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Is there a method to determine the values for the resistances? below shows when R1=R2 << R2=R3 and the capacitance is 10uF and the circuit works perfectly.

Vibrator with high capacitance

However if I try to reduce the capacitance to say 1nF there is no output, as below,

no output

I have read that the frequency should only be dependent on R2, R3, C1 and C2 and that R1 and R4 are only to limit current through the transistors. So with me reducing the capacitance this should increase the frequency as the resistors are the same values but there is no output. I have experimented and if I increase the values of R2=R3=100k I do get an output at much higher frequency but I don't really understand why I should have to change these values. If anyone could help me understand what is going on that would be awesome.

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  • \$\begingroup\$ You do not need to enter "10nC" or "10kR". Al you need is "1nF" or "100k" for a resistor. As for the question try to reduce the R4 value to "210". \$\endgroup\$
    – G36
    Jun 5, 2022 at 14:13
  • \$\begingroup\$ Ah okay thanks for the heads up. I just tried to set R4=R1=210 but with R2=R3=10k and C1=C2=1nF and there is still no output. \$\endgroup\$
    – user488476
    Jun 5, 2022 at 14:31
  • \$\begingroup\$ Set R1 = 220 and R4 = 210 and use a pulse source ( PULSE(0 5 10u) ) instead of DC or in simulation option select "start external dc supply voltage at 0V" (.tran 1m startup) \$\endgroup\$
    – G36
    Jun 5, 2022 at 14:36
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    \$\begingroup\$ SPICE's exact symmetry for all resistors, and transistors prevents toggling. If I set base voltage of one or the other transistors (not both) to "0.6V" with a ** .IC ** option statement, oscillations begin. Also, you might decrease time span in the .tran statement \$\endgroup\$
    – glen_geek
    Jun 5, 2022 at 14:38
  • \$\begingroup\$ Ah okay, i wasn't sure how to set the base voltage of the transistors but i set the option "start external DC voltages at 0V" and this produced an output :) Thanks! \$\endgroup\$
    – user488476
    Jun 5, 2022 at 14:51

2 Answers 2

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Here is it working with 1nF capacitors. I did two things, first I modified one of the capacitors slightly but still well within the tolerance of likely real capacitors. You could change one of the resistors too, the key thing is to upset the exact symmetrical arrangement slightly.

Secondly, I had the supply voltage start from 0V. There is a checkbox in the simulation options, or you can type or, or you can use a pulse voltage source instead of a constant supply voltage.

enter image description here

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To expand on SP's answer:

The reason the circuit did not work has nothing to do with you or the circuit; it's because of LTS. Simulators often have theoretically perfect resistors and capacitors, as in perfectly identical component values and zero noise. they also have perfect voltage sources with a true zero ohm output impedance and an output that magically appears at full voltage with no ramp up.

This is a problem when simulating many types of oscillator circuits because the circuit powers up perfectly balanced - there is nothing to kick the circuit into oscillating. A slight imbalance in component values and/or having the circuit power ramp up almost always cures this problem. The voltage across the slightly larger capacitor increases slightly more slowly, so one transistor begins to conduct slightly ahead of the other, which is enough to create a slight positive feedback, and off they go.

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