Figure 163$0.18-\mu \mathrm{m}$ Technology Having$\left|V_{t}\right|=0.4 \mathrm{~V}$ $k_{n}^{\prime}=0.4 \mathrm{~mA} / \mathrm{V}^{2}$

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The Answer is $(a) For matching, \begin{aligned} k_{n} & =k_{p} \ \mu_{n} C_{o x}\left(\frac{W}{L} ight)_{N} & =\mu_{p} C_{o x}\left(\frac{W}{L} ight)_{P} \ \mu_{n} imes \frac{0.27}{0.18} & =\frac{1}{3} \mu_{n} imes \frac{W_{p}}{0.18} \ \Rightarrow W_{p} & =3 imes 0.27=0.81 \mu \mathrm{m} \end{aligned} (b) For v_{I}=1.8 \mathrm{~V}, Q_{N} will be operating in the triode region: i_{D N}=k_{n}\left[\left(v_{G S}-V_{t n} ight) v_{D S}-\frac{1}{2} v_{D S}^{2} ight] For v_{O}=0.1 \mathrm{~V} , \begin{aligned} i_{D N} & =0.4 imes \frac{0.27}{0.18}\left[(1.8-0.4) imes 0.1-\frac{1}{2} imes 0.1^{2} ight] \ & =0.081 \mathrm{~mA} \ & =81 \mu \mathrm{A} \end{aligned}$
(a) For matching,

\begin{aligned}k_{n} & =k_{p} \\mu_{n} C_{o x}\left(\frac{W}{L} ight)_{N} & =\mu_{p} C_{o x}\left(\frac{W}{L} ight)_{P} \\mu_{n} imes \frac{0.27}{0.18} & =\frac{1}{3} \mu_{n} imes \frac{W_{p}}{0.18} \\Rightarrow W_{p} & =3 imes 0.27=0.81 \mu \mathrm{m}\end{aligned}

(b) For

v_{I}=1.8 \mathrm{~V}, Q_{N}

will be operating in the triode region:

i_{D N}=k_{n}\left[\left(v_{G S}-V_{t n} ight) v_{D S}-\frac{1}{2} v_{D S}^{2} ight]

For

v_{O}=0.1 \mathrm{~V}

,

\begin{aligned}i_{D N} & =0.4 imes \frac{0.27}{0.18}\left[(1.8-0.4) imes 0.1-\frac{1}{2} imes 0.1^{2} ight] \& =0.081 \mathrm{~mA} \& =81 \mu \mathrm{A}\end{aligned}

Figure 163 $0.18-\mu \mathrm{m}$ Technology Having $\left|V_{t}\right|=0.4 \mathrm{~V}$ $k_{n}^{\prime}=0.4 \mathrm{~mA} / \mathrm{V}^{2}$

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Figure 163 0.18−μm0.18-\mu \mathrm{m}0.18−μm Technology Having ∣Vt∣=0.4 V\left|V_{t}\right|=0.4 \mathrm{~V}∣Vt​∣=0.4 V kn′=0.4 mA/V2k_{n}^{\prime}=0.4 \mathrm{~mA} / \mathrm{V}^{2}kn′​=0.4 mA/V2

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Figure 163 $0.18-\mu \mathrm{m}$ Technology Having $\left|V_{t}\right|=0.4 \mathrm{~V}$ $k_{n}^{\prime}=0.4 \mathrm{~mA} / \mathrm{V}^{2}$

Correct Answer:
Verified