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I have the following question about a cascade 555 timer circuit. The thing what I do not understand in this schematic how the second input trigger (2) is triggered for low condition. I know that the capacitor (see red arrow) will be discharged after the output (3) of the first timer goes low. But how can trigger (2) of the second timer see that output (3) of the first timer goes low, because it is connected to the negative (lower) side of this capacitor.

The first 555 chip is triggered when SW1 is depressed, taking pin 2 to ground. This action takes the output on pin 3 to high, which lights LED1. Notice, however, that pin 3 of the first 555 is connected through a small capacitor to the trigger input of the second 555. As soon as the time interval expires on the first 555, its output goes low, which turns off LED1 and at the same time triggers the second 555, which in turn lights up LED2. LED2 stays lit until C2 charges, and then it goes out. The circuit then waits to be triggered again by a press of the switch.

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The voltage on pin 2 is pulled up by the base current from an internal PNP transistor, maybe 500nA. So when the output of timer 1 is high, the other side of the capacitor is also high. When the output of timer 1 goes low, the voltage across the capacitor (call it zero for the sake of discussion) does not change instantly (that's what capacitors do) and thus the trigger input is pulled low and the second timer starts.

Incidentally, that's not very good practice, though it probably works, there should be a resistor of 20K or something like that from the trigger input to the supply. It might even sort-of work with a CMOS 555, depending on the frequency vs. leakage currents, but a resistor will make it work for sure.

Internal schematic from here:

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The capacitor charges from the base current at pin 2 (Q10). When the output of timer 1 goes high again, the capacitor is discharged again.. I think initially by two diode drops above the supply voltage through the well used to isolate the lateral PNP, then it drops a bit to supply the bias current. Note that this usage breaks the absolute maximum input voltage spec on the trigger input, but hey, lots of folks do it.

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    \$\begingroup\$ Assuming Hans got the schematic from dummies.com/how-to/content/…, a user in the comments added that he had to add the pull-up resistor to the trigger. \$\endgroup\$ – Justin Trzeciak Mar 23 '14 at 21:52
  • \$\begingroup\$ @jrtrzeciak Thanks for that addendum. +1 \$\endgroup\$ – Spehro Pefhany Mar 23 '14 at 21:53
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But how can trigger (2) of the second timer see that output (3) of the first timer goes low, because it is connected to the negative (lower) side of this capacitor.

Once the series capacitor is drained or charged, both sides will be equal. First the output (3) is High, charging the capacitor. When it is charged, the input (2) is pulled high. When (3) goes low, the capacitor discharges through the led, and when it is fully discharged, both sides are equal, triggering the second 555 and turning LED 1 off.

The capacitor is transparent to the system, when charged.

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More so: the 555 trigger input: A pull up resistor is not required only a paper disc cap, it blocks DC while allowing the AC signal.

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