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varunkant2k showed two different cases, which are you referring to? One is 1.5V pure DC, I explained the result in post #17. In fact, a diode has some (very small current) at zero diode voltage as well.
In the second one, 1V DC + 0.5 sine is applied. Here the capacitor is charged to Vpeak - diode forward voltage. The diode capacitance also couples a small AC amount.
In the second one, 1V DC + 0.5 sine is applied. Here the capacitor is charged to Vpeak - diode forward voltage. The diode capacitance also couples a small AC amount.