AC Analysis

Abstract

The AC analysis deals with the small-signal, linear and time-invariant circuits. Nonlinear circuits are first linearized using a method such as Taylor expansion. The circuit transfer functions are then calculated in s-domain from the Admittance matrix using Cramer’s method combined with the Laplace expansion. Important information such as stability, bandwidth, gain, group delay, poles and zeros are obtained from transfer functions. Using a node-elimination algorithm, circuits are simplified to two input and output nodes and the resulting 2-port model is used to extract the power gains, the maximum power, the inductive or the capacitive behavior, the quality factor and the matching characteristics of the circuits using the reduced-order admittance, impedance or scattering parameters.

Questions

Answer the following questions using materials and codes you have learned in this chapter.

4.1.1 Q.4.1

Eliminate the node 5 of the X-shaped circuit and convert it to the square-shaped circuit. Find the relationship among the components (Fig. 4.109).

Fig. 4.109

Circuit of Question 1

Answers:

${G}_a=\frac{G_1{G}_2}{G_1+{G}_2+{G}_3+{G}_4}$, ${G}_b=\frac{G_1{G}_4}{G_1+{G}_2+{G}_3+{G}_4}$, ${G}_c=\frac{G_2{G}_3}{G_1+{G}_2+{G}_3+{G}_4}$ and ${G}_d=\frac{G_3{G}_4}{G_1+{G}_2+{G}_3+{G}_{4.}}$

4.1.2 Q.4.2

Eliminate the node 5 of the X-shaped circuit and convert it to a 4-node circuit. Find the relationship among the components and draw the circuit (Fig. 4.110).

Fig. 4.110

Circuit of Question 2

Answers:

The shunt component at the eliminated node of the original circuit appears as modified shunt components at all nodes of the reduced-order circuit (Fig. 4.111).

Fig. 4.111

Circuit of Question 2 Answer

${G}_a=\frac{G_0}{5}$ and ${G}_b=\frac{G_0}{5}$

4.1.3 Q.4.3

Create the small-signal AC model of the following circuit and find the admittance matrix using KCL equations (Fig. 4.112).

Fig. 4.112

Circuit of Question 3

Answers: See Fig. 4.113.

Fig. 4.113

Small-signal AC model of Question 3

$$ {\mathrm{Y}}_{\mathrm{a}}={\mathrm{G}}_1+{\mathrm{G}}_2+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{gs}1} $$

$$ {\mathrm{Y}}_{\mathrm{b}}=\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{gd}1} $$

$$ {\mathrm{Y}}_{\mathrm{c}}={\mathrm{g}}_{\mathrm{ds}1}+{\mathrm{G}}_{\mathrm{D}}+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{db}1} $$

KCL(1): i1 == g*(v1-v2)

KCL(2): 0 == g*(v2-v1) + Ya*v2 + Yb*(v2-v3)

KCL(3): 0 == gm1*v2 + Yc*v3 + Yb*(v3-v2)

4.1.4 Q.4.4

Create the small-signal AC model of the following circuit and find the admittance matrix using KCL equations (Fig. 4.114).

Fig. 4.114

Circuit of Question 4

Answers: See Fig. 4.115.

Fig. 4.115

Small-Signal AC Model of Question 4

$$ {\mathrm{Y}}_{\mathrm{a}}={\mathrm{G}}_2+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{gs}1} $$

$$ {\mathrm{Y}}_{\mathrm{b}}={\mathrm{G}}_1+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{gd}1} $$

$$ {\mathrm{Y}}_{\mathrm{c}}={\mathrm{g}}_{\mathrm{ds}1}+{\mathrm{G}}_{\mathrm{D}}+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{db}1} $$

KCL(1): i1 == g*(v1-v2)

KCL(2): 0 == g*(v2-v1) + Ya*v2 + Yb*(v2-v3)

KCL(3): 0 == gm1*v2 + Yc*v3 + Yb*(v3-v2)

4.1.5 Q.4.5

Create the small-signal AC model of the following circuit and find the admittance matrix using KCL equations (Fig. 4.116).

Fig. 4.116

Circuit of Question 5

Answers: See Fig. 4.117.

Fig. 4.117

Small-signal AC model of Question 5

$$ {\mathrm{Y}}_{\mathrm{a}}={\mathrm{G}}_1+{\mathrm{G}}_2+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{gd}1} $$

$$ {\mathrm{Y}}_{\mathrm{b}}=\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{gs}1} $$

$$ {\mathrm{Y}}_{\mathrm{c}}={\mathrm{g}}_{\mathrm{b}1}+{\mathrm{g}}_{\mathrm{ds}1}+{\mathrm{G}}_{\mathrm{S}}+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{sb}1} $$

KCL(1): i1 == g*(v1-v2)

KCL(2): 0 == g*(v2-v1) + Ya*v2 + Yb*(v2-v3)

KCL(3): 0 == gm1*(v3-v2) + Yc*v3 + Yb*(v3-v2)

4.1.6 Q.4.6

Create the small-signal AC model of the following circuit and find the admittance matrix using KCL equations (Fig. 4.118).

Fig. 4.118

Circuit of Question 6

Answers: See Fig. 4.119.

