Schematics: schematics

I'm working on improvement of my CCIR->OIRT downconverter based on LA1185 IC. Previously I've faced strange problem that on certain frequencies I'm getting just a loud silence and sometimes (depending on temperature changes, I think) it just doesn't work at all. I suspect that this can be related to the quartz crystals which I use - they are working on their 3rd harmonic. I've decided to change them to fundamental ones.

I discovered, however, that the most of the crystals from Mouser (22 MHz, 30 MHz and 36 MHz, for example) have load capacitance. If I understand this right this is the capacitance which I should match by adding two caps of that value in parallel to the both pins of my crystals (as I usually do with AVRs) to the ground so the crystal itself is designed for parallel use. However it's clearly seen from schematics that Y1-Y3 crystals are just in series between ground and OSC pin of LA1185.

There are few crystals on Mouser which are labeled as "Series" so I can't actually get full set of them. And I doubt if I can use crystals with Load Capacitance in such a way according to the schematics.

My questions are:

  1. Can I just use any crystal of right frequency but with Load Capacitance without any modifications to the current schematics/PCB?
  2. If no, is it possible to use them anyway but with, for example, adding some caps between crystal pins and ground (-5V in this schematics)?
  3. If previous is true, what will be the proper way to add such a caps to my crystals?

Thank you!

EDIT 1 Regarding question 1, let's consider that Y1 crystal have stated Load Capacitance of 10 pF (I assume it's designed to be used in parallel), what is the proper way of connecting it?

Series: series

Parallel with one cap: parallel 1

Parallel with two caps: parallel 2

I understand that frequency will vary a bit but as I said down in the comments I don't think that this change will matter a lot because even an error of a few kilohertz should be ok for downconverting the whole CCIR into OIRT (that's why I use three crystals selectable - just to be able to select necessary subrange). More interesting question is how it will impact on the stability? What are the potential pitfalls of using improper connection such as no oscillation at all, temperature dependence etc?

EDIT 2 I've found this useful AN. According to the chapter 5 I should select C1 and C2 in that way so their series combination will be close to the Load Capacitance: (C1 * C2) / (C1 + C2) = C_load. For AVRs it's around 11 pF (22 * 22 / (22 + 22)).


1 Answer 1


From the datasheet it looks like the LA1185 is intended to be used with a parallel resonant LC. There doesn't seem to be any data indicating the capacitance at the pin.

To use it with a crystal I would suggest using the same circuit but replace the LC circuit with the crystal. Leave the 30pF capacitor in place, or possibly reduce it to 22pF.

Most, but not all, parallel resonant crystals are intended for use with 20pF shunt capacitance. If the capacitance is not the same as the intended value the crystal will oscillate at a slightly different frequency.

Alternatively you could create your own oscillator and directly inject the output of the oscillator into pin 8 on the LA1185. Some experimentation will be needed to determine the correct injection voltage.

Pin 7 is described as "Local Osc Monitor" on some data sheets so presumably you could look at the signal there to determine that the signal had been injected correctly.

enter image description here

  • \$\begingroup\$ Actually this schematics is not made by me. It's pretty common scheme which can be found by searching "ccir oirt downconverter". Nevertheless thanks for explanation. But I still can't figure out how to properly shunt crystal with cap(s). Should I just connect single capacitor in parallel with the crystal or should I add two caps to both pins and connect other ends of caps to the ground? Also I think because the whole purpose of this project is to roughly downconvert CCIR spectrum to OIRT (tens of MHz) frequency mismatch will not be so large and will not harm the whole performance \$\endgroup\$ Commented Oct 2, 2020 at 16:53

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.