Semiconductor diode

Every phone charger, every laptop adapter hides the same tiny piece of crystal doing the same one job: letting current flow one way and refusing to let it flow the other. That crystal is a semiconductor diode — nothing more than a single p-n junction, grown as one continuous piece of silicon, brought out to two wires. No moving parts, no filament, no vacuum tube — just a boundary between p-type and n-type silicon, and the electric field that boundary builds for itself. What makes the diode worth a whole topic is not its construction — you already met that as the p-n junction — but its behaviour as a circuit element. Connect it one way ("forward") and it conducts almost like a closed switch once you cross a small threshold voltage. Connect it the other way ("reverse") and it behaves almost like an open switch, letting through only a current so small it is measured in microamperes. Plot current against voltage for both directions on the same graph and you get one of the most recognisable curves in electronics: flat, then a sudden steep knee on one side; flat and nearly zero on the other. That single curve — the diode's I-V characteristic — is the reason a diode can rectify alternating current into direct current, the application waiting in the next node. This article stays entirely inside what CBSE actually examines: how forward and reverse bias reshape the depletion region and the barrier potential, why that reshapes the I-V curve the way it does, and how to read threshold voltage, breakdown, and static/dynamic resistance off that curve. It does not touch Zener regulation, LEDs, photodiodes, solar cells, or transistors — all of those are outside the 2025-26 CBSE Class 12 syllabus, however familiar the names sound.

Key takeaways

A semiconductor diode is a two-terminal device made from one p-n junction; the anode is the p-side terminal, the cathode is the n-side terminal.
Forward bias (positive terminal to p-side, negative to n-side) opposes the internal barrier field, narrows the depletion region, and lets majority carriers cross freely once the applied voltage exceeds a threshold/knee voltage — about $0.7\ \text{V}$ for silicon, $0.3\ \text{V}$ for germanium.
Reverse bias (positive terminal to n-side, negative to p-side) adds to the barrier field, widens the depletion region, and permits only a tiny, almost voltage-independent reverse saturation current ( $\mu\text{A}$ range) from minority carriers — until breakdown at a large reverse voltage.
The I-V characteristic is strongly non-linear: nearly flat below threshold, then a steep exponential-like rise in forward bias; nearly flat and near-zero in reverse bias. This non-linearity is exactly why a diode is not an ohmic resistor.
Two useful resistances can be read from the curve: static (DC) resistance $R = V/I$ at one operating point, and dynamic (AC) resistance $r_d = \Delta V/\Delta I$ , the local slope's reciprocal — these differ because the curve is curved, not straight.
Because forward current at a given voltage magnitude is many orders larger than reverse current, the diode acts as a near-ideal one-way valve for conventional current — the property exploited when it is used as a rectifier.
Key idea: a p-n junction packaged as a two-terminal diode conducts freely once forward bias past a threshold (about 0.7 V for silicon) collapses the barrier, but blocks almost all current in reverse bias where the barrier widens and only a tiny, saturated minority-carrier current survives — the strong asymmetry recorded in its non-linear I-V characteristic.

The core insight

A semiconductor diode's I-V characteristic is sharply asymmetric — negligible current below a threshold voltage of about $0.7\ \text{V}$ (silicon) in forward bias, then a steep rise, versus a nearly constant, tiny reverse saturation current in reverse bias — because forward bias narrows the depletion region's barrier while reverse bias widens it.

Where it fits in the syllabus

This is the device-level topic of Unit IX, "Semiconductor Electronics," coming directly after the formation of the p-n junction and directly before the diode's use as a rectifier — the three form a tight teaching sequence. The 2025-26 CBSE syllabus keeps this topic at exactly the level treated here: reading and explaining the forward- and reverse-bias I-V characteristic, the threshold/knee voltage, and the qualitative idea of breakdown, built on the (also qualitative) depletion-region picture from the p-n junction. It drops the wider device family older textbooks bundled in — Zener diodes and voltage regulation, LEDs, photodiodes, solar cells, and any transistor or digital-logic content are not part of the current prescription, so none of that appears below.

