# Common Base CB BJT

Discussion in 'Homework Help' started by Vicros, Jan 25, 2012.

1. ### Vicros Thread Starter New Member

Jan 22, 2012
13
0
Hi all,

Back again with another BJT configuration, this time it is a Common base question.

I am asked to final suitable values for $V_{cc}$ and $R_E$ with a quiescent point $I_{CQ} = 1mA$

I have tried doing Thévenin on each of the terminals but end up with both $V_{cc}$ and $R_E$ in the equation. As well as using my regular method of

$V_{cc} = (R_C + R_E)I_C + V_{CE}$

Im having a mental block and cant seem to find $V_{cc}$ on its own first.

Please find attached the question and my attempt at solving it

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Given Vb is 0V then Ve will be -0.7V.

With Vceq=5V then Vc=+4.3V. This will lead you to Vcc since you are given Ic=1mA & Rc=2k2.

3. ### Vicros Thread Starter New Member

Jan 22, 2012
13
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Brilliant, thank you

I think i did it right, I ended up with Vcc as 6.5V when i did:

Vcc = Ic*Rc + Vc

I assumed that it was correct, so I moved on and continued to do most of the questions. Turns out that the common base AC analysis is quite similar to the common emitter i think.

I did what i believe is an alpha model of the question...

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4. ### Vicros Thread Starter New Member

Jan 22, 2012
13
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Ok, I have finished off the question now

The two major things that i would like some one to look over is the AC analysis circuit that i drew and the last section where it ask to calculate the transresistance gain $v_o / i_i$

What i did was replaced $i_i$ with $v_i / R_i$ so that the equation came down to $A_V * R_i$

but my answer is very large which doesnt look right

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
I would dispute both your values of voltage gain and input resistance.

To find Av one would use Rc||Rload=2.2k||4.7k=1.5k

Also keep in mind that Vo and Vi will be in phase for the CB configuration - at midband frequencies.

Also the effective input resistance at the emitter is dominated by re which is quite small. RE and re are in parallel rather than in series as your analysis suggests.

6. ### Vicros Thread Starter New Member

Jan 22, 2012
13
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thanks for pointing that out, i can now see that i did forget to include the load resistance.

I changed my values. i now have smaller values for both voltage and current gain.

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