## Q. 2.3

Consider a NMOS transistor biased with $\mathrm{V}_{\mathrm{GS}}=0.65 \mathrm{~V}$ and with the device parameters listed for the 0.18-μm CMOS process in Table 1.5. How much does the drain current change with a 100 mV increase in $V_{t n}$, 5% decrease in $C_{o x}$, and 10% decrease in $\mu_n$?

 Table 1.5  MOSFET parameters repre sentative of v arious C MOS t echnologies and used for rough hand calculations in this text. 0.8 μm 0.35 μm 0.18 μm 45 nm Technology NMOS PMOS NMOS PMOS NMOS PMOS NMOS PMOS $\mu \mathrm{C}_{\mathrm{ox}}\left(\mu \mathrm{A} / \mathrm{V}^2\right)$ 92 30 190 55 270 70 280 70 $\mathrm{V}_{\mathrm{t} 0}(\mathrm{~V})$ 0.80 -0.90 0.57 -0.71 0.45 -0.45 0.45 -0.45 $\lambda \cdot \mathrm{L}(\mu \mathrm{m} / \mathrm{V})$ 0.12 0.08 0.16 0.16 0.08 0.08 0.10 0.15 $\mathrm{C}_{\mathrm{ox}}\left(\mathrm{fF} / \mu \mathrm{m}^2\right)$ 1.8 1.8 4.5 4.5 8.5 8.5 25 25 $\mathrm{t}_{\mathrm{ox}}(\mathrm{nm})$ 18 18 8 8 5 5 1.2 1.2 n 1.5 1.5 1.8 1.7 1.6 1.7 1.85 1.85 $\theta(1 / \mathrm{V})$ 0.06 0.135 1.5 1.0 1.7 1.0 2.3 2.0 m 1.0 1.0 1.8 1.8 1.6 2.4 3.0 3.0 $\mathrm{C}_{\mathrm{ov}} / \mathrm{W}=\mathrm{L}_{\mathrm{ov}} \mathrm{C}_{\mathrm{ox}} (\mathrm{fF} / \mu \mathrm{m})$ 0.20 0.20 0.20 0.20 0.35 0.35 0.50 0.50 $\mathrm{C}_{\mathrm{db}} / \mathrm{W} \approx \mathrm{C}_{\mathrm{sb}} / \mathrm{W} (\mathrm{fF} / \mu \mathrm{m})$ 0.50 0.80 0.75 1.10 0.50 0.55 0.45 0.50

## Verified Solution

Note that since Table 1.5 indicates a threshold voltage of $\mathrm{V}_{\mathrm{tn}}=0.45 \mathrm{~V}, \mathrm{a} \mathrm{V}_{\mathrm{t}}$ increase of  100 mV represents a 50% decrease in $\mathrm{V}_{\mathrm{eff}},\text { from }$ $200 \mathrm{mV} \text { to }$ $100 \mathrm{mV} \text {. }$ Adopting a simple square law device model, a great simplification considering the large process variations present on λ, the nominal device drain current is given by

$\mathrm{I}_{\mathrm{D}}=\frac{1}{2} \mu_{\mathrm{n}} \mathrm{C}_{\mathrm{ox}} \frac{\mathrm{W}}{\mathrm{L}} \mathrm{V}_{\mathrm{eff}}^2$

Accounting for all of the process variations, the new drain current will be

$\mathrm{I}_{\mathrm{D}, \mathrm{new}}=\frac{1}{2}\left(0.9 \mu_{\mathrm{n}}\right)\left(0.95 \mathrm{C}_{\mathrm{ox}}\right) \frac{\mathrm{W}}{\mathrm{L}}\left(0.5 \mathrm{~V}_{\mathrm{eff}}\right)^2=\left(0.9 \cdot 0.95 \cdot 0.5^2\right) \mathrm{I}_{\mathrm{D}}=0.214 \cdot \mathrm{I}_{\mathrm{D}}$

representing a 79% decrease in drain current. Much of this is a result of the $V_t$ variation.