The physical picture

Recall the two ingredients from the p-n junction: a depletion region with almost no free carriers, and a barrier potential (built-in field) across it that opposes further diffusion. Everything about the diode's behaviour follows from what an external battery does to that barrier when connected two different ways. In forward bias, the battery's positive terminal connects to the p-side and its negative terminal to the n-side. The applied field points opposite to the internal barrier field, so it partially cancels the barrier, and the depletion region narrows. Below the threshold voltage $V_0$ , the applied voltage has not yet overcome enough of the barrier for significant numbers of majority carriers (holes from the p-side, electrons from the n-side) to cross — current stays small. Past $V_0$ , the barrier is largely defeated, and majority carriers pour across in large numbers for small further increases in voltage: current rises steeply. In reverse bias, the positive terminal connects to the n-side and negative to the p-side. The applied field now adds to the internal barrier field, widening the depletion region and strengthening the barrier majority carriers face, so their current is suppressed almost completely. A small current still flows, carried by minority carriers that the strengthened field sweeps across the instant they are thermally generated nearby. Since this current is limited by the rate of thermal generation, not by field strength, it quickly saturates to a small, nearly constant value — the reverse saturation current, $I_0$ , typically only a few microamperes for silicon. Increase the reverse voltage far enough and a sudden large current sets in at the breakdown voltage, a regime this syllabus only needs you to know exists, not to analyse. Putting forward and reverse branches on one $I$ -versus- $V$ graph, with forward current plotted upward and reverse voltage plotted to the left (standard convention), gives the characteristic curve:

I_F \approx 0 \ \text{for} \ 0 < V_F < V_0, \qquad I_F \ \text{rises steeply for} \ V_F > V_0, \qquad I_R \approx -I_0 \ (\text{nearly constant}) \ \text{for} \ V_R < V_{\text{breakdown}}. \qquad (1)

where: $I_F$ and $V_F$ are the forward current and voltage across the diode; $V_0$ is the threshold (knee) voltage, about $0.7\ \text{V}$ for silicon and $0.3\ \text{V}$ for germanium; $I_R$ is the reverse current, conventionally plotted negative and vastly smaller than typical forward currents; $I_0$ is the reverse saturation current, microampere-range for silicon; $V_{\text{breakdown}}$ is the reverse voltage where current rises sharply again, outside normal operation. The physical reading is "almost nothing, then suddenly a lot" on one side, and "almost nothing, staying almost nothing" on the other — an extreme asymmetry no ordinary resistor shows.

Figure

A current versus voltage graph for a silicon diode showing a nearly flat forward branch up to about zero point seven volts followed by a steep rise, and a nearly flat near zero reverse branch that stays small until a distant breakdown voltage

Static and dynamic resistance

Because the characteristic is curved, a diode does not have one fixed resistance the way a resistor does. The static (DC) resistance at an operating point is the ratio of the total voltage to total current there:

R_{\text{DC}} = \frac{V}{I}. \qquad (2)

where: $V$ and $I$ are the voltage and current at one specific point on the curve — useful for a quick DC estimate, but it changes from point to point since the curve is not a straight line through the origin. The dynamic (AC) resistance is the reciprocal of the local slope — how much current changes for a small change in voltage right around an operating point:

r_d = \frac{\Delta V}{\Delta I}. \qquad (3)

where: $\Delta V$ and $\Delta I$ are small changes about the chosen operating point. On the flat part of either branch, the slope is nearly zero, so $r_d$ is very large; beyond the knee the forward curve is steep, so $r_d$ becomes small — this falling dynamic resistance is the graphical signature of "the diode has turned on."

Forward and reverse bias — the derivation

Derivation AReading the threshold voltage and forward/reverse asymmetry from the barrier potential

The goal is to show, using only the depletion-region and barrier-potential picture already built for the p-n junction — no band diagrams, no exponential diode equation, both beyond this syllabus's qualitative treatment — why the forward branch of the I-V curve has a threshold followed by a steep rise, while the reverse branch stays flat and tiny until breakdown, and why these two facts together make the diode strongly asymmetric (non-linear, non-ohmic) rather than behaving like a fixed resistor in either direction.

Step 1. Start from the unbiased junction. With no external voltage, diffusion of majority carriers is exactly balanced by drift of the few minority carriers pulled back by the built-in barrier field. Net current is zero, with a fixed depletion width and barrier potential $V_B$ at equilibrium — the barrier any external voltage must fight or reinforce.

Step 2. Apply forward bias and track the depletion width. Connecting the p-side to the higher potential drives the external field opposite to the internal field, so the two partially cancel. The effective barrier majority carriers see is reduced, and the depletion region narrows. At small forward voltages the reduction is still small compared to $V_B$ , so few carriers yet have enough energy to cross: current stays close to zero.

Step 3. Cross the threshold voltage $V_0$ . Once the applied forward voltage approaches the built-in barrier potential (about $0.7\ \text{V}$ for silicon), the barrier is nearly cancelled and the depletion region has narrowed drastically. Majority carriers — holes from p to n, electrons from n to p — now cross in large numbers for little further increase in voltage. This is the "knee": current negligible until now rises steeply, because the barrier has essentially been removed, not merely reduced a little at a time.

