Chapter 4, Problem 4.71P

For the circuit in Figure P4.71, the transistor parameters are: $K_{n1}=K_{n2}=4{\text{mA/V}}^{\text{2}}$ , $V_{TN1}=V_{TN2}=2\text{V}$ , and ${\lambda }_1={\lambda }_2=0$ . (a) Determine $I_{DQ1},I_{DQ2},V_{DSQ1}$ , and $V_{DSQ2}$ . (b) Determine $g_{m1}$ and $g_{m2}$ . (c) Determine the overall small−signal voltage gain $A_{\upsilon }={\upsilon }_o/{\upsilon }_i$ .

Figure P4.71

To determine

The drain current and drain to source voltages of the transistors for the given data.

Answer to Problem 4.71P

  $I_{DQ1}=0.7569\text{ mA}$ , $I_{DQ2}=0.7569\text{ mA}$ , $V_{DSQ1}=12.431\text{ V}$ , $V_{DSQ2}=8.6465\text{ V}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  $V_{TN1}=V_{TN2}=2\text{ V,}{K}_{n1}\text{=}{K}_{n2}\text{=4 }\text{mA}/{\text{V}}^{\text{2}},{\lambda }_1\text{=}{\lambda }_2\text{=0}$

The given circuit is shown below.

  

Calculation:

The drain current in transistor 1 is

  $\begin{array}{l}{I}_{DQ1}=K_{n1}{\left(\text{}{V}_{GSQ1}-V_{TN1}\right)}^2=4{\left(\text{}{V}_{GSQ1}-2\right)}^2\\ I_{DQ1}=4\left(V_{GSQ1}^2-4V_{GSQ1}+4\right)\end{array}$

The gate voltage of the M₁ transistor is

  $\begin{array}{l}{V}_{G1}=I_{DQ1}{R}_{S1}+V_{GSQ1}+V^{-}\\ I_{DQ1}=\frac{10-V_{GSQ1}}{10}\end{array}$

Then by equating the above equations,

  $\begin{array}{l}\frac{10-V_{GSQ1}}{10}=4\left(V_{GSQ1}^2-4V_{GSQ1}+4\right)\\ 40V_{GSQ1}^2-159V_{GSQ1}+150=0\\ V_{GSQ1}=\text{2}\text{.434}\end{array}$

Then,

  $I_{DQ1}=4{\left(2.435-2\right)}^2$

  $I_{DQ1}=0.7569\text{ mA}$

  $I_{DQ2}=0.7569\text{ mA}$

From a KVL equation around the drain to source loop of transistor 1,

  $\begin{array}{c}{V}^{+}=V_{DSQ1}+I_{DQ1}{R}_{S1}+V^{-}\\ V_{DSQ1}=20-0.7569\times 10\text{ V}\\ \text{=12}\text{.431}\text{V}\end{array}$

From a KVL equation around the drain to source loop of transistor 2,

  $\begin{array}{c}{V}^{+}=V_{DSQ2}+I_{DQ1}\left(R_{S2}+R_{D2}\right)+V^{-}\\ V_{DSQ2}=20-0.7569\times 15\text{ V}\\ =\text{8}\text{.6465}\text{V}\end{array}$

To determine

The transconductance of the two transistors.

Answer to Problem 4.71P

  $g_{m1}=\text{3}\text{.48 mA}/\text{V}$ , $g_{m2}=\text{3}\text{.48 mA}/\text{V}$

Explanation of Solution

Given Information:

The given values are:

  $V_{TN1}=V_{TN2}=2\text{ V,}{K}_{n1}\text{=}{K}_{n2}\text{=4 }\text{mA}/{\text{V}}^{\text{2}},{\lambda }_1\text{=}{\lambda }_2\text{=0}$

The given circuit is shown below.

  

Calculation:

Form part (a), $I_{DQ1}=0.7569\text{ mA}$ , $I_{DQ2}=0.7569\text{ mA}$

The transconductance of transistor 1 is

  $g_{m1}=2\sqrt{K_{n1}{I}_{DQ1}}=2\sqrt{4\times 0.7569}\text{ mA}/\text{V}$

  $g_{m1}=\text{3}\text{.48 mA}/\text{V}$

The transconductance of transistor 2 is

  $g_{m2}=2\sqrt{K_{n2}{I}_{DQ2}}=2\sqrt{4\times 0.7569}\text{ mA}/\text{V}$

  $g_{m2}=\text{3}\text{.48 mA}/\text{V}$

To determine

The overall small-signal voltage gain.

Answer to Problem 4.71P

  $A_v=2.416$

Explanation of Solution

Given Information:

The given values are

  $V_{TN1}=V_{TN2}=2\text{ V,}{K}_{n1}\text{=}{K}_{n2}\text{=4 }\text{mA}/{\text{V}}^{\text{2}},{\lambda }_1\text{=}{\lambda }_2\text{=0}$

The given circuit is shown below.

  

Calculation:

The small-signal equivalent circuit is in the below figure.

  

Form part (b), $g_{m1}=\text{3}\text{.48 mA}/\text{V}$ , $g_{m2}=\text{3}\text{.48 mA}/\text{V}$

The output voltage is

  $V_o=-g_{m2}{V}_{gs2}\left(R_{D2}\parallel R_L\right)$

Then,

  $\begin{array}{l}{V}_{gs2}=\left(-g_{m1}{V}_{gs1}-g_{m2}{V}_{gs2}\right)\left(R_{S1}\parallel R_{S2}\right)\\ V_{gs2}\left[1+g_{m2}\left(R_{S1}\parallel R_{S2}\right)\right]=-g_{m1}{V}_{gs1}\left(R_{S1}\parallel R_{S2}\right)\end{array}$

The input voltage is,

  $\begin{array}{l}{V}_i=V_{gs1}-V_{gs2}\\ V_{gs1}=V_i+V_{gs2}\end{array}$

Then

  $\begin{array}{l}{V}_{gs2}\left[1+g_{m2}\left(R_{S1}\parallel R_{S2}\right)\right]=-g_{m1}\left(V_i+V_{gs2}\right)\left(R_{S1}\parallel R_{S2}\right)\\ V_{gs2}=\frac{-g_{m1}{V}_i\left(R_{S1}\parallel R_{S2}\right)}{\left[1+\left(g_{m1}+g_{m2}\right)\left(R_{S1}\parallel R_{S2}\right)\right]}\end{array}$

The output voltage will be

  $\begin{array}{l}{V}_o=-g_{m2}\left\{\frac{-g_{m1}{V}_i\left(R_{S1}\parallel R_{S2}\right)}{\left[1+\left(g_{m1}+g_{m2}\right)\left(R_{S1}\parallel R_{S2}\right)\right]}\right\}\left(R_{D2}\parallel R_L\right)\\ \frac{V_o}{V_i}=\frac{3.48\times 3.48\left(10\parallel 10\right)\left(5\parallel 2\right)}{\left[1+\left(3.48+3.48\right)\left(10\parallel 10\right)\right]}\\ A_v=\frac{\left(12.1104\right)\left(5\right)\left(1.4286\right)}{\left[1+\left(6.96\right)\left(5\right)\right]}\end{array}$

So, the small-signal voltage gain is

  $A_v=2.416$