How this instrument works
Forward current through a diode is carriers diffusing across the junction, and the count able to make the crossing at a given instant follows Boltzmann statistics: raise the forward voltage and the population with enough energy to climb the barrier grows exponentially, not linearly. That is the entire shape of the equation — I = Iₛ·(exp(V⁄nV_T) − 1). The −1 keeps the model honest at both ends: it forces zero current at zero volts, and under reverse bias it caps the result at −Iₛ, the small steady leakage a real junction shows instead of never bottoming out.
William Shockley derived this relation in a 1949 Bell Labs paper by solving the minority-carrier diffusion equation on either side of the junction and matching boundary conditions at its edge, work that also laid the groundwork for the junction transistor. The ideality factor n he left as a free parameter is doing real work: values near 1 mean diffusion current dominates, matching the derivation cleanly, while values climbing toward 2 mean recombination inside the depletion region is taking over instead — common in ordinary silicon diodes at low bias and across much of an LED's operating range.
The formula only ever describes the ideal junction. It leaves out the diode's own bulk silicon and lead resistance, so at higher forward currents a real device's voltage climbs faster than the pure exponential predicts, and the measured curve bends visibly below this one — datasheets show that bend past a few tens of milliamps on a typical small-signal part. It also treats Iₛ as a fixed input, when in a real device that saturation current is itself steeply temperature-dependent, roughly doubling for every 5 to 10°C the junction warms.
- Enter Reverse saturation current, Is, in amperes — a datasheet or curve-fit figure, often on the order of a picoamp for small silicon diodes.
- Enter the Forward voltage across the diode, the bias point in volts you want the current for.
- Set Ideality factor, n — 1 for a diffusion-dominated junction close to Shockley's ideal case, up to about 2 when recombination current dominates.
- Set Junction temperature; the instrument converts your entry to kelvin internally when it builds the thermal voltage.
- Read Diode current, the forward current the equation predicts at that voltage, temperature, and ideality factor.
Worked example — a small silicon diode at 0.6 volts
Take a small-signal silicon diode with a reverse saturation current of 1 pA, so Is = 1×10⁻¹² A, forward-biased to Vd = 0.6 V, an ideality factor n = 1, and a junction sitting at 27°C, which is 300.15 K. The thermal voltage V_T = kT⁄q works out to 25.865 mV, so the exponent V⁄(nV_T) equals 0.6 ⁄ 0.025865, about 23.197.
Raise e to that power and multiply by Iₛ: I = 1×10⁻¹² × (e^23.197 − 1) = 0.0118718694192 A, about 11.87 mA. That is enough current to light a small indicator LED comfortably or set the base-emitter bias of a small-signal transistor, and it fell out of a single exponential term.
Questions
Why does the equation subtract 1?
So the model behaves correctly at both ends of the bias range. At V = 0 the exp() term equals 1, and subtracting it forces I to exactly 0 — a diode with no voltage across it carries no current. Under reverse bias the exponential collapses toward zero, so the bracket settles near −1 and current approaches −Iₛ, the small steady leakage a real junction actually shows rather than a current that vanishes entirely or diverges.
What does the ideality factor n actually represent?
It measures how closely a real junction matches Shockley's pure diffusion theory. A value near 1 means forward current is dominated by carriers diffusing across the junction, the textbook case his 1949 derivation describes directly. Values climbing toward 2 mean recombination current inside the depletion region has taken over instead, which is typical of ordinary silicon diodes at low forward bias and of many LEDs across their entire working range.
Where does the thermal voltage V_T come from?
It is kT⁄q — Boltzmann's constant times absolute temperature, divided by the electron charge — and it sets the voltage scale of the whole exponential. At 300.15 K it comes to 25.865 mV. It is not a fitted number; it falls directly out of the Boltzmann distribution describing how many carriers have enough energy to cross the junction's potential barrier at a given bias.
Why does a small voltage change swing the current so much?
Because current sits inside an exponential rather than on a straight line. At room temperature, each extra 59 to 60 millivolts of forward bias multiplies the current roughly tenfold, so moving from 0.6 V to 0.66 V predicts about ten times as much current. That sensitivity is exactly why diodes make poor linear elements but excellent rectifiers and switches.
Does this formula include the diode's own series resistance?
No — it models only the ideal junction. At higher forward currents, the real bulk silicon and lead resistance inside the device add their own voltage drop, so the actual current for a given applied voltage runs a little below what the pure exponential predicts. That is the bend datasheets show in the true I-V curve at high current, and why practical SPICE diode models add a separate series-resistance term this equation leaves out.
What is a realistic value for reverse saturation current, Is?
Small-signal silicon switching diodes, such as the common 1N4148, typically run around 1 to 5 picoamps at room temperature. Power rectifiers are larger, often into the nanoamp range. Is is itself strongly temperature-dependent, roughly doubling every 5 to 10°C the junction warms, though this instrument treats it as the fixed figure you supply for a single operating temperature.