No, you need diodes.Thinking about connecting two outputs of the 4017 to one LED makes me wonder if it’s okay to effectively connect two outputs together
No, you need diodes.Thinking about connecting two outputs of the 4017 to one LED makes me wonder if it’s okay to effectively connect two outputs together
THAT is why I suggested buffers AND then diodes, as a "diode OR" arrangement. The buffers would increase the available drive current if you used either 9 or 12 volt power.No, you need diodes.
OK, I counted wrong. It was early, before morning coffee. And the pulse generator would be the two inverters not used as buffers, used as an oscillator. and use counts 1, 2, and then 6, 7#22 schematic - Connecting the diodes to Q1, 3, 5, and 7 will make a symmetrical blinking pattern, rather than an extra 1-count delay between the end of blue and the start of red.
Ah, but wait - The asymmetrical pattern might be more attention-getting. Hmmm...
Because the 555 is clocking a counter, the clock waveform duty cycle does not affect the output pulse widths. You can eliminate R2 and D3. Connect pins 2 and 6 together, and the other end of R1 to pin 3.
ak
No.Thinking about connecting two outputs of the 4017 to one LED
No.if it’s okay to effectively connect two outputs together.
Yes.Maybe, instead of connecting the LED directly (with one current limiting resistor to ground) the outputs should be connected to the LED via a resistor at each output,
You have arrived.or maybe a diode for each output and one resistor from the LED to ground.
Very clever, I like the fact that one can use smaller caps with this circuit.Assuming this circuit is driving generic red and green LEDs, here is a different approach. This is set for 4 blinks left and 4 blinks right. I added R3-C3 to shorten the on time for a more snappy look.
ak
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