Posted on May 17, 2011 at 12:51Hi jdmallett;
There is an important parameter that is often neglicted or unknown, it's the gain margin of the internal inverter. This parameter determines if the oscillator will be start or not. A gain margin of 5 can be considered as a minimum to ensure an efficient start-up of the oscillations.
gain margin = gm / gm_min
gm is the transconductance of the internal inverter. Refer to the table 22 and 23 in the datasheet:
For the HSE: gm = 25mA/V and for the LSE: gm = 5µA/V
the gm_min formula is given by: gm_min = 4*ESR*(2*PI*F)²(C0 + CL)²
Where: ESR is the series resistor (specified by the crystal manufacturer)
C0 is the shunt capacitance (specified by the crystal manufacturer)
CL is the load capacitance (specified by the crystal manufacturer)
F is the frequency of the crystal
let's take an example for HSE part:
let's take the part number of HOSONIC ELECTRONIC a 8MHz quartz: HC-49S-8MHz which has the following caracterstics:
CL = 10pF, C0 = 7pF, ESR = 80ohm
gm_min = 4*80*(2*3.14*8000000)²*(7E-12+10E-12)² = 2.33E-4 A/V
if we calculate the gain margin of this oscillator:
gain margin = 25E-3/2.33E-4 = 107 >> 5 ==> the oscillator has a sufficient gain to make the oscillation to start-up.
Now, let's take the example of the LSE part:
if we take MC306C12O part number from JFVNY manufacturer: its caracterstics are:
CL = 12.5pF, C0 = 2pF, ESR = 50kohm
gm_min = = 4*50E3*(2*3.14*32768)²*(12.5E-12+2E-12)² = 1.78E-6 A/V
Let's now calculate the gain margin of this oscillator:
since gm = 5E-6 A/V (refer to the table 23 in the datasheet)
gain margin = 5E-6/1.78E-6 = 2.8 < 5 ==> the oscilator won't startup since the gain is insufficient to make the oscilations to start up.
I hope that I answered the question
🙂 B.R.
M3allem
