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Semiconductor questions in JEE Main and NEET often give you a doping level and the intrinsic carrier concentration, then ask for the number of electrons or holes. The arithmetic is short. The traps are not: parts per million, cm⁻³ against m⁻³, crystals with both kinds of dopant, and the habit of dividing nᵢ by the doping level instead of squaring it. This post works through four problem types using NCERT's own value for silicon, then gives a practice routine.
Is this on the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both open their Electronic Devices unit with "Semiconductors". The CBSE physics curriculum for 2026-27 is more specific: intrinsic and extrinsic semiconductors, p and n type. The working equation is in the NCERT Class 12 chapter Semiconductor Electronics: Materials, Devices and Simple Circuits, which gives ne·nh = nᵢ² for a semiconductor in thermal equilibrium (Eq. 14.5) and works one example for arsenic-doped silicon. The copy on ncert.nic.in was still marked Reprint 2025-26 when we opened it. The JEE (Advanced) 2026 brochure does not list semiconductors in its physics syllabus. Check your own year's documents.
Which equations do you need?
Pure crystal: ne = nh = nᵢ (NCERT Eq. 14.1).
Mass-action law: ne·nh = nᵢ². It holds at a fixed temperature for pure and doped crystals alike, so doping one carrier up pushes the other down.
Heavy doping: NCERT notes that at room temperature most donor (or acceptor) atoms are ionised. When the dopant concentration is far above nᵢ, the majority concentration is roughly the dopant concentration: ne ≈ N_D for n-type, nh ≈ N_A for p-type.
Charge neutrality: NCERT stresses that the doped crystal stays neutral. Written as an equation, ne + N_A = nh + N_D. You need this only when both dopants are present or the doping is close to nᵢ.
For the examples, take NCERT's figures: silicon has about 5 × 10²⁸ atoms m⁻³ and nᵢ = 1.5 × 10¹⁶ m⁻³, so nᵢ² = 2.25 × 10³².
Worked example 1: p-type silicon from a ppm figure
Silicon is doped with boron at 0.1 ppm. Find nh and ne.
0.1 ppm means 0.1 × 10⁻⁶ of the atoms: N_A = 0.1 × 10⁻⁶ × 5 × 10²⁸ = 5 × 10²¹ m⁻³.
This is far above nᵢ, so nh ≈ 5 × 10²¹ m⁻³.
ne = nᵢ²/nh = 2.25 × 10³² / 5 × 10²¹ = 4.5 × 10¹⁰ m⁻³.
Holes outnumber electrons by about 10¹¹ to one. Notice that doping did not just add holes; it cut the electron count from 1.5 × 10¹⁶ to 4.5 × 10¹⁰.
Worked example 2: a crystal with both dopants
A silicon sample (sample numbers) contains N_D = 3 × 10²² m⁻³ of phosphorus and N_A = 1 × 10²² m⁻³ of boron. What type is it?
Donors win, so the sample is n-type. The acceptors soak up part of the donor electrons, so the net majority concentration is ne ≈ N_D − N_A = 2 × 10²² m⁻³.
nh = 2.25 × 10³² / 2 × 10²² = 1.125 × 10¹⁰ m⁻³.
A common wrong answer adds the two dopant levels. Another uses N_D alone and gets nh = 7.5 × 10⁹ m⁻³.
Worked example 3: when doping is close to nᵢ
Now suppose N_D = 2 × 10¹⁶ m⁻³ (sample number), only a little above nᵢ. The shortcut ne ≈ N_D fails, because the thermally made carriers are no longer negligible. This step goes beyond NCERT and appears in harder JEE-style questions.
Combine neutrality (ne = nh + N_D) with ne·nh = nᵢ². That gives a quadratic, ne² − N_D·ne − nᵢ² = 0, so
ne = N_D/2 + √((N_D/2)² + nᵢ²) = 1 × 10¹⁶ + √(1 × 10³² + 2.25 × 10³²) ≈ 1 × 10¹⁶ + 1.80 × 10¹⁶ = 2.80 × 10¹⁶ m⁻³
and nh = nᵢ²/ne ≈ 8.03 × 10¹⁵ m⁻³. Check: ne − nh = 2.0 × 10¹⁶, which matches N_D. The shortcut would have given nh = 1.125 × 10¹⁶, about 40% too high.
Worked example 4: changing units
A textbook gives nᵢ for silicon as 1.5 × 10¹⁰ cm⁻³. One cubic metre is 10⁶ cm³, so this is the same 1.5 × 10¹⁶ m⁻³. If a question mixes units, convert everything to one before squaring. Squaring a cm⁻³ value and dividing by an m⁻³ value gives an answer wrong by a factor of 10⁶.
Which mistakes cost the most marks?
Writing nh = nᵢ/ne. The law has nᵢ squared.
Forgetting the 10⁻⁶ in ppm. One ppm of 5 × 10²⁸ is 5 × 10²², not 5 × 10²⁸.
Thinking an n-type crystal is negatively charged. The donor ions carry the balancing positive charge. For more on the junction that forms when n and p regions meet, see our guide to p-n junction diode characteristics.
Using ne·nh = nᵢ² at a different temperature. nᵢ rises steeply with temperature, so a value given for 300 K does not hold at 400 K.
Adding nᵢ to N_D in heavily doped problems. At 5 × 10²¹ m⁻³ doping, adding 1.5 × 10¹⁶ changes nothing at the precision asked.
If you keep landing on answers that are off by a power of ten and cannot see where, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare each conversion with your own.
A practice routine for this topic
Day 1: Redo NCERT Example 14.2 without looking, then repeat it for doping levels of 0.01, 0.1 and 10 ppm. Write the electron-to-hole ratio each time.
Day 2: Three compensated-crystal questions with both dopants. Decide the type first, then calculate.
Day 3: One light-doping question solved with the quadratic, then the same question with the shortcut, so you see when the shortcut breaks.
Day 5: Mix in five questions on bands, doping and diodes from our semiconductors, diodes and logic gates guide, timed at about two minutes each.
Tonight, compute the hole concentration for silicon doped with 2 ppm arsenic and check it by multiplying back to nᵢ².
