IMO Practice Test — Semiconductor Electronics
12 Questions • 15 min • Olympiad level
15:00
Question 1 of 12
A silicon diode and a germanium diode are connected in series and forward-biased. The first to conduct as the voltage rises from zero is:
the silicon diode
the germanium diode
both at the same voltage
neither conducts
Explanation: Germanium has the lower barrier potential ($\approx 0.3\,\text{V}$ vs $0.7\,\text{V}$ for Si), so it begins conducting first.
Question 2 of 12
Doping increases the conductivity of a semiconductor mainly because it:
raises the temperature
greatly increases the number of majority charge carriers
removes all holes
widens the band gap
Explanation: A controlled impurity adds many majority carriers, far more than the intrinsic carriers, raising conductivity.
Question 3 of 12
An intrinsic semiconductor has $n_i = 1.6\times10^{16}\,\text{m}^{-3}$. After doping, $n_e = 8\times10^{22}\,\text{m}^{-3}$. The hole concentration is closest to:
$3.2\times10^{9}\,\text{m}^{-3}$
$8\times10^{22}\,\text{m}^{-3}$
$1.6\times10^{16}\,\text{m}^{-3}$
$2\times10^{6}\,\text{m}^{-3}$
Explanation: $n_h = n_i^2/n_e = (1.6\times10^{16})^2/(8\times10^{22}) = 2.56\times10^{32}/8\times10^{22} = 3.2\times10^{9}\,\text{m}^{-3}$.
Question 4 of 12
A full-wave rectifier and a half-wave rectifier use the same peak input. The ratio of their DC output voltages (full : half) is:
$1 : 1$
$2 : 1$
$1 : 2$
$\pi : 1$
Explanation: $V_{dc}$ is $2V_m/\pi$ for full-wave and $V_m/\pi$ for half-wave, a ratio of $2 : 1$.
Question 5 of 12
In a Zener regulator the input rises while the load stays fixed. To keep the output at $V_Z$, the Zener current must:
decrease
increase
stay the same
drop to zero
Explanation: Extra current from the higher input flows through the Zener (not the load), so $I_Z$ rises while the output stays at $V_Z$.
Question 6 of 12
An LED requires the band gap to be at least about $1.8\,\text{eV}$ to emit visible light because:
smaller gaps give infrared photons
smaller gaps give ultraviolet photons
the diode would not conduct
visible photons need lower energy
Explanation: Visible photons have energies of roughly $1.8$ to $3.1\,\text{eV}$; a gap below this emits lower-energy infrared light.
Question 7 of 12
A transistor has $I_C = 9.8\,\text{mA}$ and $I_E = 10\,\text{mA}$. Its value of $\alpha$ is:
$0.98$
$0.49$
$1.02$
$49$
Explanation: $\alpha = I_C/I_E = 9.8/10 = 0.98$.
Question 8 of 12
For the transistor in the previous question, the common-emitter gain $\beta$ is:
$0.98$
$49$
$98$
$100$
Explanation: $\beta = \alpha/(1-\alpha) = 0.98/0.02 = 49$.
Question 9 of 12
Two NOT gates are connected in series (output of the first feeds the input of the second). The combination behaves as:
an OR gate
an AND gate
a buffer (output equals input)
a NAND gate
Explanation: Double inversion restores the original value: $\overline{\overline{A}} = A$, so the pair acts as a buffer.
Question 10 of 12
A NOR gate has both inputs tied together. The resulting single-input gate is:
a buffer
an AND gate
a NOT gate
an OR gate
Explanation: $Y = \overline{A + A} = \overline{A}$, so a NOR with joined inputs becomes a NOT gate — showing NOR is universal.
Question 11 of 12
A common-emitter amplifier gives an output that is $180^\circ$ out of phase with the input. This phase reversal arises because:
the diode rectifies the signal
a rise in base current raises $I_C$, increasing the drop across $R_L$ and lowering the output
the capacitor charges slowly
the transistor is in cut-off
Explanation: As $I_C$ rises with the input, the larger drop across $R_L$ lowers the collector (output) voltage, inverting the signal.
Question 12 of 12
A photodiode shows almost no change in current under long-wavelength infrared light but responds to visible light. The reason is:
infrared photons have energy below $E_g$ and cannot create electron-hole pairs
infrared light is reflected
the diode is forward-biased
infrared light has too much energy
Explanation: Only photons with $h\nu > E_g$ can free carriers; long-wavelength infrared has too little energy, so no extra current flows.