The Junction Diode & its I–V Curve

The junction diode is the most-manufactured active device on Earth, and its one-way curve is the first thing every descendant relies on. Bridge rectifiers in every mains adapter use four diodes to turn AC into DC — the next article. Light-emitting diodes are forward-biased junctions whose recombination energy comes out as light; photodiodes and solar cells run the same junction in reverse or open-circuit to turn light into current. Zener diodes live deliberately in the breakdown region to pin a voltage, and fast Schottky diodes trade the $0.7\ \text{V}$ knee for a $0.2\ \text{V}$ one to switch quickly in power supplies. And the exponential I–V law is not just a diode fact: the same Boltzmann-factor physics sets the current in a transistor, so the equation you quoted here reappears at the heart of every amplifier and logic gate in the rest of this chapter. Master the curve, and the whole of electronics opens up.

A diode is nothing more than a single p–n junction with a wire (a terminal) attached to each end. The p-side terminal is called the anode, the n-side the cathode, and on the device a painted band marks the cathode. Everything the diode does follows from one question: which way does the external battery push against the junction's built-in barrier? Recall the equilibrium junction. Majority electrons from the n-side and holes from the p-side diffuse across and recombine near the boundary, leaving a thin depletion region stripped of mobile carriers and populated only by fixed, charged donor and acceptor ions. Those ions set up an internal electric field and a potential barrier — about $0.7\ \text{V}$ in silicon, $0.3\ \text{V}$ in germanium — pointing from the n-side to the p-side, which repels further diffusion. At equilibrium the diffusion of majority carriers is exactly balanced by the drift of the few minority carriers, and the net current is zero. Now bias it. In forward bias we connect the battery's positive terminal to the p-side and negative to the n-side. The battery's field opposes the junction's internal field, so it lowers the barrier. Holes on the p-side and electrons on the n-side are pushed toward the junction; the depletion region narrows. As the applied voltage climbs toward the barrier value, the barrier shrinks toward zero and majority carriers begin pouring across in enormous numbers. Below the knee almost nothing happens; just past it, the current rises almost vertically. In reverse bias we flip the battery: positive to the n-side, negative to the p-side. Now the battery's field adds to the junction's field, raising the barrier and pulling majority carriers away from the junction. The depletion region widens. Majority-carrier diffusion is choked off completely. What remains is a trickle: the thermally generated minority carriers — electrons on the p-side, holes on the n-side — that are swept across by the strong field. This is the reverse saturation current $I_0$, and it is tiny (microamps in germanium, nanoamps in silicon) and almost independent of how hard you reverse-bias, because it is limited by how fast heat generates minority carriers, not by the voltage.

So the diode's whole personality is asymmetry. One direction: open the gate, current floods, the diode barely resists. The other: shut the gate, current chokes to a trickle. To make that quantitative — and to see where the current turns on — we plot it.

The I–V characteristic is a graph of the current $I$ through the diode against the voltage $V$ across it, measured by sweeping a variable supply and reading a milliammeter (forward) or microammeter (reverse). By convention $V$ is positive in forward bias, negative in reverse. Because forward and reverse currents differ by a factor of thousands, the two halves are almost always drawn on different scales (milliamps forward, microamps reverse). In the forward region, for $V$ below the knee voltage $V_\gamma$ (also called the cut-in or threshold voltage), the current is negligible: the barrier is only partly cancelled and few carriers have enough energy to cross. Once $V$ passes the knee — about $0.7\ \text{V}$ for silicon, $0.3\ \text{V}$ for germanium — the barrier is essentially gone and the current shoots up almost vertically. Beyond the knee the diode drops only a small, nearly constant voltage; the current is then set by whatever resistance is in the external circuit. In the reverse region, the current is the tiny reverse saturation current $I_0$, nearly flat and independent of voltage — the diode is effectively "off." But if you keep increasing the reverse voltage, at the breakdown voltage $V_{BR}$ the reverse current suddenly rises very steeply (Zener or avalanche breakdown, the subject of a later article). For ordinary rectifier diodes $V_{BR}$ is tens to hundreds of volts and reaching it usually destroys the device; a diode built for breakdown (a Zener) uses it deliberately.

Two features deserve emphasis before we compute. First, the "knee" is not a hard wall: the current does not switch on abruptly at exactly $0.7\ \text{V}$; it rises exponentially the whole way, but so steeply that on a linear plot it looks like a sudden turn-on. The knee is simply where that exponential becomes visible against the milliamp scale. Second, the forward drop is not truly constant — it creeps up as current rises (roughly $60\ \text{mV}$ per tenfold increase, as we will see) — but for circuit work we approximate it as a fixed $0.7\ \text{V}$.

