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Transconductance of a differential pair

  1. Aug 26, 2009 #1
    I'm trying to figure out what the transconductance of a bipolar differential pair with a certain tail current would be, but I'm getting tangled up in thinking like "well the pair sees half the input swing, but the transconductance might be double if there are two transistors.."etc. Could someone explain to me the proper way to calculate the transconductance of such a circuit with both a resistive load and a current mirror load? Thanks!
  2. jcsd
  3. Aug 26, 2009 #2
    Maybe this will help. If the first (assume npn) transistor has a zero resistance collector load, then it is an emitter follower. If the second transistor has the base resistor(s) bypassed with a capacitor, then the second transistor is a common base circuit. So the first transistor is driving a common base transistor in parallel with the "tail" resistor. These are nice low signal amplifiers.
  4. Aug 27, 2009 #3
    By looking through some references I managed to find a really elegant derivation of the gain of a differential amplifier...I never thought the hyperbolic tangent function would show up here! I'll write it out since I need practice with LaTeX and it might help someone who is working through the same material. :biggrin:

    [tex]I_e = e^\frac{Vbe}{V_t}[/tex]


    [tex]V_{be} = V_t ln I_e[/tex]

    [tex]V_{dif} = V_{be1} - V_{be2} = V_t ln I_{e1} - V_t ln I_{e2} =
    Vt ln \frac{I_e1}{I_e2}[/tex]


    [tex]e^\frac{V_{dif}}{V_t} = \frac{I_e1}{I_e2}[/tex]

    [tex]\frac{I_{c1}}{I_{c2}} = \frac{\alpha I_{e1}}{\alpha I_{e2}} = \frac{I_{e1}}{I_{e2}}[/tex]

    The differential current ratio is:

    [tex]\frac{I_{c1} - I_{c2}}{I_{c1} + I_{c2}} = \frac{e^\frac{V_{dif}}{V_t} - 1}{e^\frac{V_{dif}}{V_t} + 1} = tanh(\frac{V_{diff}}{2V_t})[/tex] :eek:

    So by multiplying the differential current ratio top and bottom by the collector load resistance we can do this:

    [tex]\frac{R_c(I_{c1} - I_{c2})}{Rc(I_{c1} + I_{c2})} = \frac{V_{o1} - V_{o2}}{Rc(I_{c1} + I_{c2})} = \frac{V_{od}}{R_c \alpha I_o}[/tex]

    Where Io is the current through both emitters of the differential pair and alpha is the common base current gain.

    Setting the two equations equal we finally have:

    [tex]\frac{V_{od}}{R_c \alpha I_o} = tanh(\frac{V_{dif}}{2V_t})[/tex]

    [tex]V_{od} = R_c \alpha I_o tanh(\frac{V_{dif}}{2V_t})[/tex].

    For small signals, [tex]tanh(\frac{V_{dif}}{2V_t}) \approx \frac{V_{dif}}{2V_t}[/tex] and alpha can be taken to be 1, so we get for small signal gain:

    [tex] V_{od} = R_c\frac{I_o}{2V_t} [/tex].

    For large signals, the transfer function of the amplifier behaves just like the hyperbolic tangent function: it's linear in a small region around the quiescent point, but goes asymptotic as the voltage increases or decreases beyond this linear region and the input transistor saturates or goes into cutoff.
  5. Aug 28, 2009 #4
    hello bitrex-
    This a nice derivation. I do have some comments. You are using Vt which is about 26 millivolts at room temperature. It is actually kBT/q (Boltzmanns constant, temperature, electron charge). Some of your subscripts are not subscripts. This derivation I think applies to differential inputs and outputs, which is not always the case. I usually use only one (base) input and one (collector) output, for example. The large common emitter tail resistance is more important in this case. It does not appear at all in your derivation because you assume that the emitter tail current is constant (current sink). I have sometimes actually used an npn transistor in the tail to make a constant current sink.
    Bob S
  6. Aug 28, 2009 #5


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    Science Advisor

    It's quite common to use it like that Bob. BTW it can be considered as CC CB cascade in that case.
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