Clamper circuit, how to apply kirchhoff when the diode is turned around?

Question

in this clamper circuit it's easy for me to determine the output waveform:

This is how I do it:

In the positive half cycle, the diode is forward biased with a drop of 0.7 V, so the positive half output is limited to 0.7 V

In the negative half cycle, the diode is reverse biased, and the capacitor has a voltage of 3.3 V, so the negative half output, by applying Kirchhoff:

-Vout - Vin - Vc = 0 Vout = -Vin - Vc = -4-3.3 = -7.3 V

So the output waveform has a positive peak of 0.7 V and negative peak of -7.3 V

But when he asks me to turn the diode around, like this:

I face problems when I apply Kirchhoff, this is how I do it:

In the first positive half cycle, the diode is reverse biased and the capacitor has a voltage of 3.3 V, applying Kirchhoff:

-Vout + Vin - Vc = 0 Vout = Vin - Vc = 4-3.3 = 0.7 V

In the negative half cycle, the diode is forward biased, so it has a drop of 0.7 V, so the output is limited to -0.7 V

I end up with a positive peak of 0.7 V and a negative peak of - 0.7 V , but this should be wrong, the right answer should be the exact same as the first circuit but with different signs, I don't understand how though, can someone let me know, thank you.

Answer 1

The cap isn't charged until the negative half cycle, where the charge necessary is pulled in through the diode. (Currently ignoring the RC constant, since you seem to treat it as negligible.) The first half cycle, the output will track the input fairly closely. The second half cycle, it clamps the output to -0.7, and the 4V excursion on the input charges it to -3.3. Thereafter, it will stay clamped with a positive excursion of about 7.3V and a negative of 0.7.