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Setting steady electric potential differences and electric currents in an electrical circuit to choose its operating point.

Leo Esaki's 1958 report describes alloy-fabricated germanium p-n junctions, with acceptor impurity concentration $1.6\times 10^{19}{\mathrm{cm}}^{-3}$ and donor impurity concentration approximately $10^{19}{\mathrm{cm}}^{-3}$. Its readout was the current-voltage characteristic: a forward peak occurred at $0.035\pm 0.005\mathrm{V}$. The report says specimens could be reproduced; above $0.3\mathrm{V}$ the forward branch approximately followed the ordinary Shockley diode equation.^[ref]

In his 1973 lecture, Leo Esaki recalls an earlier junction whose strong room-temperature reverse conduction preceded a negative-slope forward segment revealed by cooling. Its estimated zero-bias width was about $200$ ångströms ($20\mathrm{nm}$), supported by capacitance measurements. Increasing semiconductor doping further made the negative slope visible across the measured temperatures, including room temperature. Thus preparation and temperature variation were experimental controls, not merely changes to a theoretical diagram. The low onset voltage argued against avalanche production of Electrons and electron holes; the energy band explanation connected the forward decline to loss of available final states.^[ref]

The historical readouts were junction electric current versus applied electric potential difference and a separate capacitance check; preparation and temperature were varied. The cited descriptions do not identify a meter model or give a complete wiring diagram.

For a modern repetition, measure junction voltage directly while stepping the supply and recording electric current, repeat both sweep directions, and compare temperatures and differently doped specimens. These are proposed controls, not a reconstruction of the historical apparatus. Lead electrical resistance, Joule heating, contact quality and voltage resolution can obscure the result. The tunnel-diode load-line stability calculation also predicts that an external resistor can make a negative-slope operating point inaccessible in a slow sweep. A jump in terminal readings then need not mean the underlying current-voltage characteristic is discontinuous.

A first-order Taylor expansion of an electrical circuit's response about an operating point. It describes small changes in electric potential difference and electric current; the neglected higher-order terms limit the useful signal range.

The change of output electric current per change of input electric potential difference, evaluated at an operating point with the other terminal conditions specified. In a small-signal model, $i_{\mathrm{out}}=g_m{v}_{\mathrm{in}}$ describes this contribution; transconductance has the dimensions of electrical conductance.

An original teaching model exposes the roles of input loading, output loading and the supply. Choose a small-signal model with positive input electrical resistance $r_i$, transconductance $g_m>0$, and output electrical resistance $r_o>0$. Its controlled output current source draws $g_m{v}_i$ from the output node toward the reference node. This chosen sign gives an inverting voltage gain; it is not the historical contact circuit's sign convention.

Assume a fixed operating point, negligible reverse coupling, a frequency range where coupling capacitors pass the signal, and negligible other capacitances and transit delays. A collector resistor $R_D$ connects the output to a steady ideal supply $V_{DD}$; an external electrical load $R_L$ is coupled through a capacitor, so it draws no direct current. The input source has series electrical resistance $R_s$. All voltages below are small departures from the operating point. An ideal fixed supply has zero incremental voltage and therefore becomes the reference node in the small-signal model.

Define $R_O=R_D\parallel r_o$ and $R_P=R_O\parallel R_L$, where $a\parallel b=ab/(a+b)$. Input voltage division and output Kirchhoff's current law giveConsequently the port voltage gain, overall source-to-output voltage gain, and load current gain magnitudes areThis current gain uses load electric current, not all the controlled-source electric current. It is also a different port definition from the historical emitter-to-collector current gain. Larger $R_s$ attenuates the input; smaller $R_L$ attenuates the output.

For a sinusoid input signal, use root mean square voltages. The power gain from accepted input signal power $P_i=v_i^2/r_i$ to useful load power $P_L=v_o^2/R_L$ isUnlike $|A_i|$, this tends to zero as either $R_L\to 0$ or $R_L\to \infty$. Differentiation givesThus matching the output electrical load maximizes delivered power in this unilateral resistive model. For fixed source voltage, input loading multiplies $P_L$ by $[r_i/(R_s+r_i){]}^2$; the accepted-input power gain above excludes the loss in $R_s$.

Choose $r_i=R_s=1\mathrm{k\Omega }$, $g_m=2\mathrm{mS}$, $r_o=20\mathrm{k\Omega }$, and $R_D=2\mathrm{k\Omega }$. Then $R_O=20/11\mathrm{k\Omega }$. With $R_L=2\mathrm{k\Omega }$ and $v_s=20\mathrm{mV}$ root mean square, the input is $10\mathrm{mV}$ and the inverted output has magnitude $19.0476\mathrm{mV}$ root mean square. Hence $P_i=0.100\mu \mathrm{W}$ and $P_L=0.181406\mu \mathrm{W}$: $G_P=1.81406$. These are calculated values of the chosen model, not historical measurements.

