Question

The output voltage in the following circuit is (Consider ideal diode case): 

• A. \( -5 \, \text{V} \)
• B. \( +5 \, \text{V} \)
• C. \( 10 \, \text{V} \)
• D. \( 0 \, \text{V} \)

Hint:

For ideal diodes, always remember that they conduct when forward biased and do not conduct when reverse biased. In this case, the conducting diode pulls the output voltage to 0V.

Answer 1

The Correct Option is D

To determine the output voltage in the given circuit, we need to analyze the circuit considering the ideal diode case. In an ideal diode scenario, the diode conducts perfectly when forward-biased (zero voltage drop across it) and does not conduct at all when reverse-biased (infinite resistance).

1. Initially, consider diode \( D_1 \). Since it is forward-biased with \( +5 \, \text{V} \), it will conduct, allowing current to pass through it without any voltage drop.
2. Next, examine diode \( D_2 \). The cathode of \( D_2 \) is connected to ground, making the anode potential higher than ground when the circuit is active. This results in \( D_2 \) being reverse-biased.
3. As \( D_1 \) is conducting and \( D_2 \) is not conducting, the voltage at the output \( V_{\text{out}} \) is the same as the cathode of the conducting diode \( D_1 \), causing it to be \( +5 \, \text{V} \).

Finally, observe the resistor connected to the output; it is also connected to \( +5 \, \text{V} \). With both connections at the same potential, there is no potential difference, leading the output voltage \( V_{\text{out}} \) to be \( 0 \, \text{V} \).

Thus, the correct answer is:

\( 0 \, \text{V} \)

Answer 2

In the given circuit, we are considering ideal diodes.
The behavior of an ideal diode is: 
- It conducts when forward-biased (anode is more positive than cathode). 
- It does not conduct when reverse-biased.

Let's analyze the circuit step by step:

1. Diode \( D_1 \) is forward biased because its anode is at \( +5 \, \text{V} \) and its cathode is at \( V_{\text{out}} \).
2. Diode \( D_2 \) is reverse biased because its anode is at ground potential (0V) and its cathode is at \( V_{\text{out}} \). In this configuration:
- \( D_1 \) will conduct, and the output voltage at \( V_{\text{out}} \) will be 0V, since the ideal diode has no voltage drop when it conducts.
- \( D_2 \) will not conduct as it is reverse biased.

Thus, the output voltage \( V_{\text{out}} \) is \( 0 \, \text{V} \).