What does OA1 and transistor do in this VCO circuit?

the circuit is a VCO
i know that OA2 is an integrator and the signal at Vz is trianqular
i also know that OA3 is an inversed schmitt trigger and the signal at Vy is square wave but i have some questions

what does OA1 do?
what does the transistor do (when it is on and off what happens)?
what will happen to the circuit if we remove the capacitor?

thanks

Re: help in VCO circuit

what does OA1 do?
what does the transistor do (when it is on and off what happens)?
what will happen to the circuit if we remove the capacitor?

Click to expand...

OA1 is connected is a way so that when the transistor is ON it will give a gain of -1 and -Vin will appear at the output and it will be integrated, thus the output of the integrator will start falling. When the transistor is OFF the gain of OA1 is 1 and so at its output is Vin and so the output of the integrator will increase.
The Schmitt trigger triggers when the integrator output crosses the +/- 8 volt point approximately. So Vx is like a square wave in +/- Vin, Vy a square wave of +/-8v and Vz a triangular wave with amplitudes +/- 8V but its period would depend on the value of Vin, and so would the period of Vy.
If we remove the capacitor there will be no integration function and Vz would also become a square wave but this time with amplitudes of +/- Vsat, but this tme there is no control over the period, which would depend entirely on the feedback loop delay and just keep on switching, it would then be basically a ring oscillator.
I hope this answered your questions.

help in VCO circuit

i dont understand why when transistor is off the gain is 1 and when it is on the gain is -1.can you explain it??

also what does d1 do??

Re: help in VCO circuit

When the transistor is ON the non inverting terminal of OA1 is connected to ground so it is wired up as a inverting amplifier and since R1 = R2 = 33K the gain is -1. When the transistor is off, the non inverting terminal does not have any current flowing in it, so the voltage on it is Vin and so the voltage at the inverting terminal is also Vin(due to negative feedback) and so it is just like a buffer and has a gain 1.
Diode D1 is just a voltage clamper, so that the voltage at the gate of the FET does not rise above 0.7V or so.

help in VCO circuit

ok
thank you very much