How Does Changing Load Capacitance Values Affect Crystal Oscillator Performance?

[likephysics]

How does the load capacitance affect the Crystal.
For example, IC manufacturers recommend a Crystal and also load capacitance, like 25MHZ, 16pF.
What happens if I choose a xtal with load cap of 10pF 25MHz?
I am talking about changing the load caps recommended by IC manufacturer, not the ones in xtal datasheets.Does the load cap value affect the amplifier ckt inside the IC?

 

[tech99]

How does the load capacitance affect the Crystal.
For example, IC manufacturers recommend a Crystal and also load capacitance, like 25MHZ, 16pF.
What happens if I choose a xtal with load cap of 10pF 25MHz?
I am talking about changing the load caps recommended by IC manufacturer, not the ones in xtal datasheets.Does the load cap value affect the amplifier ckt inside the IC?

The crystals with 16pF might have been intended for circuits using series resonant operation and those with 25pF for parallel. But it should not change the operation of the circuit very much and both are just pieces of Quartz.
Increasing the load capacitance lowers the parallel resonant frequency very slightly and also slightly reduces the effectiveness of the crystal as a resonator. It does not alter the basic operation of the IC oscillator circuit apart from lowering frequency and efficiency and eventually stopping oscillation.
This happens because a crystal is really a series resonant device. In many applications, however, a parallel resonance is created where the crystal is used slightly HF of its series resonant frequency to create a large inductive reactance. This then forms a parallel resonant circuit with the shunt capacitance. The losses of a crystal are physically part of the series resonant circuit, so when we use a parallel resonance, this loss resistance is transformed up by a factor Q^2 and appears as a very high resistance shunting the crystal. If we add shunt capacitance, the inductance of the crystal must be reduced, so Q (series) becomes less. If the Q of the series resoance is lowered, the shunt resistance, which depends on Q^2, is less.