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## Q. 8.2

We wish to compare the values of gm, Rin, Ro, and A0 for a CS amplifier that is designed using an NMOS transistor with L = 0.4 μm and W = 4 μm and fabricated in a 0.25-μm technology specified to have μnCox = 267 μA/V² and $V_{A}^{′}$ = 10 V/μm, with those for a CE  amplifier designed using a BJT fabricated in a process with β = 100 and VA = 10 V. Assume that both devices are operating at a drain (collector) current of 100 μA.

## Verified Solution

For simplicity, we shall neglect the Early effect in the MOSFET in determining VOV ; thus,

$I_{D} = \frac{1}{2} (μ_{n}C_{ox}) \left(\frac{W}{L}\right) V_{OV}^{2}$

$100 = \frac{1}{2} × 267 × \left(\frac{4}{0.4}\right) V_{OV}^{2}$

resulting in

VOV = 0.27 V

$g_{m} = \frac{2I_{D}}{V_{OV}} = \frac{2 × 0.1}{0.27} = 0.74 mA/V$

Rin = ∞

$r_{o} = \frac{V_{A}^{′} L}{I_{D}} = \frac{10 × 0.4}{0.1} = 40 kΩ$

Ro = ro = 40 kΩ

A0 = gmro = 0.74 × 40 = 29.6 V/V

For the CE amplifier we have

$g_{m} = \frac{I_{C}}{V_{T}} = \frac{0.1 mA}{0.025 V} = 4 mA/V$

$R_{in} = r_{π} = \frac{β}{g_{m}} = \frac{100}{4} = 25 kΩ$

$r_{o} = \frac{V_{A}}{I_{C}} = \frac{10}{0.1} = 100 kΩ$

Ro = ro = 100 kΩ

A0 = gmro = 4 × 100 = 400 V/V