How does a simple oscillator work in non-ideal conditions?

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Start date Sep 29, 2013
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Oscillator

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A simple oscillator operates in non-ideal conditions by utilizing transistors, capacitors, and resistors to create feedback loops that facilitate oscillation. The capacitor charges and discharges, affecting the base current of the transistors, which in turn influences their on/off states. Positive feedback is crucial, as it allows the circuit to maintain oscillation despite the inherent resistive losses and variations in component behavior. The interaction between the charging capacitor and the transistor base currents creates a cycle where the transistors alternately turn on and off, generating oscillations. Understanding these dynamics is essential for grasping how oscillators function in practical applications.

Oct 1, 2013

meBigGuy

Gold Member

I fully understand a bistable mult-vibrator like "oscillator" and was attempting to analyze with that in mind, especially since that was the description.
I had a problem with the off cycle.

" At this point there is not enough base current on Q1 to keep the collector of Q2 close to supply voltage. "

That's where I had a problem. My thoughts were that if R could supply enough current to turn on Q1, the current into Q1 Base would never drop enough to turn it off. I still have a small problem with that since it take very little current on Q1 to keep Q2 saturated. But, then again, it requires very little voltage change at Q2 to get Q1 fully off - maybe even noise is enough since gm is high.

The other thing I don't like is that the base of Q2 is open circuit when Q1 is off, meaning leakage current (Q2 Icb and Q1 Icb and Ice) will tend to keep it on a bit. Feels sloppy.

Looks like starving Q1 Ib with high R is what makes it work.