2
\$\begingroup\$

I've made a 32kHz crystal oscillator using the CD4060B IC. The IC is powered by a 7805 voltage regulator which in turn is connected to a 12V SMPS.

enter image description here

I've probed the Q5 output (~1024Hz) using an optocoupled MSP432 and had the following waveform:

enter image description here

The large "off" period grows larger and eventually the oscillations stop.

I've noticed that when I placed my finger at the body of the 7805, I had stable 1024Hz oscillations for some seconds, then the "off" period started growing again until the wave collapsed. I thought it could be an overheating problem, so I added a heatsink to the 7805 and probed the current using a multimeter - still had the same result (the current drawn was at most 1.1mA).

Then I've placed much larger decoupling capacitors (220uF and 470uF) - still had the same results.

Then I connected the IC directly to the 12V SMPS and had the same result...

I also tried the original circuit, but with different feedback capacitors and resistor at the crystal filter (15pF, 33pF, 56pF, and 3.3Mohm) - still no improvement...

Finally I've powered the IC directly using the 5V pin from MSP432 and probed using the same optocoupled circuit. Then I had a stable and perfect ~1024Hz square wave.

Could this be a problem with my SMPS or is it a problem with the circuit? What could possibly indicate the fact that my finger at the body of 7805 created a "temporary stable" square wave? Why does it work with the MSP432 as the power source?

Thanks in advance.

\$\endgroup\$
4
  • \$\begingroup\$ How exactly are the two grounds in this circuit connected to the MSP432 and to your oscilloscope? How is the circuit physically constructed? Please replace "the IC" with the name of the IC in question...there are three in your circuit. \$\endgroup\$
    – Elliot Alderson
    Jan 22, 2021 at 1:43
  • 2
    \$\begingroup\$ Maybe your oscillator is marginally working. Try lowering or removing the 470k resistor. That will increase the loop gain of your oscillator. \$\endgroup\$
    – user69795
    Jan 22, 2021 at 1:52
  • 3
    \$\begingroup\$ Don't make the 470 kohm zero ohms because it may damage the flimsy 32.768 kHz xtal. Drop it to 4k7 and no lower. \$\endgroup\$
    – Andy aka
    Jan 22, 2021 at 10:11
  • \$\begingroup\$ @ElliotAlderson the ground on the MSP432 is the PC ground; the ground on the oscillator is the SMPS ground - that's why I used an optocoupler (to avoid ground-loops). The "IC" always refers to the CD4060. \$\endgroup\$
    – strange_bakery
    Jan 22, 2021 at 15:56

2 Answers 2

Reset to default
5
\$\begingroup\$

I don't see anything inherently wrong with your circuit or your constant-current drive output circuit, except that you MUST connect unused inputs on the chip to a valid logic level (ground or +5 in this case). In particular, pin 12 (RESET) must be grounded or it will float around with leakage and capacitive pickup.

It's also possible the 470K is too high and your oscillator is barely able to maintain oscillation.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Grounding the reset and pulling down the other pins to GND solved the issue! Thank you very much!! :D \$\endgroup\$
    – strange_bakery
    Jan 22, 2021 at 16:59
  • 2
    \$\begingroup\$ Good to hear! It's okay to leave unused outputs open, it saves power. Just not inputs. \$\endgroup\$
    – Spehro Pefhany
    Jan 22, 2021 at 17:33
0
\$\begingroup\$

I don't really know what happens, but your output circuit seems weird.

The transistor with its emitter-resistor will introduce negative feedback. Using the 12V supply, the emitter will want to rise above 5V. The 4060 may not be able to drive the base to that level+0.6V.

So my first point of order would be: drive the coupler-LED directly from the CMOS output. No transistor, no 12V, just the series resistor to limit current.

\$\endgroup\$
2
  • 3
    \$\begingroup\$ Actually, that transistor arrangement should give a nice well-controlled collector current, at least when the 4060 is operating off of 5V. The base current will be about 1/100 of the emitter current, which is good because the CD4xxx parts are wimpy. \$\endgroup\$
    – TimWescott
    Jan 22, 2021 at 2:54
  • 1
    \$\begingroup\$ It's an unusual circuit but it's good in this application because it allows using the unregulated supply to power the optocoupler allowing the 4060 to see a more stable current. it also uses less base current that the normal low-side NPN. the emitter cannot go higher than the base. if it were to try the transistor would switch off. \$\endgroup\$
    – Jasen Слава Україні
    Jan 22, 2021 at 22:27

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.