Common Emitter Amplifier Stage Design Question

Thread Starter

derail

Joined Aug 13, 2009
2
Hi, I'm currently trying to solve a problem involving the design of a common emitter amplifier stage in an autobias configuration. The information I'm given is that

Vcc=12
Vbe=0.7
gm=0.12
hfe=120
RS=500
RL=1000
Gv=29.5dB
There should be max power transfer from source to load and max output voltage swing under AC.

So far I have calculated values for Rc and R1//R2 due to the requirement for max power transfer so that
RC=1000
R1//R2=1000

However I'm now stuck trying to find a value for RE and individual values for R1 and R2. I know it has something to do with calculating the quiescent collector current using the AC load line, and that Vce will be 0 thus eliminating it from calculation.

Essentially I need to work out how to find a value for RE. Ive attached a circuit diagram and any help would be greatly appreciated. Also i welcome any questions or discussions on my own thoughts how to solve this no matter how wrong i may be, ive been going round in circles for hours.
 

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Jony130

Joined Feb 17, 2009
5,230
The rule of a thumb for Re
Re=(0.1...0.3)Ucc/Ic
Ic=gm*26mV=3.12mA

And R1 and R2 select to achieve the quiescent collector current.
For example if we pick
Re=1V/Ic=330Ω
You must select the voltage divider R1 and R2 that they give the 1.7V on the base. And current that is flow through R1 and R2 should be large then Ib.
And remember that
Gv= Rin/(Rin+Rs) * ( gm*Rc||RL)=29.5dB=29.8V/V

Rin=R1||R2||( Hfe/gm)
 

The Electrician

Joined Oct 9, 2007
2,831
Jony130, you're not supposed to work out the problem for the OP. You're supposed to give him hints to help him do the work himself.

I think I would have started by telling him that Ic=gm*26mV, and let him proceed from there. You could tell him that if he still has difficulty, then come back to the forum and show his work up to that point, and ask for more help.
 

Thread Starter

derail

Joined Aug 13, 2009
2
Thanks for the quick replies though im not sure where you got the 26mV from, no where in any of the previous examples ive done, nor in teaching notes have i seen Ic=gm*26mV.
 

ELECTRONERD

Joined May 26, 2009
1,147
Jony130, you're not supposed to work out the problem for the OP. You're supposed to give him hints to help him do the work himself.

I think I would have started by telling him that Ic=gm*26mV, and let him proceed from there. You could tell him that if he still has difficulty, then come back to the forum and show his work up to that point, and ask for more help.
Yep, that's right. Although, once he does come up with the answers it's ok to share yours so he can check his work.
 
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