Author Topic: Have I got these crystal frequency calculations correct?  (Read 220 times)

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Offline tehtehteh

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Have I got these crystal frequency calculations correct?
« on: June 20, 2021, 03:59:25 pm »
My first question is are my calculations for this crystal frequency pulling correct?

From the MAX7044 data sheet (

The crystal oscillator in the MAX7044 is designed to present a capacitance of approximately 3pF between the XTAL1 and XTAL2 pins. If a crystal designed to oscillate with a different load capacitance is used, the crystal is pulled away from its intended operating frequency, thus introducing an error in the reference frequency. Crystals designed to operate with higher differential load capacitance always pull the reference frequency higher. For example, a 9.84375MHz crystal designed to operate with a 10pF load capacitance oscillates at 9.84688MHz with the MAX7044, causing the transmitter to be transmitting at 315.1MHz rather than 315.0MHz, an error of about 100kHz, or 320ppm. In actuality, the oscillator pulls every crystal. The crystal’s natural frequency is really below its specified frequency, but when loaded with the specified load capacitance, the crystal is pulled and oscillates at its specified frequency. This pulling is already accounted for in the specification of the load capacitance. Additional pulling can be calculated if the electrical parameters of the crystal are known. The frequency pulling is given by:

\$f_p=\frac{C_m}{2}\left(\frac{1}{{C_{case}+C_{load}}} - \frac{1}{{C_{case}+C_{spec}}}\right) * 10^6\$

\$f_p\$ is the amount the crystal frequency is pulled in ppm.
\$C_m\$ is the motional capacitance of the crystal.
\$C_{case}\$ (or \$C_o\$) is the vendor-specified case capacitance of the crystal.
\$C_{spec}\$ is the specified load capacitance.
\$C_{load}\$ is the actual load capacitance.
When the crystal is loaded as specified (i.e., \$C_{load} = C_{spec}\$) the frequency pulling equals zero.

The crystal frequency is \$f_{XTAL} = \frac{f_{RF}}{32}\$ so 433.92MHz requires a 13.56MHz crystal, and the closest I can get at this frequency is is 4pF.

The data sheet for this crystal states (


Frequency: 10.0000MHz
Plating Load: 4pF

C0   =   0.88pF
R1   =   53.82Ω
L1   =   162.02mH
C1   =   1.56fF

Frequency: 27.0000MHz
Plating Load: 4pF

C0   =   1.16pF
R1   =   11.83Ω
L1   =   9.16mH
C1   =   3.8fF

so for 10MHz:    \$\frac{0.00156}{2}\left(\frac{1}{{0.88+4}} - \frac{1}{{0.88+3}}\right) * 10^6\$   Which gives -41.1948622613 ppm error.
and for 27MHz: \$\frac{0.0038}{2}\left(\frac{1}{{1.16+4}} - \frac{1}{{1.16+3}}\right) * 10^6\$   Which gives -88.5137149672 ppm error.

I'm not sure what the negative result means as the data sheet stated that this error would increase frequency, but if I ignore the sign and extrapolate for 13.56MHz I get 51.1ppm error or 433.9422MHz.

My second question is although I could correct for this error with a capacitor, I'd like to know if this error would be considered acceptable for a car remote fob, and I could get away with no capacitor to save precious space on the PCB?

The alternative would be a slightly smaller crystal with 9pF load capacitance, and a parallel capacitor, in which case my third question is: is a 6pF parallel capacitor the correct value? (Based on this formula: \$\frac{1}{C_l}=\frac{1}{C_{series1}} + \frac{1}{C_{series2}} + \frac{1}{C_{spec} + C_{parallel}}\$)

Online bob91343

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Re: Have I got these crystal frequency calculations correct?
« Reply #1 on: June 20, 2021, 09:28:50 pm »
I think you are overengineering this.  You can't buy crystals as accurate as that, at least I don't think so.  One solution would be to try a few and see how they work.  I suspect the receiver is broad enough to tolerate some frequency error.

Offline tehtehteh

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Re: Have I got these crystal frequency calculations correct?
« Reply #2 on: June 21, 2021, 11:48:15 am »
The majority of crystals I have been looking at are 10ppm or 20ppm accuracy.

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