- #36
Jony130
- 627
- 138
But this subject is intuitive.
Let us examine this case (points A and B are not connected)
At the beginning we apply input signal (1V) to the input of our inverting amplifier with gain equal to Aol = -29V/V.
The output voltage is of-course equal to Vout = Vin * -Aol = -29V, and we have 180° phase shift between Vin and Vout.
And now, if we add a feedback network which reduces the output signal the the level of the input signal (β = 1/29) and introduces another 180° voltage shift. We will have voltage at point B equal in amplitude and in phase to the input voltage. So now if we disconnect Vin and connect point A with point B we can hope that the circuit will start the oscillations at the frequency where these two conditions (amplitude and phase) are satisfied (Aol*β = 1 at Fo = √6/(2* pi *R*C)).
But if Aol*β < 1 but we meet the phase condition. The circuit will not oscillate because we don't have enough gain in our loop (voltage at point B is smaller than Vin).
And for Aol*β > 1 and phase are identical ( 0° phase shift between Vin and voltage at point B), the circuit will also start the osculation because now we have more gain then we need (Voltage at point B is larger than Vin.). Do you get it ?
Let us examine this case (points A and B are not connected)
At the beginning we apply input signal (1V) to the input of our inverting amplifier with gain equal to Aol = -29V/V.
The output voltage is of-course equal to Vout = Vin * -Aol = -29V, and we have 180° phase shift between Vin and Vout.
And now, if we add a feedback network which reduces the output signal the the level of the input signal (β = 1/29) and introduces another 180° voltage shift. We will have voltage at point B equal in amplitude and in phase to the input voltage. So now if we disconnect Vin and connect point A with point B we can hope that the circuit will start the oscillations at the frequency where these two conditions (amplitude and phase) are satisfied (Aol*β = 1 at Fo = √6/(2* pi *R*C)).
But if Aol*β < 1 but we meet the phase condition. The circuit will not oscillate because we don't have enough gain in our loop (voltage at point B is smaller than Vin).
And for Aol*β > 1 and phase are identical ( 0° phase shift between Vin and voltage at point B), the circuit will also start the osculation because now we have more gain then we need (Voltage at point B is larger than Vin.). Do you get it ?