Electronic – Wrong load capacitance for 20 MHz, 20 pf crystal


I am 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 capacitance, which is what I planned around when I ordered external 22 pF capacitors. When using the equation described here.

$$\frac{(22+12)\cdot(22+12)}{22+12+22+12} + \sim3 = 20\,\text{pF}$$

Unfortunately, I did not do my research well enough, as it seems that XT2 (the high frequency external register) does not have internal capacitance. The XCAP bit is only for the XT1 register. My question is, will the lower capacitance be a significant problem for running the clock near 20 MHz (±500 kHz)?

$$\frac{22\cdot22}{44} + \sim3 = 14\,\text{pF}$$

That makes it 6 pF less then desired.

edit: Also, I do not know if this is how capacitors work in series, but another option is I could put two of the 22 pF ones inline on one terminal. So I could have $$\frac{(22+22)\cdot22}{22+22+22} + \sim3 \simeq 17\,\text{pF}$$ Does that work?


Best Answer

To get the specified 20 pF load for a crystal, you need two 40 pF caps, theoretically. However, these 40 pF caps in reality do include package pin capacitance (2-3 pF each pin), plus parasitic capacitance of traces and pads under the IC and under XTAL, which typically are about 5-6pF. Therefore in your case you should use about 33 pF caps. Together with pin and trace capacitance it will form the desired 40 pF.

The parameter you seem to be concerned with is called "crystal pullability", see this publication for brief explanation and typical dependence. Deviation of +- 5pF from the designated 20pF load will give you at most +- 100ppm, which is about +- 2 kHz for a 20 MHz crystal.