| Electronics > Beginners |
| Looking for a simply circuit using a 555 timer that will fade an LED in and out. |
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| T3sl4co1l:
Expanding on that further -- Note that the LEDs won't turn all the way off at high supply voltage. The reason for this is that, the transistor drops ~0.6V, and the LED drops a minimum, whatever it does -- depends on color, 1.5-3.6V for red to blue types. Meanwhile, the capacitor voltage is controlled between 1/3 to 2/3 of the supply, by the 555. If the supply is 9V, the range is 3-6V. Subtracting 0.6V for the transistor leaves 2.4V as the minimum across the LED. A blue LED will be quite dim at that voltage, if not completely off; whereas a red LED will probably still be dimly lit. A 6 or 7V supply would be more appropriate in that case, or a resistor can be connected in parallel with the LED to make it see a lower voltage. Note that the fade is approximately linear, which doesn't look right, visually. Ideally you'd have a gamma correction circuit, which causes the current to increase (towards full brightness) more rapidly, and decrease (towards cutoff) more gradually. Unfortunately, there isn't a simple way to implement this on the breadboard. Tim |
| Wimberleytech:
How about using two 555s and a comparator and drive the LED with a PWM signal? I just threw this together in LTSPice. It doubles your complexity but is going to work a lot better. I chose a random comparator from the LTSpice library...use whatever you have in your parts bin that meets the supply limitations (watch for open collector outputs). Update: I built this in the lab. It works as expected. I used an LM311 (the first comparator I could lay my hands on). |
| wilfred:
The Talkingelectronics website is always worth a look. I found this example you might find of interest. http://www.talkingelectronics.com/projects/30%20LED%20Projects/30%20LED%20Projects.html#51 |
| Jwillis:
In your design the negative side of the capacitor should be connected to ground. But you have it connected to pin 6. Pin 6 should be connected to the positive side of the capacitor . Also as mentioned before recheck the orientation of your transistor. For this circuit the collector should be connected to the positive battery supply and the emitter to the led resistor (ballast or current limiter resistor ) then to the anode or positive side of LED ( the long pin). Short pin negative or Cathode of LED connects to ground. Sorry I didn't have my glasses on. |
| Wimberleytech:
--- Quote from: Jwillis on December 06, 2019, 05:53:18 am ---In your design the negative side of the capacitor should be connected to ground. But you have it connected to pin 6. Pin 6 should be connected to the positive side of the capacitor . Also as mentioned before recheck the orientation of your transistor. For this circuit the collector should be connected to the positive battery supply and the emitter to the led resistor (ballast or current limiter resistor ) then to the anode or positive side of LED ( the long pin). Short pin negative or Cathode of LED connects to ground. Sorry I didn't have my glasses on. --- End quote --- LOL Great post! |
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