Chapter 1, Problem 1.48P

Repeat Problem 1.47 if the reverse−saturation current for each diode isy $I_S=5\times {10}^{-14}\text{A}$ . What is the voltage across each diode?

1. 

Figure P1.47

To determine

The voltage across diode.

Answer to Problem 1.48P

The diode voltage, $v_D=0.5775\text{V}$ .

Explanation of Solution

Given information:

The circuit diagram is given as:

  

The reverse saturation current for each diode is $5\times {10}^{-14}\text{A}$ .

Calculation:

The expression for the diode current is given as:

  $\begin{array}{l}{I}_D=I_S\left(e^{v_D/v_T}\right)\\ \text{Where,}\\ I_D=\text{Diode}\text{current}\\ v_D=\text{Diode}\text{voltage}\\ v_T=\text{Thermal}\text{voltage}\end{array}$

From the given circuit, evaluating the diode current:

  $\begin{array}{l}I=I_D=\frac{5-V_o}{20}\mathrm{.....}\left(1\right)\\ I=I_S{e}^{\left(v_D/0.026\right)}\left(\because v_T=0.026\text{V}\right)\\ =5\times {10}^{-14}\left[e^{\left(v_D/0.026\right)}\right]\mathrm{.......}\left(2\right)\end{array}$

From the circuit, the output voltage is same as the diode voltage.

Solving equation 1 and 2 by using the hit and trial method:

  $\begin{array}{l}{v}_D=V_o=0.5775\text{V}\\ I=0.221\text{mA}\end{array}$

To determine

The voltage across diode.

Answer to Problem 1.48P

The diode voltage, $v_D=0.579\text{V}$ .

Explanation of Solution

Given information:

The circuit diagram is given as:

  

The reverse saturation current for each diode is $5\times {10}^{-14}\text{A}$

Calculation:

The expression for the diode current is given as:

  $\begin{array}{l}{I}_D=I_S\left(e^{v_D/v_T}\right)\\ \text{Where,}\\ I_D=\text{Diode}\text{current}\\ v_D=\text{Diode}\text{voltage}\\ v_T=\text{Thermal}\text{voltage}\end{array}$

From the given circuit, evaluating the diode current:

Applying the Kirchhoff’s voltage law to the circuit from top to bottom:

  $\begin{array}{l}-5+20I+v_o+20I-5=0\\ I=\frac{10-v_o}{40}\mathrm{.......}\left(1\right)\end{array}$

Assuming the voltage across diode v_D, then applying the Kirchhoff’s voltage law from top to the output voltage:

  $\begin{array}{l}-5+20I+v_D+v_o=0\\ v_o=5-I\left(20\right)-v_D\mathrm{.......}\left(2\right)\end{array}$

From the circuit, the diode current is same as the current flowing in the circuit:

  $\begin{array}{l}I=I_D=I_S{e}^{\left(v_D/0.026\right)}\left(\because v_T=0.026\text{V}\right)\\ =5\times {10}^{-14}\left[e^{\left(v_D/0.026\right)}\right]\mathrm{.......}\left(3\right)\end{array}$

Solving equation 1, 2 and 3 by using the hit and trial method:

  $\begin{array}{l}{v}_D=0.579\text{V}\\ v_o=-0.289\text{V}\\ I=0.2355\text{mA}\end{array}$

To determine

The voltage across diode.

Answer to Problem 1.48P

The diode voltage, $v_D=0.591\text{V}$ .

Explanation of Solution

Given information:

The circuit diagram is given as:

  

The reverse saturation current for each diode is $5\times {10}^{-14}\text{A}$

Calculation:

The expression for the diode current is given as:

  $\begin{array}{l}{I}_D=I_S\left(e^{v_D/v_T}\right)\\ \text{Where,}\\ I_D=\text{Diode}\text{current}\\ v_D=\text{Diode}\text{voltage}\\ v_T=\text{Thermal}\text{voltage}\end{array}$

From the given circuit, evaluating the diode current:

Applying the Kirchhoff’s voltage law to the circuit from top to bottom:

  $\begin{array}{l}-2+5I+v_D+20I-8=0\\ I=\frac{10-v_D}{25}\mathrm{.......}\left(1\right)\end{array}$

Then applying the Kirchhoff’s voltage law from top to the output voltage:

  $\begin{array}{l}-2+5I+v_o=0\\ v_o=2-5I\mathrm{.......}\left(2\right)\end{array}$

From the circuit, the diode current is same as the current flowing in the circuit:

  $\begin{array}{l}I=I_D=I_S{e}^{\left(v_D/0.026\right)}\left(\because v_T=0.026\text{V}\right)\\ =5\times {10}^{-14}\left[e^{\left(v_D/0.026\right)}\right]\mathrm{.......}\left(3\right)\end{array}$

Solving equation 1, 2 and 3 by using the hit and trial method:

  $\begin{array}{l}{v}_D=0.5913\text{V}\\ v_o=0.1185\text{V}\\ I=0.3763\text{mA}\end{array}$

To determine

The voltage across diode.

Answer to Problem 1.48P

The diode voltage, $v_D=-5\text{V}$ .

Explanation of Solution

Given information:

The circuit diagram is given as:

  

The reverse saturation current for each diode is $5\times {10}^{-14}\text{A}$

Calculation:

Referring the given circuit, here the diode current is flowing in the reverse direction.

Hence, the reverse saturation current will be the diode current in this case.

Since the value of the reverse saturation current is given in the question.

Therefore,

  $I_S=I_D=I=5\times {10}^{-14}\text{A}$

Now evaluating the diode voltage:

Since, due to reverse direction of the current, no current will flow through the resistor. Hence the drop across the resistor will be zero.

Therefore, the value of the diode voltage will be same as the output voltage:

  $\begin{array}{l}-v_o+20I-5=0\\ v_o=-5\text{V}\left(\because I=0\right)\\ v_D=-5\text{V}\end{array}$