=D
transistor amplifier!!!
okay, remembering V
out is biased to 3V:
at cut off [Vb is too low], the transistor acts as an open switch -> no current flow. This makes the current through the collector resistor very low [neglectible], hence Vc ~ 0, Vout = 6V
at saturation [this is the tricky part]:
we take the VCE definition and make Vout go to 0 because thats at saturation the transistor amplifier acts as a closed switch -> no voltage across -> Vout is parallel to the transistor -> 0V at saturation
or we take the proper approach:
V
E + V
BE = V
BV
E = V
B - V
BE = 1 - 0.7 = 0.3V
since at saturation there's no voltage across the emitter-collector junction [closed switch], and V
out = V
CE + V
E, the lowest Vout can reach is ~0.3V
[the above method is beyong the VCE course]
[there is also better methods that are wayyy beyond this one, and will actually show that the lowest Vout can reach is about ~0.5V]
so from that, we have from ~0.3 to 6 mean at 3, after passing the capacitor we'll have +3 -2.7 [or
]