Q.Suppose a pure Si crystal has 5×1028 atoms m−3. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that ni=1.5×1016 m−3.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Intrinsic Carrier Concentration
Intrinsic carrier concentration is a foundational idea in semiconductor physics — let’s build it from the ground up, with no prior knowledge of semiconductors needed.
1. Intuition: What does "intrinsic" mean?
Imagine a pure, perfect crystal of silicon — no impurities, no defects. At absolute zero temperature (0 K), all electrons are tightly bound in the crystal lattice. No current flows.
Now, heat it up. Thermal energy shakes the atoms. Some electrons gain enough energy to break free from their bonds. When an electron leaves, it leaves behind a hole — a missing electron that behaves like a positive charge.
In this pure crystal, every free electron comes from a broken bond, and every broken bond creates one hole. So:
Number of free electrons = Number of holes
This balance is the hallmark of an intrinsic semiconductor.
2. The precise definition
Intrinsic carrier concentration (ni) is the number of free electrons (or holes) per unit volume in a pure, undoped semiconductor at thermal equilibrium.
It is denoted by ni and has units of cm−3 or m−3.
Key points:
- It depends only on the material and temperature — not on doping.
- For silicon at room temperature (300 K):
ni≈1.5×1010 cm−3
- For germanium: ni≈2.5×1013 cm−3
- For gallium arsenide: ni≈1.8×106 cm−3
3. The formula (for exams)
The precise expression is:
ni=NcNv⋅e−Eg/(2kT)
Where:
- Nc = effective density of states in the conduction band
- Nv = effective density of states in the valence band
- Eg = bandgap energy (eV)
- k = Boltzmann constant (8.617×10−5 eV/K)
- T = absolute temperature (K)
Important: The exponential term e−Eg/(2kT) dominates — a small change in Eg or T causes a huge change in ni.
4. Why does it matter?
- It sets the baseline for all semiconductor devices. Doping increases one carrier type, but the product n⋅p=ni2 always holds at equilibrium.
- Temperature sensitivity: ni roughly doubles for every 10∘C rise in silicon. This is why circuits fail in heat. …
Why this formula?
Why Intrinsic Carrier Concentration Has That Formula
The intrinsic carrier concentration ni is the number of electrons (or holes) per unit volume in a pure, undoped semiconductor at thermal equilibrium. The formula you see in every textbook is:
ni=NcNve−Eg/2kT
where Nc and Nv are the effective density of states in the conduction and valence bands, Eg is the bandgap energy, k is Boltzmann's constant, and T is absolute temperature.
This isn't pulled from thin air. It comes from a simple physical balance: in an intrinsic semiconductor, every electron in the conduction band leaves behind a hole in the valence band. So the electron concentration n must equal the hole concentration p, and both equal ni.
Step 1: The electron and hole concentrations individually
Electrons in the conduction band follow Fermi-Dirac statistics. For non-degenerate semiconductors (which intrinsic ones are, since the Fermi level lies near midgap), the distribution approximates the Maxwell-Boltzmann tail:
n=Nce−(Ec−EF)/kT
Similarly, holes in the valence band:
p=Nve−(EF−Ev)/kT
Here Ec is the conduction band edge, Ev is the valence band edge, and EF is the Fermi level. The effective densities Nc and Nv come from integrating the density of states times the Boltzmann factor — they depend on the effective masses of electrons and holes and on temperature.
Step 2: The intrinsic condition
In an intrinsic semiconductor, there are no dopants. Every electron that jumps to the conduction band creates exactly one hole. So:
n=p
Set the two expressions equal:
Nce−(Ec−EF)/kT=Nve−(EF−Ev)/kT
Take natural logs and solve for EF:
−(Ec−EF)+lnNc=−(EF−Ev)+lnNv
EF=2Ec+Ev+2kTlnNcNv
The Fermi level in an intrinsic semiconductor sits very close to the middle of the bandgap, shifted slightly by the ratio Nv/Nc. For most practical purposes, it's at midgap.
Step 3: Multiply to eliminate EF
Now here's the clever part. Instead of solving for EF directly, multiply n and p:
np=NcNve−(Ec−EF)/kTe−(EF−Ev)/kT
The EF terms cancel:
np=NcNve−(Ec−Ev)/kT=NcNve−Eg/kT
This product np is a constant for a given material at a given temperature — it does not depend on the Fermi level. This is the law of mass action for semiconductors.
