Concept Core
From energy bands to transistors and logic gates β the semiconductor framework.
Energy Bands & Types of Semiconductors
| Material | Band Gap | Property |
|---|
| Conductor | No gap (bands overlap) | Always conducts |
| Semiconductor | ~1 eV (Si: 1.1 eV, Ge: 0.7 eV) | Conducts at higher T or with doping |
| Insulator | >3 eV (diamond: 5.5 eV) | Rarely conducts |
Intrinsic semiconductor: Pure Si or Ge. Equal holes and electrons: n = p = nα΅’.
n-type: Doped with pentavalent (P, As, Sb). Majority: electrons. Minority: holes.
p-type: Doped with trivalent (B, Al, In). Majority: holes. Minority: electrons.
p-n Junction Diode
At the junction: depletion region forms (no free charge carriers). Built-in potential barrier ~0.3V (Ge) or 0.7V (Si).
Forward bias: p side to +ve terminal. Barrier reduces β current flows easily.
Reverse bias: n side to +ve terminal. Barrier increases β only small leakage current flows.
Diode allows current in one direction β rectification.
Transistor Types & Operation
BJT (Bipolar Junction Transistor): n-p-n or p-n-p. Three regions: Emitter, Base, Collector.
Ξ± = I_C/I_E (current gain, CE to EB) Ξ± < 1
Ξ² = I_C/I_B (current gain, CB to CE) Ξ² >> 1
Relation: Ξ² = Ξ±/(1βΞ±) Ξ± = Ξ²/(1+Ξ²)
I_E = I_B + I_C
Ξ² (h_FE) is typically 20β500. Transistor amplifies when in active region (emitter forward, collector reverse biased).
Logic Gates
| Gate | Symbol | Output Y |
|---|
| AND | Β· | Y = AΒ·B (1 only if both 1) |
| OR | + | Y = A+B (1 if either 1) |
| NOT | Β¬ | Y = Δ (inverts input) |
| NAND | | Y = Μ(AΒ·B) (AND + NOT) |
| NOR | | Y = Μ(A+B) (OR + NOT) |
| XOR | β | Y = AβB (1 if different) |
NAND and NOR are universal gates β any logic circuit can be built from only NAND gates (or only NOR gates).
Rectifiers
Half-wave rectifier: One diode. Output is one half of AC cycle. Ripple factor = 1.21.
Full-wave rectifier (bridge): 4 diodes. Both halves of AC are rectified. Ripple factor = 0.48.
Centre-tap rectifier: 2 diodes. Similar to full-wave.
Special Diodes
Zener diode: Operates in reverse breakdown. Used as voltage regulator. Maintains constant voltage across it regardless of load changes.
LED: Emits light when forward biased. Energy released as photon when electron recombines with hole: E = hΞ½.
Photodiode: Works in reverse bias. Current increases when light falls on it.
Formula Vault
Semiconductor device formulas and truth tables for EAPCET.
Current Gain Ξ±
Ξ± = I_C/I_E
Common base; always <1
Current Gain Ξ²
Ξ² = I_C/I_B
Common emitter; Ξ² = 20β500
Ξ±-Ξ² Relation
Ξ² = Ξ±/(1βΞ±); Ξ± = Ξ²/(1+Ξ²)
Convert between configurations
Current Relation
I_E = I_B + I_C
KCL at transistor node
Voltage Gain (CE amp)
A_v = Ξ² Γ R_C/R_i
R_C = collector resistance
AND Gate
Y = A Β· B
High output only when both inputs high
OR Gate
Y = A + B
High output when any input high
NAND Gate
Y = Β¬(A Β· B)
Universal gate
Worked Examples
5 problems β diode biasing, transistor Ξ², logic gates, Zener, and a trap.
EasyFind Ξ² if I_B = 50 ΞΌA and I_C = 5 mAβΎ
A transistor has base current I_B = 50 ΞΌA and collector current I_C = 5 mA. Find Ξ² and Ξ±.
1
Ξ² = I_C/I_B = 5Γ10β»Β³ / 50Γ10β»βΆ = 100
2
Ξ± = Ξ²/(1+Ξ²) = 100/101 β 0.99
3
I_E = I_B + I_C = 50 ΞΌA + 5 mA β 5.05 mA
β Ξ² = 100, Ξ± β 0.99
EasyTruth table for NAND gate (A=1, B=0)βΎ
For a NAND gate, find the output when A=1 and B=0.
