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The ideal diode model sets the conducting electric potential difference to zero and the blocking electric current to zero. With the stated orientation,^[ref]

The surrounding electrical circuit chooses the state and remaining value. The two branches meet at $V_D=I_D=0$. Their product vanishes, so the ideal diode dissipates no power: $P_D=V_D{I}_D=0$.

The ideal diode model omits forward drop, reverse leakage, diode reverse breakdown and switching delay.

At equilibrium the diffusion currents and drift currents cancel separately for each carrier species. Electron diffusion is from n to p, while the electric field drives electrons from p to n; hole diffusion is from p to n, while hole drift is from n to p. The net current is zero, although these opposing microscopic fluxes are present. In conventional-current signs, both majority diffusion contributions point p to n and the balancing drift contributions point n to p.

Forward bias, $V_{\mathrm{ext}}>0$, reduces the barrier and enhances majority-carrier diffusion and minority carrier injection. The ideal low-injection Shockley diode equation giveswhere positive current is p to n. For reverse bias, majority diffusion is suppressed and the field sweeps thermally supplied minority carriers across the junction, giving approximately $J=-J_s$. This saturation is limited by carrier supply rather than a current proportional to the field. Depletion-layer generation, recombination and eventual diode reverse breakdown modify the ideal result.

An ideal diffusion-dominated p-n junction obeys the Shockley diode equationwith diode ideality factor $\eta =1$ in the simplest model. Forward bias gives exponential current growth; reverse bias gives a nearly constant $-I_s$ until diode reverse breakdown, outside this ideal expression.

For a long abrupt p-n junction in low carrier injection, the diode saturation current isHere $\mathcal{A}$ is area, $N_D,N_A$ are ionized dopant densities, and $D,L$ are minority-carrier diffusion coefficients and diffusion lengths in the appropriate regions. At each temperature, extract $I_s$ either from the reverse-saturation plateau below breakdown or from a forward exponential fit using its measured diode ideality factor.

Extracting a semiconductor band gap from diode saturation current requires correcting for the transport prefactor:A plot against $1/T$ has slope $-E_g/k_B$ when the gap and effective masses are nearly temperature-independent. Alternatively divide the measured $I_s$ by a known $C(T)$ to recover the full $n_i^2(T)$. The prefactor cannot generally be assumed constant over an arbitrarily broad range. Junction-generation or recombination-dominated current can scale as $n_i$ instead, so the diffusion regime should be identified from the current-voltage behaviour.

For a p-n junction at fixed temperature, the Shockley diode equation models its electric current asHere $I_S>0$ is the diode saturation current, $n>0$ the diode ideality factor, $V_T$ the thermal voltage, $k_B$ the Boltzmann constant, and $q>0$ the elementary charge.^[ref]

For an original model example, choose $I_S=1\mathrm{nA}$, $n=2$, and $T=300\mathrm{K}$. Using $k_B=1.380649\times 10^{-23}J/K$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ gives $V_T=25.8520\mathrm{mV}$. Then $V_D=0.600\mathrm{V}$ gives $I_D=0.109591\mathrm{mA}$, whereas $V_D=0.800\mathrm{V}$ gives $I_D=5.24450\mathrm{mA}$. These are predictions of the declared model. There is no universal turn-on electric potential difference: solving the Shockley diode equation for the value needed at a chosen electric current givesAt $I_D\gg I_S$, increasing $V_D$ by $0.100\mathrm{V}$ multiplies the electric current by approximately $e^{0.100/(n{V}_T)}=6.91772$. At $V_D=-0.300\mathrm{V}$ the same model gives $I_D=-0.996979\mathrm{nA}$. Its reverse branch approaches $-I_S$; it cannot describe diode reverse breakdown.

