1
\$\begingroup\$

I know, another question about crystal capacitor values.

I would like to use this crystal in the 20 MHz version to clock an ATmega1284.

According to the crystal's datasheet, it has a load capacitance of 20 pF.

Using the formula \$C = 2(C_L - C_S)\$ and assuming a stray capacitance of 2 pF or 5 pF, I get a value for each capacitor of 36 pF or 30 pF, respectively.

However, this is outside the recommended range for the capacitors of 12-22 pF according to page 46 of the ATmega1284 datasheet.

So, can I even use this crystal, and if so, with what capacitors?

\$\endgroup\$

2 Answers 2

3
\$\begingroup\$

It will probably work with 22pF capacitors, however the frequency will be a bit off (high). Maybe 50ppm or 75ppm, not enough to worry about the maximum frequency of the MCU.

Better to buy a part that is calibrated with a lower load capacitance.

\$\endgroup\$
6
  • \$\begingroup\$ What would happen if I used 36 pF? \$\endgroup\$ Jan 11, 2018 at 2:06
  • \$\begingroup\$ Might have trouble starting when cold or hot, things like that. \$\endgroup\$ Jan 11, 2018 at 2:07
  • \$\begingroup\$ If I were to use either 22 or 36 pF, which would you recommend? (This is for a hobbyist project. Not frying my chips is my main concern.) \$\endgroup\$ Jan 11, 2018 at 2:10
  • 1
    \$\begingroup\$ Suggest 22pF... \$\endgroup\$ Jan 11, 2018 at 2:11
  • 2
    \$\begingroup\$ As Microchip says: Care should be used in selecting values of (load capacitors). Large values increase frequency stability but decrease the loop gain and may cause oscillator start-up problems. \$\endgroup\$ Jan 11, 2018 at 3:20
1
\$\begingroup\$

They are implying you should choose a part with low load capacitance and not 20pF. This also helps reduce power consumption.

A=Series
E=10pF
G=12pF
U=13pF
L=18pF
M=20pF
N=22pF
Q=30pF
R=32pF
S=33pF

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.