The smooth exponential shape of the forward curve, and the flat saturation of the reverse, are both captured by a single relation, the diode equation (also the Shockley equation, after William Shockley, who derived it in 1949):

where: $I$ is the diode current, $V$ the voltage across it (positive forward, negative reverse), $I_0$ the reverse saturation current, $n$ the ideality factor (between $1$ and $2$; take $n=1$ for an ideal junction), and $V_T = kT/e$ the thermal voltage. At room temperature ($T = 300\ \text{K}$, Boltzmann constant $k = 8.617\times10^{-5}\ \text{eV/K}$),

Deriving equation (1) properly requires counting how electrons and holes are distributed in energy — Fermi–Dirac statistics — and how they diffuse across the depletion region, which is genuinely beyond school algebra and needs calculus. We therefore quote it, exactly as every Class 12 text does, and only read off its shape. And its shape says everything: For a forward bias more than a few $V_T$ (i.e. $V \gtrsim 0.1\ \text{V}$), the exponential is huge and the "$-1$" is negligible: $I \approx I_0\,e^{V/nV_T}$ grows explosively — that is the near-vertical rise past the knee. For a reverse bias of more than a few $V_T$, the exponential collapses to nearly zero, so $I \approx I_0(0 - 1) = -I_0$: a small constant current in the reverse direction, exactly the saturation plateau. And at $V = 0$, $I = I_0(1-1) = 0$, as it must. One equation, both halves of the curve. The physical reading is the barrier picture again: the fraction of carriers with enough thermal energy to climb a barrier of height $eV$ follows a Boltzmann factor $e^{-eV/kT}$, and forward bias lowers the barrier, multiplying the current that crosses by $e^{V/V_T}$. The steepness of the diode is the steepness of that exponential.

A diode is almost never used alone; it sits in series with a resistor and a supply. The whole art of "using" a diode is finding its operating point — the one $(V, I)$ pair that satisfies both the diode's curve and the circuit at once. Apply Kirchhoff's voltage law to a supply $V_S$ driving a diode (voltage $V_D$) in series with a resistor $R$:

where: $V_S$ is the source voltage, $V_D$ the voltage across the diode, $I$ the common current, and $R$ the series resistance. Rearranged, $I = (V_S - V_D)/R$ — a straight line on the I–V plane called the load line, with slope $-1/R$. The diode must live on its own curve and on this line; where they cross is the operating point. That is the exact, graphical answer. For hand calculation we simplify.

The constant-drop model is why, in every later circuit — rectifiers, clippers, logic gates — you may simply subtract $0.7\ \text{V}$ (Si) or $0.3\ \text{V}$ (Ge) per conducting diode and treat the rest as a resistor problem. But the model hides a subtlety worth one more derivation: how much the diode actually resists small changes.

Board level — diode and resistor, silicon versus germanium (CBSE and ICSE pattern)

A silicon diode is connected in series with a $220\ \Omega$ resistor across a $5.0\ \text{V}$ battery, correctly forward biased. Take the diode's forward drop as $0.7\ \text{V}$. Find (a) the current in the circuit and (b) the voltage across the resistor. (c) Repeat for a germanium diode ($0.3\ \text{V}$ drop).

Step 1. Forward-biased and conducting, so use the constant-drop model $I = (V_S - V_\gamma)/R$. With $V_S = 5.0\ \text{V}$, $V_\gamma = 0.7\ \text{V}$, $R = 220\ \Omega$: the voltage left for the resistor is $5.0 - 0.7 = 4.3\ \text{V}$.

Step 2. Current: $I = 4.3/220 = 0.01955\ \text{A} \approx \boxed{19.5\ \text{mA}}$.

Step 3. Voltage across the resistor: $V_R = IR = 4.3\ \text{V}$ (which is just $V_S - V_\gamma$, as it must be by KVL). Answer (a): $19.5\ \text{mA}$; (b): $4.3\ \text{V}$.

Step 4. Germanium: drop is $0.3\ \text{V}$, so $V_R = 5.0 - 0.3 = 4.7\ \text{V}$ and $I = 4.7/220 = 0.02136\ \text{A} \approx \boxed{21.4\ \text{mA}}$. The germanium diode passes slightly more current because it "steals" less of the supply — a direct consequence of its lower knee voltage.

