Solve Multistage Amplifier Homework | Av=Av1*Av2

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The discussion revolves around solving a multistage amplifier circuit using the formula Av=Av1*Av2, where Av1 and Av2 represent the gains of the two stages. Participants clarify that each stage can be analyzed separately, particularly when assuming beta is infinite, which simplifies calculations by allowing the neglect of small currents like Ib2. The calculated values for Av1 and Av2 were -5.86 and -1.97, respectively, leading to a total gain of 11.56, with a noted error attributed to approximations. There is a consensus that in well-designed circuits, it is common practice to consider beta as infinite to streamline the analysis. The conversation concludes with an acknowledgment of the importance of understanding the underlying principles of transistor behavior in circuit design.
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Homework Statement


I want to solve the circuit using the formula Av=Av1*Av2 where Av1 and Av2 are the gains(volts) of the 2 stages and i have some questions.
V2 is equal with -(βIb1+Ib2)*R3=(β+1)Ib2*R5+Ib2*rbe2
Now when i calculate Av1 and Av2 i take each stage separately from circuit (example the 3rd circuit in pic)?

The Attempt at a Solution


The result must be 12.
My solution
Av1=-βR3/(rbe1+(β+1)R4))=-5.86
Av2=-βR6/(rbe2+(β+1)R5))=-1.97
Av=11.56
The error of 0.44 is because of aproximations or i did somthing wrong?
For the second transistor the emitter is to R6 resistor,npn.
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Looks from your diagram like
R3 = 1K
R4 = 0.5K
R5 = 2K
R6 = 1K
from which the gain is approximately (R3//R2)(R5/R6) = 4
I must be misreading either R3 or R4.
Anyway, the gain computes to an even number only if you assume beta = infinity and r_be1 and r_be2 = 0. This is what people normally do.
 
R3=3k
R1=50k
R2=10k
Thanks for what did you say, if i limit beta to infinity it gives R3/R2*R5/R6, didnt know this:cool: .
Is it correct to take each stage separately? I ask this because if i take the first stage, i don't have the current which comes from the 2nd stage -Ib2 but because Ib2 is very small current we can ignore it?
 
Last edited:
Drao92 said:
R3=3k
R1=50k
R2=10k
Thanks for what did you say, if i limit beta to infinity it gives R3/R2*R5/R6, didnt know this:cool: .
Is it correct to take each stage separately? I ask this because if i take the first stage, i don't have the current which comes from the 2nd stage -Ib2 but because Ib2 is very small current we can ignore it?

Exactly right. ib2 = 0 if beta is infinite. I will go on a limb and say always consider beta = infinity IN A PROPERLY-DESIGNED CIRCUIT. Unfortunately, you will often come across non-correctly-designed circuits in textbooks, and for a good reason - they want you to include a finite beta and/or rbe at times just to make sure you understand how to handle these parameters. Once you're working you forget them, typically.

Especially beta. Beta varies so widely FROM ONE TRANSISTOR TO THE NEXT OF THE SAME TYPE that you have to design to beta = infinity. rbe is more predictable but in a properly designed circuit is usually negligible compared to other error sources.
 
rude man said:
Looks from your diagram like
R3 = 1K
R4 = 0.5K
R5 = 2K
R6 = 1K
from which the gain is approximately (R3//R2)(R5/R6) = 4

Shouldn't the gain be (R3/R4)(R5/R6)?
 
The Electrician said:
Shouldn't the gain be (R3/R4)(R5/R6)?

Yes. My boo-boo.
 
I have an ugly writing :wink:.
Thanks everyone for help.
 

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