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Crystal Load Capacitance Calculator

Enter C1, C2 and stray capacitance to get the load capacitance.

How it is calculated

For a Pierce oscillator the crystal sees the two capacitors C1 and C2 in series, plus the stray capacitance of the board and the chip pins in parallel: CL = (C1 × C2) ÷ (C1 + C2) + Cstray. All three inputs are in picofarads (pF) and the result is shown in pF with two decimals. C1 and C2 must be greater than zero. Stray capacitance is optional: if you leave it empty it counts as 0, and a negative value gives no result. The tool does not know your actual stray value, so it must come from your layout or from the microcontroller datasheet. To design backwards from a crystal's rated load, with C1 = C2 use C1 = 2 × (CL − Cstray).

Worked example

An oscillator uses two 22 pF capacitors and you estimate 3 pF of stray capacitance. The series part is (22 × 22) ÷ (22 + 22) = 484 ÷ 44 = 11 pF, which is always half of C1 when C1 = C2. Adding the stray capacitance: 11 + 3 = 14.00 pF. If the crystal's datasheet asks for 12 pF, this is too high; with C1 = C2 = 2 × (12 − 3) = 18 pF you would land on 12 pF.

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Common mistakes

  • Setting C1 and C2 equal to the crystal's rated load capacitance. The two capacitors are in series, so they must be roughly twice the rated value minus the stray part.
  • Leaving stray capacitance at zero. Traces and pins add a few pF, so the real load is higher than the capacitors alone suggest.
  • Mixing units, such as entering 0.022 for a 22 nF part or values in farads. All fields are in pF.
  • Using the sum C1 + C2 as if the capacitors were in parallel, which gives a load that is several times too large.
  • Forgetting that the printed capacitor tolerance and PCB variation shift the load, so a few kHz of frequency error does not always mean the formula is wrong.

About this calculator

A quartz crystal is only accurate at its rated frequency when the oscillator circuit presents it with the exact load capacitance (CL) printed on its datasheet. In the common Pierce oscillator topology, two capacitors C1 and C2 connect the crystal pins to ground, and the load they present is CL = (C1·C2)/(C1+C2) + Cstray, where Cstray accounts for PCB trace and pin capacitance, typically 2-5 pF.

Embedded and hardware engineers work this formula backward — starting from the crystal's specified load capacitance — to pick C1 and C2 values that keep the oscillator's actual frequency within its tolerance, since a mismatched load capacitance causes frequency pulling or unreliable startup.

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Comments (1)

  • Editorial team

    Capacitors come in standard values, so the 2 x (CL - Cstray) result rarely lands on a real part. Compute with the nearest candidates and see which gets closer. For a 12.5 pF crystal with 4 pF stray, 17 pF is ideal, but 18 pF parts give 13.00 pF and 15 pF parts give 11.50 pF. The stray figure is only an estimate, so treat the result as a starting point and confirm on the board. Where did your stray estimate come from?

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