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Discrete Differential Amplifier

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  1. Jun 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculating the open loop output Impedance at 1kHz and the open loop differential gain

    Circuit.jpg
    3. The attempt at a solution
    Here is the approach.

    INPUT STAGE: Q111 and Q112

    Av1 = Rtot/2re, where Rtot is the total resistance of R117 and the input resistance of Q78 which is Rin.
    So the gain Av = Rtot/500 ohms.

    If hFE for Q78 is 250,

    Rin = hFE*(R115+re) = 250(155) = 38750 ohms.
    So Rtot = R117||Rin = 14441ohms = 14000 ohms.

    Using this value the gain G1 = 14000/500 = 28.

    If the load is 10 ohms

    hFE for Q78 and Q94 are 250 and 100 respectively.

    Load impedance RL= (250)(100)(600 ohms) = 15000000 ohms.

    SECOND STAGE: Q78

    Output resistance of Q78:
    Using Early voltage of BC557 is 50V,

    ro = (VA + VCE)/IC.

    ro = 14360 ohms.
    Early Effect Resistance Rout = 14360 ohms


    The early effect resistance and load impedance are parallel.
    So Rtot2 = RL|| Rout = 14346 ohms.

    Gains at stage 2: G2 = 14346/155 = 93.

    OUTPUT STAGE: Q94 and Q126

    For Q94, IE = 10.3mA = 10mA
    So re = VT/IE - 25mV/10mA = 2.5 ohms

    Rtot3 = R174 + re = 68 ohms + 2.5 ohms = 70.5 ohms

    Both halves are in parallel, So

    Output Impedance Zout = 70.5 ohms || 70.5 ohms = 35.25 ohms = 35 ohms

    Gain in the output stage G3 = RL/(RL + Zout) = 600 ohms/(600 ohms + 35 ohms)= 0.95

    Thus open loop gain G = G1*G2*G3 = (28)(93)(0.95) = 2473.8

    Converting to dB : Gain in dB = 20log(2473.8) = 68 dB

    I have calculated the open loop gain considering a load of 600 ohms. I ended up with an open loop gain of about 2473.8 which is about 68dB.

    Is this feasible?
     
    Last edited by a moderator: Jun 26, 2016
  2. jcsd
  3. Jun 27, 2016 #2

    rude man

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    Homework Helper
    Gold Member

    Disregard beta and re. Beta is infinite and re is zero. This problem is complicated enough without including those two parameters.
    I would also disregard the capacitors, hoping that at 1 KHz they are ignorable. But if you're really ambitious, include them.
    The circuit is mirror-identical between the npn and pnp input stages so work with the npn only, assuming HRI+ > HRI- so the pnp's are not in the gain stages.
     
  4. Jun 27, 2016 #3
    You have the following expression: Rin = hFE*(R115+re) = 250(155) = 38750 ohms.

    Where is R115?
     
  5. Jun 28, 2016 #4
    Sorry, it is a mistake, I wanted to write R153
     
  6. Jun 28, 2016 #5
    I also see this calculation: So Rtot = R117||Rin = 14441ohms = 14000 ohms.

    If R117 = 15k and Rin = 38750 ohms, we should have 15000||38750 = 10814 ohms, not 14441 ohms.
     
  7. Jun 28, 2016 #6
    You are correct. Using the value Rtot = 10814, the gain at this stage G1 is 22

    Making the total gain to be

    The open loop Gain G = G1*G2*G3 = (22)(157)(0.95) = 3281.3
    Converting to dB: GaindB = 20 log(3281.3) ≈ 70.32dB
     
  8. Jun 28, 2016 #7
    Another Mistake, If the load is 10 ohms

    The load is 600 ohms and not 10 ohms
     
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