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In which cases for higher accurancy the LFXO crystal must be used and must int's load capacitors have exact 12pF values?

Hi,

We develop two BLE projects on NRF51822 and NRF52832 chips.

In both chips, for higher accuracy (better than ±250ppm) exists the option to connect the LFXO 32kHz crystal to pins P0.00 & p0.01 in NRF52832 chip and to p0.26 & p0.27 in NRF51822 chip.

  1. According to manual this crystal is used for higher accuracy better than ±250ppm, but in which cases this accuracy is required?
  2. In the examples of connection diagram, between each crystal terminal and ground are connected the two load capacitors with the values of 12pF. But in anyone of these examples anything about required value isn’t written. To purchase the capacitors with exact 12pF value is very hard. Much more easy to purchase the capacitors with 10pF or 15pF values. Must these capacitors have the exact 12pF values or it is possible to connect the capacitors with different values. Which capacitances are more suitable for this crystal - 10pF or 15pF? Which capacitance range is recommended?
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  • 1. A crystal will have better accuracy than the RC oscillator. This will make it possible to use a narrower receive window because the timing is more accurate. Hence the time the receiver needs to be on is shorter and the average current draw is less. 

    2. The value of the load capacitors depends on the CL value of the crystal. The pin capacitance of the nRF51 and nRF52 is 4 pF so you must subtract this. The formula for the load caps then becomes: Ccap = CL * 2 - 4 pF

    12 pF is not a hard value to get.

Reply
  • 1. A crystal will have better accuracy than the RC oscillator. This will make it possible to use a narrower receive window because the timing is more accurate. Hence the time the receiver needs to be on is shorter and the average current draw is less. 

    2. The value of the load capacitors depends on the CL value of the crystal. The pin capacitance of the nRF51 and nRF52 is 4 pF so you must subtract this. The formula for the load caps then becomes: Ccap = CL * 2 - 4 pF

    12 pF is not a hard value to get.

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