semiconductor conductivity diode and transistor

Semiconductors such as silicon and germanium lie between conductors and dielectrics: their band gap is small, $\Delta E \approx 1\ \text{eV}$. Heating or illumination moves some electrons from the valence band to the conduction band, leaving holes; this intrinsic conductivity obeys $\sigma = e(n\mu_n + p\mu_p)$ with $n_i = p_i$.

Doping with donors produces an $n$-type semiconductor whose majority carriers are electrons, while doping with acceptors produces a $p$-type semiconductor whose majority carriers are holes. At the boundary between $p$- and $n$-regions a depletion layer and a potential barrier form: under forward bias the barrier is lowered and the diffusion current rises sharply, whereas under reverse bias only a small minority-carrier current remains.

A diode contains one $p$–$n$ junction and therefore conducts mainly in one direction; a transistor contains two junctions and a thin base through which a small current $I_B$ controls a large collector current $I_C \approx \beta I_B$. Charge conservation gives $I_E = I_B + I_C$, which underlies amplification and switching.

A silicon sample at room temperature has an electron concentration $n = 1{,}0\cdot10^{22}\ \text{m}^{-3}$ and a negligible hole concentration. The electron mobility is $\mu_n = 0{,}135\ \text{m}^2/(\text{V}\cdot\text{s})$. Find the conductivity $\sigma$.

A diode is connected in series with a resistor $R = 330\ \Omega$ and a source $U_0 = 5{,}0\ \text{V}$. The forward voltage drop across the diode is $U_D = 0{,}7\ \text{V}$. Determine the current in the circuit. How will the current change if the polarity of the source is reversed?

In a common-emitter circuit $U_{CC}=9{,}0\ \text{V}$, $R_C=470\ \Omega$, $R_B=100\ \text{k}\Omega$, $\beta=120$, $U_{BE}=0{,}7\ \text{V}$. Find $I_B$, $I_C$, and $U_{CE}$. Check that the transistor is not saturated.

Simulation (safe, no mains voltage). In an online simulator (for example, Falstad Circuit Simulator or PhET), build two circuits powered by a battery of at most $9\ \text{V}$: (1) a diode, a $330\ \Omega$ resistor, and a source; (2) an $n$–$p$–$n$ bipolar transistor with base and collector resistors. Record the diode’s current–voltage characteristic by reversing polarity and gradually changing the voltage; put the data in a table and plot $I(U)$. For the transistor, vary the base current through $R_B$ and record $I_B$ and $I_C$; plot $I_C(I_B)$ and estimate $\beta$ from the slope. Do not build a real mains-powered circuit—work only in the simulator or from a battery, observing correct polarity.

To show how the band structure determines conductivity and how a $p$–$n$ junction and a transistor control current.

A three-panel diagram: (a) the semiconductor band diagram with $E_v$, $E_c$, and an electron transition that leaves a hole; (b) a $p$–$n$ junction at equilibrium and under forward/reverse bias, showing the depletion region and barrier; (c) the circuit symbol and currents of an $n$–$p$–$n$ transistor with $I_B$, $I_C$, and $I_E$.

A phone charger plugged into a socket at $230\ \text{V}$ and $50\ \text{Hz}$ delivers $5\ \text{V}$ DC at $2\ \text{A}$ to the phone (a power of $10\ \text{W}$). Inside, four silicon diodes in a bridge pass current through the load on both halves of the AC cycle. Each conducting diode loses about $0.7\ \text{V}$, and two conduct in series at any instant, so roughly $1.4\ \text{V}$ is subtracted from the rectified voltage.

This exercises the one-way conduction of a $p$–$n$ junction: with forward bias (plus on the $p$-side) the depletion region narrows and current flows, while under reverse bias the current is nearly zero, so the diode behaves as a valve. The bridge uses both polarities of the input, whose period is $1/50\ \text{s} = 20\ \text{ms}$.

A bicycle headlight runs from a lithium-ion cell at $3.7\ \text{V}$. The LED needs $0.35\ \text{A}$ at $3.2\ \text{V}$, but a microcontroller pin supplies only about $3.5\ \text{mA}$ at $3.3\ \text{V}$. An $n$–$p$–$n$ transistor with current gain $\beta = 100$ bridges the gap: a base current of $0.35\ \text{A} / 100 = 3.5\ \text{mA}$ switches the full LED current on, with a base resistor of about $(3.3 - 0.7)\;\text{V} / 3.5\ \text{mA} \approx 740\ \Omega$.

This is the transistor effect in action: the emitter–base junction is forward biased (a drop of about $0.7\ \text{V}$), the collector–base junction is reverse biased, and a small base current controls a collector current $\beta$ times larger. The three terminals and the two internal $p$–$n$ junctions are the part of the theory this circuit exercises.

A digital kitchen thermometer senses temperature with a small NTC thermistor bead. At $25\ ^\circ\text{C}$ its resistance is $10\ \text{k}\Omega$; dipped into boiling water at $100\ ^\circ\text{C}$ it drops to about $1.2\ \text{k}\Omega$ — nearly an eightfold fall. The circuit turns that change into a reading of $100\ ^\circ\text{C}$ to within $\pm 0.5\ ^\circ\text{C}$.

This shows the temperature dependence of semiconductor conductivity: heating frees more and more electron–hole pairs, the charge-carrier concentration rises, and the resistance falls. In a metal, where the carriers are already free, the picture is the opposite — resistance rises on heating.