An AA flashlight cell whose Thevenin equivalent is a voltage source of 1.5V and a resistance of 1Ω is connected to the terminals of an ideal diode. Describe two possible situations that result what are the diode current and terminal voltage when

a. The connection is between the diode cathode and the positive terminal of the battery.

b. The anode and the positive terminal are connected.

A. Current = 0 A, Voltage = 1.5 V

B. Current = 1.5 A, Voltage = 0 V

Explanation:

A.

The cathode is the negative terminal or n-side of the diode. So, in this case the n-side of diode is connected to the positive terminal of battery. This is, the condition for reverse bias, when no current passes through diode and diode acts as an open circuit. In such case:

Current through diode = Id = 0 A

Voltage Across Diode = Vd = 1.5 V

B.

The anode is the positive terminal or p-side of the diode. So, in this case the p-side of diode is connected to the positive terminal of battery. This is, the condition for forward bias, when the current passes through diode and diode acts as an closed circuit. In such case ideal diode acts as a short circuit and has following characteristics:

Current through diode = Id = V/R = 1.5 V/1 Ω = 1.5 A

Voltage Across Diode = Vd = 0 V