Fig. 4.119

Small-signal AC model of Question 6

$$ {\mathrm{Y}}_{\mathrm{a}}={\mathrm{G}}_{\mathrm{S}}+\mathrm{j}\upomega \left({\mathrm{C}}_{\mathrm{gs}1}+{\mathrm{C}}_{\mathrm{sb}1}\right) $$

$$ {\mathrm{Y}}_{\mathrm{b}}={\mathrm{g}}_{\mathrm{ds}1} $$

$$ {\mathrm{Y}}_{\mathrm{c}}={\mathrm{G}}_{\mathrm{D}}+\mathrm{j}\upomega \left({\mathrm{C}}_{\mathrm{gd}1}+{\mathrm{C}}_{\mathrm{db}1}\right) $$

KCL(1): i1 == g*(v1-v2)

KCL(2): 0 == g*(v2-v1) + Ya*v2 + Yb*(v2-v3) + (gm1 + gb1)*v2

KCL(3): 0 == − (gm1 + gb1)*v2 + Yb*(v3-v2) + Yc*v3

4.1.7 Q.4.7

Create the small-signal AC model of the following circuit and find the admittance matrix using KCL equations (Fig. 4.120).

Fig. 4.120

Circuit of Question 7

Answers: See Fig. 4.121.

Fig. 4.121

Small-signal AC model of Question 7

$$ {\mathrm{Y}}_{\mathrm{a}}={\mathrm{G}}_{\mathrm{S}}+\mathrm{j}\upomega \left({\mathrm{C}}_{\mathrm{gs}1}+{\mathrm{C}}_{\mathrm{sb}1}\right) $$

$$ {\mathrm{Y}}_{\mathrm{b}}={\mathrm{g}}_{\mathrm{ds}1} $$

$$ {\mathrm{Y}}_{\mathrm{c}}={\mathrm{G}}_{\mathrm{D}}+{\mathrm{G}}_1+\mathrm{j}\upomega \left({\mathrm{C}}_{\mathrm{gd}1}+{\mathrm{C}}_{\mathrm{db}1}\right) $$

KCL(1): i1 == g*(v1-v2)

KCL(2): 0 == g*(v2-v1) + Ya*v2 + Yb*(v2-v3) + (gm1 + gb1)*v2

KCL(3): 0 == − (gm1 + gb1)*v2 + Yb*(v3-v2) + Yc*v3

4.1.8 Q.4.8

Create the small-signal AC model of the following circuit and find the admittance matrix using KCL equations (Fig. 4.122).

Fig. 4.122

Circuit of Question 8

Answers: See Fig. 4.123.

Fig. 4.123

Small-signal AC Model of Question 8

$$ {\mathrm{Y}}_{\mathrm{a}}={\mathrm{G}}_1+{\mathrm{G}}_2 $$

$$ {\mathrm{Y}}_{\mathrm{b}}=\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{gs}1} $$

$$ {\mathrm{Y}}_{\mathrm{c}}={\mathrm{G}}_{\mathrm{S}}+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{sb}1} $$

$$ {\mathrm{Y}}_{\mathrm{d}}={\mathrm{g}}_{\mathrm{d}\mathrm{s}1} $$

$$ {\mathrm{Y}}_{\mathrm{e}}={\mathrm{G}}_{\mathrm{D}}+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{db}1} $$

$$ {\mathrm{Y}}_{\mathrm{f}}=\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{gd}1} $$

KCL(1): i1 == g*(v1-v2)

KCL(2): 0 == g*(v2-v1) + Ya*v2 + Yb*(v2-v3) + Yf*(v2-v4)

KCL(3): 0 == gm1*(v3-v2) + gb1*v3 + Yb*(v3-v2) + Yc*v3 + Yd*(v3-v4)

KCL(4): 0 == gm1*(v2-v3)-gb1*v3 + Yd*(v4-v3) + Ye*v4 + Yf*(v4-v2)

4.1.9 Q.4.9

Create the small-signal AC model of the following circuit and find the admittance matrix using KCL equations (Fig. 4.124).

Fig. 4.124

Circuit of Question 9

Answers: See Fig. 4.125.

Fig. 4.125

Small-signal AC model of Question 9

$$ {\mathrm{Y}}_{\mathrm{a}}={\mathrm{G}}_2 $$

$$ {\mathrm{Y}}_{\mathrm{b}}=\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{gs}1} $$

$$ {\mathrm{Y}}_{\mathrm{c}}={\mathrm{G}}_{\mathrm{S}}+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{sb}1} $$

$$ {\mathrm{Y}}_{\mathrm{d}}={\mathrm{g}}_{\mathrm{d}\mathrm{s}1} $$

$$ {\mathrm{Y}}_{\mathrm{e}}={\mathrm{G}}_{\mathrm{D}}+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{db}1} $$

$$ {\mathrm{Y}}_{\mathrm{f}}={\mathrm{G}}_1+\mathrm{j}\upomega {\mathrm{C}}_{\mathrm{gd}1} $$

KCL(1): i1 == g*(v1-v2)

KCL(2): 0 == g*(v2-v1) + Ya*v2 + Yb*(v2-v3) + Yf*(v2-v4)

KCL(3): 0 == gm1*(v3-v2) + gb1*v3 + Yb*(v3-v2) + Yc*v3 + Yd*(v3-v4)

KCL(4): 0 == gm1*(v2-v3)-gb1*v3 + Yd*(v4-v3) + Ye*v4 + Yf*(v4-v2)