Understanding the biased diode clipper

[Aeternam]

I'm having a hard time understanding how a biased diode clipping circit works.

I think I understand what happens with D2 (ignoring D1 for the moment): on the positive portion of the AC signal (*) Vkathode > Vanode, so the diode does not conduct and we get the full signal at Vout. On the negative portion of the AC signal, Vkathode < Vanode so the diode conducts and since the output is parallel to D2 we read its operating voltage (0.6V) at Vout.

Now we bring D1 into play. As long as the input signal is below V2 (5V in the example) the diode does not conduct, we read the input signal at Vout, so far so good. Here comes the part I don't understand: as soon as the input signal goes above 5V Vkathode < Vanode, so D1 starts to conduct. Say the input is at 10V. The simulation says the output should be at 5.6V. How does that work? Where do the other 5V go? Where does the +0.6V come from (it's the diode drop allright, but why is it present at Vout)?

I'm sure the answer is obvious but I'm drawing blanks here

(*) I should probably mention that in the example, the input signal is a -10V to 10V sine.

[trophosphere]

D1 starts to conduct at 5.6v (not 5v) because D1 has a Vf of 0.6v. When D1 conducts, the output becomes clamped at (ground + 5v source + 0.6v Vf of D1). The "extra voltage" is (in relative terms - actually current limited by the resistor) shunted to ground so that the output is now maintained at the point where D1 just about turns on.

[xavier60]

The clipping will be asymmetrical. Coupling capacitors should be placed in series with the input and output.