Determine Zi, Zo, and Vo for the network of Fig. 8.77 if Vi = 4 mV.FIG. 8.77

Calculate drain current \(\left(I_{D}\right)\) at \(V_{G S}=\frac{V_{p}}{2}\).

\(\begin{aligned} V_{G S} &=\frac{V_{p}}{2} \\ &=-1.4 \mathrm{~V} \\ I_{D} &=(8 \mathrm{~mA})\left[1-\frac{\left(\frac{-2.8}{2}\right)}{(-2.8)}\right]^{2} \\ &=2 \mathrm{~mA} \end{aligned}\)

Calculate drain current \(\left(I_{D}\right)\) at \(V_{G S}=0.3 V_{P} .\)

\(V_{G S}=0.3 V_{P}\)

\(\begin{aligned} &=-0.84 \mathrm{~V} \\ I_{D} &=(8 \mathrm{~mA})\left[1-\frac{(0.3)(-2.8)}{(-2.8)}\right]^{2} \\ &=3.92 \mathrm{~mA} \\ & \cong 4 \mathrm{~mA} \end{aligned}\)

The transfer curve for the device is shown in Figure 1:

Figure 1: Required schematic

To sketch load line locates two coordinates in figure 1 and draw load line as follows:

Calculate gate source voltage \(\left(V_{G s_{Q}}\right)\)

\(V_{G S}=-I_{D} R_{s} \ldots \ldots\) (1)

Here,

Source resistance \(\left(R_{S}\right)\) is \(1.5 \mathrm{k} \Omega\)

Let \(I_{D}=0\)

Substitute \(1.5 \mathrm{k} \Omega\) for \(R_{S}\) and 0 for \(I_{D}\) in equation (1).

\(V_{G S}=-(0)(1.5 \mathrm{k} \Omega)\)

\(=0 \mathrm{~V}\)

Therefore the gate source voltage corresponding to \(0 \mathrm{~mA}\) drain current is \(0 \mathrm{~V}\).

Again solve equation (1) for \(I_{D}=4 \mathrm{~A}\).

$$ \begin{aligned} V_{G S} &=-(4 \mathrm{~mA})(1.5 \mathrm{k} \Omega) \\ &=-6 \mathrm{~V} \end{aligned} $$

Therefore the gate source voltage corresponding to \(4 \mathrm{~mA}\) drain current is \(-6 \mathrm{~V}\)

The point of intersection of the load line and a transverse curve is defined as \(Q\) -point.

Here, the load line cuts the transverse curve at \(V_{G S_{Q}} \cong-1.75 \mathrm{~V}\).

The required graph is shown in Figure 2:

Figure 2 Required schematic.

Calculate transconductance \(\left(g_{m o}\right):\)

$$ g_{m o}=\frac{2 I_{D S S}}{\left|V_{P}\right|} $$

Here,

Drain saturation current \(\left(I_{D S s}\right)\) is \(8 \mathrm{~mA}\)

Pinch-off voltage \(\left(V_{P}\right)\) is \(-2.8 \mathrm{~V}\)

Substitute \(-2.8 \mathrm{~V}\) for \(V_{P}\) and \(8 \mathrm{~mA}\) for \(I_{D s s}\) in the expression of transconductance \(\left(g_{m o}\right)\)

$$ \begin{aligned} g_{m o} &=\frac{2(8 \mathrm{~mA})}{|-2.8 \mathrm{~V}|} \\ &=\frac{16 \mathrm{~mA}}{2.8 \mathrm{~V}} \\ &=5.714 \mathrm{mS} \end{aligned} $$

