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For your consideration, I have this Astable-multivibrator via a classic Op-Amp Schmitt Trigger delivering a square signal (here 1.82Hz).

  • I am fine with the circuit with SW1 open as C2 is basically in Series with C1 making CT = 5uF.

  • However, I am not sure about the effect of SW1 being closed.

Can someone enlighten me?

Figure 1. Switch Open

enter image description here

Figure 2. Switch Closed enter image description here

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1 Answer 1

Reset to default
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With the switch on, it would basically increase the RC time constant and increase the time. It could be considered C2 shorted and the vibrator would continue with the frequency formed from R3, R1 and C1

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  • \$\begingroup\$ thanks @laptop2d, makes total sense, but then would this mean iCircuit is incorrect when showing zip on the Scope when I close the switch (Figure 2. above)? \$\endgroup\$
    – Jeruinsky
    Apr 9, 2019 at 19:27
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    \$\begingroup\$ The simulation does not look accurate. It should show a lower frequency when the switch is closed. Just for the heck of it, why not try moving the switch to be across the lower capacitor and see how the simulation fares. \$\endgroup\$
    – Dwayne Reid
    Apr 9, 2019 at 19:46
  • \$\begingroup\$ The simulation is most likely not working, use a better simulator like lt spice \$\endgroup\$
    – Voltage Spike ♦
    Apr 9, 2019 at 19:52

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