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

A transition whose frequency is stationary to first order in a specified perturbation at an operating point. For magnetic sensitivity, the derivative shown vanishes, but second and higher derivatives can remain nonzero. A clock transition need not have zero individual energy slopes: equal slopes suffice. Optical light shifts and other perturbations require their own sensitivity analysis.

An electronic amplifier uses an electrical power supply to increase a signal's amplitude or available output power. The input controls the transfer of supply energy to the load; it does not supply all the amplified output energy itself.

A voltage amplifier, a transimpedance amplifier and a charge amplifier relate different input and output quantities. To characterize one, vary the input around its operating point and measure gain, frequency response, noise and the range before saturation. A large gain does not by itself improve signal-to-noise ratio, since input noise and amplifier noise can both be amplified.

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.

Electronic gain expresses how an electronic circuit scales a specified input signal quantity into an output quantity. For a linearized voltage amplifier around its operating point, the voltage gain is $A_v=\delta v_{\mathrm{out}}/\delta v_{\mathrm{in}}$. At a sinusoidal frequency it can be a complex phasor ratio, specifying both amplitude ratio and phase shift. Power gain instead compares the average power delivered to the load with the average signal power entering the amplifier:State the input/output quantities, frequency, load and operating conditions when specifying gain. A transimpedance amplifier relates output electric potential difference to input current and has a dimensional transfer ratio in ohms.

For an original circuit example, use an inverting operational amplifier with input resistor $R_1=10\mathrm{k\Omega }$, feedback resistor $R_2=100\mathrm{k\Omega }$ and load $R_L=1\mathrm{k\Omega }$. In its ideal negative-feedback linear regime,An input of $50\mathrm{mV}$ root mean square therefore gives $0.50\mathrm{V}$ root mean square at the output, with a phase inversion. The manufacturer's primary explanation derives this resistance-ratio gain using the input and feedback currents.^[ref] Suitable supplies, output-current capability and amplifier output range are assumptions of this worked model, not specifications of a particular device.

Here the input resistance is $R_1$ in the ideal model. For resistive ports and root-mean-square signal amplitudes,The example input power is $0.25\mu \mathrm{W}$ and load power $0.25\mathrm{mW}$, giving $G_P=1000$. The additional output energy comes from the amplifier's supply, controlled by the input signal. The impedance condition for converting voltage gain to power gain is essential: voltage magnitude alone does not specify load power.

Voltage gain is conventionally expressed as $20{\log }_{10}|A_v|$ decibels, while power gain uses $10{\log }_{10}{G}_P$. This example has voltage gain $20\mathrm{dB}$ and power gain $30\mathrm{dB}$; the two expressions agree when the resistive input and load impedances are equal. A negative voltage gain indicates phase inversion; its magnitude enters the logarithm.

A direct test drives a small sine wave into the input and measures the actual terminal signals with an oscilloscope. Sweeping frequency reveals the frequency response; increasing amplitude reveals amplifier clipping or other departure from the assumed linear gain. Keep the load and measurement impedances specified. Gain can amplify input noise along with signal, so the signal-to-noise ratio requires its own measurement.

A qubit encoded in two resolved hyperfine levels of the same electronic ground state. Such states avoid optical excited-state decay and can support coherent radiofrequency or microwave control through a magnetic dipole transition. Technical magnetic noise and differential shifts can still dephase them; first-order field insensitivity requires additional choices of states and operating point.

For a slender winding around a toroid with $N$ turns, core area $A$, mean magnetic path length $\ell$ and approximately constant magnetic permeability $\mu$, Ampère's law gives $H\ell =NI$. The linked magnetic flux is $N\Phi =N\mu HA$, so $L=N\Phi /I=\mu N^{2}A/\ell$. This approximation neglects flux leakage and variation across the core and assumes operation below magnetic saturation. With a nonlinear magnetic core, an incremental inductance instead depends on differential permeability at the operating point.

