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In most CMOS applications, the use of resistive source degeneration to reduce noise and improve matching is restricted due to headroom constraints.

A) Could anyone elaborate on this statement extracted from "Ultra High-Compliance CMOS Current Mirrors for Low Voltage Charge Pumps and References" ? Besides, how do we obtain the equation of output noise current = (16/3)KT*gm in the same document ?

B) For http://www.ele.uri.edu/courses/ele447/Slides/CurMir/Chapter_07.pdf#page=78 , the Thevenin resistance would be 1/gmREF in series wth Rb, right ? If so, why there is a factor of (1+gmREF*Rb)2 in the denominator ?

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    \$\begingroup\$ Do you know how resistive degeneration works? If not look up degeneration in current mirrors. This is mostly done in mirrors using BJTs, much less in CMOS (where we scale the transistors instead). Degeneration has the disadvantage of increasing the voltage drop needed at the mirror output, this decreases headroom. When degenerating a CMOS mirror, the gm of the MOS is partly replaced by a resistor. Resistors have less noise than 1/gm of a MOS. So total noise becomes lower. \$\endgroup\$ Commented Oct 19, 2017 at 13:29
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    \$\begingroup\$ Regarding B) the formula is a bit more complex than just the 1/gm and Rb since there's also the capacitor making this it an RC lowpass filter. All the terms are squared (\$x^2\$) since we're dealing with noise currents and voltages. \$\endgroup\$ Commented Oct 19, 2017 at 13:49

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how do we obtain the equation of output noise current = (16/3)KT*gm in the same document

The thermal noise in the drain current in a MOSFET is given by:
$$\overline{\frac{i_{nd}^2}{\Delta f}} = 4kT\gamma g_{d0}$$ For transistor in saturation region \$\gamma = \frac{2}{3}\$ and \$g_{d0} \approx g_m\$, thus noise current (PSD) in each of the transistor of the current mirror is given by \$\frac{8}{3}kTg_{m}\$. Since the current in M1 (in the document you are referring) is mirrored at the output, noise current will similarly get mirrored. Since the two noise sources are uncorrelated their PSD (variances) will add giving a total noise current of \$\frac{16}{3}kTg_{m}\$ at the output.

why there is a factor of (1+gmREF*Rb)2 in the denominator

The input impedance of the \$M_{REF}\$ seen from its drain/gate is \$\frac{1}{g_{mREF}}\$. The voltage at the capacitor will be given by a simple voltage divider equation:
$$V_{C} = \frac{\frac{1}{sC}}{R_B + \frac{1}{g_mREF} + \frac{1}{sC}} = \frac{g_{mREF}}{s(1+g_{mREF}R_B)C_B + g_{mREF}}$$ Thus noise sources from reference current and \$R_B\$ will be filtered by the above transfer function and then will be amplified by \$g_{m1}\$ before getting to the output. The noise from the transistor \$M1\$, though, appears directly at the capacitor output and will not be filtered by this transfer function. Again all the noise sources are uncorrelated so we add the PSD's.

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