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