I just started learning about chips and signals using the book Make: Electronics, in one of the circuits, used to generate a pulse in Monostable mode Here

a potentiometer is used to vary the pulse length. What I'm having trouble understanding is why the right pin of the pot is connected back to the wiper and not to ground, I built the circuit on a breadboard and indeed when I connect it to ground the circuit does not function properly.


1 Answer 1


The circuit needs a variable resistance there (AKA a rheostat). It does not need a potentiometer, but potentiometers are what you can buy in the store. You use a potentiometer as a rheostat by connecting it as shown, or by leaving one leg unconnected (it's usually better to connect it as shown, in case the wiper loses contact with the resistive element).

Note that potentiometers are not always connected between power and ground, or power and minus power, etc. -- they show up connected every which way; from power (or some reference, or input) to ground is the most common, but by no means the only way.

  • \$\begingroup\$ Just to clarify for the OP: "... in case the wiper loses contact with the resistive element." With the wiper connected to the other end the circuit will still work although fixed at maximum resistance. This is usually a better outcome than an open-circuit. \$\endgroup\$
    – Transistor
    Nov 17, 2018 at 18:25
  • \$\begingroup\$ Thank you so much this makes sense now, I guess that why the 20k pot is used as a voltage divider for the trigger pin and not as a rheostat. \$\endgroup\$ Nov 17, 2018 at 18:25
  • \$\begingroup\$ I'm actually not sure why they're using a pot on the trigger pin, unless they're trying to vary the duration of the output pulse. It's not a normal feature of a 555 circuit. \$\endgroup\$
    – TimWescott
    Nov 17, 2018 at 18:30
  • \$\begingroup\$ The pot and the 10k resistor are the charging resistor for the 15uF capacitor. It is the voltage on this capacitor that triggers the 555. \$\endgroup\$ Nov 17, 2018 at 20:47
  • \$\begingroup\$ the trigger pot is a strange feature, trigger voltage is only compared against the control voltage signal and that's fixed in this circuit (unless the circuit is being used with varying supply) \$\endgroup\$
    – Jasen
    Nov 17, 2018 at 22:40

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