Step 4: Apply the intrinsic condition
Since n=p=ni in an intrinsic semiconductor:
ni2=NcNve−Eg/kT
Take the square root:
ni=NcNve−Eg/2kT …
Concept: Intrinsic Carrier Concentration — In an extrinsic semiconductor, the product n⋅p=ni2 always holds at equilibrium.
Step 1: Find donor concentration.
1 ppm means 1 As atom per 106 Si atoms.
ND=1065×1028=5×1022 m−3
Step 2: Approximate electron concentration.
Since ND≫ni, nearly all donors ionise:
n≈ND=5×1022 m−3
Step 3: Find hole concentration using mass action law. …
The key idea is that doping with a pentavalent impurity (As) adds donor electrons, making the crystal n-type. The electron concentration becomes approximately equal to the donor concentration, and the hole concentration is found using the mass-action law np=ni2. The final values are n≈5×1022 m−3 and p≈4.5×109 m−3.
Why this approach works
In a pure (intrinsic) semiconductor, the number of electrons equals the number of holes — both are ni. But when we dope with a pentavalent atom like arsenic (As), which has five valence electrons, four of them bond with neighbouring silicon atoms and the fifth becomes a free electron. This makes the crystal n-type, where electrons are the majority carriers and holes are the minority carriers.
The key principle is charge neutrality: the total positive charge must equal the total negative charge. In an n-type semiconductor at room temperature, almost all donor atoms are ionised, so the electron concentration n is essentially equal to the donor concentration ND. Then, using the mass-action law (np=ni2), we can find the hole concentration p.
Step-by-step calculation
1. Find the donor concentration from the doping level
We are told the crystal has 5×1028 Si atoms per cubic metre, and it is doped with 1 ppm (parts per million) of As. This means for every million Si atoms, there is one As atom.
So the donor concentration ND is:
ND=1061×(5×1028)=5×1022 atoms/m3
Since each As atom donates one free electron, ND is also the concentration of donor electrons (assuming full ionisation, which is valid at room temperature).
"1 ppm" here means 1 atom of impurity per 106 atoms of Si. Always check the context — in some problems ppm means parts per million by mass, but here it clearly refers to atomic concentration.
2. Determine the electron concentration
In an n-type semiconductor at room temperature, the electron concentration n is approximately equal to the donor concentration because the intrinsic carrier concentration ni is negligible compared to ND:
n≈ND=5×1022 m−3
Why "approximately"? Because a tiny fraction of electrons come from intrinsic generation, but ni=1.5×1016 m−3 is six orders of magnitude smaller than ND, so the approximation is excellent. …
Method: Mass Action Law for Extrinsic Semiconductors
This problem uses the mass action law combined with the charge neutrality condition for an n-type semiconductor.
Step 1 — Find the donor concentration
1 ppm means 1 atom of As per 106 atoms of Si. So:
ND=1065×1028=5×1022 m−3
Step 2 — Identify the majority carrier
Pentavalent As donates an extra electron. At room temperature, nearly all donor atoms are ionised. Since ND≫ni, the electron concentration is essentially equal to the donor concentration:
n≈ND=5×1022 m−3
Do not add ni to ND here — ni is 1.5×1016, which is six orders of magnitude smaller than ND. Adding it would be meaningless.
Step 3 — Apply the mass action law
For any semiconductor in thermal equilibrium:
n⋅p=ni2
So:
p=nni2=5×1022(1.5×1016)2 …
Common Mistakes & How to Avoid Them
Mistake 1: Confusing ppm with a percentage
Many students treat 1 ppm as 1% — that is, they multiply the atom density by 0.01 instead of 10−6. This gives a donor concentration that is four orders of magnitude too large, which then throws off every subsequent calculation.
How to avoid: Remember that "ppm" means parts per million, i.e., 1 ppm=1×10−6. So the donor concentration is:
ND=(5×1028)×(1×10−6)=5×1022 m−3
Never treat ppm as a percentage. Write 10−6 explicitly beside "ppm" in your working until it becomes automatic.
Mistake 2: Forgetting that ni is given in m−3, not cm−3
The problem gives ni=1.5×1016 m−3. Some students, used to the common textbook value 1.5×1010 cm−3, automatically convert or substitute the wrong number. This leads to a completely wrong hole concentration.
How to avoid: Always check the units of every given quantity before plugging into a formula. If the problem states m−3, keep everything in m−3. Do not "correct" the given data.
