Semiconductors formulas
Master Semiconductors through 30 JEE Advanced-level formulas, systematically structured with every variable spelled out. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.
Semiconductors, every formula
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Intrinsic carrier concentration
Q1MCQIntrinsicIn an intrinsic semiconductor, the electron and hole concentrations satisfy:- A$n_e=n_h=n_i$
- B$n_e\gg n_h$
- C$n_h\gg n_e$
- D$n_e=0$
- A
Mass-action law
all types
Q1MCQMass actionFor any semiconductor, $n_e\,n_h$ equals:- A$n_i^{2}$
- B$n_i$
- C$2n_i$
- D$0$
- A
p-type semiconductor
trivalent doping
Q1MCQp-typeA p-type semiconductor is formed by doping with a:- Atrivalent atom
- Bpentavalent atom
- Ctetravalent atom
- Dnoble gas
- A
n-type semiconductor
pentavalent doping
Q1MCQn-typeAn n-type semiconductor is formed by doping with a:- Apentavalent atom
- Btrivalent atom
- Ctetravalent atom
- Dnoble gas
- A
Total current
Q1MCQTotal currentThe total current in a semiconductor is:- A$i_e+i_h$
- B$i_e-i_h$
- C$i_e$
- D$i_h$
- A
Conductivity
Q1MCQConductivityThe conductivity of a semiconductor is:- A$e(n_e\mu_e+n_h\mu_h)$
- B$e n_e\mu_e$
- C$e n_h\mu_h$
- D$e(n_e-n_h)$
- A
n-type conductivity
Q1MCQn-type conductivityFor an n-type semiconductor, the conductivity is dominated by:- A$e n_e\mu_e$
- B$e n_h\mu_h$
- C$e(n_e+n_h)$
- D$0$
- A
p-type conductivity
Q1MCQp-type conductivityFor a p-type semiconductor, the conductivity is dominated by:- A$e n_h\mu_h$
- B$e n_e\mu_e$
- C$e(n_e+n_h)$
- D$0$
- A
Mobility
Q1MCQMobilityThe mobility of a charge carrier is:- A$\dfrac{v_d}{E}$
- B$v_d E$
- C$\dfrac{E}{v_d}$
- D$eE$
- A
Carrier generation
Q1MCQCarrier generationThe number of carriers excited across the gap varies as:- A$T^{3/2}e^{-E_g/2kT}$
- B$e^{E_g/2kT}$
- C$T^{2}$
- D$e^{-kT}$
- A
Dynamic resistance of a diode
Q1MCQDynamic resistanceThe dynamic resistance of a diode is:- A$\dfrac{\Delta V}{\Delta I}$
- B$\dfrac{\Delta I}{\Delta V}$
- C$VI$
- D$\dfrac{V}{I}$
- A
Emitter current
Q1NumericalEmitter currentIn a transistor, $i_b=0.1$ mA and $i_c=9.9$ mA. The emitter current (mA) is:CB dc current gain
0.95–0.99
Q1NumericalαFor $i_c=9.9$ mA and $i_e=10$ mA, the dc current gain $\alpha$ is (decimal):CB ac current gain
Q1MCQCB ac gainThe CB ac current gain is:- A$\dfrac{\Delta i_c}{\Delta i_e}$
- B$\dfrac{\Delta i_c}{\Delta i_b}$
- C$\dfrac{i_c}{i_b}$
- D$\dfrac{i_e}{i_c}$
- A
CE dc current gain
Q1NumericalβFor $i_c=9.9$ mA and $i_b=0.1$ mA, the dc current gain $\beta$ is:CE ac current gain
Q1MCQCE ac gainThe CE ac current gain is:- A$\dfrac{\Delta i_c}{\Delta i_b}$
- B$\dfrac{\Delta i_c}{\Delta i_e}$
- C$\dfrac{i_e}{i_c}$
- D$\dfrac{i_b}{i_c}$
- A
β in terms of α
Q1MCQβ from αThe relation between $\beta$ and $\alpha$ is:- A$\beta=\dfrac{\alpha}{1-\alpha}$
- B$\beta=\dfrac{1-\alpha}{\alpha}$
- C$\beta=1-\alpha$
- D$\beta=\alpha$
- A
α in terms of β
Q1MCQα from βThe relation for $\alpha$ in terms of $\beta$ is:- A$\alpha=\dfrac{\beta}{1+\beta}$
- B$\alpha=\dfrac{1+\beta}{\beta}$
- C$\alpha=1+\beta$
- D$\alpha=\beta$
- A
Voltage gain (CE)
Q1MCQVoltage gainThe voltage gain of a CE amplifier is:- A$\beta_{ac}\times\text{resistance gain}$
- B$\alpha_{ac}\times\text{resistance gain}$
- C$\beta_{ac}^{2}$
- D$\text{resistance gain}$
- A
Power gain (CE)
Q1MCQPower gainThe power gain of a CE amplifier is:- A$\beta_{ac}^{2}\times\text{resistance gain}$
- B$\beta_{ac}\times\text{resistance gain}$
- C$\beta_{ac}$
- D$\alpha_{ac}^{2}$
- A
Transconductance
Q1MCQTransconductanceThe transconductance is:- A$\dfrac{\Delta i_c}{\Delta V_{EB}}$
- B$\dfrac{\Delta V_{EB}}{\Delta i_c}$
- C$\dfrac{i_c}{i_b}$
- D$A_v R_L$
- A
OR gate
Q1MCQOR gateThe output of an OR gate is:- A$A+B$
- B$A\cdot B$
- C$\bar A$
- D$A\oplus B$
- A
AND gate
Q1MCQAND gateThe output of an AND gate is:- A$A\cdot B$
- B$A+B$
- C$\bar A$
- D$\overline{A\cdot B}$
- A
NOT gate
Q1MCQNOT gateThe output of a NOT gate is:- A$\bar A$
- B$A$
- C$A+B$
- D$A\cdot B$
- A
NAND gate
Q1MCQNAND gateThe output of a NAND gate is:- A$\overline{A\cdot B}$
- B$A\cdot B$
- C$\overline{A+B}$
- D$A+B$
- A
NOR gate
Q1MCQNOR gateThe output of a NOR gate is:- A$\overline{A+B}$
- B$A+B$
- C$\overline{A\cdot B}$
- D$A\cdot B$
- A
XOR gate
Q1MCQXOR gateThe output of an XOR gate is:- A$A\oplus B$
- B$A+B$
- C$A\cdot B$
- D$\bar A$
- A
De Morgan's first theorem
Q1MCQDe Morgan 1De Morgan's first theorem states:- A$\overline{A+B}=\bar A\cdot\bar B$
- B$\overline{A+B}=\bar A+\bar B$
- C$\overline{A\cdot B}=\bar A\cdot\bar B$
- D$\overline{A+B}=A\cdot B$
- A
De Morgan's second theorem
Q1MCQDe Morgan 2De Morgan's second theorem states:- A$\overline{A\cdot B}=\bar A+\bar B$
- B$\overline{A\cdot B}=\bar A\cdot\bar B$
- C$\overline{A+B}=\bar A+\bar B$
- D$\overline{A\cdot B}=A+B$
- A
Absorption law
Q1MCQAbsorption lawThe Boolean absorption law states $A+AB$ equals:- A$A$
- B$B$
- C$AB$
- D$A+B$
- A
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