An NMOS transistor fabricated in a $0.18-\mu \mathrm{m}$ CMOS technology has $L=0.54 \mu \mathrm{m}$ and $W=$ $10.8 \mu \mathrm{m}$. The technology is specified to have $\mu_{n} C_{o x}=400 \mu \mathrm{A} / \mathrm{V}^{2}, V_{t n}=0.5 \mathrm{~V}$, and $V_{A}^{\prime}=$ $5 \mathrm{~V} / \mu \mathrm{m}$.
(a) If the device is operated in saturation with an overdrive voltage of $0.2 \mathrm{~V}$, find the required values of $I_{D}$ and $V_{G S}$, along with the resulting values of $g_{m}$ and $r_{o}$. (b) If the gate width is doubled but the value of $V_{O V}$ is maintained, what do the values of $I_{D}, g_{m}$, and $r_{o}$ become?

Question

Quizplus · Accepted Answer

The Answer is (a) Saturation with

V_{O V}=0.2 \mathrm{~V}

,

\begin{aligned}I_{D} & =\frac{1}{2}\left(\mu_{n} C_{O x} ight)(W / L) V_{O V}^{2} \& =\frac{1}{2} imes 0.4 imes(10.8 / 0.54) imes 0.2^{2} \& =0.16 \mathrm{~mA} \V_{G S} & =V_{t n}+V_{O V}=0.5+0.2=0.7 \mathrm{~V}\end{aligned}

\begin{aligned}g_{m} & =\frac{2 I_{D}}{V_{O V}}=\frac{2 imes 0.16}{0.2}=1.6 \mathrm{~mA} / \mathrm{V} \r_{o} & =\frac{V_{A}}{I_{D}}=\frac{V_{A}^{\prime} L}{I_{D}}=\frac{5 imes 0.54}{0.16}=16.9 \mathrm{k} \Omega\end{aligned}

(b) If

W

is doubled while

V_{O V}

is maintained, we see from Eq. (1) that

I_{D}

doubles to

I_{D}=0.32 \mathrm{~mA}

From

g_{m}=\frac{2 I_{D}}{V_{O V}}

we see that

g_{m}

doubles to

g_{m}=3.2 \mathrm{~mA} / \mathrm{V}

Finally, from

r_{o}=\frac{V_{A}}{I_{D}}

we see that

r_{O}

is halved to

r_{o}=8.45 \mathrm{k} \Omega

An NMOS Transistor Fabricated in A 0.18−μm0.18-\mu \mathrm{m}0.18−μm CMOS Technology Has L=0.54μmL=0.54 \mu \mathrm{m}L=0.54μm

An NMOS Transistor Fabricated in A $0.18-\mu \mathrm{m}$ CMOS Technology Has $L=0.54 \mu \mathrm{m}$