Mistake 3: Assuming n=ND without checking the doping regime
Students often write n=ND and p=ni2/ND without first verifying that ND≫ni. If ND were comparable to ni, the full quadratic equation would be needed.
How to avoid: Compare ND and ni explicitly. Here:
ND=5×1022 m−3,ni=1.5×1016 m−3
Since ND is about 3 million times larger than ni, the approximation n≈ND is excellent. Write this comparison in your solution — examiners look for it.
A quick check: if ND>100ni, the approximation is safe for most exam problems.
Mistake 4: Using the wrong mass-action law sign or forgetting it entirely
Some students try to find p by subtracting ni from ND, or by using n+p=constant. The only correct relation is:
n⋅p=ni2
How to avoid: Write the mass-action law before you calculate anything. It is the bridge between electron and hole concentrations in any doped semiconductor.
Mistake 5: Arithmetic errors in the final division …
- COMEDK 2026Set 2026-M1 markMCQQ.A silicon sample is doped simultaneously with donor impurity phosphorus at a concentration of ND=3×1022 m−3 and acceptor impurity boron at a concentration of ND=2.8×1022 m−3. The intrinsic carrier concentration of silicon at room temperature is ni=1.5×1016 m−3. Assuming complete ionization, the hole concentration is : (A) 1.125×1010m−3 (B) 1.125×1011 m−3 (C) 2.125×1010 m−3 (D) 2.125×1011m−3
›Reveal solutionSolution
Net donor doping ND−NA=2×1021m−3 fixes the electron density; mass-action then gives p=ni2/n=1.125×1011m−3.
Net doping (the sample is n-type)
Donor (phosphorus) ND=3×1022m−3 exceeds acceptor (boron) NA=2.8×1022m−3, so under complete ionization the majority electron concentration is:
n≈ND−NA=(3−2.8)×1022=2×1021m−3
Minority (hole) concentration via the mass-action law
np=ni2⇒p=nni2 …
- COMEDK 2025Set 2025-E1 markMCQQ.An intrinsic semiconductor has equal concentrations of hole and electron which is equal to 4×108 m−3. During its conversion to extrinsic semiconductor, concentration of hole increases to 8×1010 m−3. The new electron concentration is: (A) 4×108 m−3 (B) 2×108 m−3 (C) 2×106 m−3 (D) 4×1010 m−3
›Reveal solutionSolution
The key idea is the law of mass action for semiconductors: in thermal equilibrium, the product of electron and hole concentrations is constant (ni2). Given the intrinsic carrier concentration ni=4×108m−3 and the new hole concentration p=8×1010m−3, the new electron concentration is n=ni2/p=2×106m−3, which corresponds to option (C).
Concept and Intuition
In an intrinsic (pure) semiconductor, every electron excited to the conduction band leaves a hole in the valence band, so the electron concentration n equals the hole concentration p, and both equal the intrinsic carrier concentration ni. When we dope the semiconductor to make it extrinsic, we add impurities that greatly increase one type of carrier. However, the material remains in thermal equilibrium, and a fundamental principle—the law of mass action—holds: the product n×p is constant at a given temperature, equal to ni2. So if doping raises the hole concentration, the electron concentration must drop to keep the product constant. This is not a guess; it's a direct consequence of the balance between generation and recombination of electron-hole pairs.
Step-by-Step Solution
- Identify the intrinsic carrier concentration. The problem states that in the intrinsic semiconductor, the concentrations of holes and electrons are equal and given as 4×108m−3. Therefore, the intrinsic carrier concentration is
ni=4×108m−3.
- Recall the law of mass action. For any non-degenerate semiconductor in thermal equilibrium, the product of the electron concentration n and the hole concentration p is constant:
n⋅p=ni2.
This holds for both intrinsic and extrinsic semiconductors at the same temperature.
- Apply the law to the extrinsic case. After doping, the hole concentration becomes
p=8×1010m−3.
Let the new electron concentration be n. Then:
n⋅(8×1010)=(4×108)2.… - COMEDK 2025Set 2025-M1 markMCQQ.A Si and a Ge diode has identical physical dimensions. The band gap in Si is larger than that in Ge. On applying identical reverse bias across these diodes, (A) The reverse current in Ge is lesser than that in Si . (B) The relative magnitudes of reverse currents cannot be determined from the given data. (C) The reverse current is identical in both cases. (D) The reverse current in Ge is larger than that in Si .
