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In Analog CMOS integrated circuit book, page 696, edition 2, it is stated that:

In order to restore the dynamic range, the transconductance of the transistors must be increased by a factor of square of x because thermal noise and currents scaled with root of gm(transconductance).

I have heard that the lower limit is thermal noise and the higher limit is supply voltage. Our supply voltage has been decreased by x, So how does dynamic range increase according to the above statement? Do I misunderstand the concept of dynamic range?

UPDATE

A more detailed quote:

The greatest impact of scaling on analog circuits is the reduction of the supply voltage. With ideal scaling, the maximum allowable voltage swings decrease by a factor of α, lowering the dynamic range1 of the circuit. For example, if the lower end of the dynamic range is limited by thermal noise, then scaling VDD by α decreases the dynamic range by the same factor because gm and hence thermal noise remain constant. Of course, since for analog circuits (VDD/α)(IDD/α) = VDD.IDD/(α)^2, the power dissipation drops by α^2. In order to restore the dynamic range, the transconductance of the transistors must be increased by a factor of α^2 because thermal noise voltages and currents scale with √gm. Thus, since voltage scaling requires that VGS − VT H decrease by a factor of α, we note from gm = 2×ID/(VGS − VTH ) that ID must increase by the same factor, leading to a power dissipation of (VDD/α)(α×ID) = VDD×ID. Also, from gm = μCox×(W/L×(VGS−VTH ), we conclude that if Cox is scaled up by α and L and VGS−VTH are scaled down by α, then W must increase by α (whereas in ideal scaling, it would decrease by this factor). That is, for a constant (thermal-noise limited) dynamic range, ideal scaling of linear analog circuits requires a constant power dissipation and a higher device capacitance, e.g., (α×W)(L/α)(α×Cox ) = αWLCox . Interestingly, if the lower end of the dynamic range is determined by kT/C noise, then to maintain a constant slew rate in switched-capacitor circuits, the bias current must scale up by a factor of α^2, resulting in an increase in the power dissipation. (Problem 17.17.3)

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    \$\begingroup\$ can you give a longer quote ? I dont see how that quote implies that if the supply decreases by x then the dynamic range increases \$\endgroup\$
    – Rahmany
    Oct 10, 2022 at 9:24
  • \$\begingroup\$ Hello @Rahmany, done! \$\endgroup\$ Oct 10, 2022 at 9:49

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Thank you for the detailed quote.

Our supply voltage has been decreased by x. So, how does dynamic range increase according to the above statement?

I think you haven't understood the paragraph well.

Let's define the dynamic range loosely as \$ DR = \frac{V_{DD}}{V_{OS}}\$ where \$ V_{DD} \$ is the supply voltage and \$ V_{OS}\$ is the noise level.

When \$ V_{DD} \$ is scaled by \$ \alpha \$ then \$ DR \$ is also scaled by \$ \alpha \$ as long as the transistor hasn't changed its parameters. As it states:

For example, if the lower end of the dynamic range is limited by thermal noise, then scaling VDD by α decreases the dynamic range by the same factor because gm and hence thermal noise remain constant

Then paragraph states that if we want to preserve the old dynamic range of \$\frac{V_{DD}}{V_{OS}}\$ , then the noise level should, also, be scaled by \$ \alpha \$. And since \$ V_{OS}\$ scales with \$\sqrt{g_m}\$, If we scale \$ g_m\$ by \$ \alpha^2 \$. Then the new dynamic range \$ DR_{new} \$ becomes:

\$ DR_{new} = \frac{V_{DD}'}{V_{OS}'} = \frac{\alpha V_{DD}}{\alpha V_{OS}}=\frac{V_{DD}}{V_{OS}}=DR\$

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  • \$\begingroup\$ Thank you @Rahmany , not only I do not the definition of dynamic range, but also I did not know the relationship between transconductance and thermal noise. \$\endgroup\$ Oct 10, 2022 at 12:23

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