Board level — sizing the series resistor (CBSE and A-Level pattern)

A silicon diode must run at $10\ \text{mA}$ from a $6.0\ \text{V}$ supply. What series resistance is required?

Step 1. By KVL the resistor must drop everything the diode does not: $V_R = V_S - V_\gamma = 6.0 - 0.7 = 5.3\ \text{V}$.

Step 2. Ohm's law for the required current: $R = V_R/I = 5.3/0.010 = \boxed{530\ \Omega}$.

Step 3. A quick sanity check: use the nearest standard value ($560\ \Omega$) and the current becomes $5.3/560 = 9.5\ \text{mA}$ — close, and safely on the low side. This is exactly how you choose the resistor for an LED, which is just a diode with a higher knee (treated in the special-diodes article).

Competitive level — the Shockley equation, forward and reverse (JEE Main and Advanced pattern)

A silicon diode has reverse saturation current $I_0 = 1.0\ \mu\text{A}$ and ideality factor $n = 1$ at $300\ \text{K}$, where $V_T = 0.026\ \text{V}$. Using $I = I_0\big(e^{V/nV_T}-1\big)$, find the current at (a) a forward bias $V = +0.20\ \text{V}$ and (b) a reverse bias $V = -0.20\ \text{V}$. (c) By how much must the forward voltage increase to raise the current tenfold?

Step 1. Forward. The exponent is $V/V_T = 0.20/0.026 = 7.7$, so $e^{7.7} \approx 2.2\times10^{3}$. Then $I = 1.0\times10^{-6}\,(2.2\times10^{3} - 1) \approx 1.0\times10^{-6}\times2.19\times10^{3} = 2.19\times10^{-3}\ \text{A} \approx \boxed{2.2\ \text{mA}}$. From $1\ \mu\text{A}$ of saturation to milliamps: the exponential in action.

Step 2. Reverse. Now $V/V_T = -7.7$, so $e^{-7.7} = 4.6\times10^{-4}$, essentially zero. Then $I = 1.0\times10^{-6}\,(4.6\times10^{-4} - 1) \approx -1.0\times10^{-6}\ \text{A} = \boxed{-1.0\ \mu\text{A}}$ — the current saturates at $-I_0$, independent of how hard we reverse bias. This is the flat reverse plateau.

Step 3. Tenfold rise. To multiply $I$ by $10$ we need $e^{\Delta V/V_T} = 10$, so $\Delta V = V_T\ln 10 = 0.026\times2.303 = 0.0599\ \text{V} \approx \boxed{60\ \text{mV}}$. This famous "$60\ \text{mV}$ per decade" is why the forward curve looks vertical: a factor of $1000$ in current costs only about $180\ \text{mV}$. Timing note: about two minutes. The only real work is one exponential; keep $V_T = 0.026\ \text{V}$ memorised so $V/V_T$ is a single division. Recognising that reverse bias forces $I \to -I_0$ by inspection (the exponential dies) saves the whole of part (b), and the "$60\ \text{mV}$ per decade" rule is worth carrying as a standard result — it turns part (c) into one line.

Competitive level — static and dynamic resistance (NEET and JEE pattern)

From a silicon diode's forward characteristic, the current is $30\ \text{mA}$ at $0.75\ \text{V}$, and it rises from $20\ \text{mA}$ to $40\ \text{mA}$ as the voltage goes from $0.73\ \text{V}$ to $0.77\ \text{V}$. Find (a) the static (DC) resistance at $0.75\ \text{V}$ and (b) the dynamic (AC) resistance there.

Step 1. Static resistance is simply voltage over current at the point: $R_{dc} = V/I = 0.75/0.030 = \boxed{25\ \Omega}$.

Step 2. Dynamic resistance is the change in voltage over the change in current across the point: $\Delta V = 0.77 - 0.73 = 0.04\ \text{V}$, $\Delta I = 40 - 20 = 20\ \text{mA} = 0.020\ \text{A}$. So $r_d = \Delta V/\Delta I = 0.04/0.020 = \boxed{2\ \Omega}$.

Step 3. The dynamic resistance ($2\ \Omega$) is about a twelfth of the static ($25\ \Omega$) — the hallmark of the steep exponential. Cross-check with the quoted $r_d = V_T/I = 0.026/0.030 = 0.87\ \Omega$: the same order of magnitude, the difference coming from the finite step size and a real diode's series resistance. Answers: $R_{dc} = 25\ \Omega$, $r_d = 2\ \Omega$. Timing note: under a minute. Do not confuse the two — static uses the point $(V/I)$, dynamic uses the interval $(\Delta V/\Delta I)$. Read which the question wants: "resistance of the diode" usually means static; "resistance to a small signal" or "AC resistance" means dynamic.

