Q.Draw the output signal in a p-n junction diode when a square input signal of 10 V as shown in the figure is applied across it.

Why a p-n junction rectifies

A p-n junction diode is a one-way valve for current. When forward-biased (p-side positive relative to n-side), the barrier drops and current flows freely; when reverse-biased, the barrier widens and current is negligible. Connected in series with an ac source, the diode "chops off" half the waveform — the half that would reverse-bias it — leaving only the forward half-cycles to appear across the load. This is the essence of rectification: converting alternating current into a unidirectional (though still pulsating) current.

The circuit here has one diode, so it can only conduct during one polarity of the ac cycle. That makes it a half-wave rectifier.

Step-by-step reasoning

1. Identify the input

   The ac source delivers a sinusoidal voltage $V(t) = V_0 \sin(\omega t)$, which alternates between positive and negative half-cycles.

2. Diode behaviour during the positive half-cycle

   When $V(t) > 0$, the diode is forward-biased (assuming the diode's anode is connected to the positive terminal of the source). The diode conducts, and current flows through the load resistor $R_L$. The output voltage across $R_L$ follows the input (minus a small diode drop $\approx 0.7\,\text{V}$, often neglected in ideal analysis):

3. Diode behaviour during the negative half-cycle When $V(t) < 0$, the diode is reverse-biased. It blocks current (ideally zero reverse current), so no current flows through $R_L$. The output voltage is

4. The resulting waveform

   Over one full period $T = \frac{2\pi}{\omega}$, the output consists of:

   • A positive sinusoidal hump (the forward half-cycle),
   • followed by a flat zero-voltage interval (the blocked reverse half-cycle).

   This pattern repeats every cycle: positive hump, zero gap, positive hump, zero gap, … — the signature of half-wave rectification. …

Why this works — the core idea

A p-n junction diode is not a linear resistor. Its behaviour depends entirely on the polarity of the applied voltage:

• Forward bias (p-side at higher potential than n-side): the depletion region narrows, majority carriers flow freely, and the diode acts like a closed switch — current passes, and voltage appears across the load.
• Reverse bias (n-side at higher potential): the depletion region widens, only a tiny leakage current flows (negligible here), and the diode acts like an open switch — no current, no voltage across the load.

This property is called rectification: converting an alternating (or bidirectional) signal into a unidirectional one. The simplest circuit using this is a half-wave rectifier — a single diode in series with a load resistor.

Step-by-step solution

1. Identify the circuit configuration

The problem shows a p-n junction diode connected in series with a load resistor $R_L$. The square wave input of amplitude 10 V is applied across the series combination. The output voltage $V_o$ is taken across $R_L$.

2. Understand the input signal

The input is a square wave that alternates between $+10\ \text{V}$ and $-10\ \text{V}$ with equal time periods. There is no zero-voltage interval — it jumps instantly from one extreme to the other.

3. Analyse the forward-bias half-cycle

When the input is $+10\ \text{V}$, the p-side of the diode is at a higher potential than the n-side. This is forward bias. The diode turns on, and its voltage drop is small (≈0.7 V for silicon, but for an ideal diode we take it as zero). The entire input voltage appears across $R_L$:

4. Analyse the reverse-bias half-cycle

When the input switches to $-10\ \text{V}$, the p-side is now at a lower potential than the n-side. This is reverse bias. The diode turns off, behaving as an open circuit. No current flows through $R_L$, so no voltage develops across it:

5. Describe the output waveform …