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Sep 25, 2013 at 2:24 vote accept Thiago
Sep 22, 2013 at 21:56 history edited Alfred Centauri CC BY-SA 3.0
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Sep 22, 2013 at 21:54 comment added Vasiliy I'm missing your point. I'll try to read this again later - maybe then I'll understand. Thx
Sep 22, 2013 at 21:36 history edited Alfred Centauri CC BY-SA 3.0
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Sep 22, 2013 at 21:21 comment added Alfred Centauri @Vasily, the linked document on clipping levels is quite straightforward. Setting R3 to zero, it is clear that the most negative clipping voltage across the load is precisely: \$v^-_O = (I_E R_E) + V_{CEsat} - (V^+ - I_C R_C)\$ The first term on the right is the DC voltage across the emitter bypass capacitor. The last term on the right is the DC voltage across the output coupling capacitor. For clipping level calculations, it is assumed that the coupling capacitors can be replaced by batteries, i.e., that they are AC short circuits with a DC voltage across.
Sep 22, 2013 at 20:59 comment added Vasiliy DC voltage yes, but regarding voltage swing you mean AC voltages, no?
Sep 22, 2013 at 20:55 comment added Alfred Centauri @Vasiliy, \$R_E \$ must appear in the equation since the voltage across the emitter bypass capacitor is precisely \$I_E R_E \$.
Sep 22, 2013 at 20:39 comment added Vasiliy I think that \$R_E\$ should not appear in the equation for negative clipping since it is completely bypassed by the capacitor. In the reference you provided the battery does replace the capacitor, but the presence of \$R_3\$ is not analogous to OP's circuit.
Sep 22, 2013 at 19:29 history edited Alfred Centauri CC BY-SA 3.0
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Sep 22, 2013 at 19:15 history undeleted Alfred Centauri
Sep 22, 2013 at 19:11 history edited Alfred Centauri CC BY-SA 3.0
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Sep 22, 2013 at 13:26 history edited Alfred Centauri CC BY-SA 3.0
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Sep 22, 2013 at 12:19 history deleted Alfred Centauri via Vote
Sep 22, 2013 at 12:18 history answered Alfred Centauri CC BY-SA 3.0