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I'm using a PIC24FJ256GB110 connected via SPI to a CAN Bus Controller chip MCP2515. The MCP2515 requires a crystal connected to it but has a Clock Output pin for driving a connected MCU. So I was going to connect the Clkout from the MCP2515 to the MCU.

My question is that in the osillator data sheet for the PIC24F family of processors it shows this external clock connection with a NOT Logic gate in the circuit. I usually get parts from farnell but struggling with what it is I need for this NOT gate. Can it be any general NOT Gate or is it something more specific then that, like a buffer or some sort of impedance magic? Or is it as simple as a single NOT gate used to "clean" up the clock signal?

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It's just a gate to clean up the waveform (or just to indicate that there is a logic output driving the input)- it would not generally be used if you were employing a clock oscillator nearby the MCU such as this one (photo from Digikey.com).

enter image description here

You could use a ST gate such as this one, but usually it's not necessary.

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These (Pierce) oscillators are commonly built around CMOS ports for their high impedance. Usually a 74HC04, 4069 will do the job. Use the remaining ports as output buffer so do do not unnecessarily load the oscillator itself.

A Schmitt Trigger is not useful for a Pierce oscillator.

Use of the CMOS Unbuffered Inverter in Oscillator Circuits - Texas Instruments Especially check the chapter on Practical Design Tips

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    \$\begingroup\$ That download link was very slow for me, an alternative is ti.com/litv/pdf/szza043 \$\endgroup\$
    – alexan_e
    Commented Feb 19, 2014 at 20:19
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    \$\begingroup\$ The 74HC04 is NOT a good candidate for Pierce oscillators. Use 74HCU04 (buffered versions can do strange things because of the extra delays). The CD4069 is okay because it has only a single stage. \$\endgroup\$ Commented Feb 19, 2014 at 20:36
  • \$\begingroup\$ @SpehroPefhany I was reading about (un)buffered in the TI document I linked, but I was unable to figure out which devices go with that. Turns out the HC U was the answer. \$\endgroup\$
    – jippie
    Commented Feb 19, 2014 at 20:38
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    \$\begingroup\$ @alexan_e changed the link. \$\endgroup\$
    – jippie
    Commented Feb 19, 2014 at 20:39

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