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In circuits where there is no input signal shown explicitly as in the figure shown below, how do we determine the behaviour of the circuit? I understand that the power supply given to the opamps drives the circuit, but don't know exactly how we consider their effects initially. I'd like to know how we take initial conditions into the equations at different nodes(say Vout, here) enter image description here

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  • \$\begingroup\$ Consider a resistor that is grounded with one terminal, I apply a voltage (input) at the other terminal, a current flows and that is my output. I could also force a current into the resistor (input) and measure the voltage (output). My point: information (and input and output) can be a voltage but also a current. So given one terminal, there are still two entities that can be determined. \$\endgroup\$ – Bimpelrekkie Mar 4 at 15:07
  • \$\begingroup\$ What do you mean with "behaviour"? The circuits reaction after power switch-on? Can be answered without calculation - look at the lower opamp. What will be the result of the feedback path? \$\endgroup\$ – LvW Mar 4 at 16:47
  • \$\begingroup\$ Yes ,it is clear now \$\endgroup\$ – Pranav Kumar Mar 4 at 17:50
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Clues:

Have a look at the configuration of OA2. (You should have labeled all the components.)

  • Do you recognise the configuration?
  • If not, is there anything unusual about the feedback?
  • If that gives you a clue then what will be the nature of the output of this part of the circuit.

Then,

  • What is OA1 doing?
  • Do you recognise its configuration?

When you put the two together what happens?

Assume some non-zero Vout initially and see what the maths tells you.

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  • \$\begingroup\$ Thank you for accepting my "answer". Did it prompt you enough to answer the question yourself? If you like you can post your own answer and accept it yourself. I think that if you can explain it you'll prove that you've understood and we can give some more pointers if something is not clear. \$\endgroup\$ – Transistor Mar 5 at 13:43
  • \$\begingroup\$ Yes it did.I realised that the Schmitt Trigger output had to be in either of its two states due to some inequality in voltages at the terminals which exists(minute, but big enough to cause saturation).From there, we work out what the integrator does. \$\endgroup\$ – Pranav Kumar Mar 14 at 18:26
  • \$\begingroup\$ Good. Thanks for letting me know. Watch out for simulations of circuits like these. Sometimes they start in perfect balance and you need to inject a little "noise" to get something like the Schmitt trigger to go one way or the other. \$\endgroup\$ – Transistor Mar 14 at 18:38
  • \$\begingroup\$ Noted.Thank you. \$\endgroup\$ – Pranav Kumar Mar 15 at 8:49

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