Why are Pins 7 and 5 floating for this 555 IC and how does an NPN and PNP transistor when connected to E B C act a switch? Without the use of Pin 7 surely the capacitor cannot discharge which would render it useless? Circuit Diagram

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    \$\begingroup\$ The output pin 3 is being used for discharging the capacitor when a transistor is connected. I'd recommend to simulate the circuit in LTspice or any other tool and check what happens. You can find the use of pin 5 on the internet. \$\endgroup\$ – Huisman May 1 '19 at 13:17
  • \$\begingroup\$ 10wonga, Are you still not understanding how the capacitor is discharged? Pin 7 is an open-collector output capable of sinking some current. Pin 3 is a push-pull output and able to sink still more current than pin 7. Pin 7 will be active only when pin 3 is LO. So they both can sink current at the same time (only pin 3 can source current), except that pin 3 can sink more than pin 7. \$\endgroup\$ – jonk May 2 '19 at 17:20

Capacitor connected to pins six and two is charged and discharged by the output pin 3.

That is a common charge/discharge method for producing a 50% duty cycle.

Pin 5 can be left open like it is. Normally a capacitor is connected from pin 5 to ground. Pin 5 is the control voltage reference for pins 2 and 6. So a bypass capacitor might be needed for better stable reference voltage.

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