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design of a common emitter at cut off .. ?
A CE amplifier circuit and char. cures are shown in the sollowing link :
http://www.electronics-tutorials.ws/amplifier/amp_...
if I intend to design a common emitter that is biased at cut off , then VCE must equal VCC and
IC nearly 0 A , my question is that according to the characteristic curves and the load line is it true that the negative section of the input signal will be amplified and the positive section would be clipped ?
and if i want the negative section be clipped and get the positive section of the input amplified then VCE must be near Vsat ?
am I thinking right ?
thanks
3 Answers
- billrussell42Lv 79 years agoFavorite Answer
No. You have everything backwards.
Negative part of the input is the positive part of the output, and as the output bias point has the collector close to Vcc, that part cannot be amplified.
What happens is the positive part of the input (the negative part of the output) is amplified, and the negative part of the input (positive part of the output).
If you want the negative part of the input to be clipped, you need to bias the transistor near Vsat.
- ?Lv 79 years ago
First the common emitter amplifier both amplifies and inverts the input signal.
A common emitter amplifier using a PNP transistor biased at cutoff will amplify the negative portion of the input signal. The collector of the transistor will be at Vcc which is a negative voltage when the transistor is cutoff and will go positive when the input goes negative
If the common emitter amplifier uses a NPN transistor biased at cutoff, it will amplify the positive voltage of input. The collector will be at Vcc (positive voltage) and will go negative when the input goes positive.
A common emitter amplifier using a PNP transistor biased at saturation will amplify the positive portion of the input signal, the collector output voltage will go negative as the input goes positive.
Finally a common emitter amplifier using a NPN transistor biased at saturation will amplify the negative portion of the input, the collector will go positive as the input goes negative.
- RickLv 79 years ago
The answer to your first question is no, because when the input goes further negative, the output will remain unchanged. For your second queston, the output voltage will decrease when the input goes positive.
This is not a good way to design a clamping circuit. You could just use a diode depending on the input signal amplitude.