Energy bands in solids (qualitative)
In one atom, an electron can have only some fixed energies (levels). In a solid, crores of atoms sit close together. Their levels spread into wide bands.
- Valence band: the top band that is full of bound electrons.
- Conduction band: the band above it. An electron here is free to move and carry current.
- Energy gap (Eg): the empty energy space between the two bands. No electron can stay there.
Three kinds of solids
- Conductor (metal): bands overlap, Eg = 0. Many free electrons. Conducts well.
- Semiconductor: small gap, Eg less than about 3 eV (Si 1.1 eV, Ge 0.7 eV). At room temperature a few electrons jump. Conducts a little.
- Insulator: big gap, Eg more than 3 eV (diamond about 5.4 eV). Electrons cannot jump. Does not conduct.
Heating a semiconductor gives more electrons the energy to jump, so its resistance falls with temperature. A metal behaves the opposite way.
Intrinsic semiconductor
Intrinsic means pure. In pure Si or Ge each atom has 4 outer electrons and shares them with 4 neighbours (covalent bonds).
Heat breaks a bond. The electron becomes free. The empty place it leaves is a hole. A hole acts like a positive charge, because a nearby electron can jump into it, and then the hole seems to move the other way.
Free electrons and holes are always made in pairs, so ne = nh = ni (ni = intrinsic carrier concentration). Total current = electron current + hole current.
Extrinsic semiconductor: n-type and p-type
Doping = adding a very small amount (about 1 atom in a million) of another element. The result is an extrinsic semiconductor. It conducts much better than pure silicon.
n-type
Add a pentavalent atom (5 outer electrons: P, As, Sb). Four electrons make bonds; the fifth is loosely held and becomes free easily. The dopant is called a donor. Majority carriers = electrons; minority carriers = holes.
p-type
Add a trivalent atom (3 outer electrons: B, Al, In, Ga). One bond is left incomplete: a hole. The dopant is called an acceptor. Majority carriers = holes; minority carriers = electrons.
Key rule
In any semiconductor at a steady temperature, ne × nh = ni². More electrons means fewer holes.
Important: an n-type or p-type crystal is still electrically neutral. Each extra electron came with its own atom.
Formation of a p-n junction
A p-n junction is one crystal with p-type on one side and n-type on the other.
- Diffusion: n side has many electrons, p side has many holes. Each spreads to where there are fewer. Electrons cross to p, holes cross to n.
- They meet and cancel. Near the border only fixed ions remain: negative ions on the p side, positive ions on the n side.
- This thin carrier-free strip is the depletion layer (about a micrometre thick).
- The fixed ions make an electric field from n to p. It pushes back further diffusion and causes a drift current the other way.
- When diffusion current = drift current, the junction is in balance. The p.d. across the layer is the barrier potential: about 0.7 V for Si, 0.3 V for Ge.
Diode I-V characteristics: forward and reverse bias
A diode is a p-n junction with two leads. Symbol: an arrow (p side, anode) touching a bar (n side, cathode). The arrow shows the easy current direction.
Forward bias
p to +, n to −. The battery pushes against the barrier, so the depletion layer gets thinner and the barrier falls to (V0 − V). Below the threshold (knee) voltage (≈ 0.7 V Si) the current is tiny. Above it, current rises very fast (milliamperes).
Reverse bias
p to −, n to +. The barrier rises to (V0 + V) and the layer gets wider. Only minority carriers cross: a tiny, almost constant reverse saturation current (microamperes). At a large reverse voltage, breakdown happens and current rises suddenly.
The graph
The I-V curve is not a straight line, so a diode is non-ohmic. Dynamic resistance rd = ΔV / ΔI, read from the slope near a point. It is small in forward bias and very large in reverse bias.
Try it (practical): in the 3D at step 5, drag the bias slider from 0 to 1 V slowly and watch the red dot climb the I-V curve; then drag to −3 V and see the current almost vanish. At home with a multimeter on diode mode: red probe on the anode shows about 0.6–0.7; swap probes and it shows OL.
Diode as a rectifier
Rectification = changing AC (direction flips) into DC (one direction). The diode is a one-way door, so it can do it.
Half-wave rectifier
One diode in series with the load. In the positive half the diode is forward biased and current flows; in the negative half it is reverse biased and nothing flows. Output = only the positive halves. Output ripple frequency = input frequency (50 Hz). Half the energy is wasted.
