Question 10.9: Consider a common-gate amplifier with gm = 1.25 mA/V, ro = 2...

Figure 10.27 shows the CG amplifier circuit at midband frequencies. We note that

v_o = iR_L

v_sig = i(R_sig + R_in)

Thus, the overall voltage gain is given by

G_{v} = \frac{v_{o}}{v_{sig}} = \frac{R_{L}}{R_{sig}  +  R_{in}}

The value of R_in is found from Eq. (10.101) as

R_{in} = \frac{r_{o}  +  R_{L}}{1  +  g_{m} r_{o}} \simeq \frac{r_{o}  +  R_{L}}{g_{m} r_{o}}                          (10.101)

R_{in} = \frac{r_{o}  +  R_{L}}{1  +  g_{m} r_{o}}

= \frac{20  +  20}{1  +  (1.25  ×  20)} = 1.54  kΩ

Thus, G_v can now be determined as

G_{v} = \frac{20}{10  +  1.54} = 1.73  V/V

Observe that as expected G_v is very low. This is due to the fact that the CG amplifier draws a large input current, equal, in fact, to the load current i.

To obtain an estimate of the 3-dB frequency f_H, we first determine R_gs and R_gd using Eqs. (10.100) and (10.102),

R_gs = R_sig || R_in                  (10.100)

R_gs = R_sig || R_in = 10 || 1.54 = 1.33 kΩ

R_gd = R_L || R_o                    (10.102)

where R_o is given by Eq. (10.103),

R_o = r_o + R_sig + (g_mr_o)R_sig                                (10.103)

= 20 + 10 + 25 × 10 = 280 kΩ

Thus,

R_gd = 20 || 280 = 18.7 kΩ

Now we can compute the sum of the open-circuit time constants, τ_H,

τ_H = C_gsR_gs + (C_gd + C_L) R_gd

τ_H = 20 × 10⁻¹⁵ × 1.33 × 10³ + (5 + 25) × 10⁻¹⁵ × 18.7 × 10³

= 26.6 × 10⁻¹² + 561 × 10⁻¹²

= 587.6 ps

and the upper 3-dB frequency f_H can be obtained as

f_{H} = \frac{1}{2π  τ_{H}} = \frac{1}{2π  ×  587.6  ×  10^{-12}} = 270.9  MHz

Observe that f_H is indeed much higher than (about twice) the corresponding value for the CS amplifier found in Example 10.8. Another important observation can be made by examining the two components of τ_H: The contribution of the input circuit is 26.6 ps, while that of the output circuit is 561 ps; thus the limitation on the high-frequency response is posed by the output circuit.