Reverse-biased diode

Turn a p-n junction diode around — join its p-side to the negative terminal of a battery and its n-side to the positive terminal — and something almost the opposite of forward bias happens. Instead of a flood of current, you get a current so small it barely registers on an ordinary meter: a few microamperes, roughly a millionth of what the same diode carries when connected the "right way." This deliberately blocking connection is called reverse bias, and it is not a failure mode or a wiring mistake — it is the diode's other, equally essential personality. Reverse bias is what makes a diode a genuinely one-way device rather than just a resistor with a funny label. Push the depletion region's ions further apart with an external field pointing the same way as the internal built-in field, and majority carriers are shoved back from the junction rather than pulled across it. What little current does flow comes from an entirely different population of charge carriers — minority carriers, born from ordinary thermal jostling of the crystal lattice — and it behaves strangely: it barely changes as you crank the reverse voltage up, yet it climbs noticeably if you simply warm the diode. This article builds the reverse-bias picture on top of forward bias: same junction, same depletion region, same built-in field, but the external battery now points the other way. We track how the depletion region widens, why the reverse current "saturates," why it is carried by minority carriers instead of majority carriers, and how a diode's forward and reverse branches together sketch the full I-V characteristic the CBSE syllabus requires you to read off a graph — the very characteristic that makes the diode useful as a rectifier.

Key takeaways

Reverse bias connects the battery backwards: positive terminal to the n-side, negative terminal to the p-side, so the external field points the same way as the internal built-in field across the depletion region.
The depletion region widens and the potential barrier at the junction increases, because the external field adds to, rather than opposes, the internal field.
Majority carriers are blocked almost completely — they are pushed away from the junction, not pulled across it — so they contribute essentially nothing to reverse current.
A small reverse saturation current still flows, carried entirely by minority carriers generated by thermal activity in the crystal, typically in the microampere range.
This current "saturates": below the breakdown voltage, it stays nearly constant as reverse voltage increases, because it is limited by the rate of carrier generation, not by field strength.
Reverse current is strongly temperature-sensitive (it roughly doubles for every $10\,^\circ\text{C}$ rise in typical silicon diodes) even though it is nearly voltage-independent — the opposite behaviour to forward current.
Key idea: reversing the battery reinforces the internal field, widens the depletion region, blocks majority carriers almost entirely, and leaves only a small, temperature-sensitive, voltage-independent minority-carrier saturation current until breakdown.

The core insight

Reverse bias reinforces the diode's internal field, widening the depletion region and blocking majority carriers, so the only current that flows is a small, nearly voltage-independent, temperature-sensitive minority-carrier saturation current.

Where it fits in the syllabus

Within CBSE Class 12 Unit IX, "Semiconductor Electronics," reverse bias is the second half of the I-V characteristic the syllabus asks you to know: forward bias in the earlier article, reverse bias here, and together they explain why the diode is used as a rectifier in the article that follows. The syllabus keeps this qualitative — you are expected to describe how reverse biasing affects the depletion region and barrier potential, identify a reverse-biased connection from a circuit diagram, and read the reverse branch of the I-V graph, including the near-flat reverse saturation current below breakdown. Note that the mechanism of breakdown itself, the Zener diode, and any device beyond the plain p-n junction diode were removed from the CBSE 2025-26 rationalisation — this article stays inside the diode's ordinary reverse-bias operation, well below breakdown.

The physical picture

Start from the p-n junction at equilibrium, before any external battery is connected. Diffusion of majority carriers across the junction — electrons from the n-side, holes from the p-side — leaves behind a thin depletion region stripped of mobile carriers, with exposed positive ions on the n-side and exposed negative ions on the p-side. This charge separation sets up a built-in electric field across the depletion region, pointing from the n-side to the p-side, together with a built-in potential barrier that opposes further diffusion. Now connect the battery's positive terminal to the n-side and its negative terminal to the p-side — reverse bias. The externally applied field also points from n to p, in the same direction as the built-in field, so the two fields add rather than cancel.

