Diamond, silicon and germanium each have four valence electrons per atom. Which band-gap ordering is correct near room temperature?
(a)Eg(diamond)>Eg(Si)>Eg(Ge)
(b)Eg(Si)>Eg(diamond)>Eg(Ge)
(c)All three gaps are equal
(d)Eg(Ge)>Eg(Si)>Eg(diamond)
Show answer
(a) Eg(diamond)>Eg(Si)>Eg(Ge)
Equal valence count does not determine equal crystal band gaps: approximately 5.4,1.1 and 0.7 eV respectively.
Question 2
A student says diamond must conduct as well as silicon because both atoms are tetravalent. Which response is correct?
(a)Silicon conduction requires all covalent bonds to vanish
(b)Diamond atoms have no electrons
(c)Their equal valence counts make their band gaps and conductivities equal
(d)Their crystal band gaps differ greatly, so thermal carrier populations differ
Show answer
(d) Their crystal band gaps differ greatly, so thermal carrier populations differ
Band structure depends on bonding in the solid, not valence count alone. Diamond has a much wider gap.
Electron-hole pairs, conventional currents and temperature
Question 3
In an intrinsic semiconductor, electrons drift left and holes drift right under an electric field. How do their conventional-current contributions combine?
(a)Both point left and add
(b)Both point right and add
(c)Only electrons can contribute current
(d)They cancel because the particles move in opposite directions
Show answer
(b) Both point right and add
Conventional electron current is opposite electron velocity; hole current follows hole velocity.
Question 4
What effective charge is assigned to a hole in a silicon valence band?
(a)Zero
(b)+e
(c)−e
(d)+2e
Show answer
(b) +e
A hole is the vacancy left by an electron and behaves as an effective positive charge.
Donors, acceptors, neutrality and carrier equilibrium
Question 5
An n-type crystal contains far more conduction electrons than holes. What is its overall charge in the usual neutral bulk condition?
(a)Negative because electron density exceeds hole density
(b)It alternates sign each second
(c)Zero; donor ion cores balance the extra electron charge
(d)Positive because all dopants remove electrons
Show answer
(c) Zero; donor ion cores balance the extra electron charge
Mobile carrier imbalance is compensated by the fixed ionised dopant cores.
Question 6
Which dopant type produces p-type silicon with holes as majority carriers?
(a)Pentavalent donors such as phosphorus
(b)Only neutral vacancies with no impurity
(c)Trivalent acceptors such as boron
(d)Tetravalent silicon added to silicon
Show answer
(c) Trivalent acceptors such as boron
A trivalent substitution leaves an available acceptor state and an associated valence-band hole.
Diffusion, depletion ions and equilibrium drift
Question 7
Why can a depletion region create an electric field even though it lacks most mobile majority carriers?
(a)Every hole in it becomes a free proton
(b)Immobile positive donor and negative acceptor ions remain
(c)It contains no charge of any kind, yet creates the field
(d)The electric field is produced by moving dopant atoms
Show answer
(b) Immobile positive donor and negative acceptor ions remain
Carrier depletion leaves fixed space charge: positive on n and negative on p.
Question 8
In an unbiased p-left, n-right junction, which fixed-charge arrangement is present near the junction?
(a)No electric charges on either side of the depleted region
(b)Only mobile neutral atoms carry the barrier charge
(c)Negative ions on p; positive ions on n
(d)Positive ions on p; negative ions on n
Show answer
(c) Negative ions on p; positive ions on n
Ionised acceptors remain on p and ionised donors on n after mobile carriers diffuse and recombine.
Bias polarity, barrier width and reverse current
Question 9
What happens to barrier height and depletion width when forward bias is increased within the usual operating range?
(a)Height increases while width is unchanged
(b)Both decrease
(c)Both become independent of applied voltage
(d)Both increase
Show answer
(b) Both decrease
Forward bias opposes the built-in potential and reduces the depleted barrier.
Question 10
A junction has built-in barrier 0.7 V. A forward bias 0.2 V is applied. In the elementary barrier model, what is the remaining barrier?
(a)1.4 V
(b)0.5 V
(c)0.9 V
(d)0.7 V
Show answer
(b) 0.5 V
Forward bias subtracts from the built-in barrier:0.7−0.2=0.5 V.
Diode characteristics and circuit models
Question 11
The graph gives a locally straight part of a diode’s forward characteristic. What is the approximate dynamic resistance near 15 mA?
(a)0.01 Ω
(b)50 Ω
(c)10 Ω
(d)100 Ω
Show answer
(c) 10 Ω
Use the local changes:ΔV=0.1 V and ΔI=10 mA=0.010 A, giving 10 Ω. The dc ratio at 0.75 V and 15 mA is 50 Ω and is a different quantity.
Question 12
A diode’s reverse voltage reaches breakdown while an external resistor limits current safely. What happens to reverse current?
(a)It must become forward current with no polarity change
(b)It destroys charge conservation
(c)It remains zero because reverse bias always blocks current
(d)It rises sharply
Show answer
(d) It rises sharply
The reverse-open approximation fails at breakdown; the current-limiting circuit controls the resulting large reverse current.
Question 13
Why must an ordinary diode rectifier have a reverse-voltage rating above the applied reverse peak?
(a)The rating makes the semiconductor intrinsic
(b)The rating sets the ac frequency
(c)No such rating matters because reverse current is always exactly zero
(d)Otherwise reverse breakdown may cause large current and overheating
Show answer
(d) Otherwise reverse breakdown may cause large current and overheating
A sufficiently large reverse voltage causes breakdown, so ignoring reverse current indefinitely is unsafe circuit reasoning.
Half-wave and full-wave rectification and capacitor filtering
Question 14
A student places the only smoothing capacitor in series between rectifier output and load. Which correction permits the capacitor to supply the load between charging peaks?
(a)Replace the rectifier with a wire
(b)Remove the load so the supply can never deliver current
(c)Move the capacitor across the load terminals
(d)Keep it in series because a smoothing capacitor must carry all dc load current
Show answer
(c) Move the capacitor across the load terminals
The smoothing capacitor should be connected in parallel across the load terminals, with the correct polarity if it is electrolytic.
Near a rectified-voltage peak, the source charges the capacitor. Between peaks, the capacitor discharges through the load and helps maintain the load voltage.
A capacitor in series would block steady direct current instead of serving as a reservoir across the load. The correction is to move it across the load terminals.
Question 15
Where should a smoothing capacitor be connected in a rectifier with a resistive load?
(a)Only across a disconnected battery
(b)In parallel with the load
(c)In series with the transformer primary only
(d)In series with the load
Show answer
(b) In parallel with the load
Across the load it charges near peaks and supplies load current between peaks.
Common misconception in Semiconductor Electronics: Materials, Devices and Simple Circuits
Does a diode conduct as soon as any positive voltage appears?
Wrong
A silicon diode behaves like a short circuit at any forward voltage.
Why it feels right. The circuit symbol suggests a simple one-way switch.
Right
Real forward current grows strongly with forward voltage. The ideal switch is an approximation.
Worked example
This qualitative current–voltage curve shows very small current at low forward bias and much larger current near 0.7 V.
Compare a typical silicon diode near 0.1 V and near 0.7 V forward bias.
The current can be negligible in the first case and appreciable in the second, depending on the diode and circuit.
Cookies
We use cookies, including Meta’s and, if a ChatGPT advert brought
you, OpenAI’s, to see how visitors find us.
They are never used on your child’s screens in the app. You can turn them
off here, and change your mind any time.
Privacy policy