If in a p-n junction diode, a square input signal of 10 V is applied as shown.
(a) Option A
(b) Opti

Q: Why doesn't the diode consume any voltage itself during the positive cycle?

In many basic circuit problems, unless a specific 'cut-in' or 'barrier' voltage (like $0.7\text{ V}$ for Silicon) is mentioned, we assume the diode is 'ideal'. An ideal diode has zero resistance in forward bias, meaning it doesn't consume any voltage.

Q: What would happen if we flipped the diode around in the circuit?

If the diode were reversed, it would be forward-biased during the negative half-cycle ($-5\text{ V}$) and reverse-biased during the positive half-cycle ($+5\text{ V}$). The output would then be $0\text{ V}$ for the first half and $-5\text{ V}$ for the second half.

Q: Why is the output exactly zero during the negative cycle? Couldn't there be a tiny bit of voltage?

In reality, a real diode has a very tiny 'reverse saturation current', which would cause a microscopic voltage drop across the resistor. However, this current is so small (usually in microamps or nanoamps) that for all practical purposes in this context, we consider the current, and thus the output voltage, to be exactly zero.

Q: Is this circuit a rectifier?

Yes! This exact setup is the most basic form of a rectifier, specifically a 'half-wave rectifier'. Its job is to convert alternating current (AC), which goes positive and negative, into direct current (DC), which only flows in one direction.

Q: What if the input was a smooth sine wave instead of a blocky square wave?

The principle remains exactly the same. The diode would allow the positive 'humps' of the sine wave to pass through unchanged, but it would completely flatten out the negative 'valleys' to zero volts.

Question

If in a p-n junction diode, a square input signal of 10 V is applied as shown.

Sigma Percentile · Accepted Answer

### The One-Way Valve of Electronics

Imagine you have a pipe with a special valve that only lets water flow in one direction. If you try to push water backward, the valve slams shut. A p-n junction diode is exactly that—a one-way valve for electrical current.

In this problem, we are feeding a square wave signal into a simple circuit containing a diode and a load resistor, $R_L$. The input signal, $V_i$, is a bit bipolar; it jumps between a positive state of $+5	ext{ V}$ and a negative state of $-5	ext{ V}$. Our mission is to figure out what the voltage across the resistor, $V_o$, looks like.

### The Positive Cycle: The Open Gate

Let's break the signal down. During the first half of the cycle, the input voltage is $+5	ext{ V}$. This means the p-side of the diode (the flat back of the triangle symbol) is connected to a higher potential than the n-side (the line at the tip).

When the p-side is more positive than the n-side, the diode is in a state called **forward bias**. In an ideal scenario, a forward-biased diode acts just like a closed switch. It offers zero resistance. Because the diode lets the current pass freely, the entire $+5	ext{ V}$ from the source drops across the load resistor $R_L$.

$$V_o = V_i = +5	ext{ V}$$

So, for the positive half of the cycle, the output graph perfectly mimics the input graph.

### The Negative Cycle: The Blockade

Now, the plot twists. In the second half of the cycle, the input voltage suddenly drops to $-5	ext{ V}$. Now, the p-side is at a lower potential compared to the n-side.

This puts the diode into **reverse bias**. Remember our one-way valve? The water is trying to flow backward, and the valve slams shut. A reverse-biased ideal diode acts as an open circuit, offering infinite resistance. It completely blocks the flow of current.

If there is no current flowing through the circuit ($I = 0$), then according to Ohm's Law, the voltage drop across the resistor must also be zero.

$$V_o = I 	imes R_L = 0 	imes R_L = 0	ext{ V}$$

During the negative half-cycle, the output voltage is flatlined at zero.

### The Final Verdict: Half-Wave Rectification

As the input signal continues to oscillate, the diode keeps repeating its routine: open gate, closed gate, open gate, closed gate. It effectively "clips" off the entire negative portion of the input signal.

This process is known as **half-wave rectification**, because only half of the input wave makes it to the output. Looking at the given options, the graph that shows a $+5	ext{ V}$ pulse followed by a $0	ext{ V}$ flatline is option (d).

Animated Solution for Physics - Semiconductors: If in a p-n junction diode, a square input signal of 10 V is applied as shown.

Select Answer:

Visualized Solution

Analyzing the Setup

Positive Half Cycle

Forward Bias Condition

Negative Half Cycle

Reverse Bias Condition

Final Output Waveform

The Sigma Insight: P-N Junction Diode

The One-Way Valve of Electronics

The Positive Cycle

The Negative Cycle

The Final Verdict

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