Chapter 2, Problem 2.51P

Assume $V_{\gamma }=0.7\text{V}$ for each diode in the circuit in Figure P251. Plot ${\upsilon }_O$ versus ${\upsilon }_I$ for $-10\le {\upsilon }_I\le +10\text{V}$ .

Figure P2.51

To determine

To plot: The output voltage, $v_0$ versus the input voltage, $v_1$ .

Explanation of Solution

Given:

  $V_{\gamma }=0.7\text{V}$

  $-10\text{V}\le v_1\le +10\text{V}$

  

Calculation:

Case 1:

Assume that diodes $D_1$ and $D_4$ are reverse biased and diodes $D_2$ and $D_3$ are forward blased as shown in Figure 1.

  

Figure 1

Apply the Kirchhoff’s voltage to the outer branch of the circuit shown in figure 1.

  $-10+(10K)I_1+0.7+(10k)I_1=0$

Therefore,

  $\begin{array}{l}{I}_1=\frac{10-0.7}{10k+10k}\\ =\frac{9.3}{20k}\\ =0.465\text{mA}\end{array}$

The voltage at node $4$ of the circuit shown in figure 1 is,

  $\begin{array}{l}{v}_0=I_1(10k)\\ =(0.465\times {10}^{-3})(10k)\\ =4.65\text{V}\end{array}$

The voltage at node $1$ of the circuit shown in figure $1$ is,

  $\begin{array}{l}{v}_1=0.7+I_1(10k)\\ =0.7+(0.465\times {10}^{-3})(10k)\\ =5.35\text{V}\end{array}$

Apply the Kirchhoff’s voltage to the input branch of the circuit shown in figure $1$ .

  $-v_1+0.7+(10k)I_2-10=0$

Therefore,

  $\begin{array}{l}{I}_2=\frac{10-0.7+v_1}{10k}\\ =\frac{9.3+v_1}{10k}\end{array}$

The voltage at node $2$ is,

  $\begin{array}{l}{v}_2=I_2(10k)-10\\ =\left(\frac{9.3+v_1}{10k}\right)(10k)-10\\ =9.3+v_1-10\\ =v_1-0.7\end{array}$

The diode $D_1$ gets reverse biased if,

  $\begin{array}{l}{v}_1-v_3<0.7\\ 5.35-v_1<0.7\\ v_1>4.65\text{V}\end{array}$

The diode $D_4$ gets reverse biased if,

  $\begin{array}{l}{v}_0-v_2<0.7\\ 4.65-(v_1-0.7)<0.7\\ 4.65-v_1+0.7<0.7\\ v_1>4.65\text{V}\end{array}$

Therefore, diodes $D_1$ and $D_4$ gets reverse biased for $v_1>4.65\text{V}$ .

The output voltage is constant for $v_1>4.65\text{V}$ .

  $v_0=4.65\text{V}$ .

Case 2:

Assume that diodes $D_1$ and $D_4$ are forward biased and diodes $D_2$ and $D_3$ are reverse biased as shown in Figure 2.

  

Figure 2

Apply the Kirchhoff’s voltage to the outer branch of the circuit shown in figure 2.

  $\begin{array}{l}\left(10k\right)I_4+0.7+\left(10k\right)I_4-10=0\\ I_4=\frac{10-0.7}{10k+10k}\\ =\frac{9.3}{20}\text{mA}\\ =0.465\text{mA}\end{array}$

The voltage at node $3$ of the circuit shown in figure 2 is,

  $\begin{array}{l}{v}_0=-I_4(10k)\\ =-\left(0.465\times {10}^{-3}\right)(10k)\\ =-4.65\text{V}\end{array}$

The voltage at node $4$ of the circuit shown in figure 2 is,

  $\begin{array}{l}{v}_4=I_4(10k)-10\\ =\left(0.465\times {10}^{-3}\right)(10k)-10\\ =-5.35\text{V}\end{array}$

Apply the Kirchhoff’s voltage to the input branch of the circuit shown in figure 2.

  $-10+(10k)I_3+0.7+v_1=0$

Therefore,

  $\begin{array}{l}{I}_3=\frac{10-0.7-v_1}{10k}\\ =\frac{9.3-v_1}{10k}\end{array}$

The voltage at node $2$ of the circuit shown in figure 2 is,

  $\begin{array}{l}{v}_2=-I_3(10k)+10\\ =\left(\frac{9.3-v_1}{10k}\right)(10k)+10\\ =-9.3+v_1+10\\ =v_1+0.7\end{array}$

The diode $D_2$ gets reverse biased if,

  $\begin{array}{l}{v}_2-v_0<0.7\\ (v_1+0.7)-(-4.65)<0.7\\ v_1<-4.65\text{V}\end{array}$

The diode $D_3$ gets reverse biased if,

  $\begin{array}{l}{v}_1-v_4<0.7\\ v_1-(-5.35)<0.7\\ v_1+5.35<0.7\\ v_1<-4.65\text{V}\end{array}$

Therefore, diodes $D_2$ and $D_3$ gets reverse biased for $v_1<-4.65\text{V}$

Therefore, the output voltage is constant for $v_1<-4.65\text{V}$ .

  $v_0=-4.65\text{V}$ .

Case 3:

Assume that diodes $D_1$ and $D_2$ are forward biased and diodes $D_3$ and $D_4$ are reverse biased as shown in Figure 3.

  

Figure 3

Apply the Kirchhoff’s current law at node $2$ of the circuit shown in figure 3.

  $\begin{array}{l}{I}_5+I_6=\frac{10-v_2}{10k}\\ I_5+\frac{v_2-0.7}{10k}=\frac{10-v_2}{10k}\\ I_5=\frac{10-v_2-v_2+0.7}{10k}\\ =\frac{10.7-2v_2}{10k}\end{array}$

The diode $D_1$ gets forward biased if,

  $\begin{array}{l}{I}_5\ge 0\\ \frac{10.7-2v_1}{10k}\ge 0\\ v_1\le 5.35\text{V}\end{array}$

Apply the Kirchhoff’s voltage law to the input branch of the circuit shown in figure 3.

  $\begin{array}{l}-v_2+0.7+v_1=0\\ v_2=0.7+v_1\end{array}$ …….(1)

The voltage across diode $D_2$ is,

  $\begin{array}{l}{v}_2-v_0=2.7\\ v_0=v_2-0.7\end{array}$

Consider equation(1).

Rewrite as,

  $\begin{array}{l}{v}_0=\left(0.7+v_1\right)-0.7\\ =v_1\end{array}$

Therefore, the output voltage is,

  $v_0=\{\begin{array}{l}-4.65\text{V},v_1<-4.65\text{V}\\ v_1,-4.65\text{V}\le v_1\le 4.65\text{V}\\ 4.65\text{V},v_1>4.65\text{V}\end{array}$

Therefore, plot $v_0$ versus $v_1$ as shown in figure 4.

  

Conclusion:

Therefore, plot of $v_0$ versus $v_1$ is drawn as shown in Figure 4.