Q.A $220\,\text{V}$ (rms) A.C. supply is connected between points A and B. From A, an ideal diode (arrow pointing from A into the circuit) charges a capacitor C, whose other plate connects to B. There is no load resistor across the capacitor. What is the steady potential difference $V$ across the capacitor?

P-N Junction Rectification: Turning AC into DC

Imagine you have a water pipe with a one-way valve. Water can flow freely in one direction, but if you try to push it the other way, the valve slams shut and nothing moves. That's exactly what a p-n junction does — but with electric current instead of water.

The Intuition: Why Does It Let Current Flow Only One Way?

A p-n junction is made by joining two pieces of semiconductor: one p-type (with extra "holes" — think of them as positive charge carriers) and one n-type (with extra electrons). At the junction, something interesting happens.

Electrons from the n-side diffuse into the p-side, and holes from the p-side diffuse into the n-side. They meet and recombine, leaving behind a region with no free charge carriers — the depletion region. This region acts like a tiny battery, creating an internal electric field that points from n to p.

Now here's the key: this internal field opposes the flow of majority carriers. It's like a spring that's been compressed — it wants to push things back.

Forward Bias: Opening the Gate

Connect the p-side to the positive terminal of a battery and the n-side to the negative terminal. This is forward bias.

The external battery pushes holes from p toward n, and electrons from n toward p. They both march toward the depletion region. If the battery voltage is large enough (about 0.7 V for silicon), it overcomes the internal field. The depletion region shrinks, and current flows easily.

Reverse Bias: Locking the Gate

Now swap the battery: p-side to negative, n-side to positive. This is reverse bias.

The battery pulls holes away from the junction on the p-side, and electrons away on the n-side. The depletion region widens — the spring stretches. No current flows (except a tiny leakage current from minority carriers, which we ignore for now).

The Precise Statement

Mathematically, the current-voltage relationship is given by the Shockley diode equation:

Where:

• $I$ = diode current
• $I_S$ = reverse saturation current (tiny, typically $10^{-12}$ to $10^{-6}$ A)
• $V$ = applied voltage (positive for forward bias, negative for reverse)
• $n$ = ideality factor (usually 1 for ideal, 1–2 for real diodes)
• $V_T$ = thermal voltage $\approx 25.85$ mV at room temperature (300 K)

For forward bias ($V > 0$), the exponential term dominates, so $I \approx I_S e^{V/(nV_T)}$ — current grows rapidly.

For reverse bias ($V < 0$), $e^{V/(nV_T)} \approx 0$, so $I \approx -I_S$ — a tiny constant leakage current.

How Rectification Works in Practice …