V_{\text{barrier, reverse}} = V_0 + V_{\text{applied}}

where: $V_0$ is the built-in potential barrier already present at equilibrium (a fraction of a volt, set by the semiconductor material), and $V_{\text{applied}}$ is the magnitude of the external reverse voltage. The physical reading is direct: the effective barrier a majority carrier must climb to cross the junction has grown, not shrunk, and it keeps growing as you increase the reverse voltage. Because a stronger net field must be supported by more exposed ion charge on each side, the depletion region itself widens to match — wider depletion region, taller barrier, both consequences of the same reinforced field.

Figure

A p-n junction under reverse bias showing a widened depletion region with exposed ions, the battery connected with its positive terminal to the n-side and negative terminal to the p-side, the internal and external fields both pointing from n to p, and a tiny minority-carrier current crossing the junction while majority carriers are pushed away

With majority carriers pushed away and a taller barrier ahead of them, essentially none cross the junction. But the depletion region and its neighbourhood are never entirely free of charge carriers: ordinary thermal energy continually breaks a few covalent bonds, creating electron-hole pairs even where they are minorities — electrons on the p-side, holes on the n-side. The reinforced field readily sweeps any such minority carrier that wanders near the junction across to the other side, producing a small but nonzero reverse saturation current $I_0$ , typically a few microamperes for a silicon diode — roughly a million times smaller than a typical forward current at a comparable voltage magnitude.

Derivation AWhy reverse current saturates instead of growing with voltage

The goal here is to explain, using only the physical picture already built — no calculus beyond a rate argument — why the reverse current stops depending on the applied voltage once that voltage is large enough, producing the flat "saturation" plateau the syllabus requires you to read off the I-V graph, and why it depends instead on temperature. This matters because it is the single most commonly mis-stated fact about diode behaviour: students routinely (and wrongly) expect more voltage to mean more current, as it does on the forward side.

Step 1. Identify what actually limits the reverse current. Once a reverse voltage is applied, however small, the field across the depletion region already points strongly enough from n to p to sweep across essentially every minority carrier that reaches the junction's edge. The rate at which minority carriers are supplied — not how hard the field pulls them — is what sets the current, because the field's job (sweeping them across) is already saturated at "collect everyone who shows up."

Step 2. Identify what controls the supply rate. Minority carriers are generated purely by thermal breaking of covalent bonds throughout the crystal, at a rate that depends on temperature $T$ and the material's properties, not on the externally applied voltage $V_{\text{applied}}$ . Write this supply rate as $G(T)$ , carriers generated per second per unit volume near the junction.

Step 3. Combine the two facts. Since the field already sweeps across all carriers supplied by $G(T)$ , increasing $V_{\text{applied}}$ further cannot increase the current — there are simply no additional minority carriers waiting to be swept. The reverse current settles at a value set only by $G(T)$ , independent of $V_{\text{applied}}$ over a wide range: this is the reverse saturation current $I_0(T)$ , and it is the flat plateau seen on the reverse branch of the I-V graph below breakdown.

Step 4. Explain the temperature sensitivity. Because $G(T)$ increases sharply with temperature (more thermal energy breaks more covalent bonds per second), $I_0$ also increases sharply with temperature — roughly doubling for every $10\,^\circ\text{C}$ rise in a typical silicon diode — even though it stays almost flat against voltage. This is the opposite dependence pattern from forward current, which is strongly voltage-dependent (through the barrier-lowering effect) and only mildly temperature-dependent by comparison.

Worked examples

Board level - reading the reverse saturation current from an I-V graph

A silicon diode is reverse biased with a $9\ \text{V}$ battery, and a sensitive ammeter in series reads a reverse current of $2\ \mu\text{A}$ . The reverse voltage is then increased to $15\ \text{V}$ , well below the diode's rated breakdown voltage, and the ammeter reads almost the same current, about $2.1\ \mu\text{A}$ . Explain why the current barely changed despite the considerable voltage increase, and state what the ammeter reading represents.