Step 4. Apply reverse bias and track the same width. Connecting the n-side to the higher potential adds the external field to the internal field, so the barrier increases and the depletion region widens, pushing majority carriers further away — that current is suppressed almost entirely, the opposite of Step 2. What remains is the trickle of minority carriers generated thermally near the junction, swept across as soon as they appear. Since this trickle is set by a generation rate, not by how hard the field pulls, more reverse voltage does not raise the current much — it saturates at $I_0$ , orders of magnitude below typical forward currents.

Step 5. Combine the two branches. The forward branch is barrier-limited up to $V_0$ , then carrier-supply-limited (steep); the reverse branch is generation-limited throughout (flat, tiny) until breakdown (outside this syllabus). Because the limiting process is completely different in the two directions, the I-V curve cannot be a single straight line through the origin the way an ohmic resistor's is — the entire content of "the diode is a non-linear, one-way device," and exactly what makes rectification possible in the next topic.

Worked examples

Board level - reading threshold voltage and static resistance from a forward I-V table

A student records forward voltage and current for a silicon diode: at $V = 0.6\ \text{V}$ , $I = 0.5\ \text{mA}$ ; at $V = 0.7\ \text{V}$ , $I = 5\ \text{mA}$ ; at $V = 0.8\ \text{V}$ , $I = 40\ \text{mA}$ . (a) Identify the approximate threshold voltage. (b) Find the static resistance at $V = 0.7\ \text{V}$ . (c) Find the dynamic resistance between $V = 0.7\ \text{V}$ and $V = 0.8\ \text{V}$ .

Step 1. Locate the threshold from the table. Current jumps tenfold (to $5\ \text{mA}$ ) between $V = 0.6\ \text{V}$ and $V = 0.7\ \text{V}$ , and eightfold again to $V = 0.8\ \text{V}$ . The steep rise is clearly under way by $0.7\ \text{V}$ , consistent with the standard silicon threshold voltage.

Step 2. Static resistance at $V = 0.7\ \text{V}$ using equation (2): $R_{\text{DC}} = V/I = 0.7\ \text{V} / 5\times10^{-3}\ \text{A} = 140\ \Omega$ .

Step 3. Dynamic resistance between the two nearby points using equation (3): $\Delta V = 0.8 - 0.7 = 0.1\ \text{V}$ , $\Delta I = 40\ \text{mA} - 5\ \text{mA} = 35\ \text{mA} = 35\times10^{-3}\ \text{A}$ , so $r_d = \Delta V/\Delta I = 0.1 / 0.035 \approx 2.9\ \Omega$ .

Step 4. Compare and interpret. Answer: threshold $V_0 \approx 0.7\ \text{V}$ ; static resistance at $0.7\ \text{V}$ is $140\ \Omega$ ; dynamic resistance from $0.7\ \text{V}$ to $0.8\ \text{V}$ is about $2.9\ \Omega$ . Dynamic resistance is far smaller here because the curve is climbing steeply — a small extra push in voltage buys a disproportionately large extra current, the "barrier nearly gone" picture from Step 3 of the derivation.

Competitive level - comparing forward and reverse operating points on the same diode

The same silicon diode from the board-level example is now reverse biased at $V_R = 5\ \text{V}$ , and the reverse saturation current is measured as $I_0 = 2\ \mu\text{A}$ , essentially unchanged up to $V_R = 9\ \text{V}$ where $I_0$ is still $2.1\ \mu\text{A}$ . (a) Find the static resistance in reverse bias at $5\ \text{V}$ . (b) Compare it with the forward static resistance at $0.7\ \text{V}$ found above, as a ratio. (c) Comment on what this ratio means for the diode's use as a one-way valve.

Step 1. Reverse static resistance using equation (2): $R_{\text{DC,rev}} = V_R/I_0 = 5\ \text{V} / 2\times10^{-6}\ \text{A} = 2.5\times10^{6}\ \Omega = 2.5\ \text{M}\Omega$ .

Step 2. Ratio to the forward static resistance from Step 2 above ( $140\ \Omega$ ): $\dfrac{R_{\text{DC,rev}}}{R_{\text{DC,fwd}}} = \dfrac{2.5\times10^{6}}{140} \approx 1.8\times10^{4}$ .

