5
\$\begingroup\$

The MC44144 is a gated phase locked loop intended for video applications that is described in its datasheet as "sensitive to shunt capacitance" and that "crystal switching should be avoided".

What issues might I expect to encounter if I were to attempt crystal switching such as in the schematic below? I understand this would introduce around 5pf to ground in series with the selected crystal, and 5pf in parallel. Would I get non-oscillation, unstable oscillation, drift, ?

Are there any measures I might take to mitigate these effects?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
6
  • 1
    \$\begingroup\$ An aside from the question you asked: To put M1 in "on" state, you're likely going to have to drive SELECT to a couple volts above the VCC used by the microcontroller. \$\endgroup\$
    – The Photon
    Aug 8, 2019 at 20:26
  • \$\begingroup\$ You placed M2 correctly above whatever you are turning on/off, and you placed M1 incorrectly, you should've placed M1 below so it was connected to ground. Then at least you will connect the transistors correct. Remember, it is the voltage difference between source and gate that turns a transistor on. Look at where the source is at M1(incorrectly placed) and M2(correctly placed). \$\endgroup\$ Aug 8, 2019 at 20:48
  • \$\begingroup\$ I'd just use two MC44144 each with its own crystal. The surrounding circuit around the IC is so simple. \$\endgroup\$
    – carloc
    Aug 9, 2019 at 12:12
  • \$\begingroup\$ Thanks @HarrySvensson I will review the circuit before attempting to build anything. \$\endgroup\$
    – Batperson
    Aug 11, 2019 at 9:51
  • \$\begingroup\$ Updated the schematic so source of both mosfets is connected to ground. @carloc I'm not showing the other components required by the MC44144, decoupling, AC-coupling capacitor etc. \$\endgroup\$
    – Batperson
    Aug 12, 2019 at 4:41

1 Answer 1

3
\$\begingroup\$

What issues might I expect to encounter if I were to attempt crystal switching such as in the schematic below?

The VCO will run at an incorrect frequency, and is likely to fail to lock either video standard.

Are there any measures I might take to mitigate these effects?

Use a different part -- preferably one that's still in production. This part would have transitioned from Motorola to On Semi when Motorola's analog division was spun off in 1999, but On's web site doesn't even acknowledge that the part ever existed.

If that weren't the case, or if a similar issue arose with a substitute part, I would recommend using two sync separators, each with their own crystal, and switching between their outputs.

\$\endgroup\$
5
  • \$\begingroup\$ Care to elaborate why the VCO will fail to lock or run incorrectly? ie. There is a specific clock startup sequence on power-up that is required and you can't arbitrarily jump ship in the middle. \$\endgroup\$
    – DKNguyen
    Aug 8, 2019 at 21:17
  • 1
    \$\begingroup\$ @DKNguyen My concerns would center around the MOSFETs not behaving as ideal switches and introducing unwanted capacitance, and/or coupling the wrong crystal into the circuit. The datasheet describes the VCXO as "sensitive to shunt capacitance", and cautions that "layout and crystal positioning are critical"; it seems highly unlikely that it would tolerate a MOSFET being used here. \$\endgroup\$
    – user39382
    Aug 8, 2019 at 22:22
  • \$\begingroup\$ Thanks @duskwuff. There must have been alternatives back in the days of analog broadcast television, I'm hoping some may still be around. If you or anybody else know of any, please comment. \$\endgroup\$
    – Batperson
    Aug 9, 2019 at 3:44
  • \$\begingroup\$ What features of the MC44144 are you depending on? Color burst detection is typically integrated into larger NTSC decoders nowadays, rather than being a discrete part. \$\endgroup\$
    – user39382
    Aug 9, 2019 at 4:28
  • \$\begingroup\$ @duskwuff I'm working on an application (colour overlay) that requires a clock in phase with the colourburst. \$\endgroup\$
    – Batperson
    Aug 11, 2019 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.