Is it correct to say: Saying an n-type semiconductor is negatively charged?

Each donor atom is neutral; its extra electron is balanced by the positive donor ion left behind. The crystal as a whole stays neutral.

Is it correct to say: Using n_e = n_h for a doped semiconductor?

Equality holds only for intrinsic material. For doped material use n_e n_h = n_i², with the majority density ≈ dopant concentration.

Is it correct to say: Thinking holes are real positive particles?

A hole is a vacancy left by a missing valence electron. It behaves like a charge +e and moves when neighbouring electrons fill it, so holes drift opposite to electrons.

Is it correct to say: Giving 100 Hz as the output frequency of a half-wave rectifier on 50 Hz mains?

Half-wave: output frequency = input frequency (50 Hz). Full-wave: output frequency = 2 × input (100 Hz).

Is it correct to say: Assuming the depletion layer contains free charge carriers?

The depletion region is emptied of free carriers. It holds only immobile donor and acceptor ions, whose charge creates the barrier potential.

Is it correct to say: Mixing up how bias changes the barrier?

Forward bias lowers the barrier to V₀ − V and narrows the depletion layer; reverse bias raises it to V₀ + V and widens the depletion layer.

📄 Formula sheet (PDF)

Semiconductor Electronics: Materials, Devices and Simple Circuits

Energy bands, intrinsic and extrinsic semiconductors, the p-n junction, diode characteristics and the junction diode as a rectifier — NCERT Class 12 Physics Ch 14

📐 13 formulas✏️ 4 examples🎯 6 practice⚖️ Unit IX: 7 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Draw and label energy-band diagrams for a conductor, an insulator and a semiconductor, and for n-type and p-type material showing the donor and acceptor levels.
→Explain depletion-layer formation in two steps: diffusion of majority carriers across the junction, then drift caused by the built-in field, until the two currents balance.
→For rectifiers, draw the circuit and the input and output waveforms. Half-wave output frequency equals the input frequency; full-wave output frequency is twice the input.
→Diode V–I characteristics use different scales: mA for forward current, μA for reverse current. Label both axes and mark the threshold voltage.
→In any doped semiconductor n_e n_h = n_i² still holds, so the minority-carrier density is n_i² divided by the majority-carrier density.

📐 Formulas(13)

Energy Band Gap★ Board fav

E_{g} = E_{C} - E_{V}

Symbol	Meaning
$E_{g}$	Energy band gap (eV)
$E_{C}$	Energy at the bottom of the conduction band (eV)
$E_{V}$	Energy at the top of the valence band (eV)

Intrinsic Semiconductor

n_{e} = n_{h} = n_{i}

Symbol	Meaning
$n_{e}$	Free-electron density (m⁻³)
$n_{h}$	Hole density (m⁻³)
$n_{i}$	Intrinsic carrier concentration (m⁻³)

Current in a Semiconductor

I = I_{e} + I_{h}

Symbol	Meaning
$I_{e}$	Current due to free electrons (A)
$I_{h}$	Current due to holes (A)

Mass-Action Law★ Board fav

n_{e} n_{h} = n_{i}^{2}

n-type Semiconductor★ Board fav

n_{e} \approx N_{D} ≫ n_{h}, n_{h} = \frac{n _{i}^{2}}{N _{D}}

Symbol	Meaning
$N_{D}$	Donor atom concentration (m⁻³)

p-type Semiconductor

n_{h} \approx N_{A} ≫ n_{e}, n_{e} = \frac{n _{i}^{2}}{N _{A}}

Symbol	Meaning
$N_{A}$	Acceptor atom concentration (m⁻³)

Both Donors and Acceptors Present

n_{e} \approx N_{D} - N_{A}, n_{h} = \frac{n _{i}^{2}}{N _{D} - N _{A}} (N_{D} > N_{A})

Effective Barrier Height Under Bias

Forward bias: V_{0} - V, Reverse bias: V_{0} + V

Symbol	Meaning
$V_{0}$	Barrier potential at equilibrium (V)
$V$	Applied bias voltage (V)

Threshold (Cut-in) Voltage

V_{t h} \approx 0.7 V (Si), V_{t h} \approx 0.2 V (Ge)

Dynamic Resistance of a Diode

r_{d} = \frac{Δ V}{Δ I}

Symbol	Meaning
$r_{d}$	Dynamic resistance (Ω)
$Δ V$	Small change in voltage (V)
$Δ I$	Corresponding change in current (A)

Forward-Biased Diode in Series with a Resistor

I = \frac{V - V _{t h}}{R}

Symbol	Meaning
$V$	Battery voltage (V)
$R$	Series resistance (Ω)

Half-Wave Rectifier Output Frequency

ν_{o u t} = ν_{in}

Full-Wave Rectifier Output Frequency★ Board fav

ν_{o u t} = 2 ν_{in}

✏️ Solved Examples

1Solved Exampleeasy2 steps

For a silicon diode, the forward current is 10 mA at 0.70 V and 20 mA at 0.72 V. Find its dynamic resistance in this range.