Calculate transconductance \(\left(g_{m}\right)\)

$$ g_{m}=g_{m o}\left(1-\frac{V_{G S_{\rho}}}{V_{P}}\right) $$

Substitute \(-2.8 \mathrm{~V}\) for \(V_{P}\) and \(-1.75 \mathrm{~V}\) for \(V_{G S_{Q}}\) and \(5.714 \mathrm{mS}\) for \(g_{m o}\) in the expression of transconductance \(\left(g_{m}\right) .\)

$$ \begin{aligned} g_{m} &=(5.714 \mathrm{mS})\left[1-\frac{(-1.75 \mathrm{~V})}{(-2.8 \mathrm{~V})}\right] \\ &=(5.714 \mathrm{mS})(1-0.625) \\ &=(5.714 \mathrm{mS})(0.375) \\ &=2.14 \mathrm{mS} \end{aligned} $$

Check approximation condition

$$ r_{d} \geq 10 R_{D} $$

Here,

Small signal input impedance \(r_{d}\) is \(40 \mathrm{k} \Omega\)

Drain resistance \(R_{D}\) is \(3.3 \mathrm{k} \Omega\)

Substitute \(3.3 \mathrm{k} \Omega\) for \(R_{D}\) and \(40 \mathrm{k} \Omega\) for \(r_{d}\) in approximation condition

$$ \begin{array}{l} 40 \mathrm{k} \Omega \geq 10(3.3 \mathrm{k} \Omega) \\ 40 \mathrm{k} \Omega \geq 33 \mathrm{k} \Omega \end{array} $$

Calculate input impedance \(\left(Z_{i}\right)\)

$$ \begin{aligned} Z_{i}=& R_{S} \| \frac{1}{g_{m}} \\ =& \frac{\left(R_{S}\right)\left(\frac{1}{g_{m}}\right)}{R_{S}+\frac{1}{g_{m}}} \end{aligned} $$

Substitute \(1.5 \mathrm{k} \Omega\) for \(R_{S}\) and \(2.14 \mathrm{mS}\) for \(g_{m}\) in the expression of input impedance \(\left(Z_{i}\right)\)

$$ \begin{aligned} Z_{i} &=\frac{(1.5 \mathrm{k} \Omega)(467 \Omega)}{1.5 \mathrm{k} \Omega+(467 \Omega)} \\ &=356.3 \Omega \end{aligned} $$

Hence, the input impedance \(\left(Z_{i}\right)\) is \(356.3 \Omega\).

As \(r_{d} \geq 10 R_{D}\), therefore output impedance \(\left(Z_{o}\right)\) is equal to drain resistance \(\left(R_{D}\right)\).

\(Z_{o}=R_{D}\)

Substitute \(3.3 \mathrm{k} \Omega\) for \(R_{D}\) in the expression of output impedance \(\left(Z_{o}\right)\).

\(Z_{o}=3.3 \mathrm{k} \Omega\)

Hence, the output impedance is \(3.3 \mathrm{k} \Omega\).

As \(r_{d} \geq 10 R_{D}\), therefore calculate voltage gain \(\left(A_{v}\right)\)

\(A_{v}=g_{m} R_{D}\)

Substitute \(3.3 \mathrm{k} \Omega\) for \(R_{D}\) and \(2.14 \mathrm{mS}\) for \(g_{m}\) in the expression of voltage gain \(\left(A_{v}\right)\).

\(\begin{aligned} A_{v} &=(2.14 \mathrm{mS})(3.3 \mathrm{k} \Omega) \\ &=7.062 \end{aligned}\)

Hence, the voltage gain \(\left(A_{v}\right)\) is \(7.062\).

Calculate output voltage \(\left(V_{o}\right)\)

\(V_{o}=A_{v} V_{i}\)

Here,

Input voltage \(\left(V_{i}\right)\) is \(4 \mathrm{~V}\).

Substitute \(4 \mathrm{~V}\) for \(V_{i}\) and \(7.062 \mathrm{~V}\) for \(A_{v}\) in the expression of \(\left(V_{o}\right)\)

\(V_{o}=(7.06)(4 \mathrm{~V})\)

\(=28.248 \mathrm{~V}\)

Hence, the output voltage \(\left(V_{o}\right)\) is \(28.248 \mathrm{~V}\).