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Establish the relationship between $g_m$ & $R_{on}$ for MOSFET

Subject: CMOS VLSI Design

Topic: CMOS analog building blocks

Difficulty: Medium

cvd • 1.0k  views
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${I_{D_{sat}}=\frac{1}{2}\frac{W}{L}\mu_n C_{ox}(V_{GS}-V_{Th})^2}$ ${\\}$

${I_{D_{un}}=\frac{1}{2}\frac{W}{L}\mu_n C_{ox}[2(V_{GS}-V{TH})V_{DS}-V_{DS}^2]}$ ${\\}$

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If ${V_{DS}}$ $\lt\lt$ $ {2(V_{GS}-V_{TH})}$ i.e. ${V_{DS}}$ $\lt\lt$ ${2V_0-}$ MOSFET$\Large{-}$ deep triode region.

${\therefore}$ ${I_D \approx \frac{1}{2}\frac{W}{L}\mu_n C_{ox}[(V_{GS}-V_{TH})]V_{DS}}$

${R_{on}=\frac{V_{DS}}{I_D}=\frac{V_{DS}}{\frac{W}{L}\mu_n C_{ox}[(V_{GS}-V_{TH})]V_{DS}}}$

${R_{on}=\frac{1}{\frac{W}{L}\mu_n C_{ox}[(V_{GS}-V_{TH})]}}$

${R_{on}=\frac{\partial V_{DS}}{\partial I_D}}$

As,${g_m=\frac{\partial I_D}{\partial V_{DS}}={\frac{W}{L}\mu_n C_{ox}[(V_{GS}-V_{TH})]}}$

${\therefore R_{on}=\frac{1}{g_m}}$

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