Full-wave rectifier (centre-tap)
A centre-tapped transformer and two diodes. In one half D1 conducts, in the other half D2 conducts; both send current through the load in the same direction. Output = all humps upward. Ripple frequency = 2 × input = 100 Hz.
Filter
The output is one-way but bumpy (pulsating DC). A capacitor across the load charges at each peak and slowly gives charge in the gaps, so the output becomes almost steady.
Key formulas and definitions
- n_e × n_h = n_i²
- Intrinsic: n_e = n_h = n_i
- Forward bias: barrier = V₀ − V; Reverse bias: barrier = V₀ + V
- Dynamic resistance r_d = ΔV / ΔI
- Current in series diode circuit: I = (V − V_knee) / R
- Half-wave: f_out = f_in; Full-wave: f_out = 2 f_in
- Photon from gap: λ = hc / Eg ≈ 1240 / Eg(eV) nm
Worked examples
1. Pure silicon at 300 K has n_i = 1.5 × 10¹⁶ m⁻³. How many holes per m³ does it have?
Step 1: Pure means intrinsic, so n_h = n_e. Step 2: n_h = n_i = 1.5 × 10¹⁶ m⁻³.
2. Silicon (n_i = 1.5 × 10¹⁶ m⁻³) is doped with arsenic so that n_e = 5 × 10²² m⁻³. Find n_h. Is it n-type or p-type?
Step 1: Use n_e × n_h = n_i². Step 2: n_h = (1.5 × 10¹⁶)² / (5 × 10²²) = 2.25 × 10³² / 5 × 10²² = 4.5 × 10⁹ m⁻³. Step 3: Electrons ≫ holes, and arsenic has 5 outer electrons, so it is n-type.
3. A Si diode (knee 0.7 V) is in series with a 200 Ω resistor and a 5 V battery, forward biased. Find the current.
Step 1: The diode takes about 0.7 V. Step 2: Voltage left for the resistor = 5 − 0.7 = 4.3 V. Step 3: I = 4.3 / 200 = 0.0215 A = 21.5 mA.
4. The same circuit, but the diode is turned round (reverse bias). Find the current roughly.
Step 1: Reverse bias: only a tiny reverse saturation current (microamperes). Step 2: So I ≈ 0 (a few µA). The resistor gets almost no voltage; the full 5 V is across the diode.
5. On a diode's forward curve, voltage rises from 0.70 V to 0.72 V and current rises from 10 mA to 20 mA. Find the dynamic resistance.
Step 1: ΔV = 0.72 − 0.70 = 0.02 V. Step 2: ΔI = 20 − 10 = 10 mA = 0.01 A. Step 3: r_d = ΔV / ΔI = 0.02 / 0.01 = 2 Ω.
6. In reverse bias the voltage changes from 10 V to 20 V and the current changes from 1 µA to 1.5 µA. Find the dynamic resistance.
Step 1: ΔV = 10 V; ΔI = 0.5 µA = 5 × 10⁻⁷ A. Step 2: r_d = 10 / (5 × 10⁻⁷) = 2 × 10⁷ Ω = 20 MΩ. Step 3: Compare: forward r_d was a few ohms. That huge difference is why a diode works as a one-way door.
7. Input AC is 50 Hz. What is the output ripple frequency of (a) a half-wave and (b) a full-wave rectifier?
Step 1: Half-wave passes one hump per cycle: 50 Hz. Step 2: Full-wave passes two humps per cycle: 2 × 50 = 100 Hz.
8. A semiconductor has Eg = 2.0 eV. Can light of wavelength 700 nm free electrons across the gap?
Step 1: Photon energy E = 1240 / λ(nm) eV = 1240 / 700 ≈ 1.77 eV. Step 2: 1.77 eV < 2.0 eV, so the photon is too weak. Step 3: No. Light shorter than 1240/2 = 620 nm is needed.
Common mistakes
- Saying n-type is negatively charged. It is neutral: each extra electron came with a positive donor atom.
- Thinking holes are real particles. A hole is a missing electron in a bond; it just behaves like a + charge.
- Forgetting the 0.7 V drop in circuit sums: I = (V − 0.7)/R, not V/R.
- Mixing up the ripple frequency: full-wave output is 2 × input frequency (100 Hz for 50 Hz mains), not 50 Hz.