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MSP430F5529: Wrong load capacitance for 20 MHz, 20 pf crystal

Part Number: MSP430F5529

I'm new to microcontroller development, and I wanted to try running the MSP430f5529 off of an external crystal of 20 MHz that requires 20 pf of capacitance. I read on the datasheet that the pins can support up to 12 pf of internal capacitence, which is what I planned around when I ordered external 22 pf capacitors. When using the equation described here ((22 + 12) * (22 + 12))/(22 + 12 + 22 + 12) + ~3 = 20 pf. 

Unfortunately, I didn't do my research well enough, as it seems that XT2 (the high frequency external) dosen't have internal capacitance. The XCAP bit is only for XT1. My question is, will the lower capacitance be a significant problem for running the clock near 20 MHz (+-500 KHz)?

((22)*(22))/44 + ~3 = 14 pf

Thanks,

Ashwin Gupta

  • Hello Ashwin,

    Thank you for your detailed post. It's helpful to understand your thought process and what you've learned. To answer your question, the effective load capacitance is very important because it allows the external crystal to provide the proper frequency. Otherwise, the crystal will operate at a different frequency and could exhibit instability.

    Our MSP430™ 32-kHz Crystal Oscillators app note contains a lot of important discussion about how to chose and use a crystal oscillator. For example, Figure 5 shows the impact of effective load capacitance on the oscillation frequency for a selected 32kHz crystal.

    You're also correct about the XT2 interface not having dedicated internal load capacitors, but I did find that it has an integrated effective load capacitance of approximately 1pF (see Section 5.16 in the datasheet). This means that each pin has approximately 2pF parasitic capacitance. Using a form of Equation 3 in the MSP430™ 32-kHz Crystal Oscillators app note, you can find out the required capacitance for each of the two load capacitors.

    CLoad = (CL1 + Cparasitic) / 2, assuming CL1 = CL2                                                                 (Eq. 3)

    CL1 = CL2 = (2 x CLoad) - Cparasitic

    CL1 = CL2 = (2 x 20pF) - 2pF, which equals CL1 = CL2 = 38pF

    With your 22pF capacitors, the effective load capacitance equals (22pF + 2pF) / 2 = 12pF instead of the required 20pF. For a correct setup, the effective load capacitance should always match the specification of the used crystal.

    However, Footnote 1 on page 29 in the datasheet mentions that effective load capacitances up to 18pF are supported, so if you have any issues using 38pF load capacitors, you may want to get a crystal with a smaller effective load capacitance.

    Hopefully this helps!

    Regards,

    James

    MSP Customer Applications

  • Thanks James. That helps tremendously, I will look at purchasing new capacitors! For the time being, I will attempt my project with the wrong ones, as a little variance (a couple KHz) is actually acceptable to me. We will see how that goes. 

    Now, hopefully digikey ships these new ones fast! 

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