Chapter 9, Problem 9.63P

Let $R=10k\Omega$ in the differential amplifier in Figure P9.63. Determine thevoltages $v_X,v_Y,v_O$ and the currents $i_1,i_2,i_3,i_4$ for input voltages of(a) $v_1=1.80V,v_2=1.40V$ ; (b) $v_1=3.20V,v_2=3.60V$ ; and (c) $v_1=-1.20V,v_2=-1.35V$ .

To determine

The value of the voltages $v_X$ , $v_Y$ and $v_O$ . Also determine the value of the current $i_1$ , $i_2$ , $i_3$ and $i_4$

for the input voltage $v_1=1.80\text{V}$ and $v_2=1.40\text{V}$ .

Answer to Problem 9.63P

The value of the voltage $v_O$ is $-4\text{V}$ , $v_x$ is $1.273\text{V}$ , $v_y$ is $1.273\text{V}$ , $i_2$ is $0.0527\text{mA}$ , $0.0527\text{mA}$ , $i_3$ is $0.0127\text{mA}$ and $i_4$ is $0.0127\text{mA}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The expression for the value of the voltage $v_x$ is given by,

  $v_x=v_y$

Apply KCL at the node $v_y$ .

  $\begin{array}{l}\frac{v_2-v_y}{R}=\frac{v_y}{10R}\\ v_y=\frac{10v_2}{11}\end{array}$ ....... (1)

Substitute $v_x$ for $v_y$ in the above equation.

  $v_x=\frac{10v_2}{11}$

Apply KCL at the node $v_x$ .

  $\begin{array}{l}\frac{v_1-v_x}{R}=\frac{v_x-v_O}{10R}\\ v_O=11v_x-10v_1\end{array}$

Substitute $v_y$ for $v_x$ in the above equation.

  $v_O=11v_y-10v_1$ ….. (2)

Substitute $\frac{10v_2}{11}$ for $v_y$ in the above equation.

  $\begin{array}{c}{v}_O=11\left(\frac{10v_2}{11}\right)-10v_1\\ =10\left(v_2-v_1\right)\end{array}$

The expression for the current $i_1$ is given by,

  $i_1=\frac{-v_1-v_x}{R}$

Substitute $v_y$ for $v_x$ in the above equation.

  $i_1=\frac{-v_1-v_y}{R}$

Substitute $\frac{10v_2}{11}$ for $v_y$ and $10\text{k}\Omega$ for $R$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{-v_1-\left(\frac{10v_2}{11}\right)}{10\text{k}\Omega }\\ =\frac{11v_1-10v_2}{110\text{k}\Omega }\end{array}$

Substitute $i_2$ for $i_1$ in the above equation.

  $i_2=\frac{11v_1-10v_2}{110\text{k}\Omega }$ ….. (3)

The expression for the current $i_3$ is given by,

  $i_3=\frac{v_2-v_y}{R}$

Substitute $\frac{10v_2}{11}$ for $v_y$ and $10\text{k}\Omega$ for $R$ in the above equation.

  $\begin{array}{c}{i}_3=\frac{v_2-\frac{10v_2}{11}}{10\text{k}\Omega }\\ =\frac{v_2}{110\text{k}\Omega }\end{array}$

Substitute $i_4$ for $i_4$ in the above equation.

  $i_4=\frac{v_2}{110\text{k}\Omega }$ ....... (4)

Substitute $1.40\text{V}$ for $v_2$ in equation (1).

  $\begin{array}{c}{v}_y=\frac{10\left(1.40\text{V}\right)}{11}\\ =1.273\text{V}\end{array}$

Substitute $v_x$ for $v_y$ in the above equation.

  $v_x=1.273\text{V}$

Substitute $1.40\text{V}$ for $v_2$ and $1.80\text{V}$ for $v_2$ in equation (2)

  $\begin{array}{c}{v}_O=10\left(1.40\text{V}-1.80\text{V}\right)\\ =-4\text{V}\end{array}$

Substitute $1.40\text{V}$ for $v_2$ and $1.80\text{V}$ for $v_2$ in equation (3)

  $\begin{array}{c}{i}_2=\frac{11\left(1.80\text{V}\right)-10\left(1.40\text{V}\right)}{110\text{k}\Omega }\\ =0.0527\text{mA}\end{array}$

Substitute $i_1$ for $i_2$ in the above equation.

  $i_1=0.0527\text{mA}$

Substitute $1.40\text{V}$ for $v_2$ and $1.80\text{V}$ for $v_2$ in equation (4)

  $\begin{array}{c}{i}_4=\frac{1.40\text{V}}{110\text{k}\Omega }\\ =0.0127\text{mA}\end{array}$

Substitute $i_4$ for $i_3$ in the above equation.

  $i_3=0.0127\text{mA}$

Conclusion:

Therefore, the value of the voltage $v_O$ is $-4\text{V}$, $v_x$ is $1.273\text{V}$ , $v_y$ is $1.273\text{V}$ , $i_2$ is $0.0527\text{mA}$ , $0.0527\text{mA}$ , $i_3$ is $0.0127\text{mA}$ and $i_4$ is $0.0127\text{mA}$ .