Step 3. Sanity-check using the near-constancy of $I_0$ : even at $V_R = 9\ \text{V}$ , current only crept up from $2\ \mu\text{A}$ to $2.1\ \mu\text{A}$ — direct numerical evidence of saturation, confirming reverse current is set by generation rate, not field strength.

Step 4. Interpret. Answer: reverse resistance at $5\ \text{V}$ is $2.5\ \text{M}\Omega$ , roughly eighteen thousand times the forward resistance at $0.7\ \text{V}$ . A ratio this large, between the two directions of the same device, is exactly what makes a diode a near-ideal one-way valve for current. Timing note: about two minutes. The whole problem is two applications of $R = V/I$ plus one ratio; the only trap is unit care — keep reverse current in microamperes and convert to amperes before dividing.

Common traps

Treating the diode as an ohmic (linear) resistor with one fixed resistance — its I-V graph is markedly non-linear, especially past the knee; $V = IR$ with a constant $R$ never describes the whole curve, only a resistance defined at one point.
Confusing static resistance ( $V/I$ at a point) with dynamic resistance ( $\Delta V/\Delta I$ , the local slope's reciprocal) — they can differ by orders of magnitude, as the worked example above shows ( $140\ \Omega$ versus $2.9\ \Omega$ ).
Assuming zero current flows in reverse bias — a small but non-zero reverse saturation current always flows; it only becomes negligible compared to forward current, not literally zero.
Confusing anode and cathode, or forward and reverse polarity — the anode is the p-side; forward bias connects the battery's positive terminal to the anode and negative to the cathode.
Believing larger reverse voltage always produces proportionally larger reverse current — below breakdown it is nearly saturated, the opposite of the steep rise seen in forward bias beyond threshold.
Quoting the threshold voltage without naming the material — $0.7\ \text{V}$ is specifically for silicon; germanium has a threshold near $0.3\ \text{V}$ .

You should be able to...

Identify the anode (p-side) and cathode (n-side) of a diode and state which terminal connects to the battery's positive terminal in forward bias versus reverse bias.
Sketch and interpret the I-V characteristic in both forward and reverse bias, correctly showing the threshold/knee voltage and the near-flat reverse branch.
Explain, using the depletion region and barrier potential, why forward current is negligible below threshold and rises steeply beyond it.
Explain why reverse current is small and nearly constant (saturated) below breakdown, in terms of minority-carrier generation rate rather than applied field strength.
Compute static resistance ( $V/I$ ) and dynamic resistance ( $\Delta V/\Delta I$ ) at a given operating point from tabulated or graphed I-V data, and explain why they differ.
Explain why a diode is called a non-linear (non-ohmic) device, and connect this asymmetry to its usefulness as a rectifier.

Ask the tutor: Ready to see this same I-V asymmetry put to work — ask your AI tutor how a single forward-conducting, reverse-blocking diode converts an alternating input into a one-directional (rectified) output.

Why is the threshold voltage about 0.7 V for silicon specifically — where does that number come from?

It comes from silicon's own built-in barrier potential at the junction, set by the material's band gap and doping — the same barrier you met when the junction first forms. Applied forward voltage has to cancel most of that barrier before carriers cross freely, and for silicon that takes about $0.7\ \text{V}$ . Germanium's smaller band gap gives a lower threshold, near $0.3\ \text{V}$ . You don't need to derive the number — the syllabus wants you to know and use it.

If reverse current is basically constant, why do textbooks even show a slope in the reverse branch?

Real reverse branches are drawn with a very slight downward creep as reverse voltage increases, because a small extra fraction of minority carriers get generated at higher fields, plus a tiny leakage contribution. It is genuinely close to flat compared to the forward branch's steepness, which is why "nearly constant" or "saturated" is the correct description for this syllabus, not perfectly flat.

Is dynamic resistance the same thing as the resistance value printed on an actual resistor?

No — an ordinary resistor's resistance is one fixed number because its I-V graph is a straight line through the origin (Ohm's law), so static and dynamic resistance are identical everywhere on it. A diode's curve is not straight, so the two differ and change from point to point. Dynamic resistance matters when a diode carries a small AC signal riding on a DC bias — the small-signal behaviour follows the local slope, not the overall $V/I$ ratio.

Why doesn't this article cover Zener diodes or LEDs — aren't they diodes too?

They are diodes structurally, but the 2025-26 CBSE rationalisation removed Zener voltage regulation, LEDs, photodiodes and solar cells from this chapter, along with transistors and digital logic. This article sticks to exactly what remains prescribed — the ordinary junction diode's I-V characteristic in forward and reverse bias — so nothing here implies those devices are still on your board syllabus, even if you meet them in an older textbook.