I have seen countless times the following diagram for a Schmitt trigger oscillator.

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As shown in the picture, the slow-rise wave-form generated by the charging and discharging capacitor is translated to a square wave-form at the output, as a result of the Schmitt trigger. However, I have no idea how the capacitor even charges up in the first place. I don't see any "input voltage" other than the upper and lower bounds on the actual Schmitt trigger. How is the capacitor charging up in the first place?

I'm sorry if this is a very dumb question.


The schmitt trigger inverter is what generates the signal that charges and discharges the capacitor. Assume the input on the left starts at 0V; the schmitt inverter will therefore output +5V. This +5V will charge Ct via Rt until the voltage crosses the schmitt trigger's rising voltage threshold. At that point, the output will change to low, and start discharging the same capacitor until the schmitt trigger's falling voltage threshold is crossed, starting the cycle over again.

Power for the schmitt trigger's output voltage comes from its power supply - marked +5V in the diagram.


Basically it works because the output of the Schmitt trigger inverter is either 0 or 1 (low or high). Imagine you look at the circuit at some random time. By its nature, the Schmitt trigger inverter output is either 0V or 5V (or transitioning between them but we can ignore that). If the output is 0V then the capacitor output is higher than the Schmitt Trigger inverter output, therefore the capacitor is discharged through the resistor (the "falling" part of the capacitor waveform). The capacitor discharges until the input to the Schmitt inverter is low enough that it trips it to high (5V). Now, the output of the Schmitt Trigger is higher than the capacitor voltage so the current flows the other way into the cap (the rising part of the caapcitor waveform). This cycle repeats.

The key that makes this work is the "hysteresis" in the Schmitt Trigger. Basically this means the inverter trip point is different depending on whether we are coming from a high voltage or a low voltage.


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