Chapter 4, Problem 4.1P

To determine

The two possible situations that will happen for the given conditions.

The diode current and terminal voltage across diode for two situations.

Answer to Problem 4.1P

The value of diode current and diode terminal voltage is:

For situation 1:

V_D = 1.5 V

i_D = 0 A

For situation 2:

  $\begin{array}{l}{V}_D\text{ = 0 V}\\ i_D\text{ = }1.5\text{ A}\end{array}$

Explanation of Solution

Given:

Thevenin equivalent of AA Flashlight cell with:

Thevenin voltage (V_th) = 1.5 V

Thevenin resistance (R_th) = 1 Ω

Positive side of battery is connected to the cathode of the ideal diode.

Calculation:

The two possible situations can be:

Situation 1: Cathode terminal of the diode is connected to the positive terminal of the battery.

Situation 2: Anode terminal of the diode is connected to the positive terminal of the battery.

Case 1:

When the cathode is connected to the positive terminal of battery.

The Thevenin equivalent circuit with diode connected as given can be drawn as:

  

The diode is operating in a reversed biased region as the positive side of the battery is connected to the cathode terminal of diode.

Hence, the ideal diode will behave as an open circuit.

So,

  $i_D\text{ = }0\text{ A}\cdots \cdots \text{(1)}$

Terminal voltage can be calculated by applying mesh law in the circuit:

  ${\text{V}}_{\text{th}}{\text{ = R}}_{\text{th}}{i}_D\text{ + }{V}_D$

By putting the given values and from equation (1):

  $\begin{array}{l}{\text{V}}_{\text{th}}\text{ = }{V}_D\\ V_D\text{ = 1}\text{.5 V}\end{array}$

Case 2:

When anode terminal of the diode is connected to the positive terminal of battery.

The Thevenin equivalent circuit with diode connected as given can be drawn as:

  

The diode is operating in forward biased region as the positive side of the battery is connected to the anode terminal of diode.

As it is ideal diode, the cut-in voltage across it will be 0.

Hence,

  $V_D\text{ = 0 V}\cdots \cdots \text{(2)}$

Current can be calculated by applying mesh law in the circuit:

  ${\text{V}}_{\text{th}}{\text{ = R}}_{\text{th}}{i}_D\text{ + }{V}_D$

By putting the given values and from equation (2):

  $\begin{array}{l}\text{1}\text{.5 = 1}{i}_D\\ i_D\text{ = }\frac{\text{1}\text{.5}}{1}\text{ A}\\ i_D\text{ = }1.5\text{ A}\end{array}$