I'm having issues with this very simple circuit. I'm trying to identify the different values that the output Vs can take considering that the voltage threshold of the diode is 0.7V. Ve is also a sine (Vpp = 10V).

I started by stating :

If Ve(t) > 0, the diode is an open circuit so we're dealing with a simple voltage divider : Vs = (Rc/(R1 + Rc))*Ve(t)

Then i get confused :

if -0.7<(Rc/(R1 + Rc))*Ve(t)<0 : it's still the same because the diode is not conducting so Vs = (Rc/(R1 + Rc))*Ve(t).

if (Rc/(R1 + Rc))*Ve(t) < -0.7 : the diode is conducting current so Vs = 0.7V.

I'm not verry about the second case in which Ve(t) < 0 si if someone can help me, that would be great.

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  • \$\begingroup\$ Start by Vdiode = 0.7 V and not from Ve. \$\endgroup\$ Nov 17 at 19:42
  • \$\begingroup\$ You're right as far as I can see, assuming a piecewise linear diode model. With these values, it should kick in at Ve<0.777... \$\endgroup\$ Nov 17 at 19:52

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