Changes in voltage and current

Δ V = 0.72 - 0.70 = 0.02 V, Δ I = 20 - 10 = 10 mA = 0.010 A

2Solved Exampleboard3 steps

Pure silicon has 5×10²⁸ atoms m⁻³ and an intrinsic carrier concentration of 1.5×10¹⁶ m⁻³. It is doped with boron at 1 atom per 10⁷ silicon atoms. Find the hole and electron densities and state the type of semiconductor formed.

Boron is trivalent (acceptor), so each boron atom gives one hole

N_{A} = \frac{5 \times 1 0 ^{28}}{1 0 ^{7}} = 5 \times 1 0^{21} m^{- 3} \Rightarrow n_{h} \approx 5 \times 1 0^{21} m^{- 3}

3Solved Exampleboard3 steps

A silicon sample with n_i = 1.5×10¹⁶ m⁻³ is doped with 4.0×10²¹ m⁻³ arsenic atoms and 1.0×10²¹ m⁻³ indium atoms. Find n_e and n_h and decide whether the material is n-type or p-type.

Arsenic is a donor and indium an acceptor. The acceptors cancel part of the donors

n_{e} \approx N_{D} - N_{A} = 4.0 \times 1 0^{21} - 1.0 \times 1 0^{21} = 3.0 \times 1 0^{21} m^{- 3}

4Solved ExampleHOTS3 steps

A silicon diode (forward drop 0.7 V) is connected in series with a resistor R across a 6.0 V battery so that it is forward biased. The diode can safely carry at most 20 mA. (a) Find the minimum value of R. (b) What current flows if the battery terminals are reversed?

Loop rule: the battery voltage is shared between the diode drop and the resistor

6.0 = 0.7 + I R

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Saying an n-type semiconductor is negatively charged
✓Each donor atom is neutral; its extra electron is balanced by the positive donor ion left behind. The crystal as a whole stays neutral.
2
Using n_e = n_h for a doped semiconductor
✓Equality holds only for intrinsic material. For doped material use n_e n_h = n_i², with the majority density ≈ dopant concentration.
3
Thinking holes are real positive particles
✓A hole is a vacancy left by a missing valence electron. It behaves like a charge +e and moves when neighbouring electrons fill it, so holes drift opposite to electrons.
4
Giving 100 Hz as the output frequency of a half-wave rectifier on 50 Hz mains
✓Half-wave: output frequency = input frequency (50 Hz). Full-wave: output frequency = 2 × input (100 Hz).
5
Assuming the depletion layer contains free charge carriers
✓The depletion region is emptied of free carriers. It holds only immobile donor and acceptor ions, whose charge creates the barrier potential.
6
Mixing up how bias changes the barrier
✓Forward bias lowers the barrier to V₀ − V and narrows the depletion layer; reverse bias raises it to V₀ + V and widens the depletion layer.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
The input to a half-wave rectifier is 50 Hz AC. What is the output frequency? What would it be for a full-wave rectifier?
Q2
A semiconductor with n_i = 1.0×10¹⁶ m⁻³ is doped with 1.0×10²² m⁻³ acceptor atoms. Find the electron density.
Q3
Diamond has a band gap of 5.4 eV. Is it a conductor, a semiconductor or an insulator?
Q4
A p-n junction has a barrier potential of 0.7 V. Find the effective barrier height when (a) a forward bias of 0.3 V and (b) a reverse bias of 2.0 V is applied.
Q5
On a diode's forward characteristic, a 0.10 V increase in voltage raises the current by 25 mA. Find its dynamic resistance.
Q6
A silicon diode (forward drop 0.7 V) and a 100 Ω resistor are connected in series with a 3.0 V battery, with the diode forward biased. Find the current.

📝 Notes

Semiconductor Electronics

This chapter builds from energy bands to doped semiconductors, then to the p-n junction diode and its use as a rectifier.

Bands decide the material

The size of the gap $E_g$ between the valence and conduction bands sorts solids into three groups:

Conductors: bands overlap, so plenty of free electrons.
Semiconductors: small gap ( $E_g < 3$ eV); a few electrons cross it at room temperature.
Insulators: large gap ( $E_g > 3$ eV); practically no conduction.

Doping and carrier densities

Pure (intrinsic) material has $n_e = n_h = n_i$ . Adding a pentavalent impurity makes it n-type ( $n_e \approx N_D$ ); a trivalent impurity makes it p-type ( $n_h \approx N_A$ ). In every case $n_e n_h = n_i^2$ , so the minority density is found by dividing $n_i^2$ by the majority density. Doped material stays electrically neutral.

The p-n junction and the diode

At the junction, electrons and holes diffuse across and recombine, leaving a depletion layer of fixed ions and a barrier potential $V_0$ . Under forward bias the barrier falls to $V_0 - V$ and current rises sharply above the threshold voltage (about 0.7 V for Si). Under reverse bias the barrier rises to $V_0 + V$ and only a tiny current flows until breakdown. The slope of the V–I curve gives the dynamic resistance $r_d = \Delta V/\Delta I$ .

Diode as a rectifier

Because a diode conducts in one direction only, it turns AC into pulsating DC:

Half-wave (one diode): output on one half-cycle; $\nu_{out} = \nu_{in}$ .
Full-wave (two diodes, centre-tap transformer): output on both half-cycles; $\nu_{out} = 2\nu_{in}$ .