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When we use this circuit, the LED blinks. I know an inverter transforms high to low voltage and low to high voltage, but here the voltage source is connected to the inverter all the time so the inverter receives high voltage all the time, right? Then why does the inverter ever have a high output?

Also, what's the role of the capacitor in all of this? Here I have the same question as before: it's connected to the voltage source all the time, so when does it ever discharge?

I'm trying to understand this circuit picturing the electron flow as that is currently the only way of understanding it with my limited background knowledge.

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marked as duplicate by Marcus Müller, Andy aka, Dmitry Grigoryev, dim, awjlogan Nov 27 '18 at 12:10

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migrated from physics.stackexchange.com Nov 8 '18 at 17:26

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    \$\begingroup\$ Are you missing a component? \$\endgroup\$ – TimWescott Nov 8 '18 at 17:31
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    \$\begingroup\$ This circuit will not make the LED blink. You are missing a resistor from the inverter output to the inverter input. \$\endgroup\$ – Janka Nov 8 '18 at 17:31
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    \$\begingroup\$ "I don't understand how it does the latter..." -- do you mean you do not understand how the inverter itself works? \$\endgroup\$ – TimWescott Nov 8 '18 at 17:32
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    \$\begingroup\$ I'm voting to close this question as off-topic because it's based on the false claim that this circuit will oscillate. It won't. \$\endgroup\$ – Marcus Müller Nov 8 '18 at 17:56
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    \$\begingroup\$ This has been discussed so many times: Schmitt Trigger Oscillator - How does it work? \$\endgroup\$ – Dmitry Grigoryev Nov 22 '18 at 14:15
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This curcuit might work without the otherwise required feedback resistor - if and only if the input is not buffered.

In this case the (internal) current for hysteresis could affect the voltage of the capacitor at the input.

Note that relying on parasitics currents is usually a rather bad idea - especially when you can just simply put a feedback resistor between the output and the input.

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    \$\begingroup\$ … I mean, that'd be true if the leakage current of the cap wouldn't pull it low, while we're starting to consider parasitics. And electrolytics are not especially known for low leakage current. \$\endgroup\$ – Marcus Müller Nov 8 '18 at 18:13
  • \$\begingroup\$ I've seen oscillations like these with normal TTL inverters and large capacitors. The characteristics of their input current can be enough to bias them precisely in the middle of their noise band. \$\endgroup\$ – Edgar Brown Nov 8 '18 at 20:47
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The key is that the inverter has a Schmitt trigger input, which has two threshold voltages. When the input voltage rises above the upper threshold voltage the output abruptly changes from high to low. The resistor will then discharge the capacitor until the voltage on it falls below the lower threshold, at which point the output will abruptly change from low to high and the cycle will begin again. So the capacitor and resistor set the timing. There is a curved sawtooth waveform at the input and a square waveform at the output. You might consider building or simulating it to gain some insight. 4000 series CMOS will tolerate 9V.

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