Chapter 2, Problem 2.59P

Each diode cut−in voltage in the circuit in Figure P2.59 is 0.7 V. Determine $I_{D1},I_{D2},I_{D3}$ , and ${\upsilon }_O$ for (a) ${\upsilon }_I=0.5\text{V}$ , (b) ${\upsilon }_I=1.5\text{V}$ , (c) ${\upsilon }_I=3.0\text{V}$ , and (d) ${\upsilon }_I=5.0\text{V}$ .

Figure P2.59

To determine

The values of $I_{D1},I_{D2},I_{D3},v_o$ .

Answer to Problem 2.59P

  $\begin{array}{l}{I}_{D1}=I_{D2}=I_{D3}=0\text{A}\\ v_o=0.5\text{V}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{v}_I=0.5\text{V}\\ v_{\gamma }=0.7\text{V}\end{array}$

Calculation:

For $v_I=0.5\text{V}$ , diode D₁ is in cut off region and acts as open circuit because the voltage difference between positive and negative terminal is less than 0.7 V. The diode D₂,D₃ are in reverse bias mode and act like open circuit. Hence, the currents through the diodes are zero.

  $I_{D1}=I_{D2}=I_{D3}=0\text{A}$

The input voltage is dropped across the output because the current in the circuit is zero.

  $v_o=0.5\text{V}$

To determine

The values of $I_{D1},I_{D2},I_{D3},v_o$ .

Answer to Problem 2.59P

  $\begin{array}{l}{I}_{D1}=0.0667\text{mA}\\ I_{D2}=I_{D3}=0\text{A}\\ v_o=1.234\text{V}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{v}_I=1.5\text{V}\\ v_{\gamma }=0.7\text{V}\end{array}$

Calculation:

For $v_I=1.5\text{V}$ , the diode D₁ is forward biased and in active mode. The diode D₂ is in cut off mode and acts as open circuit because the voltage difference between positive and negative terminal is less than 0.7 V. The diode D₃ is in reverse bias region and acts like open circuit. Hence, the currents through the diodes D₂ and D₃ are zero.

  $I_{D2}=I_{D3}=0\text{A}$

The value of diode current $I_{D1}$ is determined as follows:

The modified circuit is:

  

Applying Kirchhoff’s voltage law:

  $\begin{array}{l}-v_I+I_{D1}{R}_1+I_{D1}{R}_2+0.7=0\\ I_{D1}=\frac{v_I-0.7}{R_1+R_2}\\ I_{D1}=\frac{1.5-0.7}{\left(4+8\right)\times {10}^3}\\ I_{D1}=0.0667\text{mA}\end{array}$

The value of output voltage $v_o$ is:

  $\begin{array}{l}{v}_o=I_{D1}{R}_2+0.7\\ v_o=\left(0.0667\times {10}^{-3}\right)\left(8\times {10}^3\right)+0.7\\ v_o=1.234\text{V}\end{array}$

To determine

The values of $I_{D1},I_{D2},I_{D3},v_o$ .

Answer to Problem 2.59P

  $\begin{array}{l}{I}_{D1}=0.171\text{mA},I_{D2}=0.0615\text{mA}\\ I_{D3}=0\text{mA,}{v}_o=2.069\text{V}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{v}_I=3\text{V}\\ v_{\gamma }=0.7\text{V}\end{array}$

Calculation:

Assuming the diodes are in forward bias and in active mode.

Applying Kirchhoff’s current law at output node:

  $\begin{array}{l}\frac{v_o-3}{4}+\frac{v_o-0.7}{8}+\frac{v_o-1.7}{6}+\frac{v_o-2.7}{4}=0\\ v_o\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{6}+\frac{1}{4}\right)=\frac{5.7}{4}+\frac{0.7}{8}+\frac{1.7}{6}\\ v_o\left(\frac{1}{2}+\frac{1}{8}+\frac{1}{6}\right)=\frac{34.2+2.1+6.8}{24}\\ v_o\left(\frac{12+3+4}{24}\right)=\frac{43.1}{24}\\ v_o=\frac{43.1}{19}\\ v_o=2.268\text{V}\end{array}$

From above calculation diodes D₁,D₂ are in forward bias active mode but the diode D₃ is in cut off mode because the voltage difference between positive and negative terminal is less than 0.7 V. The current through diode D₃ is zero.

  $I_{D3}=0\text{A}$

Hence, the assumption is incorrect.

The modified figure is:

  

Applying Kirchhoff’s current law at output node:

  $\begin{array}{l}\frac{v_o-3}{4}+\frac{v_o-0.7}{8}+\frac{v_o-1.7}{6}=0\\ v_o\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)=\frac{3}{4}+\frac{0.7}{8}+\frac{1.7}{6}\\ v_o\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{6}\right)=\frac{18+2.1+6.8}{24}\\ v_o\left(\frac{6+3+4}{24}\right)=\frac{26.9}{24}\\ v_o=\frac{26.9}{13}\\ v_o=2.069\text{V}\end{array}$

The diode currents $I_{D1},I_{D2}$ are determined as follows:

  $\begin{array}{l}{I}_{D1}=\frac{v_o-0.7}{\text{8}\text{kΩ}}\\ I_{D1}=\frac{2.069-0.7}{8\times {10}^3}\\ I_{D1}=0.171\text{mA}\\ I_{D2}=\frac{v_o-1.7}{\text{6}\text{kΩ}}\\ I_{D2}=\frac{2.069-1.7}{6\times {10}^3}\\ I_{D2}=0.0615\text{mA}\end{array}$

To determine

The values of $I_{D1},I_{D2},I_{D3},v_o$ .

Answer to Problem 2.59P

  $\begin{array}{l}{I}_{D1}=0.275\text{mA},I_{D2}=0.2\text{mA}\\ I_{D3}=0.05\text{mA,}{v}_o=2.9\text{V}\end{array}$

Explanation of Solution

Given Information:

The given circuit is shown below.

  

  $\begin{array}{l}{v}_I=5\text{V}\\ v_{\gamma }=0.7\text{V}\end{array}$

Calculation:

For $v_I=5\text{V,}$ all diodes are forward biased and in active mode.

Applying Kirchhoff’s current law at output node:

  $\begin{array}{l}\frac{v_o-5}{4}+\frac{v_o-0.7}{8}+\frac{v_o-1.7}{6}+\frac{v_o-2.7}{4}=0\\ v_o\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{6}+\frac{1}{4}\right)=\frac{7.7}{4}+\frac{0.7}{8}+\frac{1.7}{6}\\ v_o\left(\frac{1}{2}+\frac{1}{8}+\frac{1}{6}\right)=\frac{46.2+2.1+6.8}{24}\\ v_o\left(\frac{12+3+4}{24}\right)=\frac{55.1}{24}\\ v_o=\frac{43.1}{19}\\ v_o=2.9\text{V}\end{array}$

The values of diode currents are:

  $\begin{array}{l}{I}_{D1}=\frac{v_o-0.7}{\text{8}\text{kΩ}}\\ I_{D1}=\frac{2.9-0.7}{8\times {10}^3}\\ I_{D1}=0.275\text{mA}\\ I_{D2}=\frac{v_o-1.7}{\text{6}\text{kΩ}}\\ I_{D2}=\frac{2.9-1.7}{6\times {10}^3}\\ I_{D2}=0.2\text{mA}\\ I_{D3}=\frac{v_o-2.7}{4\text{kΩ}}\\ I_{D3}=\frac{2.9-2.7}{4\times {10}^3}\\ I_{D3}=0.05\text{mA}\end{array}$