How does a Quartz Crystal in say a Colpitts Crystal Oscillator provide enough Current to Bias the transistor?
Is the Transistor Base just on the edge of being active and the small fluctuation from the Crystal pushes the transistor to conduct?
In a Circuit like this picture, except RL is an LC tank Circuit.

 

R1 and R2 provide the bias for the transistor.
C2 and C3 feed back some signal from the emitter to base, sufficient to maintain oscillations.
The crystal, tuned by C1, filters the feed back to attenuate frequencies away from the cystal frequency and so control the oscillation frequency.

 

Gotcha. So the crystal is excited when the voltage from Emitter hits it and since the crystal is an RLC it resonates at its Fr, biasing the transistor..etc
I'm more curious in the initial start-up of the circuit. The very first nSec that power is switched on for the circuit, the transistor has to turn on to allow power to the crystal to start that oscillation effect, correct?

I may be complicating this more than it should be but my main confusion stems from how such a small signal from a crystal can help bias the transistor. I was under the impression that you can not put much any power onto/into a crystal.

 

You're mis-using the term "biasing." Biasing refers to establishing a DC operating point for the circuit, with voltages and currents somewhere in the transistor's linear operating region; the resistors in your circuit provide the biasing.

The crystal has ABSOLUTELY NOTHING to do with biasing. Rather, it is a resonant element that determines the frequency of oscillation, which is sustained by feedback through C2 and C3.

I'm more curious in the initial start-up of the circuit. The very first nSec that power is switched on for the circuit, the transistor has to turn on to allow power to the crystal to start that oscillation effect, correct?

The initial turn-on transient at power-up, Johnson noise in the resistors, and noise in the transistor all provide a starting point for oscillation, which starts out small and builds up until the amplitude reaches steady-state.

 

Okay I'm understanding.
I've been trying to understand the sole affect that the crystal had on the transistor instead of considering the Crystals affect on the rest of the circuit that influences the transistor.
Take the Crystal out and the Transistor is simply On, put the crystal back in and the circuit has a passive element to pull and push with.

 

See how the operating point changes (average voltage at the emitter):