Question 10.6: Consider an IC CS amplifier fed with a source having Rsig = ...

The low-frequency gain A_M is given by

A_{M} = −g_{m}R_{L}^{′} = −g_{m} (r_{o} || r_{o})

Thus,

A_{M} = −\frac{1}{2} g_{m}r_{o} = −\frac{1}{2} × 1.25 × 20

= −12.5 V/V

The 3-dB frequency f_H can be found using Eq. (10.67),

f_{H} = \frac{1}{2π (C_{L}  +  C_{gd}) R_{L}^{′}}                                (10.67)

= \frac{1}{2π (25  +  5) × 10^{−15} × 10 × 10^{3}}

= 530.5 MHz

and the unity-gain frequency, which is equal to the gain–bandwidth product, can be determined as

f_t = |A_M| f_H = 12.5 × 530.5 = 6.63 GHz

The frequency of the zero is obtained using Eq. (10.66) as

f_{Z} = \frac{g_{m}}{2π  C_{gd}}                        (10.66)

f_{Z} =\frac{1}{2π} \frac{g_{m}}{C_{gd}}

=\frac{1}{2π} \frac{1.25 × 10^{−3} }{5 × 10^{−15}} \simeq 40  GHz

Now, to double V_OV , I_D must be quadrupled. The new values of g_m and R_{L}^{′} can be found as follows:

g_{m} = \frac{I_{D}}{V_{OV}/2} = 2.5  mA/V

R_{L}^{′} = \frac{1}{4} × 10 = 2.5  kΩ

Thus the new value of A_M becomes

A_{M} = −g_{m}R_{L}^{′} = −2.5 × 2.5 = −6.25  V/V

That of f_H becomes

f_{H} = \frac{1}{2π (C_{L}  +  C_{gd}) R_{L}^{′}}

\frac{1}{2π (25  +  5) × 10^{−15} × 2.5 × 10^{3}}

= 2.12 GHz

and the unity-gain frequency (i.e., the gain–bandwidth product) becomes

f_t = 6.25 × 2.12 = 13.3 GHz

We note that doubling V_OV results in reducing the dc gain by a factor of 2 and increasing the bandwidth by a factor of 4. Thus, the gain–bandwidth product is doubled—a good bargain!

Finally, the frequency of the transmission zero f_Z will be doubled, becoming 80 GHz.