Max Input Signal: 0.3V for 1.5mA Collector Current

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Now, the input circuit is also biased as the transistor to work in the active region and without the biasing of the input circuit , the input circuit may get reversed biased during the negative half cycle of the signal voltage. Due to this bias, there should be some base current say$I_{bB}$.The minimum value of $I_{bB}$ should be such that (taking the maximum current due to the signal voltage in positive half cycle as $I_{signal}$ ) $I_{signal} + I_{bB} = 30~μA$ and in the negative half cycle $-I_{signal} + I_{bB} = 0$ .
This gives $I_{bB} = I_{signal} = 15μA$
Similarly, the collector current due to the signal voltage alone is 1.5 mA.

Now, it is given that 1V change in base voltage,i.e. 1V of signal voltage gives 5 mA of collector current.
So, for giving 1.5 mA of collector current, the signal voltage should be $\frac{1.5} 5 = 0.3~V$.
Is this correct?
Why does the question say that $V_{BE}$ gets changed by 1V.
Isn't $V_{BE}$ always 0.7V for sillicon transistor?

Is it necessary that I shoud specify input signal into voltage, not current?

 

[cnh1995]

Why does the question say that VBEVBEV_{BE} gets changed by 1V.
Isn't VBEVBEV_{BE} always 0.7V for sillicon transistor?

No, not in small signal analysis. In this case,
a linear approximation of the exponential diode equation is used, which gives i_d=kV_d.
(This approximation holds as long as the signal is small.)

 

[Pushoam]

So, for giving 1.5 mA of collector current, the signal voltage should be 1.55=0.3 V1.55=0.3 V\frac{1.5} 5 = 0.3~V.
Is this correct?

What about this?
My book says that the answer is 0.6V.

 

[rude man]

I think they meant the peak input voltage. So the answer makes no sense since 0.6V < 1V knee.
And if they meant rms input voltage that = √2(0.6) = 0.85V which is still below the knee.

So assuming the former case I think they meant the answer to be 1.6V (or 0.6V above the knee) since that would give a dc collector current of (1.6 - 1.0) * 5mA/V = 12V/4K = 3mA.
Why do you think you were given beta BTW?