# Homework Help: Transistor : Output voltage of the amplifier

1. Dec 27, 2017

### Jahnavi

Ah ! Finally .

But the irony is that in the exams I need to stick with the diagram in post #31 however flawed it might be . ( It is the standard reference text )

Thank you for your patience .Please bear with me for some more time . I have few more relevant enquiries .

2. Dec 27, 2017

### Jahnavi

In the picture in post#38 , the DC voltage from battery and applied AC input signal get added ?

Is the net input voltage in the Base sinusoidal but the peak voltage now oscillating between VBB+Vo to VBB-Vo ?

3. Dec 27, 2017

### Delta²

Ok @Jahnavi maybe in the text of the book mentions something like that all DC voltage sources are assumed to have a (small) internal resistance?

4. Dec 28, 2017

### Staff: Mentor

Yes.
Yes, it's an AC signal with a DC offset. Mathematically you could write it as:

$V_{BE}(t) = V_{BB} + V_o sin(\omega t)$

5. Dec 28, 2017

### Jahnavi

Thanks .

What happens to the net input voltage in the original problem (post #1 where there is no coupling capacitor ) ?

I am referring to the case when AC input signal is applied across the branch having resistor and DC battery .

6. Dec 28, 2017

### cnh1995

Base-emitter voltage will be equal to the input ac voltage and the base current will have a half wave rectified waveform.
Plus, the battery will add a dc component in the ac source current.
Edit: I was assuming BE junction to be ideal (0V drop).If the BE junction is assumed to have 0.7V drop and if the ac source is of 1mV, then, as gneill said, it won't be able to turn the transistor on.

Last edited: Dec 28, 2017
7. Dec 28, 2017

### Staff: Mentor

The base DC bias circuit (battery and resistor branch) is rendered ineffective by the placement of the input source, so an equivalent circuit as far as the transistor is concerned would be:

Now, $1\;mV$ is not enough to forward bias the transistor, which needs about $700\;mV$ for a silicon transistor. Without a proper base bias to place the transistor operating point in the appropriate location on its $IV$ characteristic curves, the transistor would never turn on and no signal would pass.

8. Dec 28, 2017

### Jahnavi

Does that mean there would be no current in the branch consisting of the 2kΩ resistor and DC battery ?

Why does AC input voltage dominate over the DC battery so as to make it ineffective ?

9. Dec 28, 2017

### Staff: Mentor

No, the AC source will provide a path for it. KVL around the loop will show that the majority of the bias will be dropped across the resistor.
Any ideal source, AC or DC, would do the same. The point is, what ever source you place in that position will set the potential difference between the two ends of that "bias" branch.

10. Dec 28, 2017

Thanks !