With the same parameters but $R_L=100\Omega$, $|A_i|=1.89573$ while $G_P=0.359381<1$. At $R_L=10\mathrm{k\Omega }$, $|A_v|=3.07692$ while $G_P=0.946746<1$. Neither current gain nor voltage gain alone establishes power gain.

For the direct current load line in the figure, choose $V_{DD}=6\mathrm{V}$, $I_Q=1\mathrm{mA}$ and $V_Q=4\mathrm{V}$, consistent with $V_Q=V_{DD}-R_D{I}_Q$. Locally extend the model as $I=I_Q+g_m{v}_i+(V-V_Q)/r_o$. With the external electrical load isolated at direct current, its intersection with $I=(V_{DD}-V)/R_D$ is $V=V_Q-g_m{R}_O{v}_i$. Keeping $|v_o|$ small compared with the available voltage headroom is essential; extrapolation to arbitrarily large signals would be unphysical.

At the chosen operating point the collector supply delivers $V_{DD}{I}_Q=6\mathrm{mW}$: $2\mathrm{mW}$ is dissipated in $R_D$ and $4\mathrm{mW}$ in the output device. Under a small signal, that energy flow is redistributed to provide load power. The incremental controlled current source hides the supply's steady energy flow; it does not create energy. A real complete amplifier also needs its input biasing network and has further losses. This calculation establishes neither its total efficiency nor a universal law for point-contact transistors.

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.

A current-voltage characteristic determines possible steady operating points; it does not alone determine their linear stability. Consider an ideal voltage source $V_s$ feeding a tunnel diode through a resistor $R>0$, with a positive constant capacitance $C$ across the junction. Assume a quasistatic device relation $I(V)$ and neglect inductance, transit delays, noise and heating. Kirchhoff's current law gives the original one-state modelA steady operating point $V_{\ast }$ satisfies $I(V_{\ast })=(V_s-V_{\ast })/R$: the device curve intersects the descending supply line. This is the source-side counterpart of a resistive load line. A small perturbation $V=V_{\ast }+\delta V$ obeysThus the equilibrium point is locally attracting if $1/R+I^{\prime }(V_{\ast })>0$ and unstable if the inequality reverses. Equality requires higher-order analysis. Even on a negative differential resistance segment, a sufficiently small supply electrical resistance can stabilize this model. The assumption about the electrical circuit matters: an inductive or delayed circuit has additional state variables and may behave differently.

Choose an original toy forward characteristic, with no fitted device parameters:where $V_0=0.10\mathrm{V}$, $I_0=1.0\mathrm{mA}$ and $0\le x\le 3.6$. For every $x>0$, $i(x)>0$, so this entire forward curve absorbs positive total power. NeverthelessThe model therefore separates negative differential resistance from negative total $V/I$. The corresponding peak and valley are at approximately $0.1184\mathrm{V}$ and $0.2816\mathrm{V}$; these voltages deliberately differ from the historical peak.

With $R=100\Omega$ and $V_s=0.60\mathrm{V}$, the supply line is $I/I_0=6-x$. The intersection equation factors exactly:The three operating points are $(0.10\mathrm{V},5\mathrm{mA})$, $(0.20\mathrm{V},4\mathrm{mA})$ and $(0.30\mathrm{V},3\mathrm{mA})$. Their values of $1/R+I^{\prime }(V_{\ast })$ are respectively $20$, $-10$ and $20\mathrm{mS}$. The outer two attract nearby trajectories; the middle one repels them. This is bistability in the specified one-state electrical circuit.

For the second panel, choose $R=25\Omega$ and $V_s=0.30\mathrm{V}$. Now $I/I_0=12-4x$ andThere is exactly one real operating point, again $(0.20\mathrm{V},4\mathrm{mA})$. Here $1/R+I^{\prime }(V_{\ast })=20\mathrm{mS}>0$, despite the device's own slope $I^{\prime }(V_{\ast })=-20\mathrm{mS}$. With $C=10\mathrm{nF}$, the perturbation decays with time constant $C/(20\mathrm{mS})=0.50\mu \mathrm{s}$; the unstable middle point of the first panel has growth time $1.0\mu \mathrm{s}$.

The one-state autonomous model supports switching between attracting operating points after a perturbation or parameter change; it cannot produce a nonconstant periodic orbit. A proposed oscillator needs additional dynamics. This limitation is useful: a load-line intersection establishes a possible steady reading, the slope test establishes local attraction in a specified model, and neither calculation alone predicts an oscillation frequency for a real tunnel diode.