# colpitts crystal oscillator loaded Q factor

#### calebpinter

Joined Feb 6, 2017
2
Hello,

if I have a crystal oscillator defined as the image shown below the Q factor can be defined as $$Q = \frac{2\pi\omega_o L_1}{R1}$$.

If a resistor is added in parallel similar to what would happen in a colpitts oscillator, how would the Q factor be reduced?

#### calebpinter

Joined Feb 6, 2017
2
My first attempt was to find the Q of the series RLC branch,
transform R1 into a parallel resistor:
$$R_p = R_2*Q^2$$

find the equivalent resistance between R1 and R2:
$$R_{eq} = R1||R2$$

and then calculate the new Q of entire network as if it were a RLC network:
$$Q = \frac{R_{eq}}{2\pi\omega_oL1}$$
but if I plug in values such that the series resonant frequency is on the scale of 10 MHz and R2 is a 20 or 30 kohm, Q factor is close to 0 which doesn't make sense.