# Homework Help: Transistor : Output voltage of the amplifier

1. Dec 27, 2017

The input voltage is often a weak ac voltage or current source, e.g. a photodiode or a microphonic element for an audio (sound) system. If you attach it directly to the point $V_{BE}$, it could simply get fried, if it is a low resistance element. $V=1.0$ volts is often far too much for that sort of thing. $\\$ The purpose of amplifier circuits such as this is to amplify the (small) signals, so that the signal is large enough to easily work with, and use. Sometimes additional amplification is then employed, e.g. to drive speakers in a sound system.

Last edited: Dec 27, 2017
2. Dec 27, 2017

### Staff: Mentor

If the circuit is meant to represent an ac small signal equivalent, then the VBE would be zero (it's part of the DC bias circuit).
It doesn't short the battery, it shorts the branch containing the battery and its series 2 kΩ resistor. Applying an ideal source between two nodes fixes the potential difference between those nodes. In the given example the "input source" establishes the potential difference between the emitter and the base, rendering any bias circuit irrelevant.

3. Dec 27, 2017

I think the OP would do well to consider the recommendations of post 12 and 14. I myself have sometimes tried to work with books that weren't 100% in college. A good book can be worth its weight in gold, while a not-so-good book is worth about as much as you get get for it by recycling the paper.

4. Dec 27, 2017

### Jahnavi

Are you saying that potential difference between Base and Emitter is equal to the input signal voltage ( in the original problem where no coupling capacitor is present ) ?

5. Dec 27, 2017

### cnh1995

Yes, and the battery is redundant.

6. Dec 27, 2017

### Jahnavi

@gneill , please see the attached picture . This is from another reference book .Do you find it reasonable ?

7. Dec 27, 2017

### Staff: Mentor

Yes. Note that capacitor $C_1$ DC-isolates the input signal from the bias network consisting of $V_{BB}$ and $R_B$. $R_B$ also contributes to the input impedance for the amplifier.

8. Dec 27, 2017

### Jahnavi

OK.

Are you saying that the two voltages , DC from battery and applied AC input signal get added ?

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

9. Dec 27, 2017

### Staff: Mentor

Sorry, I need to retract my previous statement (post #32). In the circuit shown in post #31 the source $V_{BB}$ is going to clamp the potential at the end of $R_B$, so the input signal will have no effect. The input signal will be developed across the capacitor and the transistor will never see it.

I'm finding that your reference materials have rather dubious quality...

10. Dec 27, 2017

### Jahnavi

Suppose the lower terminal of battery VBB is earthed . Are you saying that the potential at left end of RB is at a constant voltage VBB irrespective of the input AC signal ?

11. Dec 27, 2017

### Staff: Mentor

Yes, assuming that by "earthed" you mean that the bottom rail is taken to be the reference node.

12. Dec 27, 2017

### Delta²

Ok sorry for me intervening out of nowhere BUT I think $V_{BB}$ will have some internal resistance so I think the transistor will see the input voltage after all...

13. Dec 27, 2017

### Jahnavi

@gneill ,This picture is from another reference book . Here the author has put AC input signal in series with the DC battery . Does this look reasonable ?

14. Dec 27, 2017

### Staff: Mentor

Yes, that's a better representation. Note though that it doesn't depict an input resistor for the base in this case. That might not be a problem with careful choice of $V_{BB}$.

15. Dec 27, 2017

### Staff: Mentor

Unless the problem states that the components are not ideal, it's not advisable to make that assumption. Further, you'd have no idea what size of resistance to ascribe to that internal resistance, so analyzing the circuit would be problematical.

16. 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 .

17. 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 ?

18. 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?

19. Dec 28, 2017

### Staff: Mentor

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

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

20. 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 .

21. 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
22. 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.

23. 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 ?

24. 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.

25. Dec 28, 2017

Thanks !