1
NAND: Y = Β¬(AΒ·B) = Β¬(1Β·0) = Β¬0 = 1
β Y = 1
MediumIdentify the circuit: two NAND gates, inputs both connected togetherβΎ
Two inputs A and A are connected to a NAND gate (both inputs tied together). What is the output?
1
With both inputs tied: A = A. Y = Β¬(AΒ·A) = Β¬A
2
This is a NOT gate β NAND with tied inputs acts as NOT.
3
This demonstrates NAND is a universal gate.
β Output = Δ (NOT gate) β NAND with tied inputs = NOT
EAPCET LevelFind output of logic circuit: Y = AB + ΔBβΎ
Simplify the Boolean expression Y = AB + ΔB.
2
Since A + Δ = 1 (Boolean identity): Y = B Γ 1 = B
3
The circuit simplifies to just passing B β output equals B regardless of A.
β Y = B
Trap Questionp-type semiconductor has positive charge β True or False?βΎ
A student says: 'p-type semiconductor is positively charged because it has extra holes.' Evaluate.
1
The trap: p-type semiconductor is created by adding trivalent impurity atoms, which are electrically NEUTRAL.
2
The trivalent atom has 3 valence electrons but accepts one from the lattice, creating a hole (missing electron) in the lattice.
3
However, the impurity atom gains an electron β becomes negatively charged ion. The hole is mobile, not the ion.
4
Net result: the p-type semiconductor as a whole is electrically neutral β the extra negative ions are balanced by mobile holes.
β False β p-type semiconductor is electrically neutral overall; doping adds equal + and β charges
Mistake DNA
3 semiconductor errors from EAPCET distractor analysis.
π
Reverse Biased Diode Conducts No Current
A reverse-biased diode conducts a very small leakage current (minority carrier current), not exactly zero.
β Wrong
Reverse biased diode:
I = 0 exactly β
(ideal approximation only)
β Correct
Reverse bias: very small
leakage current flows β
(minority carriers) β
Approximated as ~0 in circuits
In practice, a reverse-biased diode passes a tiny leakage current due to minority charge carriers. For circuit analysis, this is usually negligible (treated as open circuit), but conceptually it's not zero.
β‘
Ξ± > 1 for a Good Transistor
Ξ± = I_C/I_E. Since I_C < I_E (some current goes to base), Ξ± is always less than 1.
β Wrong
Good transistor: Ξ± > 1 β
(more collector than
emitter current?)
β Correct
Ξ± = I_C/I_E < 1 always β
Ξ± β 0.95 to 0.99 β
Ξ² >> 1 (not Ξ±)
Ξ± < 1 because I_E = I_B + I_C means I_C < I_E. The quantity that is much greater than 1 is Ξ² = I_C/I_B, since I_B is very small compared to I_C.
π’
n-type Has Excess Positive Charge
n-type semiconductor (doped with pentavalent) has excess free electrons but is electrically neutral overall.
β Wrong
n-type: excess electrons
β negatively charged β
(it's neutral!)
β Correct
n-type: neutral overall β
Pentavalent donor atom
gives extra eβ» but becomes
+ve ion β net neutral β
When a pentavalent atom substitutes into the lattice, it provides one extra electron but becomes a fixed positive ion. The semiconductor remains neutral β free electrons and fixed positive ions balance.
Chapter Intelligence
Semiconductors bridges modern physics and electronics β a clean chapter with predictable questions.
EAPCET Weightage (2019β2024)
Transistor Ξ², Ξ± calculations~7 Logic gates and Boolean algebra~7 n-type vs p-type properties~4 Zener diode and special diodes~3 High-Yield PYQ Patterns
Find Ξ² from I_B and I_CΞ± from Ξ²: Ξ±=Ξ²/(1+Ξ²)Logic gate truth tableForward vs reverse biased diodeNAND/NOR as universal gatesBoolean algebra simplificationZener diode application
Exam Strategy
- Transistor: Ξ² = I_C/I_B (large, 20β500). Ξ± = I_C/I_E (less than 1, close to 1). Relation: Ξ² = Ξ±/(1βΞ±).
- Logic gates: learn truth tables for AND, OR, NOT, NAND, NOR, XOR. NAND and NOR are universal (can implement any logic).
- NAND with both inputs tied = NOT gate. NOR with both inputs tied = NOT gate. These are frequently tested configurations.
- Forward bias β current flows (barrier reduced). Reverse bias β negligible current (barrier increased). Zener is used in reverse breakdown for voltage regulation.
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