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I'm new to electronics and I'm learning some circuits with op amps. Now I have been shown this astable multivibrator.

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I understand how it works.

  1. The op amp is saturated at one side or the other
  2. Voltage is divided and a portion of it goes to non-inverting input.
  3. Voltage also charges the capacitor until it gets equal to the voltage on the non-inverting output
  4. At that point the op amp swaps its output to the other saturation point.

Why is the op amp saturated at point 1 to begin with?

If I remove the capacitor (either shorting it, opening it or replacing it with a resistance), the op amp will sit at zero volts, since both inputs are zero, and that is totally what I would expect.

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Why, by plugging in an uncharged capacitor, does the op amp suddenly decides to saturate?

The uncharged capacitor should not create any change in the difference of voltage between inverting and non inverting inputs that the op amp could amplify, so I would expect that in a perfect simulation this circuit would not start oscillating in the first place unless the capacitor does hold a tiny bit of charge.

Why does it start oscillating?

EDIT: so it turns out that in the simulator the capacitor actually had an initial voltage, and indeed if I set it to zero then the simulation will not oscillate.

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    \$\begingroup\$ It is starting due to thermal noise.Many oscillators start from thermal noise. \$\endgroup\$
    – Miss Mulan
    Apr 10, 2021 at 16:32
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    \$\begingroup\$ Thermal noise starts the oscillations and it gets amplified until it is not.It is not about ideality of components. \$\endgroup\$
    – Miss Mulan
    Apr 10, 2021 at 16:38
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    \$\begingroup\$ Yes but you will have noise to the output as well and because this is a closed loop configuration(with feedback) and the voltage dividers and the circuit is asymmetric (bottom you have a voltage divider , top dont have) you will get a voltage between the inverting and the non-inverting terminal. \$\endgroup\$
    – Miss Mulan
    Apr 10, 2021 at 16:54
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    \$\begingroup\$ I am not talking about falstad , i am talking about the real circuit if you build it on a breadboard why the oscillation starts. \$\endgroup\$
    – Miss Mulan
    Apr 10, 2021 at 17:01
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    \$\begingroup\$ Yes it will saturate. \$\endgroup\$
    – Miss Mulan
    Apr 10, 2021 at 17:16

2 Answers 2

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The uncharged cap should not create any change in the difference of voltage between inverting and non inverting inputs that the op-amp could amplify, so I would expect that in a perfect simulation this circuit would not start oscillating in the first place unless the capacitor does hold a tiny bit of charge.

You are correct and there are many questions on this site from people wondering why their simulation doesn't oscillate.

I suspect that your simulator (Falstad?) is introducing a little offset error or noise to more realistically simulate a real device specifically for applications such as yours. I think that a smart solution (for them) would be to introduce it in the power-up of the op-amp so that the output is pulled a little more than the other and once fully powered the component continues with ideal performance.

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    \$\begingroup\$ You where right!, after checking it's properties it turns out the cap was not actually discharged on initial conditions in the simulator, probably for this reason. After setting it to zero initial charge and restarting the simulator, the oscilator does not start. However IRL, does the capacitor hold a little bit of charge caused by external noise so that it creates a tiny difference in the inputs and starts the oscilation? \$\endgroup\$
    – Joaquin Brandan
    Apr 10, 2021 at 16:56
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    \$\begingroup\$ No, generally the capacitor would be discharged but the op-amp would be non ideal and will have some offsets on the inputs and won't power up symmetrically. \$\endgroup\$
    – Transistor
    Apr 10, 2021 at 16:59
  • \$\begingroup\$ So it would actually output a small voltage that would charge a tiny bit the capaciitor and start the whole thing. \$\endgroup\$
    – Joaquin Brandan
    Apr 10, 2021 at 17:00
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    \$\begingroup\$ Maybe. A real op-amp will have tiny input bias currents which might charge the capacitor a bit but notice that the feedback resistor would discharge it. A more likely scenario is that there is a slight offset voltage between the two inputs - say in the order of nV or uV - and when this is multiplied by the open-loop gain it will result in the output voltage deviating from zero enough to get things going. \$\endgroup\$
    – Transistor
    Apr 10, 2021 at 17:10
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    \$\begingroup\$ @Miss, the circuit has a positive feedback loop which gives a Schmitt trigger action. \$\endgroup\$
    – Transistor
    Apr 10, 2021 at 17:26
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If you give the op-amp even a tiny bit of offset voltage when you connect the uncharged capacitor the output will slam to one rail or the other and oscillation begins.

Similarly, even if perfectly balanced on knife-edge, a microvolt of change in offset voltage will be amplified by perhaps 1,000,000 and the output again saturates, since that 1V becomes 500mV at the non-inverting input.

It’s possible for simulations that use too-ideal components to fail to start oscillating. Giving them a minute ‘kick’ or preventing them from seeking a perfect balance in the first place may help.

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  • \$\begingroup\$ Notes about using (real) op amps for comparator-like applications could be included ("However, the best advice on using an op amp as a comparator is very simple - don't!"). \$\endgroup\$
    – Peter Mortensen
    Apr 11, 2021 at 7:07

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