CE Transistor Amp - Quiescent point

Hey guys,

I'm trying to do some past paper questions, but am stuck on this one:



Can anyone give some advice to help me get started?

Thanks!

Dear grapeman
Hi
What is your problem in that ?

goldsmith said:

Dear grapeman
Hi
What is your problem in that ?

Click to expand...

Hi,

ye sorry I forgot to say: I don't know how to determine the quiescent point

It is simple . quiescent point , consists from Vce and Ic and IB . so for Ic you can consider a loop from Vdc and biasing resistors up to ground . ( R1 and BE and then RE , thus you'll find IE and then you can find VEC easily )
Best Wishes
Goldsmith

I guess you are right to be confused... there is a missing parameter concerning Vbe of Q1. Usually it is assumed around 0.7 to 0.8V here. Even if the exponential formula of the PN junction (of BE), I versus V, has to be used, the formula constants for Q1 are also missing.
When you are solving the V/I equations, I guess you can assume Vbe = 0.75 V and you will be fine.

KerimF said:

I guess you are right to be confused... there is a missing parameter concerning Vbe of Q1. Usually it is assumed around 0.7 to 0.8V here. Even if the exponential formula of the PN junction (of BE), I versus V, has to be used, the formula constants for Q1 are also missing.
When you are solving the V/I equations, I guess you can assume Vbe = 0.75 V and you will be fine.

Click to expand...

Hey - thanks for the reply

Here is what I've done:

Vb = 15000/15000+5000 * 15 = 11.25V
Ve = 11.25 - 0.7 = 10.55V
Ic = 10.55 / 1000 = 10.55mA
15 = Vce + 3000*Ic + 1000*Ie
Ic ~ Ie
15 = Vce + 31.65 + 10.55
Vce = -27.2V

Now Vce obviously cannot be a negative voltage, so I assume there is a large amount of current being drawn as Ib which changes Vb. When I try to apply Thevenin though, Vce goes into the thousands!

What am I doing wrong?

Hi again
Are you think that , the CE voltage can be about -27 volts when you have just +15 volts , supply voltage ??!!! are you familiar with principles of circuit analysis ? of course it is completely wrong .
Best Wishes
Goldsmith

grapeman said:

Now Vce obviously cannot be a negative voltage...
What am I doing wrong?

Click to expand...

You're right to be confused. It looks like there's a mistake in the question paper.

With the values given, the transistor is saturated and won't do anything useful. If you swap the values of Rb1 and Rb2, you'll get sensible answers (and an amplifier that works).

---------- Post added at 02:06 ---------- Previous post was at 01:58 ----------

grapeman said:

I'm trying to do some past paper questions....

Click to expand...

There's a couple of other questions you might like to try in this thread: https://www.edaboard.com/threads/253524/.

It's nice to see a student working for a change.
IMO, there's too many "Please do my homework for me" threads around here.

Hi grapeman,

Your work is right since it was assumed that the transistor is linear (given beta=Ic/Ib=300). But if we find out that the result cannot be real then this leads us to think that something we have assumed wasn't a good choice in the first place (hence it wasn't your fault).

In this special case (circuit), we consider the transistor being saturated. In other words, Vce (or Vsat) is close to zero (about 50 to 400 mV and this depends on Ic/Ib. For instance, decreasing Ic/Ib decreases Vce but at a much slower rate).

The new equations would be:
Vcc - Vsat = Ic*Rc + Ie*Re

Ie = Ic + Ib
Note: Here Ib is not negligible

Vb= Vbe + Ie*Re

Ib = (Vcc - Vb)/Rb1 - Vb/Rb2

Unkown: Ic, Ib, Ie and Vb
Known: Vcc, Vsat, Rc, Re, Vbe, Rb1 and Rb2
And we have 4 equations

Good luck.

Kerim

Added:
Obviously this circuit, as it is, doesn't work as it was supposed to do... as an AC amplifer.

godfreyl said:

You're right to be confused. It looks like there's a mistake in the question paper.
With the values given, the transistor is saturated and won't do anything useful. If you swap the values of Rb1 and Rb2, you'll get sensible answers (and an amplifier that works).

Click to expand...

Yes, godfreyl is right. Rb1 and Rb2 are to be swapt.
In this case, the solution is rather simple.
* Assuming the base current can be neglected in comparison to the current through Rb1-Rb2, the base voltage is 3.75 volts.
* With Vbe approx 0.75 volts the drop across Re is 3 volts resulting in Ic=Ie=3 mA.
* With beta=300 Ib=10µA which confirms the above assumption (I through Rb1-Rb2 approx 750µA).

Of course, it is only an approximation but this is quite normal and the only practical way because of large uncertainties regarding active tolerances.

Hey guys,

Thanks for the help! :-D

After a few pages of workings I managed to get similar results as a simulated model. Hopefully I wont get a saturated 'amp' in my exam!

Hi,

Wish you good luck.

For instance, in my high studies that I passed many decades ago, I got the worst grades, for different reasons, in the fields that I became professional in, just a few years after graduation till these days So no matter what you get in your exam(s), I think you will have a bright future because you like what you do and you depend on yourself while doing it.

Kerim