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I am reading about noise of cascode amplifier in Razavi book, Design of Analog CMOS Integrated Circuits.

However, I couldn't understand the yellow highlighted sentence below. Could anyone explain it?

Why the noise from M3, M4, M5 and M6 are negligible at low frequencies?

Thank you.

enter image description here

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  • \$\begingroup\$ Is it due to lower channel operating conductance of the cascode current sources? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 2 '16 at 22:41
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    \$\begingroup\$ Not authoritative : but I suspect it's because the amplifier works by acting as a current source to (whatever load impedance) and the cascode devices operate at unity current gain, so there's no amplification of their internal noise. \$\endgroup\$ – Brian Drummond Nov 2 '16 at 23:04
  • \$\begingroup\$ I think it is because M3 to M6 are biased nonlinear loads. Their nonlinearity increases with frequency as well as M1 and M2 input transistors. Frequency response of M1 and M2 delays with increasing frequency, depending on the response of M1 and M2, M3 and M4 also show a delayed response. As you see, nonlinearity actually increase by mutliply as number of the cascoded transistors. But for low frequency, frequency response of all transistors are fine. \$\endgroup\$ – Alper91 Nov 7 '16 at 6:48
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They do not contribute the the noise because of the degeneration. Having emitter/source degeneration in effect introduces negative feedback, greatly improving the performance of the transistors. For a more quantitative discussion see the end of this presentation.

You end up with the following equation for the noise: $$\frac{i_D}{1 + g_m r_o} \approx 0$$ Where \$r_o\$ is the degeneration. For high frequencies \$C_{gs}\$ provides an alternative current path, reducing the degeneration.

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  • \$\begingroup\$ Thanks for the explanation. I have one question. The noise current source don't have any polarity but the presentation calculation seems to assume that it has a specific polarity. So does that effect the result? \$\endgroup\$ – anhnha Nov 8 '16 at 4:14
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    \$\begingroup\$ @anhnha it's just drawn that way. You can change the polarity, if you want. After all, 1V of noise is the same as -1V. \$\endgroup\$ – user110971 Nov 8 '16 at 4:18
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    \$\begingroup\$ @anhnha of course you have to keep it consistent with the transistor equations \$\endgroup\$ – user110971 Nov 8 '16 at 4:22
  • \$\begingroup\$ Thanks. That makes sense. Let's me check it again with reverse polarity. \$\endgroup\$ – anhnha Nov 8 '16 at 4:37

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