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I have designed this monostable (one-shot) vibrator using the 555 timer, yet I am getting this type of error on Pspice for the output voltage only. Where have I gone wrong?

The monostable circuit

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    \$\begingroup\$ What do you mean "no output voltage". Do you mean that the output voltage stays at 0 V? How are you measuring it? With scope? With a meter? With just a simulation? \$\endgroup\$ Oct 10 at 17:14
  • \$\begingroup\$ I'm getting an "invalid expression for Vout" whenever I run the Pspice simulation. The other voltages are working well. \$\endgroup\$
    – Projkt88
    Oct 10 at 17:15
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    \$\begingroup\$ Then please change the title of your question, because it doesn't reflect your actual issue. Change it to: "Why does Pspice say "invalid expression for Vout" for this 555 circuit?" \$\endgroup\$ Oct 10 at 17:17
  • \$\begingroup\$ Yes, you can edit the title. No need to resubmit the question. \$\endgroup\$ Oct 10 at 17:19
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    \$\begingroup\$ Thanks for the heads-up, done \$\endgroup\$
    – Projkt88
    Oct 10 at 17:20
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In the standard monostable circuit, the 555 triggers on the negative-going edge. With Vcc = 9 V, the Trigger input has to sit above 3 V in the idle state, then go below 3 V to trigger the output.

But ...

The 555 is not a normal monostable. If the trigger input still is low when the circuit times out (in around 2 s in your case), the output will stay high until the Trigger input goes high. If the narrow part of your Vtrigger signal is the high part, the circuit will not work because the low part of the trigger signal (2.45 s)is longer than the output period (2 s).

You want Vtrigger to rest high, and go low for 50 ms.

If the trigger signal cannot be adjusted, a solution is to add a differentiator to the Trigger input. This is a series capacitor between Vtrigger and the Trigger input, and a shunt resistor from the Trigger input to GND. The time constant should be shorter than 50 ms so it can recover curing the 50 ms Vtrigger period.

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  • \$\begingroup\$ My assignment's given is as follows: Design a monostable (one-shot) circuit using the 555 IC with the following characteristics: T=2s Vtrigger=Vpulse with (PW=50ms, PER=2.5s, TD=0, TR=TF=1ms, V1=4V, V2-0V) Plot (V0, Vc, Vtrigger) versus time. Run to 2.5s. What am I supposed to do? \$\endgroup\$
    – Projkt88
    Oct 10 at 17:33
  • \$\begingroup\$ Turns out I forgot the pullup resistor. Thank you very much \$\endgroup\$
    – Projkt88
    Oct 10 at 17:36
  • \$\begingroup\$ I'm now getting a 0V output after adding a 10k resistor to the output \$\endgroup\$
    – Projkt88
    Oct 10 at 17:41
  • \$\begingroup\$ You still have not said whether V1 is the amplitude of the 50 ms pulse, or the amplitude of the signal for the rest of the 2.5 s period. Also, a 555 does not need a load of any kind to function properly. It has a totem pole output, and can pull high or low all by itself. \$\endgroup\$
    – AnalogKid
    Oct 10 at 19:30

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