[SOLVED] Common Emitter Designing.

[Shayaan_Mustafa]

Hello experts!

I had given a question in exam. The question is given below.

Question
Sketch a circuit diagram of a CE amplifier. Determine the values of \[{R}_{1}\],\[{R}_{2}\] and \[{R}_{E}\], if the required operating point is \[{I}_{C}\]=2mA and \[{V}_{CE}\]=4v. Whereas current through the series network of \[{R}_{1}\] and \[{R}_{2}\] is \[{I}_{1}\]=10\[{I}_{B}\] (\[{\beta}_{dc}\]=50, \[{V}_{BE}\]=0.7v, \[{R}_{C}\]=2.2K\[\Omega\],\[{V}_{CC}\]=9v).

Attempted Solution

For \[{R}_{E}\]:

Applying KVL to the collector-emitter side,
\[{V}_{CC}\]=\[{I}_{C}\]\[{R}_{C}\]+\[{V}_{CE}\]+\[{I}_{E}\]\[{R}_{E}\]----(1)
As we know
\[{I}_{E}\]=\[{I}_{C}\]+\[{I}_{B}\]
and
\[{I}_{E}\]=\[{I}_{C}\]+\[\frac{{I}_{C}}{{\beta\]}_{dc}}\]
now substituting values gives \[{I}_{E}\] as,
\[{I}_{E}\]=2.04mA

Now solving eq(1) for \[{R}_{E}\]
\[{R}_{E}\]=\[\frac{{V}_{CC}\]-\[{I}_{C}\]\[{R}_{C}\]-\[{V}_{CE}\]}{\[{I}_{E}}\]
Now substituting values and solving for \[{R}_{E}\]

\[{R}_{E}\]=294.11\[\Omega\]

As you can see I have solved for \[{R}_{E}\]

But I want to confirm about my result. Have I found this \[{R}_{E}\] value is correct?

And how to start calculations for \[{R}_{1}\] and \[{R}_{2}\]?

Thanks in advance. And sorry for LAteX. If you find any problem then kindly solve that by yourself.

 

[mister_rf]

Yes, that’s the correct value for Re. :-D

---------- Post added at 02:00 ---------- Previous post was at 01:39 ----------

U_b= U_e + 0.7V
U_e = IeR_e= 0.6V
U_b= 1.3V
But also U_b= U_R2 = 1.3V
Use KVL
U_R1+U_R2= Vcc
U_R1= Vcc- U_R2 = 9 – 1.3 = 7.7V
We have
U_R2= R2(I1-Ib)
and
U_R1= R1(I1), but
I1= 10 Ib
And Ib= Ic/50= 2/50 mA
I1= Ic/5 = 2/5 mA
R1= U_R1/I1= 7.7/(2/5)= 19.25 k
R2= U_R2/(I1-Ib)= 1.3/(10Ib- Ib)= 1.3/9Ib= 1.3/(9* 2/50)= 1.3*50/18= 3.611k

 

[Shayaan_Mustafa]

Thanks a lot. I will post another. And I have some more question. And I want you as well as other members to solve it. Thanks once again.