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BJT current gain

Discussion in 'Electronics Homework Help' started by OrangeArav, Dec 6, 2015.

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  1. OrangeArav

    OrangeArav

    38
    0
    Jul 14, 2015
    Hello guys, I need help with the problem below:
    [​IMG]
    I'm having difficulty finding Iout in terms of small-signal ib_3 of Q3. My work is attached below under spoiler. It seems that the dependent current source for Q3 is open (Off), so then looking at my pg 3 I'm ambiguous about finding Iout in terms of ib_3, since ib_3 would be 0?
    [​IMG]
    [​IMG]
    [​IMG]

    Please help in finding the Iout/Iin
     
  2. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    I think it would be easier to split the all thing into 3 stages which are cascaded.
    There is a requirement to use voltage dividers.
    For each separate block calculate Av,Rin,Rout while assuming it is driven by a voltage source .
    Now,
    Vout=
    Vin*[Rin1/(R2+Rin1)]*Av1 *[Rin2/(Rout1+Rin2)]*Av2 *[Rin3/(Rout2+Rin3)]*Av3*[RL/(Rout3+RL)]

    Iout=Vout/RL
    Iin=Vin/R2

    Ai=Iout/Iin=
    R2*[Rin1/(R2+Rin1)]*Av1 *[Rin2/(Rout1+Rin2)]*Av2* [Rin3/(Rout2+Rin3)]*Av3*[(Rout3+RL)]


    2015-12-05_220334.jpg cascade.JPG
     
  3. Ratch

    Ratch

    1,094
    334
    Mar 10, 2013
    Well, let's see. Hmmm. Just about all the Iin will be present in the Q1 base-emitter-C1 path, because that conduction path has a very low impedance compared to the 1k of the base resistor. Q1 will betatize that current and send it through R3, because R3 has a much lower impedance than R7||R8 seen through the base of Q2. A voltage of Iin*R3 will appear across R7 and R8. The current of R8 (Iin*R3/R8) will just about be equal to the collector current of Q3. Using the current divider rule, we see that the current present in RL will be 2/3 the collector current Q3 ( 400/(400+200) ). That makes the small signal current gain ( (2/3)*Iin*R3/R8)/Iin = ((2/3)*R3/R8 . I have ignored re,rb,0.07, etc. Notice that the three transistors represent the common-emitter, common-collector, and common-base configurations of a BJT.

    Ratch
     
    Last edited: Dec 6, 2015
    Arouse1973 likes this.
  4. OrangeArav

    OrangeArav

    38
    0
    Jul 14, 2015
    dorke,

    Still, not sure what to do with that third stage on pg3? The small-signal current i_b3 is zero from my calculations. Then, the only way to calculate Iout is to assume that r_e + R8 are connected to R10 and Rload via ground nodes? That's a bit confusing. Also, how would I calculate this without knowing the current going into the emitter of Q3? Please check if my work below is correct for the first two stages.
    [​IMG]

    [​IMG]

    [​IMG]

    Ratch,

    I'm a bit confused on your post there, not sure if I see the schematic layout to which to apply your calculations... I prefer everything drawn and written out
     
  5. Ratch

    Ratch

    1,094
    334
    Mar 10, 2013
    What is your confusion? The schematic layout is your own, and my description of the calculations reference your component designations. Start from my first sentence explanation of the current (amperage) value, and let me know what you do not understand. Stop at the first point of confusion and wait for me to explain it.

    Ratch
     
    Martaine2005 likes this.
  6. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    OrangeArav,
    Put aside your calculations and try to think about the all thing as composed of cascaded basic building blocks!

    The last stage is a C.B amplifier Q3 with RC=R10 .
    for that you already have the Av,Rin,Rout formulas "ready made"(you can develop that if you want,but absolutely don't have to).
    As I have drown the blocks,"my Rin of block" would be Rin(C.B)|| R8.

    Use that approach to all 3 stages and your life would be much simpler...;)
     
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