amplifiers

[Tea.bag]

In the 2007 examination...

Question 2 for electronics section...why is the clipping at +3 and -3??

[aksman]

because it is a correctly biased amplifier circuit, the V_out when V_in=0 is 3.0V. which means that the output signal can have a mazimum oscillation till 6V and minimum till 0V, hence up 3V and down 3V (because 6V is constant voltage supply, you cannot exceed this). when the signal is decoupled by the capacitor, it comes back down to 0V from 3.0V. but the oscillation is still at maximum of up 3V and down 3V
therefore the signal is clipped at +3V and -3V!

hope that helped

good luck for tommorow!

[Mao]

=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_B
V_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 ]

[Tea.bag]

wow thnx...that really helped.

any tips on drawing the amplified graphs?