# Common Emitter Amplifier Design Problem

Discussion in 'Homework Help' started by testing12, Feb 20, 2012.

1. ### testing12 Thread Starter Member

Jan 30, 2011
80
2
Hello Everyone,
I am working on my homework and am stuck with problem below. I have a set of rules by my prof to place a transistor close to the center of the active region.

They are:
VBB = Vcc/3
IcRc= Vcc/3
0.1 Ie < Ib < Ie

I have tried a variety of things but am stuck now. (I have erased a few attempts but below is the setup of my latest) Can someone please walk me through the problem, I'd really like to understand how to approach this.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Your prof's rules seem inconsistent with the design goals.

If the drop across Rc is Vcc/3=4V then Ic = 4/5000=800uA.

In the linear region one would expect Ib=Ic/β=800uA/100=8uA.

But your rule set includes a condition that Ib must be greater than 0.1Ie which is 0.1*808uA or 80.8uA. This makes no sense. Is it more likely Ib is this case refers to the base bias voltage divider branch current rather than the base current?

In any event I doubt you can achieve the design goals specified in the printed question under the constraints of your prof's set of rules.

Last edited: Feb 21, 2012
testing12 likes this.
3. ### testing12 Thread Starter Member

Jan 30, 2011
80
2
Agreed, I arrive at a contradiction when I use my profs guidelines. Since the base current ends up being 8 uA which is out of the range he specified.
I would like to use the beta relation as this governs this current through the transistor. This time ignoring my profs guidelines, how can I approach the problem?

4. ### crutschow Expert

Mar 14, 2008
13,501
3,375
The purpose of the bias circuit is to minimize the effect of beta variation (which can typically vary 3:1 or more between the same transistor type) on the bias point. Thus you do not want to use the beta value as a way to determine the bias.

testing12 likes this.
5. ### testing12 Thread Starter Member

Jan 30, 2011
80
2

Ok, but the bias currents are related to beta as follows:

Ic = B Ib
Ie = (b+1) Ib
I could not find a way for me to have Ib in the range specified above.

Dec 26, 2010
2,147
300
Yes, the expected range of beta values can only serve as a basis for deciding a bias divider chain current, which will normally be much more than Ic/βmin.

Given that many transistors may have a typical β of about 100, and probably not much less than a few tens, a chain current of Ic/10 may be reasonable.

A working β as low as 10 would not be so common except in saturated operation, or perhaps for some larger power or high voltage transistors.

testing12 likes this.
7. ### testing12 Thread Starter Member

Jan 30, 2011
80
2
I am going to try this again ignoring the guidelines above.
I have a few questions though

- Can i select Vcc arbitrarily?
- Is Rc 5K? (the wording of the question confuses me)
- How should I Proceed?

Thank you all for the assistance.

8. ### hobbyist Distinguished Member

Aug 10, 2008
773
62
You could proceed by considering " 0.1 IE < Ib < IE" to be used as the base voltage divider current, instead of the base current itself.

Then continue to choose arbitrary value for VCC, and work from there.

A CE stage usually uses around 1/2VCC to be across RC, but continue to use the constraint given which was "1/3 ICRC".

Remember VBB is constrained to 1/3VCC also, so accounting for Vbe, you will be able to calculate required VE needed to give RE value.

Then by using Ib above for the dicvider current, you could solve for RB2 and RB1 respectively.

Then do a check on the Zin using your given value for Beta @ 100.
And see if it is greater than requirement given.

testing12 likes this.
9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
There are a couple of issues in the design 'spec' that require consideration.

The input resistance is 20kΩ [presumably at β=100] & $v_\pi=5mV$ peak with the specified source conditions.

I'm assuming $v_\pi=5mV$ is the same as the small-signal vbe.

If the source is 100mV peak with a 20kΩ resistance and a matching 20kΩ input resistance then the [peak] ac source current would be 0.1V/40kΩ=2.5uA.

Part of this 2.5uA small signal ac flows into the base and the remainder into the parallel combination of the base bias resistors RB1 & RB2. The base peak ac voltage would also be 50mV.

Suppose the emitter DC current has been set to 1mA - then the dynamic emitter resistance [re] is of the order of 25Ω. So the emitter resistance RE would be determined from the relationships

vbe=ie*re=5mV

based on $v_\pi=5mV$ & where re=25Ω [With IE=1mA]

(1+β)*ib*(re+RE)=ie*(re+RE)=50mV

Knowing ib will give the small signal ac current into the parallel combination of RB1 & RB2 = 2.5uA - ib

Last edited: Feb 21, 2012
testing12 likes this.
10. ### hobbyist Distinguished Member

Aug 10, 2008
773
62
testing12,

Please disregard my (hobbyist) above post,
as ,,t n k states, there are some factors in the specifications that need consideration.

My suggestion, did not take any of those specs into account.

Look over what t n k posted and work at it from those perspectives.

testing12 likes this.

Jan 30, 2011
80
2