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    Improved current mirror

    I do not understand "The circuit on the right is a feedback amplifier with loop gain T. Since all time constants in that loop are of the same order of magnitude, they create a system with several poles"

    Could anyone point me to some maths reference / equations ? Razavi book does not mention about the circuit on the right though.

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    Re: Improved current mirror

    How do I derive the feedback loop gain T = (gm1)*(Rin) ?



    •   Alt12th October 2017, 17:06

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    Re: Improved current mirror

    The improved Wilson current mirror is discussed in Gray/Hurst/Lewis/Meyer Analysis and Design.



    •   Alt12th October 2017, 21:28

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    Re: Improved current mirror

    Thanks. Let me start with 3T Wilson mirror in BJT implementation.

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    A feedback path is thus formed that regulates IC3 so that it is nearly equal to the input current, reducing the systematic gain error caused by finite βF.
    systematic gain error due to finite βF ?



    •   Alt13th October 2017, 05:30

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    Re: Improved current mirror

    if you use a simple bipolar current mirror, the output current is smaller than the input current because the base currents are subtracted from the input current to generate input bias currents for the 2 transistors. The sys gain error is 1-Iout/Iin, and if you can reduce the base currents the gain error will decrease. You can reduce the base currents by increasing the BF, but for bipolar devices it is finite. And gain error will also occur by the Early effect. At this circuit the explanation is similar, but more complicated because of the feedback.



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    Re: Improved current mirror

    The sys gain error is 1-Iout/Iin
    How would I get to this error expression ?



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    Re: Improved current mirror

    Gain Error = Ideal Gain (1) - Real Gain (Iout/Iin)

    For 1:1 current mirror. For 1:M current mirror it is M-Iout/Iin.

    And about Wilson current mirror you can read more here: https://wiki.analog.com/university/c...ext/chapter-11



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    Re: Improved current mirror

    For the loop gain T, you can image that from the drain node of M3, the impedance is 1/gm3 +


    Rin=ro1+ 1/gm3.

    For M1, from its gate to its drain, the gain is gm1*ro1, which is aproximately gm1*Rin. Also, from the drain of M1 to the gate of M3, to the source of M4, the gain is 1, so the total loop gain T is around gm1*Rin.



    •   Alt14th October 2017, 04:48

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    Re: Improved current mirror

    a) Why is the gain from the drain of M1 to the drain of M3 = 1 ?

    b) In most CMOS applications, the use of resistive source degeneration to reduce noise and improve matching is restricted due to headroom constraints. Could anyone elaborate on this statement extracted from "Ultra High-Compliance CMOS Current Mirrors for Low Voltage Charge Pumps and References" ?

    Ultra High-Compliance CMOS Current Mirrors for Low Voltage Charge Pumps and References.pdf
    Last edited by promach; 15th October 2017 at 02:41.



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    Re: Improved current mirror

    Source follower, the gain is around 1.



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