Step 1. Identify the current. Both readings are the reverse saturation current $I_0$ , carried entirely by minority carriers swept across the junction by the reinforced field, since the diode is operating well below breakdown in both cases.

Step 2. Identify what limits this current. The number of minority carriers available per second is set by the rate of thermal generation at the operating temperature, not by field strength — the field already sweeps across essentially every minority carrier generated, true even at $9\ \text{V}$ .

Step 3. Apply this to the voltage change. Since carrier supply, not field strength, is the limiting factor, raising the voltage from $9\ \text{V}$ to $15\ \text{V}$ does not create more carriers to sweep, so the current stays nearly constant — matching the observed rise from $2\ \mu\text{A}$ to only $2.1\ \mu\text{A}$ .

Step 4. State the conclusion. Answer: the reading is the reverse saturation current, about $2\ \mu\text{A}$ , essentially unchanged because it is limited by thermal carrier generation, not by the applied reverse voltage. This near-flat behaviour is exactly what the reverse branch of the diode's I-V characteristic shows.

Competitive level - combining temperature and voltage dependence in one problem

The same diode from the board-level example is reverse biased at $9\ \text{V}$ , giving $I_0 = 2\ \mu\text{A}$ at room temperature. It is then heated in an oven to a temperature $20\,^\circ\text{C}$ higher, still at $9\ \text{V}$ reverse bias, and the reading rises to about $8\ \mu\text{A}$ . A second diode, identical in construction and kept at the original room temperature, is instead reverse biased at $20\ \text{V}$ (still well below breakdown). Predict, with reasoning, roughly what current the second diode should show, and identify which of the two changes — heating or raising the voltage — actually altered the underlying carrier-generation rate.

Step 1. Analyse the heated diode. A $20\,^\circ\text{C}$ rise corresponds to two "doubling" steps in the roughly-doubles-per- $10\,^\circ\text{C}$ rule, so $I_0$ should scale by about $2 \times 2 = 4$ , giving $2\ \mu\text{A} \times 4 = 8\ \mu\text{A}$ — matching the given reading exactly, confirming the temperature-driven mechanism from the derivation above.

Step 2. Analyse the second diode. At room temperature, the thermal carrier-generation rate $G(T)$ is unchanged from the original $9\ \text{V}$ case; only the reverse voltage has increased, from $9\ \text{V}$ to $20\ \text{V}$ , still well below breakdown. Since the derivation established that reverse current is limited by carrier supply rather than field strength once the field already sweeps across all generated carriers, raising the voltage further should not meaningfully change the current.

Step 3. State the prediction. Answer: the second diode should still read approximately $2\ \mu\text{A}$ , essentially unchanged from its $9\ \text{V}$ value, because raising the reverse voltage does not alter the thermal generation rate of minority carriers.

Step 4. Identify the mechanism. Answer: only heating altered the underlying carrier-generation rate $G(T)$ ; raising the voltage at fixed temperature does not, since it acts on the field, not on how many carriers are thermally created. This is the key discriminator examiners test: voltage moves the barrier, temperature moves the carrier supply, and reverse current below breakdown responds almost entirely to the second, not the first. Timing note: under ninety seconds once you recognise the pattern. Do not compute anything from Ohm's-law-style proportionality — reverse current below breakdown is governed by the "doubles per $10\,^\circ\text{C}$ " temperature rule and by nothing else; a voltage change alone, however large, should be flagged as producing essentially no change, which is itself the answer competitive papers are checking for.

