Online Test — Semiconductor Electronics
18 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 18
The energy gap of an insulator is typically:
$0\,\text{eV}$
about $1\,\text{eV}$
greater than $3\,\text{eV}$
negative
Explanation: Insulators have a large forbidden gap (e.g. diamond $\approx 6\,\text{eV}$), so electrons cannot reach the conduction band.
Question 2 of 18
In an n-type semiconductor the dopant is:
trivalent (acceptor)
pentavalent (donor)
tetravalent
a noble gas
Explanation: A pentavalent donor (P, As, Sb) supplies an extra free electron, making the material n-type.
Question 3 of 18
For an extrinsic semiconductor, the carrier concentrations satisfy:
$n_e = n_h$
$n_e\, n_h = n_i^2$
$n_e + n_h = n_i$
$n_e\, n_h = 0$
Explanation: The law of mass action holds: $n_e\, n_h = n_i^2$ at a given temperature.
Question 4 of 18
Under forward bias, the width of the depletion region:
increases
decreases
stays the same
becomes infinite
Explanation: Forward bias opposes the barrier field, so the depletion region narrows and current flows easily.
Question 5 of 18
The barrier potential of a germanium p-n junction is about:
$0.3\,\text{V}$
$0.7\,\text{V}$
$1.1\,\text{V}$
$5\,\text{V}$
Explanation: Germanium has a barrier potential of about $0.3\,\text{V}$; silicon is about $0.7\,\text{V}$.
Question 6 of 18
The DC output of a full-wave rectifier with peak voltage $V_m$ is:
$V_m/\pi$
$2V_m/\pi$
$V_m$
$V_m/2$
Explanation: Using both half-cycles gives $V_{dc} = 2V_m/\pi$, double the half-wave value.
Question 7 of 18
A capacitor filter is connected:
in series with the load
in parallel with the load
in series with the diode only
across the AC source
Explanation: A filter capacitor sits in parallel with the load, charging at peaks and discharging through the load to smooth the output.
Question 8 of 18
A Zener diode is normally operated in:
forward bias
reverse breakdown
the active region
cut-off
Explanation: The Zener regulates voltage by operating in the reverse breakdown region where $V_Z$ stays nearly constant.
Question 9 of 18
The colour of light from an LED is determined mainly by:
the forward current
the band gap $E_g$ of the material
the reverse voltage
the size of the diode
Explanation: The emitted photon energy $h\nu \approx E_g$, so the band gap fixes the wavelength and colour.
Question 10 of 18
A solar cell operates with:
forward bias
reverse bias
no external bias
AC bias
Explanation: A solar cell needs no external bias; the junction field separates photo-generated carriers to produce a photovoltage.
Question 11 of 18
In a transistor the relation $I_E = I_B + I_C$ shows that:
the base current is the largest
the emitter current is the sum of base and collector currents
the collector current is zero
all currents are equal
Explanation: Charge conservation gives $I_E = I_B + I_C$, with $I_B$ small and $I_C$ large.
Question 12 of 18
If $\alpha = 0.95$, the common-emitter current gain $\beta$ is:
$19$
$0.95$
$95$
$1.05$
Explanation: $\beta = \alpha/(1-\alpha) = 0.95/0.05 = 19$.
Question 13 of 18
A transistor used as a switch is driven between:
active and breakdown
cut-off and saturation
forward and reverse bias only
two amplifier gains
Explanation: Switching uses cut-off (OFF) and saturation (ON), the two extreme states.
Question 14 of 18
The output of an OR gate is 0 only when:
all inputs are 1
all inputs are 0
exactly one input is 1
any input is 1
Explanation: $Y = A + B$ is 0 only when every input is 0; otherwise it is 1.
Question 15 of 18
The Boolean expression $Y = \overline{A \cdot B}$ represents a:
NOR gate
AND gate
NAND gate
OR gate
Explanation: An AND followed by a NOT is a NAND gate: $Y = \overline{A \cdot B}$.
Question 16 of 18
A common-emitter amplifier has $\beta = 50$, $R_L = 4\,\text{k}\Omega$ and $R_{in} = 1\,\text{k}\Omega$. Its voltage gain magnitude is:
$50$
$200$
$12.5$
$4$
Explanation: $|A_V| = \beta R_L/R_{in} = 50 \times 4 = 200$.
Question 17 of 18
The reverse saturation current in a diode is carried by:
majority carriers
minority carriers
donor ions
the depletion ions
Explanation: Only minority carriers are pushed across the junction in reverse bias, giving a tiny saturation current.
Question 18 of 18
As temperature increases, the resistance of a pure semiconductor:
increases
decreases
stays constant
becomes infinite
Explanation: More electrons cross the small gap as temperature rises, increasing carrier number and decreasing resistance.