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During thigh the capacitor is charging through RA from Vcc/3 to 2Vcc/3 thigh = RA * C * ln ((starting voltage across RA)/(finishing voltage across RA)) thigh = RA * C * ln ((2/3 of Vcc - Vdiode)/(1/3 of Vcc - Vdiode)) Multiply all the factors in the brackets by 3 and you get your second equation. During tlow the capacitor is discharging through RB from 2Vcc/...

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Here's what works for me. You can ignore Q3 and R51, who's base is grounded by that remote relay contact, which causes RL6 to be triggered independently. With values shown timing is from 6 seconds to many minutes depending on RP1 setting. Sorry about the two pics!

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Your 470 ohms load is asking for a current of 5V/470 ohms= 10.6mA. But the datasheet of a Cmos 555 with a 5V supply has a 1mA output current, not 10 times more that you have. Why is your schematic as large as my neighborhood with lots of space between its parts? I cropped it.

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Microcontroller with enough I/Os can do that. If you want something more elegant and don't feel like writing code, you could program a Silego Greenpak to do this. The resulting circuit would be really tiny, and you could eliminate those diodes.

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I've had issues with the NE555 in the past. The NE variant is TTL style. There is a better chip that is pin compatible, the LMC555. It is CMOS. From the datasheet: The LMC555 offers the same capability of generating accurate time delays and frequencies as the LM555 but with much lower power dissipation and supply current spikes When I switched over, all ...

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If it is just for a Led effect, why not to use a transistor ring counter. With a little trick you can move backward also.

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If you have a look at the internal circuit of the CD4017 you can get a good idea of how to make one using other CMOS chips - if you have them. You can also see why it's a good idea to use the CD4017. Figure 1. Image source: Intersil. The obvious thing to do is use a micro-controller with at least six IO pins. If you have an Arduino or similar to hand that ...

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The 20kHz oscillation hints at the whole thing working, but somehow the 555's output Q isn't making it all the way down to 0V. I suspect that the problem is whatever you've connected to that output (Q, pin 3), not anything shown here. Or a broken 555 output driver. EDIT: I'm certain that it's oscillating, because you see the 4V to 8V excursions at the ...

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...then goes to a 20 kHz oscillation between 12 V and 11.7 V. I think that then Q2 is slowly switching off. The combination of C10, R12 and D14 doesn't give a "hard" voltage to the gate of Q2 is it isn't switched on/off abruptly. When Q2 is switching off but still in a high resistance state (so it can pull some current from C13) it will help to ...

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When power first is applied, C13 is completely discharged, forcing the output high. This is independent of the state of Q2. You don't give us the Enable voltage in its low state. The 7555 datasheet says its typical threshold value is 0.75 V. If your Enable voltage is higher than this, the 555 will be trying to operate even when Enable is "low" ...

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The lower 1M preset is used to set the frequency as its resistance is common to both charge and discharge paths of the 4n7 capacitor. Then use the upper 1M pot to set the duty cycle. As the 1M preset is wound up (as its resistance is increased) it will not only reduce the output frequency but it will also increase the minimum achievable output high and ...

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It looks like the originator has added a 1M$\Omega$ rheostat, which would have the effect of limiting the range of adjustment of the PWM pot. When turned down to minimum resistance you'd have full range of adjustment, at maximum resistance you'd have very little adjustment range. You can leave it out (ie. replace it with a short) if you want. Call the PWM ...

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I am surprised at the responses from Bimpel and Tony. I think Bimpel is incorrect, while Jansen got it right. As for Tony's response, he starts out saying R1 is redundant and possibly an error, and then the rest of that same sentence says exactly why R1 is needed. I agree that one of the reasons given for the added resistor, that it pulls the capacitor ...

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Using the resistor from pin 3 (output) is best done with a CMOS 555 and with minimal loading on the output. If you have to use a 555 at all. The old bipolar versions have an output swing (even under minimal loading) that will go from near 0V (zero sink current) to a volt or more below the supply voltage (zero source current). You can fiddle the duty cycle as ...

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R1 corrects partially for the asymmetry of the bipolar 555. The CMOS 555 is also assymetric but not as bad as the bipolar. how close do 50% do you need?

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Whilst the 555's output is high, C1 is charging via R1 and the diode. Whilst the 555's output is low C1 is discharging only via R2. R1 is sized smaller than R2 to account for the voltage drop across the diode and the calculator probably assumes that the voltage drop across the diode is 0.7V approx. At such low C1 charge and discharge currents, the voltage ...

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Resistor R1 is used to ensure that the capacitor charges up fully to the same value as the supply voltage. That's actually a nonsense reason because it's not true. As soon as the voltage at pin 6 (threshold) exceeds 2/3 Vcc then the output (pin 3) will change to a low state. So the 555 doesn't even allow C to be charged to the supply voltage! The only thing ...

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