0
\$\begingroup\$

I'm searching a (transistor-based) schematic that can flash a LED regularly but where the off-on and on-off phases look slow like with an ordinary flashing circuit with a good old light-bulb.

With a micro-controller I'd use pulse-width modulation, but how to achieve the effect with as few as possible discrete components?

\$\endgroup\$
  • \$\begingroup\$ Fast attack, slow decay, 2-3 second period? \$\endgroup\$ – EM Fields Feb 14 '16 at 12:34
  • \$\begingroup\$ It will look something like a charging and discharging capacitor or like an AC wave generator. Now search the internet and do some math; this question is very closable because it's very open ended. \$\endgroup\$ – Dave Feb 14 '16 at 12:37
  • 1
    \$\begingroup\$ You can probably get close with a fairly large capacitor in parallel with the LED. \$\endgroup\$ – Brian Drummond Feb 14 '16 at 13:02
2
\$\begingroup\$

Old LEBs (light emitting bulbs) heated up and emitted light fairly quickly, but took longer to cool down and stop emitting light. You might try something around 10 ms time constant for on, and 50 ms for off.

It would be useful to have a voltage control the LED current. This can be the base of a transistor with a resistor on its emitter and the LED on the collector. That becomes a controlled current sink.

ut a cap on the base and a resistor across the cap. Those two set the off time constant. Drive the cap thru another resistor and diode or transistor. The two resistors in parallel with the cap set the on time constant.

\$\endgroup\$
  • \$\begingroup\$ I'm not exactly the fan boy type, but I just want to take a second to say that I appreciate what you do for the stack. It's like getting advice from Gandalf - and I mean that with nerdy veneration. \$\endgroup\$ – Sean Boddy Feb 14 '16 at 23:50
-1
\$\begingroup\$

A capacitor of choice in parallel to the LED and nothing else will do the ramping and fading.

\$\endgroup\$
  • \$\begingroup\$ A capacitor on its own in parallel with an LED will do nothing at all - it will charge up to the supply voltage and then appear open circuit. \$\endgroup\$ – Tom Carpenter Feb 15 '16 at 2:03

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.