[time-nuts] have 10MHz need 19.5Mhz

kevin-usenet at horizon.com kevin-usenet at horizon.com
Thu Jun 6 12:24:08 EDT 2013


I've had good luck with TI's CDC913/925/937/949 programmable PLLs.
(The middle digit is the number of PLLs, and the last digit is the
number of outputs.)

I haven't tested them for phase noise, but for an NTP application
all you need is long-term phase lock.  They're fractional-N PLLs,
so you can use deep divider ratios.

They're I2C programmable and have onboard EEPROM so you can set
a power-on configuration.  They do require a +1.8V supply, which the
Pi already has.


Now, first of all, you're misreading the schematic.  X2 is 19.2 MHz,
not 19.5.  So you need to multiply by 192/100 = 48/25.

The onboard VCO can go up to 230 MHz, so the obvious
way to do this is to use an input divisor of 5, a feedback
divisor of 48 or 96 (thus a VCO frequency of 96 or 192 MHz),
and an output divisor of 5 or 10.


Any 2-pin crystal connection to an IC is a simple Pierce oscillator.
There's an on-chip inverter, with a high-impedance feedback resistor
to keep the duty cycle balanced.

You want to disconnect the output pin XTALN, and feed the input XTALP
with an AC-coupled waveform of about 1 V p-p.  So use a capacative
divider between the PLL output and ground to get the voltage right.


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