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question about common emitter amplifier with resistive emitter degeneration

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jimito13

jimito13

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Dear forum members and experts,

I have a very basic question concerning the small-signal analysis of a common emitter amplifier with resistive emitter degeneration.I provide the circuit's schematic and it's small-signal equivalent at the attached image :


My question is :

At emitter node of the above ckt we can write kirchoff's law : ie=ib + (gm*uπ)+ir0.Up to here everything is clear to me.
We know that ie=ib+ic but now we have one more current term that is the ir0,thus the equation ie=ib+ic=ib+βib=(β+1)ib=>ie=(β+1)ib is no more valid.Am i right or am i making a mistake in my thoughts?

Thanks in advance for any helpful answer.

Best Regards,
Jimito13-Analog RFIC Designer
 

Look at the equivalent circuit diagram and you will see that the current ic (current through the C node) now consists of two parts which are summed together.
That's all. The equation ie=ib+ic is always valid since it is defined for the currents through the nodes/pins E, B and C.

---------- Post added at 13:09 ---------- Previous post was at 13:08 ----------

 
Ok LvW,thanks a lot,i got it ;-)

Sorry to ask again...I have one more supplemental question :

I agree that the equation ie=ib+ic is always valid,but...i will pose my doubt below :

Now that ic has two parts (gm*uπ , ir0) we write the kirchoff's law at the emitter node (as i wrote it at my first post) as :

ie=ib+ic=> ie=ib + (gm*uπ)+ir0

How can we claim/prove that the equation with bold letters ie=ib+ic=ib+βib=(β+1)ib=>ie=(β+1)ib is still valid since it is proved given that
ic is equal β*ib and has no other parts?
 
Last edited:

Anybody willing to give a clarification to my question?
 

jimito13 said:
Ok LvW,thanks a lot,i got it ;-)

Sorry to ask again...I have one more supplemental question :

I agree that the equation ie=ib+ic is always valid,but...i will pose my doubt below :

Now that ic has two parts (gm*uπ , ir0) we write the kirchoff's law at the emitter node (as i wrote it at my first post) as :

ie=ib+ic=> ie=ib + (gm*uπ)+ir0

How can we claim/prove that the equation with bold letters ie=ib+ic=ib+βib=(β+1)ib=>ie=(β+1)ib is still valid since it is proved given that
ic is equal β*ib and has no other parts?
Click to expand...

Therefore, β = [(gm*uπ)+ir0] / ib :wink:
 

Therefore, β = [(gm*uπ)+ir0] / ib
Click to expand...

Yes but we also have : gm*uπ=β*ib , thus ir0=0!

I know that i have a misunderstanding above and i am waiting for a clear answer,not just an equation with no explanations...
 

jimito13 said:
Yes but we also have : gm*uπ=β*ib , thus ir0=0!

I know that i have a misunderstanding above and i am waiting for a clear answer,not just an equation with no explanations...
Click to expand...

Hi Jimito,

you should know that the common equation ic=β*ib is a simplification only.
It results from one of the twoport equations describing the BJT in hybrid form (h-parameter) :

i2=h21*i1+h22*v2

In common emitter configuration: i2=ic ; i1=ib and v2=vce

From this: For v2=0: ic=h21*ib=β*ib (h21=β).

Because of the condition v2=0 the current gain h21=β is called "short circuit current gain".
That's the secret of the contradiction you might have seen.

---------- Post added at 14:01 ---------- Previous post was at 13:57 ----------

One additional remark:

The parameter h22 (output conductance) is a measure of the slope of the Ic-Vce characteristics.
Thus 1/h22=ro.
 
Last edited:
Hi LvW,

I can understand your analysis absolutely but the question still remains in my head...If this doesn't bother or take time i would like to ask you for further assistance :

For the ckt i uploaded how can i write the equation ic=β*ib in a correct way?And what ic is the one that fits this equation?I suppose from the above that the correct equation is : ic=β*ib => (gm*uπ)+ir0=β*ib .But now if we take this equation as correct then gm*uπ≠β*ib => uπ/ib ≠ β/gm => rπ ≠ β/gm !!!

Thank you in advance for any helpful answer.

Best Regards,
Jimito 13-Analog RFIC Designer
 

Jimito, Hello again !

Sorry, but I don't understand your question "For the ckt i uploaded how can i write the equation ic=β*ib in a correct way?"

As I have tried to explain: This equation is NOT correct within an amplifier circuit - and, therefore, you cannot write it in a "correct way".
It is a simplified equation that is correct only for the case Vce=0 (ac wise only).

---------- Post added at 16:16 ---------- Previous post was at 16:14 ----------

Another view: The simplified approach cannot be correct by 100% since the BJT is not an ideal current source.
 
Allright,but can somebody explain to me why in all books when they subsitute a circuit with BJT devices to it's small-signal low-frequency equivalent they write next to the voltage contolled current source either gm*uπ or β*ib (implying that those quantities are equivalent)?

If the equation gm*uπ=β*ib is NOT valid in a practical circuit then the equation that we all know rπ=β/gm is NOT valid as well!
 

Correct! It is "not valid".
However, don't be to disappointed - nothing in the electronic world is correct by 100%.
Because of other uncertainties (in case of transistors: very large tolerances) it is a good engineering practice always to neglect some minor influences.
This leads to equations that are handy and exact enough for most purposes.

For the particular case under discussion:

vbe=h11*ib + h12*vce
ic=h21*ib + h22*vce

That means: transconductance g=ic/vbe=(h21*ib + h22*vce)/(h11*ib + h12*vce)

Simplification: For vce=0 (neglecting backward influence and output conductance): g=h21/h11=β/rbe
 
Jimito, perhaps the following remark can help:

If the equivalent small-signal circuit contains a cuurent source i=ib*h21 or i=g*Vbe the simplification that I have mentioned very often is corrected by additional elements which represent the neglected terms:
ro=1/h22 in parallel to the current source and
v12=h12*vce in series with h11=rbe .
 
jimito13 said:
Allright,but can somebody explain to me why in all books when they subsitute a circuit with BJT devices to it's small-signal low-frequency equivalent they write next to the voltage contolled current source either gm*uπ or β*ib (implying that those quantities are equivalent)?

If the equation gm*uπ=β*ib is NOT valid in a practical circuit then the equation that we all know rπ=β/gm is NOT valid as well!
Click to expand...

most books all assume ro is very high and hence iro is negligible.
 

Ok guys i run through the theory and the equations as LvW guided me and now things are clarified to me.Thanks a lot for your help.
 

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