A constructed extension adds internal series electrical resistance $R_s$ to a junction whose electric potential difference is $V_j$. Applying the Kirchhoff cycle law givesFor $R_s=20\Omega$ and the same $I_S,n,T$, sustaining $I_D=5.24450\mathrm{mA}$ needs $V_D=0.904890\mathrm{V}$: $0.800\mathrm{V}$ across the junction plus $0.104890\mathrm{V}$ across its internal electrical resistance. The derivativeshows why the current-voltage characteristic becomes nearly a straight line when the internal series electrical resistance dominates. This is still a fixed-temperature model; self-heating can change $I_S$, $n$ and $V_T$. Real diodes also have leakage and charge-storage effects omitted here.^[ref]

A tunnel diode is a p-n junction with sufficiently heavy semiconductor doping on both sides to make a thin depletion region and permit substantial interband tunnelling. The neutral regions are degenerate semiconductors. Its distinctive forward current-voltage characteristic contains a peak followed by a valley: over part of the curve, increasing the applied electric potential difference decreases the electric current. The Esaki tunnel-diode experiment made this behavior experimentally concrete.

The useful quantity is the local slope, not the ratio of total voltage to total electric current. With conventional electric current $I$ entering the positive-voltage terminal, defineNegative differential resistance means $g_d<0$, hence $r_d<0$ where defined. A forward operating point can still have $V>0$, $I>0$ and absorbed power $VI>0$. A small voltage increase then gives a negative current increment; this does not make the total electric current negative or turn the device into an unlimited source of energy. The external supply sets the operating point and supplies the energy available for an electronic amplifier.

An original idealized construction explains the shape. Put the n-type semiconductor on the left and the p-type semiconductor on the right. Take $q>0$ as the elementary charge and $V=V_p-V_n$. Increasing forward bias lowers the right-hand Electron energy levels relative to the left. Denote the two reservoir electrochemical potentials by ${\mu }_n,{\mu }_p$, so that ${\mu }_n-{\mu }_p=qV$. At zero temperature, occupied left-hand conduction band states extend up to ${\mu }_n$; empty right-hand valence band states start at ${\mu }_p$. For transfer conserving Electron energy, the available interval isIt is nonempty only when its upper endpoint exceeds its lower endpoint. Both an occupied initial quantum state and an empty final quantum state must exist at the same energy. The Pauli exclusion principle prevents transfer into an already occupied final quantum state.

For the original diagram, choose arbitrary energy units with $E_{c,n}=-0.5$, ${\mu }_n=0$, ${\mu }_p=-u$ and $E_{v,p}=0.5-u$, where $u$ is $qV$ in those units. The available interval has widthThus $L(0.25)=0.25$, $L(0.50)=0.50$, $L(0.70)=0.30$, and $L(1.15)=0$. Initially forward bias opens a transfer window; further forward bias closes it by moving the right-hand valence band edge below the left-hand conduction band edge. This is a state-counting illustration, not a formula for a measured electric current.

For reverse bias, the transfer direction changes: occupied right-hand valence band states can feed empty left-hand conduction band states. In the same construction, $u=-0.70$ opens the interval $0<E<0.70$ for this reverse transfer. A narrow depletion region allows a large reverse quantum tunnelling electric current at small electric potential difference, whereas the ordinary Shockley diode equation predicts only a small reverse saturation electric current before diode reverse breakdown. At zero biasing, equal reservoir occupations give zero net electric current; quantum tunnelling does not require a net equilibrium flow.

A finite-temperature refinement replaces sharp occupation cutoffs with the Fermi-Dirac distribution. An illustrative weak-transfer expression isHere $\mathcal{W}\ge 0$ describes allowed initial/final quantum states and their coupling, and $f_n,f_p$ are Fermi-Dirac distributions with the respective electrochemical potentials. The occupation difference follows by subtracting the reverse transfer factor from the forward one: $f_n(1-f_p)-f_p(1-f_n)=f_n-f_p$. An overlap interval alone does not determine $\mathcal{W}$: transmission probability, density of states, crystal momentum selection and phonons can matter. A complete germanium calculation is more demanding than multiplying two constant density of states factors. At larger forward bias, ordinary carrier injection and diffusion supply a rising current branch even after direct band overlap is lost.