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I get that the trigger must be lower than vcc/3 in order for the comparator to show '1'.

But initially how is the trigger higher than vcc/3? In some text it says trigger is initially vcc and so comparator shows '0'. But how is trigger initially vcc?

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    \$\begingroup\$ Welcome to EE.SE. Your question is impossible to answer without the schematic for the circuit being discussed. Please add one in. There's a schematic button on the editor toolbar and there's a 555 timer component available in it. \$\endgroup\$ – Transistor May 3 at 13:15
  • \$\begingroup\$ In a monostable configuration, there's usually a pullup resistor for the trigger input. \$\endgroup\$ – Dave Tweed May 3 at 13:39
  • \$\begingroup\$ If this pull-up resistor has a capacitor in parallel in your schematic, initially, that capacitor forms a short, shorting Vcc to trigger. \$\endgroup\$ – Huisman May 3 at 16:38

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