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Homework Help: Calculate gain for differential amplifier

  1. Apr 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate gain (u_o/u_i) for a differential amplifier with symmetric output. Both transistors have the same transconductance gm. Transistor output resistance r_0 is neglected here.

    circuit: http://sv.tinypic.com/view.php?pic=14cea8p&s=8#.VSf7Evl_vxM

    2. Relevant equations

    Calculate gain A = u_o/u_i

    3. The attempt at a solution

    We rewrite the circuit to small-signal.
    Voltage sources +/- E are set to zero.
    I also split the source resistance Rs, into two parts so that we can divide the differentiator between a left and right side.

    [tex]v_{gs1} = u_{i}-g_{m}v_{gs1}\frac{R_{s}}{2} [/tex]
    [tex]v_{gs1} = \frac{u_{i}}{1+g_{m}\frac{R_{s}}{2}} [/tex]

    [tex]v_{gs2} = 0 - g_{m}v_{gs2}\frac{R_{s}}{2} [/tex]
    [tex]v_{gs2}(1 + g_{m}\frac{R_{s}}{2}) = 0 [/tex]
    [tex]v_{gs2} = 0 [/tex]

    [tex]u_{o} = v_{o1}-v_{o2}[/tex]
    [tex]v_{o1} = -R_{D1}g_{m}v_{gs1} [/tex]
    [tex]v_{o2} = -R_{D2}g_{m}v_{gs2} [/tex]

    [tex]u_{o} = -R_{D1}g_{m}v_{gs1}+R_{D2}g_{m}v_{gs2} = -R_{D1}g_{m}v_{gs1} [/tex]
    [tex]u_{o} = -R_{D1}g_{m}(\frac{u_{i}}{1+g_{m}\frac{R_{s}}{2}} ) [/tex]
    [tex]\frac{u_{o}}{u_{i}} = \frac{ -R_{D1}g_{m}}{1+g_{m}\frac{R_{s}}{2}} = \frac{ -(R_{D}-\frac{\Delta R}{2})g_{m}}{1+g_{m}\frac{R_{s}}{2}}[/tex]

    The answer above is wrong, u_o/u_i = -gm*Rd is the correct one. What am I doing wrong?
  2. jcsd
  3. Apr 11, 2015 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Don't use small-signal equivalent circuits. All you need is gm = dI/dVgs for each transistor. You still can and should use small-signal parameters.
    I suggest:
    Ground uin- . Apply small voltage uin+ . Compute ΔI1 and ΔI2.
    Then compute uut+ , uut- and finally uut+ - uut- .
    Note that ΔI1 and ΔI2 are independent of RD1 and RD2.
    Your given answer is correct if we assume differentially small ΔRD but if not then the gain I calculated is somewhat less than gmRD.
    Also, if 2gmRs >> 1 the gain also → gmRD.
    If you apply +vin/2 to the + input and -vin/2 to the - input instead of +vin to the + input and 0 to the - input, your given gain is correct at gmRD. This is because then the "common-mode voltage" (the average of the two input voltages) is always zero, the source voltage never changes, and i1 + i2 = 0.
    Last edited: Apr 12, 2015
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