (connected to the base of a transistor) will gently fade out the led...

If you want automatic a 555 can be used to "breathe" ..

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