Crystal load capacitance calculator
The two load capacitors that give a crystal its specified load capacitance, or the load an existing pair actually presents, with the closest standard values.
- Nearest E12 value
- 15 pF
- Load with 15 pF on each side
- 11.5 pF -0.50 pF from the spec
| C1 | C2 | Load | Error |
|---|---|---|---|
| 10 pF | 39 pF | 11.96 pF | -0.04 pF |
| 15 pF | 18 pF | 12.18 pF | +0.18 pF |
| 12 pF | 22 pF | 11.76 pF | -0.24 pF |
| 10 pF | 47 pF | 12.25 pF | +0.25 pF |
| 12 pF | 27 pF | 12.31 pF | +0.31 pF |
| 10 pF | 33 pF | 11.67 pF | -0.33 pF |
How this is calculated
CL = C1 · C2 / ( C1 + C2 ) + Cs C1 = C2 = 2 · ( CL − Cs )
- Stray capacitance
- Cs is the oscillator pin capacitance plus the PCB traces. Microchip AN826 says it can be assumed to be 2 to 5 pF. Short traces keep it low.
- Standard values
- E12 and E24 series per IEC 60063:2015. The table lists the closest pairs of standard values, including unequal pairs, with the load each gives.
- Direction of error
- More load capacitance than specified pulls the crystal below its nominal frequency; less pulls it above.
- Not covered
- Oscillator gain margin and drive level. Check those against the MCU or SoC datasheet for the crystal ESR you choose.
- Microchip AN826, Crystal Oscillator Basics and Crystal Selection for rfPIC and PICmicro DevicesDS00826A, 2002
- Ezurio, How to choose the right crystal and capacitors for your module design (worked 12.5 pF example)
- IEC 60063, Preferred number series for resistors and capacitorsEdition 3.0, 2015
Questions
The crystal sees C1 and C2 in series, plus the stray capacitance. Two equal capacitors in series give half their value, so a 12 pF crystal with 4 pF of stray needs 2 × (12 − 4) = 16 pF on each side, not 12 pF and not 24 pF.
Microchip AN826 uses 2 to 5 pF. Start from the oscillator pin capacitance in your MCU or SoC datasheet and add a few pF for short traces. A long trace or a socket raises it.
The crystal runs off frequency. Too much load pulls it slow, too little pulls it fast. In the Ezurio example, 25 pF capacitors on a 12.5 pF crystal with 5 pF stray present 17.5 pF, 5 pF more than specified; 15 pF capacitors give exactly 12.5 pF.
Yes. The formula holds for any pair, and an unequal pair of standard values often lands closer to the target than the nearest equal pair. The table ranks every E12 pair from 1 pF to 82 pF, or E24 pair to 91 pF, by error.