I know this question has been asked many times before, however this question differs in the use of an uneven set of capacitors.

I'm replacing the crystal on an old computer.

The original symptoms were as follows: If the crystal got hot enough the system would start up, heating up the crystal by aiming my hot air station at 100C for 2-3 seconds would heat the crystal enough for it to work a few minutes.

I replaced the 32.768kHz crystal marked with S3238E with a new one (32.768kHz, 12.5pF) however although the system is less sensitive to the crystal not being warm, it still has the same issues. The only difference is that the crystal now also stops functioning when it gets hot enough... (>65C)

Since the 25 years between this device got manufactured I assume a bunch has changed in the world of crystals and maybe its original crystal has a higher load capacitance.

The circuit being used is as follows:

Crystal circuit

The MCU in question is an MC68HC05E1 which the datasheet can be found for over here: https://www.nxp.com/docs/en/data-sheet/HC05E1GRS.pdf

  • What would the Load Capacitance of the crystal be like?
  • What does it tell about the circuit when a crystal will only work at a certain temperature?
  • \$\begingroup\$ Perhaps other components have degraded over time? Have you checked DC voltage of OSC1 and OSC2, especially when it fails to oscillate? Note that this is a high-impedance circuit, and the loading effect of a 10M multimeter may disturb the DC-operating point. Ideally, DC voltage @ OSC1 & OSC2 should be similar: half-DCsupply. \$\endgroup\$
    – glen_geek
    May 19 '19 at 14:17
  • \$\begingroup\$ High chance they have, I sadly don't have a probe suitable for this circuit as all probes will have a too high capacitance. \$\endgroup\$
    – Bart
    May 19 '19 at 14:34
  • 1
    \$\begingroup\$ Probing capacitance may well temporarily stop oscillations...my point is that if DC operating voltages are not half-supply , oscillations can fail to start in a non-probed circuit. \$\endgroup\$
    – glen_geek
    May 19 '19 at 14:41
  • \$\begingroup\$ The voltage at OSC2 is 2V while the voltage at OSC1 is 1.6V (or ~1.36V but that iirc was while the system is willing to boot so oscillation was occurring. In general having a probe on OSC1 will stop the system from booting, OSC2 experiences no problems from having a probe). Is this reason for concern? \$\endgroup\$
    – Bart
    May 19 '19 at 15:02
  • \$\begingroup\$ Might be OK if Vdd=5V. OSC1 is the really high-impedance pin, and is most susceptible to probe loading, so it is no wonder that probing OSC1 stops the oscillator. DC measurements with a multimeter can change, depending on whether it is oscillating or not oscillating. Attaching a probe can kick it into oscillating, can also kick it into non-oscillating. You might try probing OSC2, and see if its DC voltage changes from +2V when you heat-treat various components. \$\endgroup\$
    – glen_geek
    May 19 '19 at 15:46

The effective capacitance on a crystal is the series equivalent of the attached capacitances. The \$330\mathrm{k}\Omega\$ resistor isolates the crystal from the chip's output pin, so the effective capacitance is 33pF in series with the parallel combination of the 10pF cap and the microprocessor input pin. So the 10pF cap plus another 10pF of parasitic capacitance would get you to 12.5pF -- 10pF sounds high, but not unreasonable, to me.

But I suspect that a better question to ask would be the broader one of "why is this oscillator not oscillating", and the answer would be "either your board or your micro has aged" (Since you seem to have eliminated the crystal as a suspect). You do need the \$330\mathrm{k}\Omega\$ resistor there to reduce the drive to the crystal, but you might be able to reduce the resistance for greater drive (I wouldn't go below \$100\mathrm{k}\Omega\$, but that's a pure guess on my part).

  • 1
    \$\begingroup\$ The 330K also provides phase-shift, needed by Barkhausen. \$\endgroup\$ May 20 '19 at 0:05
  • 1
    \$\begingroup\$ @analogsystemsrf good point. And (now that I break out the calculator) the phase shift may be insufficient with a smaller resistor. It makes me think that his processor has gone bad, or maybe that his board is gummed up with something conductive. \$\endgroup\$
    – TimWescott
    May 20 '19 at 0:30
  • \$\begingroup\$ Yeah fair chance the processor has just gone bad due to its age, I think I'm just going to make a little circuitboard with a full oscillator on there. Disconnect OSC2 from the circuit and send the oscillator output in OSC1. Should make it more reliable . Thanks \$\endgroup\$
    – Bart
    May 20 '19 at 10:07

The load capacitance of the crystal heavily depends on the board layout. So noone can tell you the load capacitance.

Usually, the load capacitance is "tuned" by measuring the frequency of the crystal (with a <1pF probe) and then iteratively change the load capacitance until it matches the desired frequency. (Increasing C for decreasing f and vice versa).

But in your case, I think you have another problem: The circuit does not oscillate at all. This could be because it has too much resistive damping. Try shorting the 330 kOhms resistor from OSC2. This could solve your problem. (Note that when increasing the temperature, the resistance of a resistor decreases. This could be the reason for oscillation at higher temperature).

  • \$\begingroup\$ I shorted the 330kOhm resistor and the device does directly boot. I'll let it rest for a little while so make sure everything is down to room temperature and try again. That might have fixed it :) \$\endgroup\$
    – Bart
    May 19 '19 at 14:22
  • \$\begingroup\$ Sadly after a couple of attempts the device again didn't want to start anymore. Any other ideas? I don't have a low capacitance probe for my scope either \$\endgroup\$
    – Bart
    May 19 '19 at 14:31
  • 2
    \$\begingroup\$ Maybe the crystal was damaged because of too much drive current when shorting the resistor. Then you could try to use e.g. a 100kOhms resistor after inserting a new crystal. Forget the tuning, it should not be the cause of your problems. \$\endgroup\$ May 19 '19 at 14:34
  • 4
    \$\begingroup\$ The resistor (330K) and 33pf are in the OUTPUT path of the IC's amplifier. The time constant 330,000 ohms * 33 pF == 10+uSeconds and is a substantial part of the phaseshift needed for oscillating. I'd use a 100Kohm or 180Kohm, so there is some phaseshift and also there is some protection for the crystal. If your VDD is correct, then the internal transistors have aged (doping has migrated) IMHO, and different Caps or new value for resistor should help. This is a PI resonator with a voltage stepup-ratio of 33/10; try 8pF or 5pF for higher stepup. \$\endgroup\$ May 19 '19 at 17:15
  • 1
    \$\begingroup\$ @analogsystemsrf Recommend swapping the 330k for 180k and the 10pF for 8pF? Honestly I might just remove the crystal and directly connect a full oscillator to OSC1. Costs a little bit more part wise but for a repair like this I don't mind the few bucks \$\endgroup\$
    – Bart
    May 19 '19 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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