Hi, actually problem is coming up regarding an answer given as in how the voltage signal allowable is this. 8v to -3.4v
Thank you

 

I can understand your confusion. That problem seems to have incorrect answers.
With a 1mA constant-current emitter current, the transistor has no AC gain since the emitter current cannot change, and thus the collector current also cannot change, so the collector voltage is essentially constant with a change is base voltage (see simulation below).
You can get AC amplification if the emitter is bypassed with a capacitor to ground.

 

I can understand your confusion. That problem seems to have incorrect answers.
With a 1mA constant-current emitter current, the transistor has no AC gain since the emitter current cannot change, and thus the collector current also cannot change, so the collector voltage is essentially constant with a change is base voltage (see simulation below).
The only way to get AC amplification is if the emitter is bypassed with a capacitor to ground.

View attachment 104614

Seems like so, but in previous editions also, this exercise has carried on.

 

Hello,

Show figure 3.72 also.

 

Here is the fig. 3.72.

 

Here is the fig. 3.72.

Hello again,

Ok for the negative swing they must be talking about the theoretical max negative deviation given a constant saturation voltage of 0.3 volts which would be sort of typical. If we use that standard we get approximate values that match the answer sheet. It is still a little strange though that they would talk about that in that way with this exact circuit, but perhaps you can look at the rest of the papers and see if you can find ANY information that relates to a saturation voltage of 0.3 volts or close to that.
The saturation voltage is often used as a limiting factor in the output swing of transistor amplifiers because it limits how well the transistor can force the collector voltage to match the emitter voltage. With less voltage drop the swing can go down lower.