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Old 29th January 2006, 01:40 AM   #1
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Default Clock Oscillator circuit

I have a circuit on a synthesizer. It provides a clock for keyscan. I have included a schematic. The osc. is 10kHz +5V -9V. How does this circuit work?
I have replaced the IC with CD4069UBCN. The cap is 1200pf.
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Old 29th January 2006, 05:53 AM   #2
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This is a very basic oscillator. All elements are inverters.

Start with pin 10 is 0 volts. Pin 12 has 15 volts (the power supply voltage is 15 volts). The capacitor is charging. What happens next?

The 47k can be 1 M or more, just protection.
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Old 29th January 2006, 06:55 AM   #3
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attach oscilloscope to pin 8 can't see wave. Maybe i am test incorrectly because I tried two ICs. Cap is marked "122". Only one in my home. I dont know how to test the 12oopf cap.
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Old 29th January 2006, 07:21 AM   #4
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I notice from your schematic that the 4069 is fed with 25 volts in total Datasheet says 15 volts max if I remember correctly.

122 is 1200 pF (and hardly broken) but if you want to test you can take 1 uF and 1 Mohms this will create a real slow oscillator. Don't forget to tie the 3 free inputs to either ground (pin 7) or supply voltage (pin 14).
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Old 29th January 2006, 02:03 PM   #5
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Schematic has error. The IC is fed +5 -10. It is on the Main ic supply rails.
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Old 30th January 2006, 02:48 AM   #6
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What can I expect to see on my Oscilloscope from pin 8. What time division should I use?
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