output stage

Discussion in 'Homework Help' started by Rubs, Nov 11, 2011.

  1. Rubs

    Thread Starter New Member

    Jun 20, 2011
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    hey..i have to design an audio amplifier with a gain of around 3600V/V (load is a 10W, 4ohm speaker and input is 5mV(p-p) i have designed two stages and cascaded them for this gain (attached)..nw i have to design the output stage using class AB type biasing it with Vbe multiplier (using tip31c and tip32c transistors)...i designed one bt the gain drops remarkably wen i cascade it..hw shud i bias it so that the gain does not drop??
     
  2. crutschow

    Expert

    Mar 14, 2008
    13,033
    3,241
    Show the output stage you have designed and how you are connecting it to the first stages. Otherwise we are just guessing as to the problem.
     
  3. Rubs

    Thread Starter New Member

    Jun 20, 2011
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    Vbb=2Vt ln((Iq)/Is), i assumed Iq (quiescent current= 100mA):
    Vbb=2Vt ln((100*10^-3)/10^-9) =0.92V
    R1+R2=Vbb/IR (i wuznt sure hw to divide Iq between IR and IC1...i used 16.83mA and 84.2mA respectively)
    R1+R2=0.92/16.83m = 54.65
    Vbe1=Vt ln ((IC1)/(Is))
    Vbe1=Vt ln ((84.2*10^-3)/(10^-9)) =0.46V
    R1=Vbe1/IR
    R1=0.46/16.83m=27.11
    R2=54.65-27.11=27.54
    the input impedence of the output stage shud be greater than the output impedence of the cascade i attached earlier so that the gain doesnt drop..cud u plz help me achieve that..plus im not even sure of the assumptions ive used for the output stage...
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Why you choose Iq = 100mA ?
    Normally we set this current equal to
    Iq > IL_max/(Hfe_min) = 2.3A/(400) = 5.8mA

    Hfe_min --->
    from the data sheet
    http://www.bourns.com/pdfs/tip31.pdf

    I choose Iq = 10mA and we can connect your second CE stage directly to drive the output stage.
    See the example diagram.

    R1 = (Vcc - 2Vbe) / 10mA ≈ 1K ---> 2Vbe because you use Darlington stage.

    Ic4 = ( Vcc - 2Vbe)/1K = 13.8mA

    Ib4 = Ic4/Hfe = 13.8mA/200 = 70uA

    hfe ---> from the data sheet figure 11
    http://elenota.pl/pdf/Fairchild/2n3904.pdf

    R4 + R5 = (15V - Vbe) /Ib4 = 205KΩ
    R4 = R5 = 100K ---> In real life we add potentiometer in series with R5 to set the output voltage equal to 0V.
    R2/R3 = 2
    Read this file
    http://www.ece.drexel.edu/courses/ECE-E352/ClassABAmp.pdf
     
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  5. Rubs

    Thread Starter New Member

    Jun 20, 2011
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    thanx alot for ur help :) i simulated ur design and the output voltage still does drop to a very small value :(...plus y have u used the 0.1ohm resistors (just so dat i cud explain to my teacher) cud v use larger ones?
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I simulated this circuit. And the voltage gain is equal to 90KV/V
    So I need reduce the gain.
     
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  7. Audioguru

    New Member

    Dec 20, 2007
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    Why do you talk WEIRD? Please talk in English.
     
  8. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Audioguru, you are reminded that not all AAC members are natives of an English-speaking country, nor are obliged to have an impeccable English language education.

    AAC is an English speaking forum, but only to provide a common base of conversation between its members. Good use of the English language facilitates conversation, but is not absolutely required to participate in the forums.

    If you have difficulty reading some of the posts, you can always choose not to process it.
     
  9. Rubs

    Thread Starter New Member

    Jun 20, 2011
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    i cant seem to understand the purpose of R3,C2,T3,R4,R8,Re2 and Re3... could you please explain that to me... and why isnt your output stage design working with my input stage :( cant we make a few changes in my design to make it work with yours?
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    R3 plus R4 apply negative feedback from the output to the base of T3 so that the output is at 0VDC. But then a high or low hFE of T3 changes it.
    C2 filters the AC audio out of the negative feedback to keep the AC gain high.
    T1 and T3 provide voltage gain (amplification).
    R8 is the load for T5 and it can quickly turn off T7 when T5 turns off.
    Re2 and Re3 reduce the differences in vbe and hfe of T7 and T8 so that they match better and so that their idle current is controlled.

    The input stage here has a low output impedance from its T2 emitter-follower so it can drive the fairly low input impedance of T3.
    Your input stage has a fairly high output impedance because it doesn't have an emitter-follower on its output and the low impedance of T3 loads it down and reduces its output level.

    Thank you for turning off the weird "text speak used by little kids on their cell phones" and now you are speaking pretty good English.
     
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