## Chapter 8

## Q. 8.2

We wish to compare the values of g_{m}, R_{in}, R_{o}, and A_{0} 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 μ_{n}C_{ox} = 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 V_{A} = 10 V. Assume that both devices are operating at a drain (collector) current of 100 μA.

## Step-by-Step

## Verified Solution

For simplicity, we shall neglect the Early effect in the MOSFET in determining V_{OV} ; 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

V_{OV} = 0.27 V

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

R_{in} = ∞

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

R_{o} = r_{o} = 40 kΩ

A_{0} = g_{m}r_{o} = 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

R_{o} = r_{o} = 100 kΩ

A_{0} = g_{m}r_{o} = 4 × 100 = 400 V/V