I have designed a basic application of a 555 in a monostable configuration to delay the LOW/off state of an output signal. (The output drives some power electronics through an optocoupler).

As per my simulations and breadboard tests, when the foot pedal is depressed and held the output should remain high until it is released where the output remains high for a further ~0-10s depending on the value of the adjustment R5.

The idea is that C7 discharges through Q4 until the pedal is released where it charges up to 2/3VCC and then turns the output LOW. Unfortunately, when I had this circuit board printed and I populated it, it behaves as expected only for the smallest values of the potentiometer (delays up to around a second), after which, at around 100k or so the output just stays high indefinitely until I turn the knob back down below this value.

I cannot think of any theoretical reason this should be happening. Is there something I'm missing?

Any advice is greatly appreciated, thanks.

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    \$\begingroup\$ You could try using a larger capacitor and a smaller pot e.g. 100uF & 100K. You are also missing a pull down resistor on the base of Q3 and Q4. \$\endgroup\$ – HandyHowie Aug 5 '19 at 9:49
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    \$\begingroup\$ issues I see with this circuit: Don't share base resistor R7 between Q3 and Q4, give each transistor its own base resistor. The wire between J1 and R7 has a connection to the right, what is there? Could that be lifting this node up in voltage, preventing Q4 from switching off properly? At R5 = 1 Mohm you get in the range where leakage currents and input currents (TRE input of 555) start to matter. Also Q4 might not be switched off properly (there's no pulldown) and leak some current. \$\endgroup\$ – Bimpelrekkie Aug 5 '19 at 9:49
  • \$\begingroup\$ Running two base resistors instead of R7 is equivalent since each transistor just draws half the current passing through R7 so I doubt it's that. The connection to the right of J1 is just an indicator LED. I might try a bigger cap and smaller resistor. Do the transistors need a pulldown? Since they draw no current when the pedal is up then they will be completely off? If I added a pulldown then R7 and the pulldown would act as a voltage divider, wouldn't they? \$\endgroup\$ – Michael Fotopoulos Aug 5 '19 at 11:10
  • \$\begingroup\$ Yes, you need a pull down. You could add the pull down before R7, then you wont be making a potential divider. \$\endgroup\$ – HandyHowie Aug 5 '19 at 12:25

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