Question 10.10: This example illustrates the advantages of cascoding by comp...

This example illustrates the advantages of cascoding by comparing the performance of a cascode amplifier with that of a common-source amplifier in two cases:

(a) The resistance of the signal source is significant, Rsig = 10 kΩ.

(b) Rsig is negligibly small.

Assume all MOSFETs have gm = 1.25 mA/V, ro = 20 kΩ, Cgs = 20 fF, Cgd = 5 fF, Cdb = 5 fF, and CL (excluding Cdb) = 10 fF. For case (a), let RL = ro = 20 kΩ for both amplifiers. For case (b), let RL = ro = 20 kΩ for the CS amplifier and RL = Ro for the cascode amplifier. For all cases, determine Av, fH, and ft .

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

(a) For the CS amplifier:

A0 = gmro = 1.25 × 20 = 25 V/V

Av = −gm (RL || ro) = −gm (ro || ro)

= −\frac{1}{2} A_{0} = −12.5  V/V

To obtain τH we use Eq. (10.87) and note that Rsig = R_{sig}^{′} and that here CL does not include Cdb, thus

τ_{H} = b_{1} = C_{gs}R_{gs} + C_{gd}R_{gd} + C_{L}R_{C_{L}}                           (10.87)

τ_{H} = C_{gs}R_{sig} + C_{gd} [(1 + g_{m}R_{L}^{′})R_{sig} + R_{L}^{′}] + (C_{L} + C_{db}) R_{L}^{′}

where

R_{L}^{′} = r_{o}  ||  R_{L}= r_{o}  ||  r_{o} = 10  kΩ

τH = 20 × 10 + 5 [(1 + 12.5) 10 + 10] + (10 + 5)10

= 200 + 725 + 150 = 1075 ps

Thus,

f_{H} = \frac{1}{2π × 1075 × 10^{−12}} = 148  MHz

ft = |Av| fH = 12.5 × 148 = 1.85 GHz

For the cascode amplifier:

Ro = 2ro + (gmro) ro = (2 × 20) + (25 × 20) = 540 kΩ

Av = −gm (Ro || RL)

= −1.25 (540 || 20) = −24.1 V/V

R_{in2} = \frac{r_{o}  +  R_{L}}{g_{m}r_{o}} = \frac{r_{o}  +  r_{o}}{g_{m}r_{o}}  = \frac{2}{g_{m}} = \frac{2}{1.25} = 1.6  kΩ

Rd1 = ro || Rin2 = 20 || 1.6 = 1.48 kΩ

Using Eq. (10.109),

τH = Rsig [Cgs1 + Cgd1 (1 + gm1Rd1)] + Rd1 (Cgd1 + Cdb1 + Cgs2) + (RL || Ro) (CL + Cgd2)                            (10.109)

τH = Rsig [Cgs1 + Cgd1 (1 + gm1Rd1)]

+ Rd1 (Cgd1 + Cdb1 + Cgs2)

+ (RL || Ro) (CL + Cdb2 + Cgd2)

= 10 [20 + 5 (1 + 1.25 × 1.48)]

+1.48 (5 + 5 + 20)

+ (20 || 540) (10 + 5 + 5)

= 342.5 + 44.4 + 385.7

= 772.6 ps

f_{H} = \frac{1}{2π × 772.6 × 10^{−12}} = 206  MHz

ft = 24.1 × 206 = 4.96 GHz

Thus cascoding has increased both the dc gain and the 3-dB frequency, with the combined effect being an increase of ft by a factor of 2.7.
(b) For the CS amplifier:

Av = −12.5 V/V

τ_{H} = (C_{gd} + C_{L} + C_{db}) R_{L}^{′}

= (5 + 10 + 5) 10 = 200 ps

f_{H} = \frac{1}{2π × 200 × 10^{−12}} = 796  MHz

ft = 12.5 × 796 = 9.95 GHz

For the cascode amplifier:

RL = Ro = 540 kΩ

Av = −gm (Ro || RL)

= −1.25 (540 || 540) = −337.5 V/V

R_{in2} = \frac{r_{o}  +  R_{L}}{g_{m}r_{o}} = \frac{20  +  540}{1.25  ×  20} = 22.4 kΩ

Rd1 = ro1 || Rin2 = 20 || 22.4 = 10.6 kΩ

τH = Rd1 (Cgd1 + Cdb1 + Cgs2) + (RL || Ro) (CL + Cgd2 + Cdb2)

= 10.6 (5 + 5 + 20) + (540 || 540) (10 + 5 + 5)

= 318 + 5400 = 5718 ps

f_{H} = \frac{1}{2π × 5718 × 10^{−12}} = 27.8  MHz

ft = 337.5 × 27.8 = 9.39 GHz

Thus cascoding increases the dc gain from 12.5 V/V to 337.5 V/V. This increase has been obtained at the cost of a decrease in fH by approximately the same factor, resulting in the unity-gain frequency (which, in this case, is equal to the gain–bandwidth product) remaining nearly constant.

Related Answered Questions