Homework Help: Transistor : Output voltage of the amplifier

1. Dec 27, 2017

Jahnavi

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

This is a solved example given in my Physics textbook . I am not understanding a couple of things .

1) How is Iin calculated as Vin/Rin ?

Why is the book not taking potential difference between Base Emitter into consideration ?

Instead shouldn't Kirchoff's Voltage rule be applied in the Base Emitter loop ?

I think it should be Iin =( 1mV -VBE )/2KΩ

2) Shouldn't the diagram be showing Vout across 100kΩ ?

I don't understand why book considers Vout to be VCE ?

2. Dec 27, 2017

Delta²

$V_{BE}$ is equal to input voltage which probably is small, much smaller than 1mV, so $1mV-V_{BE}\approx1mV$.

3. Dec 27, 2017

Jahnavi

I don't think this is the case .

I think in a silicon PN junction potential difference between Base and Emitter is about 0.7 V .

This is quite large as compared to the bias battery .

4. Dec 27, 2017

Delta²

You might be right , I am not an EE and don't know much about transistors, just said to share my thought on this.

5. Dec 27, 2017

Jahnavi

6. Dec 27, 2017

To me, it looks like a problem that is poorly presented. e.g., to forward bias $V_{BE}$ , you normally need $V_{BE}=.7$ volts. They also don't specify the collector bias voltage. Perhaps someone else has a solution, but my recommendation is to not spend much time on this one as is. This "common emitter" electronic configuration is a good one to know, but this problem looks to me to be a poor one to use to try to learn it.

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

Jahnavi

Thanks .

Suppose instead of 1mV battery there is a 1 V battery , then do you think what I did in the original post Iin=( 1V -VBE)/2KΩ is correct ?

8. Dec 27, 2017

I looked at the problem in the "link" you had posted. For that one, with the book's example, a $V_{BE}$ of 18.5 Volts I think is totally inaccurate. I expect if you hooked it up in the lab, you might find that $I_B$ would be larger than what is computed using the value $\beta=200$, and you would find $V_{BE}=.7$ volts. And then I think you might have removed the "link": https://www.physicsforums.com/threads/find-the-state-of-transistor.935500/#post-5909999 $\\$ Edit: And on this one. a base voltage of 1 volt might work, but they still need to specify the collector voltage. Also $V_{in}$ (ac) usually gets coupled in with a capacitor, and $V_{out}$ (ac) is usually also coupled with a capacitor.

9. Dec 27, 2017

Jahnavi

Thank you .

Do you think post 7 is correct ?

10. Dec 27, 2017

That part works, but please see my edited additions to post 8.

11. Dec 27, 2017

cnh1995

I don't think this problem is practical and as Charles says, not much helpful in learning CE configuration.
The battery needs to be in series with the input ac source (theoretically). In this circuit, as per circuit theory, the battery can't dc-bias the transistor as the ac source is directly connected across the BE junction. Even if the battery were a 10V one, it couldn't dc-bias the transistor in this diagram.

12. Dec 27, 2017

See also my post 8. IMO, this textbook looks like it may contain numerous errors. I encountered that type of thing from time to time in my college days. The best solution would be for the OP @Jahnavi to learn the material from a book that presents it better and with fewer mistakes, if that is at all possible.

13. Dec 27, 2017

Jahnavi

@Charles Link , do you also think this is correct ?

14. Dec 27, 2017

cnh1995

I have seen many junior college physics books in India showing wrong biasing circuit for CE amplifier. @Jahnavi, I believe that includes the NCERT textbook as well.
I agree that you should study amplifiers from some standard electronis books (or COP by H C Verma).

15. Dec 27, 2017

The way they show it, if you use $V=1$ volt, it would work, provided the $V_{in}$ was ac coupled (with a capacitor).

16. Dec 27, 2017

cnh1995

Yes, but I wasn't including a coulping capacitor in my reasoning. If there isn't one, the battery is ineffective.

17. Dec 27, 2017

Jahnavi

But then it would be something like an input AC signal placed parallel to a DC battery (1 V)

What would then be the net effective voltage applied across Base Emitter ?

This is so confusing .

18. Dec 27, 2017

cnh1995

That Iin wouldn't be the true Iin as it won't flow through the base. It only flows through the battery. The base current will be determined by the input ac voltage, which will be the same even if you remove the battery.

19. Dec 27, 2017

Jahnavi

I was assuming AC input signal was not present .

20. Dec 27, 2017

cnh1995

The net BE voltage will be the input ac voltage. The battery is redundant as far as dc biasing is concerned.
Edit: assuming no coupling capacitor.

Ok, but when you connect the ac source, Iin≠IB.