To determine

The value of the voltages $v_X$ , $v_Y$ and $v_O$ . Also determine the value of the current $i_1$ , $i_2$ , $i_3$ and $i_4$ for the input voltage $v_1=3.20\text{V}$ and $v_2=3.60\text{V}$ .

Answer to Problem 9.63P

The value of the voltage $v_O$ is $4\text{V}$

  $v_x$ is $3.273\text{V}$ , $v_y$ is $3.273\text{V}$ , $i_2$ is $-0.00727\text{mA}$ , $-0.00727\text{mA}$ , $i_3$ is $0.0327\text{mA}$ and $i_4$ is $0.0327\text{mA}$ .

Explanation of Solution

Calculation:

Substitute $3.60\text{V}$ for $v_2$ in equation (1).

  $\begin{array}{c}{v}_y=\frac{10\left(3.60\text{V}\right)}{11}\\ =3.273\text{V}\end{array}$

Substitute $v_x$ for $v_y$ in the above equation.

  $v_x=3.273\text{V}$

Substitute $3.20\text{V}$ for $v_1$ and $3.60\text{V}$ for $v_2$ in equation (2)

  $\begin{array}{c}{v}_O=10\left(3.60\text{V}-3.20\text{V}\right)\\ =4\text{V}\end{array}$

Substitute $3.20\text{V}$ for $v_1$ and $3.60\text{V}$ for $v_2$ in equation (3)

  $\begin{array}{c}{i}_2=\frac{11\left(3.20\text{V}\right)-10\left(3.60\text{V}\right)}{110\text{k}\Omega }\\ =-0.00727\text{mA}\end{array}$

Substitute $i_1$ for $i_2$ in the above equation.

  $i_1=-0.00727\text{mA}$

Substitute $3.60\text{V}$ for $v_2$ in equation (4)

  $\begin{array}{c}{i}_4=\frac{3.60\text{V}}{110\text{k}\Omega }\\ =0.0327\text{mA}\end{array}$

Substitute $i_4$ for $i_3$ in the above equation.

  $i_3=0.0327\text{mA}$

Conclusion:

Therefore, the value of the voltage $v_O$ is $4\text{V}$

  $v_x$ is $3.273\text{V}$ , $v_y$ is $3.273\text{V}$ , $i_2$ is $-0.00727\text{mA}$ , $-0.00727\text{mA}$ , $i_3$ is $0.0327\text{mA}$ and $i_4$ is $0.0327\text{mA}$ .

To determine

The value of the voltages $v_X$ , $v_Y$ and $v_O$ . Also determine the value of the current $i_1$ , $i_2$ , $i_3$ and $i_4$ for the input voltage $v_1=-1.20\text{V}$ and $v_2=-1.35\text{V}$ .

Answer to Problem 9.63P

The value of the voltage $v_O$ is $-1.5\text{V}$

  $v_x$ is $-1.227\text{V}$ , $v_y$ is $-1.227\text{V}$ , $i_2$ is $-0.00273\text{mA}$ , $i_1$ is $-0.00273\text{mA}$ , $i_3$ is $-0.0123\text{mA}$ and $i_4$ is $-0.0123\text{mA}$ .

Explanation of Solution

Calculation:

Substitute $-1.35\text{V}$ for $v_2$ in equation (1).

  $\begin{array}{c}{v}_y=\frac{10\left(-1.35\text{V}\right)}{11}\\ =-1.227\text{V}\end{array}$

Substitute $v_x$ for $v_y$ in the above equation.

  $v_x=-1.227\text{V}$

Substitute $-1.20\text{V}$ for $v_1$ and $-1.35\text{V}$ for $v_2$ in equation (2)

  $\begin{array}{c}{v}_O=10\left(-1.35\text{V}-\left(-1.20\text{V}\right)\right)\\ =-1.5\text{V}\end{array}$

Substitute $-1.20\text{V}$ for $v_1$ and $-1.35\text{V}$ for $v_2$ in equation (3)

  $\begin{array}{c}{i}_2=\frac{11\left(-1.20\text{V}\right)-10\left(-1.35\text{V}\right)}{110\text{k}\Omega }\\ =-0.00273\text{mA}\end{array}$

Substitute $i_1$ for $i_2$ in the above equation.

  $i_1=-0.00273\text{mA}$

Substitute $-1.35\text{V}$ for $v_2$ in equation (4)

  $\begin{array}{c}{i}_4=\frac{-1.35\text{V}}{110\text{k}\Omega }\\ =-0.0123\text{mA}\end{array}$

Substitute $i_4$ for $i_3$ in the above equation.

  $i_3=-0.0123\text{mA}$

Conclusion:

Therefore, the value of the voltage $v_O$ is $-1.5\text{V}$

  $v_x$ is $-1.227\text{V}$ , $v_y$ is $-1.227\text{V}$ , $i_2$ is $-0.00273\text{mA}$ , $i_1$ is $-0.00273\text{mA}$ , $i_3$ is $-0.0123\text{mA}$ and $i_4$ is $-0.0123\text{mA}$ .