Confusing static and dynamic resistance. $V/I$ (static) and $\Delta V/\Delta I$ (dynamic) give very different numbers because the curve is not a straight line through the origin. JEE and NEET items exploit this constantly; read whether the question wants the resistance at a point or to a small change.

Using the wrong knee value. Silicon is $0.7\ \text{V}$, germanium $0.3\ \text{V}$ — mixing them up shifts every current answer. State which material you are using and subtract its knee. Most modern diodes are silicon unless told otherwise.

Treating reverse current as exactly zero. It is the small saturation current $I_0$, not zero, and it doubles roughly every $10\ ^\circ\text{C}$. On a "is it forward or reverse biased, find the current" question, reverse bias means "$\approx I_0$, essentially off," not "no current at all."

Forgetting the diode drop in the loop. In a series circuit the diode eats $0.7\ \text{V}$ (Si) before the resistor gets anything: $I = (V_S - V_\gamma)/R$, not $V_S/R$. Dropping the $0.7\ \text{V}$ is the single most common numerical slip.

Calling the $0.7\ \text{V}$ a source. The diode drop is a fall in potential in the direction of current, absorbing energy, never a battery that can drive current on its own.

Board scope. The qualitative diode — one-way conduction, forward/reverse bias, the I–V shape, the knee, the idea of breakdown — is core on CBSE, ICSE (ISC), JEE Main, JEE Advanced, NEET, IGCSE 0625, and A-Level 9702. The quantitative tools split by board: constant-drop for JEE, NEET, and A-Level 9702, dynamic resistance for JEE and NEET, but the exponential diode equation with $V_T$ for JEE only — CIE 9702 stays qualitative. IGCSE keeps it purely qualitative (recognise the symbol, know it conducts one way, sketch the characteristic — no $V_T$, no Shockley equation). AP Physics 2 does not carry semiconductor electronics at all, so it is not listed. Answer an IGCSE question in words and a sketch; save the exponential for the competitive boards.

Texts. NCERT Physics Class 12, Chapter 14 ("Semiconductor Electronics"), the junction-diode and I–V-characteristic sections · H. C. Verma, Concepts of Physics Vol. 2, "Semiconductors and Semiconductor Devices" · Boylestad & Nashelsky, Electronic Devices and Circuit Theory, for the load line and diode models · Sedra & Smith, Microelectronic Circuits, for the Shockley equation done properly.

Provenance. Constants and anchors from the lane spec: $V_T = kT/e = 8.617\times10^{-5}\times300/1 \approx 0.026\ \text{V}$; Si knee $0.7\ \text{V}$, Ge knee $0.3\ \text{V}$. Recompute: constant-drop currents $(5-0.7)/1000 = 4.3\ \text{mA}$ and $(5-0.3)/1000 = 4.7\ \text{mA}$; example currents $(5-0.7)/220 = 19.5\ \text{mA}$, $(5-0.3)/220 = 21.4\ \text{mA}$; resistor sizing $(6-0.7)/0.010 = 530\ \Omega$; static $0.72/0.015 = 48\ \Omega$ and $0.75/0.030 = 25\ \Omega$; dynamic $0.02/0.010 = 2\ \Omega$ and $0.04/0.020 = 2\ \Omega$; quoted $r_d = V_T/I$ giving $0.026/0.015 = 1.7\ \Omega$ and $0.026/0.030 = 0.87\ \Omega$; Shockley equation at $V=\pm0.20\ \text{V}$ with $I_0 = 1\ \mu\text{A}$, $V_T=0.026\ \text{V}$: forward $\approx 2.2\ \text{mA}$, reverse $\approx -1.0\ \mu\text{A}$; decade step $V_T\ln 10 = 0.060\ \text{V}$. All in verify_hs2-diode-iv-characteristics.py.

Prerequisites. The p–n Junction — the depletion region, barrier potential, and diffusion-versus-drift equilibrium that this article puts a battery across. Everything here is that junction, biased. Leads to. Rectifiers: Half-Wave & Full-Wave, which chains diodes to turn AC into DC using exactly the one-way conduction established here. For the exponential law derived from carrier statistics rather than quoted, the university seam on semiconductor diode lasers and junction device physics treats the Shockley equation, Fermi–Dirac occupation, and carrier diffusion in full (a go-deeper pointer, never an exam dependency here).