›Reveal solutionSolution
[!TLDR]
Reverse current depends on thermally generated minority carriers, which increase as the band gap decreases; Ge's smaller gap gives it the larger reverse current.
Concept
In CBSE Class 12 Semiconductor Electronics, the reverse saturation current of a diode is due to minority carriers produced by thermal generation across the band gap. The intrinsic carrier concentration follows ni2∝e−Eg/kT, so a larger band gap means far fewer carriers and a smaller reverse current.
Solution
- For identical dimensions and the same temperature and reverse bias, the only difference is the band gap Eg.
- Silicon has a larger Eg than germanium, so ni(Si)<ni(Ge); fewer thermally generated minority carriers means a smaller reverse current in Si.
- Germanium, with the smaller band gap, generates more minority carriers, so its reverse current is larger. …
- COMEDK 2024Set 2024-A1 markMCQQ.In intrinsic semiconductors at room temperature, number of electrons and holes are (A) zero (B) infinity (C) unequal (D) equal
›Reveal solutionSolution
In an intrinsic (pure) semiconductor at room temperature, every electron excited to the conduction band leaves behind a hole in the valence band, so the number of electrons and holes must be exactly equal. The correct option is (D).
Concept & Intuition
An intrinsic semiconductor is a pure crystal with no intentional impurities. At absolute zero, all electrons are tightly bound in the valence band, so there are no free charge carriers. As temperature rises, thermal energy can break some covalent bonds, promoting an electron from the valence band to the conduction band. This process creates two mobile carriers: the electron (now free to move in the conduction band) and the hole (the empty state left behind in the valence band, which behaves like a positive charge carrier). Crucially, each promotion creates one electron and one hole together — they come in pairs. No other mechanism generates carriers in a pure material. Therefore, the number of electrons (n) must always equal the number of holes (p) in an intrinsic semiconductor, at any temperature.
Step-by-step reasoning
-
Definition of intrinsic semiconductor
An intrinsic semiconductor has no dopant atoms. The only source of charge carriers is thermal excitation across the band gap. There are no impurities to donate extra electrons or accept extra holes.
-
Carrier generation mechanism
When an electron gains enough energy (≥ band gap energy Eg), it jumps from the valence band to the conduction band. This single event produces exactly one free electron and one hole. The process is called electron-hole pair generation.
-
Charge neutrality condition
The crystal as a whole must remain electrically neutral. In an intrinsic semiconductor, the total negative charge from conduction electrons equals the total positive charge from holes. Since each electron carries charge −e and each hole carries +e, this neutrality condition is:
n=p
where n is the electron concentration and p is the hole concentration.
- No other sources of carriers …
-
- COMEDK 2024Set 2024-E1 markMCQQ.The conductivity of a semiconductor increases with increase in temperature because A) number density of free current carriers increases B) relaxation time increases C) both number density of carriers and relaxation time increase D) number density of current carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density (A) C (B) D (C) A (D) B
›Reveal solutionSolution
Conductivity rises because carrier number density increases strongly; relaxation time decreases but its effect is much smaller — statement D.
In a semiconductor σ=neμ with μ∝τ. Heating:
- promotes many electrons across the gap, so n rises steeply (dominant effect);
- increases lattice vibration/scattering, so relaxation time τ (and hence mobility) falls slightly. …
- KCET 2022Set B-31 markMCQQ.The resistivity of a semiconductor at room temperature is in between (A) 106 to 108 Ωcm (B) 1010 to 1012 Ωcm (C) 10−2 to 10−5 Ωcm (D) 103 to 106 Ωcm
›Reveal solutionSolution
Place the semiconductor band on the resistivity scale between conductors and insulators, then check it against the measured room-temperature resistivity of intrinsic silicon.
Step 1 — The resistivity ladder.
Resistivity ρ=1/σ is what separates the three classes of solid, and the ranges are enormous (spanning ~20 orders of magnitude):
ρmetal∼10−8–10−6 Ωcm≪ρsemiconductor≪ρinsulator∼1013–1021 Ωcm
Why: in a metal the conduction band is partly filled, so carriers are freely available. In a semiconductor a band gap Eg≈1 eV must be crossed thermally, so only a small fraction of electrons (∝e−Eg/2kBT) contribute — far fewer carriers, far higher ρ. In an insulator Eg≳3 eV and essentially none cross.
Step 2 — Pin it down with a real number.