Common traps

Assuming reverse bias means exactly zero current — a small, nonzero reverse saturation current always flows due to minority carriers; "blocking" is a strong approximation, not an absolute one.
Believing a larger reverse voltage always produces a larger reverse current, by analogy with the forward branch or with Ohm's law — below breakdown, reverse current is nearly saturated, essentially independent of voltage.
Forgetting that reverse current is carried by minority carriers, not majority carriers — the opposite of forward current, and the reason reverse current is so much smaller.
Overlooking or reversing the temperature dependence — students often treat reverse current as fixed regardless of conditions, when it is in fact strongly temperature-sensitive even while being voltage-insensitive.
Confusing the barrier's direction of change — reverse bias increases the barrier and widens the depletion region, the opposite of forward bias, because the external field reinforces rather than opposes the internal field.

Common trapReverse current depends on temperature, not on voltage

The single biggest mark-losing habit is expecting reverse current to behave like forward current or like a resistor: "more voltage, more current." Below breakdown, that instinct is simply wrong. Once the reverse field is strong enough to sweep across essentially every minority carrier generated near the junction, the current is capped by how many carriers thermal activity supplies per second — a quantity set by temperature, not by the size of the applied voltage. Increasing the reverse voltage from a few volts to many tens of volts, while staying below breakdown, changes the reverse current only marginally; increasing the temperature by even $10$ - $20\,^\circ\text{C}$ can double or quadruple it. If a question changes the temperature, expect the current to change; if it only changes the reverse voltage (below breakdown), expect the current to stay essentially flat.

You should be able to…

State which terminal (p or n) connects to the battery's positive terminal in reverse bias, and explain why the external and internal fields add.
Explain why the depletion region widens and the potential barrier increases under reverse bias.
Explain why reverse current is carried by minority carriers, and why this makes it much smaller than forward current.
Explain, in terms of carrier-supply rate versus field strength, why reverse saturation current stays nearly constant as reverse voltage increases below breakdown.
Explain why reverse saturation current increases sharply with temperature, and contrast this with the voltage-driven mechanism controlling forward current.
Read the reverse branch of a diode's I-V characteristic graph and identify the saturation plateau and (qualitatively) the onset of breakdown.

Ask the tutor: Curious how forward bias and reverse bias together let a diode turn alternating current into one-directional current — ask your AI tutor to connect this reverse-branch picture to the diode's use as a rectifier.

Doesn't a reverse-biased diode carry exactly zero current, since it is blocking?

Not quite — "blocking" is a useful approximation, not a literal statement. The diode blocks the abundant majority carriers almost completely, but a small population of minority carriers, generated by ordinary thermal activity in the crystal, is still swept across the junction by the reinforced field. This gives the reverse saturation current, typically only a few microamperes for silicon, which is why in circuit problems you can often approximate it as zero without much error — but "approximately zero" and "exactly zero" are different statements, and exam questions sometimes test exactly this distinction.

Why does the depletion region widen under reverse bias, when it was already there at equilibrium?

At equilibrium, the depletion region's width is whatever is needed to support the built-in field alone. Reverse bias adds an external field pointing the same way as the built-in field, so the total field across the junction is now stronger. A stronger field needs more exposed fixed ion charge to sustain it (from Gauss's-law-style reasoning), and the only way to expose more ion charge is for more mobile carriers to be swept away from the junction on each side — so the depletion region widens to match the taller, reinforced field.

If reverse current is basically constant, why does the I-V graph eventually shoot up steeply?

That steep rise is breakdown, and it happens once the reverse voltage grows large enough — well beyond the ordinary operating range this article covers. The physical mechanisms behind breakdown are outside the CBSE 2025-26 syllabus (the Zener diode and voltage regulation were explicitly removed), so you only need to know that ordinary reverse-bias operation, with its flat saturation current, stays safely below this breakdown voltage, and that the graph's steep upturn marks where that safe regime ends.

Is the reverse saturation current the same physical quantity in every diode?

No — its size depends on the material, the junction's construction, its temperature, and its area, so different diodes of the same type still show somewhat different $I_0$ values in practice. What is universal, and what the syllabus expects you to know, is the behaviour: nearly flat against reverse voltage below breakdown, strongly increasing with temperature, and orders of magnitude smaller than typical forward current — not a single fixed number common to all diodes.