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What circuit does is turns the lights on sequentialy like on this gif. enter image description here

What I think is happening is that the capacitor thats connected between ground and the output pin through that 4k7 res is slowly getting charged up each time the pulse goes low and the 555 sinks the current through the output pin. The cap is charging up from 0V to some voltage that can activate first transistor and then each time it gains 0.7V more volts its can over come two then three transistors and so on. What is confusing me is that why do they turn of completely? How does the cap empty to 0V fast?

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  • \$\begingroup\$ This is a perfect opportunity to learn about simulators. Its stop simple enough to quickly simulate with each.g. ltspice \$\endgroup\$ – PlasmaHH Jul 7 '18 at 9:46
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How does the cap empty to 0V fast?

The diode across the 4k7 resistor fast discharges the capacitor that was previously slowly charged via the 4k7 resistor during the period when the 555 outputted a high pulse. The 555 output goes low after some time and rapidly discharges the capacitor via the diode.

In all other respects your simple analysis appears correct.

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  • \$\begingroup\$ Thenks for yor answer. I have one more question. I trying to understand why dont the transistors turn on imediately when the output goes high. I mean its logical that the current should flow from the output straight to the transistors and activate them because but instead it charges up the capacitor. My reasoning is that the cap is basically a short circuit to the ground so all of the current goes to it first. Is that correct? \$\endgroup\$ – Dangz1 Jul 7 '18 at 22:06
  • \$\begingroup\$ The current taken through the bases of the transistors doesn’t much affect the slow charging voltage developed on the capacitor. \$\endgroup\$ – Andy aka Jul 8 '18 at 0:07
  • \$\begingroup\$ Yes, the cap is basically the dominant low impedance and looks like zero ohms when at 0 volts and sinks all the current. (I re-read your comment). I say 0 volts but if you have operated this multiple times then the starting cap voltage might be a couple of hundred milli volts or close to 0.6 volts. It'll still be the dominant impedance though. \$\endgroup\$ – Andy aka Aug 3 '18 at 16:00
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I second Andy's answer but would like to add (not allowed to comment yet), that the capacitor does not discharge to 0V. Instead it gets discharged down to about one diode drop (~0.7V) through the upper 1N4148 when the 555 is outputting a low signal.

To ensure that the lowest transistor does not turn on with this remaining voltage in the capacitor, there is the lower 1N4148 between ground and the emitter to lift the voltage needed to turn the transistor by also about one diode drop.

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