Intrinsic (pure) silicon at 300 K:
ni≈1.5×1010 cm−3,ρSi≈2.3×105 Ωcm
That value lies squarely inside 103–106 Ωcm. …
- KCET 2020Set A-11 markMCQQ.Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg submerged if the density of ice is ρi=0.917 g cm−3? (A) 0.917 (B) 1 (C) 0.458 (D) 0
›Reveal solutionSolution
For a floating body, the submerged fraction equals the ratio of the body's density to the fluid's density — here 0.917/1.00=0.917.
Step 1 — The condition for floating.
The iceberg floats in equilibrium, so its weight is exactly balanced by the buoyant force:
W=FB
Step 2 — Write each side.
- Weight of the whole iceberg (volume V, density ρi):
W=ρiVg
- Archimedes' principle: the buoyant force equals the weight of the displaced water. Only the submerged volume Vsub displaces water:
FB=ρwVsubg
Step 3 — Equate and solve for the fraction.
ρiVg=ρwVsubg
The g cancels (the answer is the same on any planet), leaving
VVsub=ρwρi
Step 4 — Substitute the numbers.
Water: ρw=1.00 g cm−3. Ice: ρi=0.917 g cm−3 (both in the same units, so no conversion is needed).
VVsub=1.000.917=0.917
Step 5 — Physical sense-check. …
- KCET 2020Set A-11 markMCQQ.A positive hole in a semiconductor is (A) an anti-particle of electron. (B) a vacancy created when an electron leaves a covalent bond. (C) absence of free electrons. (D) an artificially created particle.
›Reveal solutionSolution
A positive hole is best understood as a vacancy created when an electron leaves a covalent bond — it behaves like a positive charge carrier, but it is not an anti-particle or an artificially created entity.
The concept of a "hole" in semiconductor physics is one of the most subtle ideas you'll encounter. Many students try to think of it as a real particle, like an electron, but with positive charge — that's a trap. The hole is a conceptual convenience, not a fundamental particle.
In a pure semiconductor crystal, atoms are bonded together by covalent bonds — each atom shares electrons with its neighbours to complete its valence shell. At absolute zero, every electron is locked in a bond, and no current flows. But when you add energy (heat, light), an electron can break free from its covalent bond and become a "free electron" that can move through the crystal.
What's left behind? That broken bond is missing an electron. The neighbouring electrons can now "jump" into this empty spot, effectively moving the vacancy around. This moving vacancy behaves exactly like a positive charge carrier — it's called a hole.
Now let's examine each option carefully.
-
Option (A): "An anti-particle of electron."
This sounds scientific, but it's wrong in this context. An anti-particle (like a positron) is a real fundamental particle with the same mass as an electron but opposite charge. A hole is not a fundamental particle — it's an absence in a crystal lattice. The hole's effective mass is different from an electron's, and it only exists inside the semiconductor. So (A) is incorrect.
-
Option (B): "A vacancy created when an electron leaves a covalent bond."
This is exactly right. When an electron absorbs enough energy to break its covalent bond, it becomes a free electron. The empty bond site is the hole. This hole can move as neighbouring electrons fill it, creating a new vacancy elsewhere. This is the standard definition used in every textbook and exam.
-
Option (C): "Absence of free electrons." …
-
- KCET 2018Set A-11 markMCQQ.The density of an electron-hole pair in a pure germanium is 3×1016 m−3 at room temperature. On doping with aluminium, the hole density increases to 4.5×1022 m−3. Now the electron density (in m−3) in doped germanium will be (A) 1×1010 (B) 2×1010 (C) 0.5×1010 (D) 4×1010
›Reveal solutionSolution
Use the mass-action (law of electrical neutrality/equilibrium) relation nenh=ni2 for a doped semiconductor.
Step 1 — The concept.
In a pure (intrinsic) semiconductor electrons and holes are created in pairs, so ne=nh=ni. When the crystal is doped, the total carrier densities change, but at thermal equilibrium the rates of thermal generation and recombination still balance, which gives the mass-action law
nenh=ni2
This is why doping to raise one carrier density automatically suppresses the other (extra holes recombine with the electrons).
Step 2 — Identify the given quantities.
"The density of an electron–hole pair in pure germanium" is the intrinsic density:
ni=3×1016 m−3
Aluminium is a trivalent (Group 13) dopant in Ge, hence an acceptor: it makes the crystal p-type, so holes become the majority carriers:
nh=4.5×1022 m−3
Step 3 — Compute the minority (electron) density. …
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