# Capacitor for crystal

C

#### ClueLess

Jan 1, 1970
0
For very slow crystals it is necessary to add a series resistance in the crystal
circuit. How do you calculate its value with respect to the two capacitors that
connect the crystal to the ground?

R

#### Rich Webb

Jan 1, 1970
0
For very slow crystals it is necessary to add a series resistance in the crystal
circuit. How do you calculate its value with respect to the two capacitors that
connect the crystal to the ground?

First I'd read the datasheets and/or application notes for the
microcontroller and for the crystal under consideration.

S

#### Spehro Pefhany

Jan 1, 1970
0
For very slow crystals it is necessary to add a series resistance in the crystal
circuit. How do you calculate its value with respect to the two capacitors that
connect the crystal to the ground?

The resistor is there to limit drive power.. so it depends on crystal
characteristics (ESR, max drive power) and other things (drive
voltage, load caps). It's not a straightforward calculation,
particularly if you want to minimize power consumption and/or drive
power. Overdriving the crystal can lead to excessive drift or failure.
Typically 32kHz-ish crystals have *maximum* drive powers in the 1uW to
100nW range.

Best regards,
Spehro Pefhany

C

#### ClueLess

Jan 1, 1970
0
The resistor is there to limit drive power.. so it depends on crystal
characteristics (ESR, max drive power) and other things (drive
voltage, load caps). It's not a straightforward calculation,
particularly if you want to minimize power consumption and/or drive
power. Overdriving the crystal can lead to excessive drift or failure.
Typically 32kHz-ish crystals have *maximum* drive powers in the 1uW to
100nW range.

Actually the micro is running slower, say about 5%, than what it should (a
thirty second operation takes place in 31.5 seconds) so I was wondering whether
the RC values have any bearing.

Perhaps I should try to change the C value by inserting a trimmer instead of the
fixed cap.

Thanks again for your time and help

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