Could use help on Diff Amp calculation

Discussion in 'Homework Help' started by laguna92651, Mar 22, 2010.

  1. laguna92651

    Thread Starter Active Member

    Mar 29, 2008
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    I posted this earlier with no response, I could really use some direction on this.

    How do I calculate the max peak to peak output swing and the max input voltage that can be applied without any distortion.

    20>=Ad<=30 (single ended output)
    Rid>=20k
    ±10v

    I'm not sure what I need to calculate and what assumptions to make to calculate.
     
  2. Jony130

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    Feb 17, 2009
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    For single ended output
    Au=Rc/( 2*(Re+re)) = 2.4KΩ/ 2*(28Ω+10.4Ω)=2.4KΩ/76.8Ω=31[V/V]

    And the voltage output swing for negative voltage is cut-off of a Q3.
    And for positive swing cut-off if a Q4.
    max input voltage = ( max peak to peak output swing)/Au
     
  3. laguna92651

    Thread Starter Active Member

    Mar 29, 2008
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    Thanks for the reply,what is the re=10.4Ω, where did that come from? Is Au a the max output peak to peak or the gain? Thanks
     
  4. t_n_k

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    Mar 6, 2009
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    re is the dynamic emitter resistance dependent upon the DC emitter current value and device temperature - plus some constants.

    re is typically calculated from the relationship 26/Ie(mA)Ω
     
  5. laguna92651

    Thread Starter Active Member

    Mar 29, 2008
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    How is this the max peak to peak output?

    For single ended output
    Au=Rc/( 2*(Re+re)) = 2.4KΩ/ 2*(28Ω+10.4Ω)=2.4KΩ/76.8Ω=31[V/V]

    This looks like a gain, so it is relative? I assume the max output is before distortion occurs, so it is an absolute number?
     
  6. Jony130

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    Feb 17, 2009
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    To calculate max negative swing remove Q3 from the circuit and calculate the output voltage.
    For max positive voltage remove Q4 and again calculate the output voltage.
    And Au is a voltage gain.
     
  7. t_n_k

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    Mar 6, 2009
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    I'd be interested to see how the re=10.4Ω value was determined. I'm not sure how one could easily predict the current source value (& subsequently re) from an analysis of the circuit - without reference to some device parameters.
     
  8. Jony130

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    Feb 17, 2009
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    The output current of a current mirror is given:

    Io=\frac{Vt}{R2}*In\frac{Ic2}{Io}

    So for R2=12.4Ω ; Vt=26mV; and IR3=(20V-0.7V)/320Ω=60.3mA
    And from datasheet we can read the "exact" value of VBE for Ic=60mA.
    And β=200.
    http://www.iele.polsl.pl/elenota/ON_Semiconductor/p2n2222a-d.pdf
    Vbe≈0.73V and finally IR3=60.21mA so Ic2≈β/(β+2)*IR3=59.61mA.
    So know to solve and find Io we must use iteration.
    So I start the first iteration with assume Io=50mA
    26mV/12.4Ω=2.096mA

    Io = 2.096mA*In(59.61mA/50mA) = 368.613uA (1)
    Io = 2.096*In(59.61mA/368.613uA) = 10.6638mA (2)
    Io = 2.096*In(59.61mA/10.6638mA) = 3.6084mA (3)
    Io = 2.096*In(59.61mA/ 3.6084mA) = 5.8804mA (4)
    Io = 2.096*In(59.61mA/5.8804mA) = 4.8565mA (5)
    Io = 2.096*In(59.61mA/4.8565mA) = 5.2576mA (6)
    Io = 2.096*In(59.61mA/5.2576mA) = 5.0912mA (7)
    Io = 2.096*In(59.61mA/5.0912mA) = 5.1586mA (8)
    Io = 2.096*In(59.61mA/ 5.1586mA) = 5.1311mA (9)
    Io = 2.096*In(59.61mA/5.1311mA) = 5.1423mA (10)
    So I end calculation here with final Io ≈ 5.1mA
    So
    re=26mV/2.55mA
    10.1Ω ( previously i calculate re for 2.5mA)
    And spice results are Ic2=59.74mA and Io=5.91mA .
    And I skip the fact that current morrow are very poor design.
     
    Last edited: Mar 25, 2010
  9. t_n_k

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    Mar 6, 2009
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    Thanks Jony130,

    I'd not seen that done before. Much appreciated for your effort.

    I wonder if the original 'question' entertained the idea that this degree of deduction was required on the part of the student.


    Cheers,

    t_n_k
     
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