"Semiconductor Electronics" — Practice

Question 1

Explain why the resistance of a copper wire rises with temperature while that of a silicon sample falls, and estimate the factor by which silicon's carrier concentration changes between $300$ K and $350$ K. Take $E_g=1.1$ eV.

Accepted Answer

In copper the carrier number is fixed, so heating only increases lattice scattering and the resistance rises.

In silicon carriers must be created across the gap. With $kT=0.0259$ eV at $300$ K and $0.0302$ eV at $350$ K:

$\dfrac{E_g}{2kT}$ falls from $21.2$ to $18.2$

$\dfrac{n_{350}}{n_{300}}=e^{3.0}	imes\left(\dfrac{350}{300}ight)^{3/2}=20.1	imes1.26=25$

About twenty-five times more carriers, which swamps the increased scattering entirely.

Question 2

A germanium sample has $n_i=2.4	imes10^{19}$ m$^{-3}$ and is doped so that the electron concentration is $4.8	imes10^{22}$ m$^{-3}$. Find the hole concentration and state the type.

Accepted Answer

$n_h=\dfrac{n_i^{2}}{n_e}=\dfrac{\left(2.4	imes10^{19}ight)^{2}}{4.8	imes10^{22}}=\dfrac{5.76	imes10^{38}}{4.8	imes10^{22}}$

$n_h=1.2	imes10^{16}$ m$^{-3}$

Electrons outnumber holes by a factor of $4	imes10^{6}$, so the material is **n-type**.

Note that doping has suppressed the hole concentration by a factor of two thousand relative to the intrinsic value.

Question 3

Intrinsic germanium has $n_i=2.4	imes10^{19}$ m$^{-3}$, $\mu_e=0.39$ and $\mu_h=0.19$ m$^{2}$ V$^{-1}$ s$^{-1}$. Find its conductivity and compare it with intrinsic silicon's value of $4.4	imes10^{-4}$ S m$^{-1}$.

Accepted Answer

$\sigma=\left(1.6	imes10^{-19}ight)\left(2.4	imes10^{19}ight)\left(0.39+0.19ight)$

$=\left(3.84ight)\left(0.58ight)=2.23$ S m$^{-1}$

That is about **five thousand times** silicon's value, almost entirely because germanium's smaller gap of $0.7$ eV gives vastly more thermal carriers.

The same small gap is why germanium devices fail at modest temperatures.

Question 4

A silicon diode has a reverse saturation current of $10$ nA at $300$ K. Estimate it at $340$ K, and explain why the value is almost independent of the reverse voltage applied.

Accepted Answer

A rise of $40$ K is four doublings:

$I_0=10	imes2^{4}=160$ nA

It is voltage-independent because it is carried by **minority carriers**, and every one that reaches the depletion region is already swept across. Raising the voltage pushes them harder but cannot create more of them.

What does create more is temperature, through the thermal generation factor $e^{-E_g/2kT}$.

Question 5

A diode carries $0.5$ mA at $0.65$ V and $5$ mA at $0.71$ V. Find its dynamic resistance in each case and comment.

Accepted Answer

At $0.5$ mA: $r_d\approx\dfrac{26}{0.5}=52\ \Omega$

At $5$ mA: $r_d\approx\dfrac{26}{5}=5.2\ \Omega$

Ten times the current gives a tenth the dynamic resistance, exactly as the inverse relation requires.

The static resistances, by contrast, are $1300\ \Omega$ and $142\ \Omega$ — quite different numbers describing the same device at the same two points.

Question 6

A diode breaks down at $8.2$ V. Identify the mechanism and state how the breakdown voltage changes when the device is heated.

Accepted Answer

At $8.2$ V, above $6$ V, the mechanism is **avalanche** breakdown by impact ionisation in a wide, lightly doped depletion layer.

Its temperature coefficient is **positive**, so heating **raises** the breakdown voltage.

A device below $5$ V would break down by direct field ionisation instead, with the opposite sign. Near $5.6$ V both act and the coefficients cancel, which is why that rating is used for voltage references.

Question 7

A full-wave rectifier delivers a peak current of $150$ mA. Find the average and root-mean-square currents, and the ripple frequency for a $50$ Hz supply.

Accepted Answer

$I_{dc}=\dfrac{2I_m}{\pi}=\dfrac{2	imes150}{3.1416}=95.5$ mA

$I_{rms}=\dfrac{I_m}{\sqrt2}=106$ mA

Ripple frequency $=2	imes50=100$ Hz.

A half-wave circuit with the same peak would give $47.7$ mA average at $50$ Hz ripple, which is both smaller and harder to smooth.

Question 8

A $12$ V supply is regulated to $6$ V by a Zener for a load taking $30$ mA, with a minimum Zener current of $10$ mA. Find the series resistance and the Zener dissipation with the load disconnected.

Accepted Answer

$R_S=\dfrac{12-6}{\left(10+30ight)	imes10^{-3}}=\dfrac{6}{0.04}=150\ \Omega$

With no load, the whole current passes through the Zener:

$I_Z=\dfrac{6}{150}=40$ mA

$P_Z=6	imes0.040=0.24$ W

The Zener must therefore be rated for at least a quarter of a watt, even though at full load it dissipates only $0.06$ W.

Question 9

An LED emits at $520$ nm. Find the band gap of the material, and explain why germanium could not be used.

Accepted Answer

$E_g=\dfrac{1240}{520}=2.38$ eV

Germanium's gap is only $0.7$ eV, which would correspond to $1771$ nm — far into the infrared.

Any material used for a visible LED needs a gap between about $1.8$ and $3.1$ eV, which excludes both germanium and silicon and is why gallium compounds are used instead.

Question 10

A transistor has $\beta=120$ and an emitter current of $6.05$ mA. Find $\alpha$, the collector current and the base current.

Accepted Answer

$\alpha=\dfrac{\beta}{1+\beta}=\dfrac{120}{121}=0.9917$

$I_C=\alpha I_E=0.9917	imes6.05=6.00$ mA

$I_B=I_E-I_C=6.05-6.00=0.05$ mA $=50\ \mu$A

Check: $\dfrac{I_C}{I_B}=\dfrac{6.00}{0.05}=120$, which is $\beta$ as required.

Question 11

A common-emitter stage has $V_{CC}=12$ V, $R_C=3\ 	ext{k}\Omega$ and $\beta=150$. Find $V_{CE}$ for a base current of $15\ \mu$A, and the base current at which the transistor saturates.

Accepted Answer

$I_C=150	imes15	imes10^{-6}=2.25$ mA

$V_{CE}=12-\left(2.25	imes10^{-3}ight)\left(3000ight)=12-6.75=5.25$ V

Saturation requires $I_C=\dfrac{12}{3000}=4$ mA, so

$I_B=\dfrac{4}{150}=26.7\ \mu$A

At $15\ \mu$A the stage sits near the middle of the load line, which is where an amplifier should be biased.

Question 12

Construct an AND gate and a NOT gate using only NOR gates, and state why NAND and NOR are called universal.

Accepted Answer

**NOT:** tie both inputs of a NOR together. The output is $\overline{A+A}=\bar A$.

**AND:** invert each input with a NOR-as-NOT to get $\bar A$ and $\bar B$, then feed both into a third NOR:

$\overline{\bar A+\bar B}=A\cdot B$ by De Morgan's relation.

Since every basic gate can be built from copies of NOR alone, and likewise from NAND alone, either gate suffices to construct any logic function whatever — which is why integrated circuits are made from one repeated gate type.

Practice — "Semiconductor Electronics"

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