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The diode I-V graph looks simple: a flat line, then a steep rise. Yet questions built on it catch many students, because the graph hides two different scales, two kinds of resistance and a threshold that is easy to misuse. This post explains what happens inside the junction, how to read both curves, and works two circuit problems.
What does the syllabus ask for?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list "I-V characteristics in forward and reverse bias" under Electronic Devices, and both include the experiment "Characteristic curves of a p-n junction diode in forward and reverse bias". The CBSE physics curriculum for 2026-27 lists the same I-V characteristics and has a practical to draw the curve. So every reader of this post needs it. For the full map of the unit, including logic gates, see our guide to semiconductors, diodes and logic gates. Check your own year's syllabus, as each body publishes a new one.
What happens at the junction before you connect a battery?
When p-type and n-type regions meet, electrons diffuse from n to p and holes from p to n. Each electron that leaves the n side exposes a fixed positive donor ion, and each hole that leaves the p side exposes a fixed negative acceptor ion. The NCERT Semiconductor Electronics chapter calls this layer of fixed charge the depletion region and gives its thickness as about one-tenth of a micrometre.
The fixed charges set up an electric field across the junction, which opposes further diffusion. The potential difference it creates is the barrier potential. At equilibrium, diffusion one way is balanced by drift the other way, and no net current flows.
Why does forward current stay tiny until about 0.7 V?
In forward bias the p side goes to the positive terminal. NCERT explains that the barrier height falls from V₀ to (V₀ − V) and the depletion layer narrows. At first the current barely rises. Past a threshold, which NCERT gives as about 0.7 V for silicon and 0.2 V for germanium, the current rises steeply, growing exponentially with voltage.
The ideal-diode equation used in university courses shows why: I = I₀(e^(eV/kT) − 1). At 300 K, kT/e is about 25.9 mV, so a rise of 60 mV multiplies the forward current by roughly ten. The "0.7 V" is not a switch-on voltage at which current jumps from zero. It is the region where the exponential has become large enough to matter on a milliampere scale. William Shockley developed the theory of p-n junctions in a 1949 Bell System Technical Journal paper, describing junction currents as carried by the diffusion of holes in n-type material and electrons in p-type material. He shared the 1956 Nobel Prize in Physics with John Bardeen and Walter Brattain for research on semiconductors and the transistor effect.
Why is the reverse current almost constant?
In reverse bias the barrier rises to (V₀ + V) and the depletion layer widens. Only minority carriers drift across, and NCERT notes this current is limited by how many minority carriers there are, not by the applied voltage. That is why the reverse curve is nearly flat, at microamperes rather than milliamperes. NCERT calls it the reverse saturation current.
At the breakdown voltage the reverse current rises sharply. NCERT warns that if the current is not limited by the external circuit, the diode is destroyed by overheating, and that ordinary diodes are not used beyond the reverse saturation region. Zener diodes are built to work there; see our Zener regulator worked problems.
Worked example 1: static and dynamic resistance
A silicon diode's forward curve passes through (0.70 V, 5.0 mA) and (0.75 V, 15.0 mA). In reverse bias at −10 V the current is −2.0 µA.
Dynamic resistance near 0.75 V: r = ΔV/ΔI = 0.05 V ÷ 10.0 mA = 5.0 Ω.
Static resistance at 0.75 V: R = V/I = 0.75 V ÷ 15.0 mA = 50 Ω.
Reverse resistance: 10 V ÷ 2.0 µA = 5.0 × 10⁶ Ω.
NCERT defines dynamic resistance as ΔV/ΔI and works a similar example. The static value is ten times the dynamic one here, because the curve is steep but does not pass through the origin.
Worked example 2: which diode conducts?
A germanium diode and a silicon diode are connected in parallel, both forward biased, and the pair is in series with a 1.0 kΩ resistor and a 5.0 V battery. Take thresholds of 0.2 V and 0.7 V.
As the voltage across the pair rises, the germanium diode reaches its threshold first.
Once it conducts, it holds the voltage across the pair near 0.2 V. That is below the silicon threshold, so the silicon diode carries almost no current.
Current = (5.0 − 0.2) V ÷ 1.0 kΩ = 4.8 mA, all through the germanium diode.
Now reverse the germanium diode. It is reverse biased and passes only a tiny reverse current, so the silicon diode conducts: (5.0 − 0.7) ÷ 1.0 kΩ = 4.3 mA.
Which mistakes cost marks?
Reading a µA reverse current as mA.
Saying the depletion layer widens in forward bias. It narrows.
Using V/I when the question wants dynamic resistance.
Putting both parallel diodes at their own thresholds at once. The lower-threshold diode conducts and fixes the voltage.
Forgetting the series resistor sets the current; the diode only sets the voltage drop.
The exponential shape is easier to understand when you can watch it form. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read, so you can link a change in voltage to a change on the graph.
A practice routine
Day 1: draw the forward and reverse curves from memory, label threshold, saturation current and breakdown, and check against NCERT's figure.
Day 2: four resistance problems, computing both static and dynamic resistance each time.
Day 3: four circuits with two or more diodes. Before solving, write which diodes are forward biased.
Day 4: a timed set of previous-year questions on diode characteristics.
Tonight, sketch the I-V curve of a silicon diode with both axes labelled in the right units, without looking at the book.
