Hi everybody
I am learning about colpitts crystal oscillator circuit. I am currently reading the document "CRYSTAL OSCILLATOR CIRCUITS" by author Robert J. Matthys.
On page 37 of the document, I see that the author wrote that the colpitts oscillator circuit has three distinct operating states:
"The amplifier is an emitter follower with a gain of 1. The transistorconducts current over only a small portion of each oscillation cycle, usually about 15-20%. The transistor starts conducting a little before (about 10%) its base reaches the most positive peak of the sinusoid and stops educated immediately after the positive peak.In the positive peak, the transistor saturates and clamps the crystal to the power supply bus
through the forward-biased base-collector junction. Positive peak saturation about about 5-10% of an oscillation cycle. The transistor shuts off and remains nonconducting over the rest (80-85%) of the cycle. Thus, three circuit conditions exist during each cycle: a short interval (10%), with the transistor repeated properly and acting as an emitter follower; a second short interval (5-lO%), with the transistor saturated and shorting out the crystal ; and a third long interval (80-85%), with the transistorshut off and nonconducting. "
I really do not understand the circuit operation according to the author's explanation and really want a clearer explanation. Look forward to the attention of everyone. Sincerely thank you

I am learning about colpitts crystal oscillator circuit. I am currently reading the document "CRYSTAL OSCILLATOR CIRCUITS" by author Robert J. Matthys.
On page 37 of the document, I see that the author wrote that the colpitts oscillator circuit has three distinct operating states:
"The amplifier is an emitter follower with a gain of 1. The transistorconducts current over only a small portion of each oscillation cycle, usually about 15-20%. The transistor starts conducting a little before (about 10%) its base reaches the most positive peak of the sinusoid and stops educated immediately after the positive peak.In the positive peak, the transistor saturates and clamps the crystal to the power supply bus
through the forward-biased base-collector junction. Positive peak saturation about about 5-10% of an oscillation cycle. The transistor shuts off and remains nonconducting over the rest (80-85%) of the cycle. Thus, three circuit conditions exist during each cycle: a short interval (10%), with the transistor repeated properly and acting as an emitter follower; a second short interval (5-lO%), with the transistor saturated and shorting out the crystal ; and a third long interval (80-85%), with the transistorshut off and nonconducting. "
I really do not understand the circuit operation according to the author's explanation and really want a clearer explanation. Look forward to the attention of everyone. Sincerely thank you
