Chapter 11, Problem 11.93P

For the transistors in the circuit in Figure $\text{P}11.93,$ the parameters are: $K_n=0.2\text{mA}/{\text{V}}^2,V_{TN}=2\text{V},$ and $\lambda =0.02{\text{V}}^{-1}.$ (a) Determine the differential-mode voltage gain $A_d=v_{o3}/v_d$ and the common-mode voltage $\mathrm{gain}{A}_{cm}=v_{o3}/v_{cm}.$ (b) Determine the output voltage $v_{o3}$ if $v_1=2.15\sin$ $\omega t\text{V}$ and $v_2=1.85$ sin $\omega t$ V. Compare this output to the ideal output that would be obtained if $A_{cm}=0$ .

To determine

The values of the differential mode voltage gain and the common mode voltage gain for the given transistor circuit.

Answer to Problem 11.93P

The differential-mode voltage gain is $A_d=-3.73$ .

The common-mode voltage gain is $A_{cm}=0.0718$ .

Explanation of Solution

Given:

The given circuit is shown below.

  

The transistor parameters are:

  $\begin{array}{l}\lambda =0.02V^{-1}\\ K_n=0.2m\text{A}/{\text{V}}^2\\ V_{TN}=2\text{V}\end{array}$

Calculation:

The small signal model of the given transistor circuit is shown below:

  

The value of the gate-source voltage of the transistor 4 is calculated as shown below:

  $\begin{array}{l}{I}_1=K_n{(}^{V_{GS4}}\\ I_1=\frac{24-V_{GS4}}{R_1}\\ I_1=\frac{24-V_{GS4}}{55\times {10}^3}\\ \frac{24-V_{GS4}}{55\times {10}^3}=K_n{(}^{V_{GS4}}\\ 24-V_{GS4}=55\times 0.2{(}^{V_{GS4}}\\ 24-V_{GS4}=11{(}^{V_{GS4}}\\ 24-V_{GS4}=11(V^2{}_{GS4}-4V_{GS4}+4)\\ 11V^2{}_{GS4}-43V_{GS4}+20\\ V_{GS4}=\frac{43\pm \sqrt{{43}^2-4\times 11\times 20}}{2\times 11}V\\ V_{GS4}=3.37V\end{array}$

The quiescent current is,

  $\begin{array}{l}{V}_{GS4}=3.37V\\ I_1=I_Q=\frac{24-V_{GS4}}{55\times {10}^3}\\ I_Q=\frac{24-3.37}{55\times {10}^3}A\\ I_Q=0.375mA\end{array}$

The value of the gate-source voltage of the transistor 3 is calculated as shown below:

  $\begin{array}{l}{I}_{D3}=K_n{(}^{V_{GS3}}\\ I_{D3}=\frac{v_{02}-V_{GS3}}{R_5}\\ v_{02}=12-\left(\frac{I_Q}{2}\right)\times 40\times {10}^{3}V\\ v_{02}=12-\left(\frac{0.375\times {10}^{-3}}{2}\right)\times 40\times {10}^{3}V\\ v_{02}=4.5V\\ \frac{4.5-V_{GS3}}{R_5}=K_n{(}^{V_{GS3}}\\ \frac{4.5-V_{GS3}}{6\times {10}^3}=0.2\times {10}^{-3}{(}^{V_{GS3}}\\ 4.5-V_{GS3}=1.2{(}^{V^2}\\ 1.2V^2{}_{GS3}-3.8V_{GS3}+0.3=0\\ V_{GS3}=\frac{3.8\pm \sqrt{{3.8}^2-4\times 1.2\times 0.3}}{2\times 1.2}V\\ V_{GS3}=3.09V\\ I_{D3}=\frac{4.5-3.09}{6}A\\ I_{D3}=0.235mA\end{array}$

The transconductance parameters for the differential gain are calculated as shown below:

  $\begin{array}{l}{g}_{m2}=2\sqrt{K_n{I}_{D2}}\\ g_{m2}=2\sqrt{0.2\times \left(\frac{0.375}{2}\right)}\frac{\text{mA}}{\text{V}}\\ g_{m2}=0.387\frac{\text{mA}}{\text{V}}\\ g_{m3}=2\sqrt{K_n{I}_{D3}}\\ g_{m3}=2\sqrt{0.2\times 0.235}\frac{\text{mA}}{\text{V}}\\ g_{m3}=0.434\frac{\text{mA}}{\text{V}}\end{array}$

The differential gain is,

  $\begin{array}{l}{A}_{d1}=\frac{1}{2}{g}_{m2}{R}_D\\ A_{d1}=\frac{1}{2}\times 0.387\times 40\\ A_{d1}=7.74\\ A_2=\frac{-(0.434)\times 4}{1+0.034\times 6}\\ A_2=-0.482\\ A_d=A_{d1}\times A_2\\ A_d=7.74\times -0.482\\ A_d=-3.73\end{array}$

The common mode gain is,

  $\begin{array}{l}{r}_{05}=\frac{1}{\lambda I_Q}\\ r_{05}=R_0\\ R_0=\frac{1}{\lambda I_Q}\\ R_0=\frac{1}{0.02\times 0.375}\Omega \\ R_0=133\Omega \\ A_{cm1}=\frac{-g_{m2}{R}_D}{1+2g_{m2}{R}_0}\\ A_{cm1}=\frac{-0.387\times 40}{1+2\times 0.387\times 133}\\ A_{cm1}=-0.149\\ A_{cm}=A_2\times A_{cm1}\\ A_{cm}=-0.149\times -0.482\\ A_{cm}=0.0718\end{array}$

To determine

The values of the output voltage for the given input voltages.

To compare: The result with the ideal case.

Answer to Problem 11.93P

The output voltage is $v_{03}=-0.975\sin \omega t\text{V}$ , $v_{03}(\text{ideal})=-1.12\sin \omega t\text{V}$

Explanation of Solution

Given:

The given circuit is shown below.

  

The circuit parameters are:

  $\begin{array}{l}{v}_1=2.15\sin \omega t\text{V}\\ v_2=1.85\sin \omega t\text{V}\\ A_d=-3.73\\ A_{cm}=0.0718\end{array}$

Calculation:

The output voltage is calculated as shown below:

  $\begin{array}{l}{v}_d=v_1-v_2\\ v_d=0.3\sin \omega t\\ v_1-v_2=0.3\sin \omega t\\ v_{cm}=\frac{v_1+v_2}{2}\\ \frac{v_1+v_2}{2}=2\sin \omega t\\ v_{03}=A_d{v}_d+A_{cm}{v}_{cm}\\ v_{03}=-3.73\times 0.3+0.0718\times 2\\ v_{03}=-0.975\sin \omega t\text{V}\end{array}$

On comparing the above output voltage with the ideal case,

  $\begin{array}{l}{A}_{cm}(\text{ideal})=0\\ v_{03}(\text{ideal})=A_d{v}_d+0\\ v_{03}(\text{ideal})=-3.73\times 0.3\\ v_{03}(\text{ideal})=-1.12\sin \omega t\text{V}\end{array}$