Input capacitance inferred from measured admittance is a parameter estimate from the quadrature component of a specified electrical port's response. In the parallel model of input capacitance, write its electrical admittance as $Y=G+i\omega C$ with the $e^{i\omega t}$ convention. A small sinusoidal excitation and simultaneous input-current measurement giveBoth amplitude ratio and relative wave phase matter. The differential versus common-mode terminals and operating point conditions must be specified, as in the manufacturer's input-path diagrams.^[ref]

For an original error calculation, suppose the amplitude ratio is exact but the measured current-to-voltage wave phase has offset $\delta$. Then $Y_{\mathrm{meas}}=Y{e}^{i\delta }$ andThe phase calibration error in an admittance-based capacitance estimate can therefore mix a conductance term into the small capacitive estimate. For small $\delta$, the leading error is $(G/\omega )\delta$, not a fixed percentage of $C$.

Choose $G=10\mu \mathrm{S}$, $C=20\mathrm{pF}$, and $f=10\mathrm{kHz}$. The true susceptance is only $\omega C\approx 1.257\mu \mathrm{S}$. A one-degree wave phase offset yields $\hat{C}\approx 22.775\mathrm{pF}$, an error of about $13.9\%$, despite perfect amplitude readings. These are chosen port-model values, not measured device specifications.

A useful measurement repeats the estimate over frequency and excitation amplitude. A nearly constant fitted $C$ and a residual consistent with $G+i\omega C$ support the assumed model; they do not prove that all parasitic effects were removed. A known fixture admittance adds to a parallel port and must be subtracted or retained in the model. At very small susceptance, uncertainty in the relative wave phase can dominate, so a high DC input impedance alone is not evidence for a precise capacitance measurement.

Input impedance is the ratio of a small applied input-voltage phasor to the resulting input-current phasor, with the output termination and operating point specified. A finite input impedance loads a voltage source. For an ideal operational amplifier the input impedance is infinite, so neither input terminal draws current.

The input offset voltage of an operational amplifier is the differential direct-current input voltage needed to make its output zero at a specified operating point. Its effect resembles a small unwanted input signal and can cause sustained output drift in an op amp integrator.

Loaded voltage gain is the ratio of an amplifier's output to input signal electric potential differences with a specified load connected. In a small-signal model, let the open-circuit output be $v_{\mathrm{oc}}=A_0{v}_{\mathrm{in}}$ and let the complete amplifier have output impedance $Z_o$. A load $Z_L$ then givesThis is the voltage divider result from Thévenin's theorem at the output terminals. The university source explains why a linear circuit can be replaced at specified terminals by a source and series impedance.^[ref] The output impedance used here must describe the complete amplifier at the stated operating point and frequency, including its feedback.

The input terminal matters too. For a generator $v_s$ with source impedance $Z_s$ driving amplifier input impedance $Z_i$, the corresponding source-to-load ratio in a unilateral model isThis includes both input and output loading; it differs from the terminal-to-terminal voltage gain defined above.

For an original resistive example, choose $A_0=12$, $Z_o=100\Omega$, $Z_L=900\Omega$, $Z_i=1\mathrm{k\Omega }$ and $Z_s=50\Omega$. The loaded terminal gain is $10.8$, while the source-to-load gain is $72/7\approx 10.286$. Reducing the load to $100\Omega$ makes the terminal gain six, even though the stipulated open-circuit gain remains twelve. These are model parameters, not device specifications.

A direct test holds the actual amplifier input signal fixed, changes a known load and measures the output with an oscilloscope whose probe impedance is included in the effective load. Use small enough signals to retain linear operation and respect the amplifier output range. If the output-to-input ratio changes with loading, an unloaded gain alone is insufficient to predict the circuit's behavior.

At nonzero frequency the impedances and gain can be complex. A reactive-load phase shift in a finite-output-impedance amplifier can change both magnitude and phase, while input-output coupling or nonlinear current limiting can invalidate this simple unilateral divider model. The ideal feedback resistance-ratio examples in the manufacturer's introduction apply within their corresponding operating assumptions.^[ref]

Spin-valve read-head noise limits the field-to-electrical response: sensitivity is a response relative to noise, not merely a large total resistance. A small free-layer volume gives larger thermally driven angular fluctuations and can lose magnetic stability when its anisotropy barrier is insufficient compared with $k_{B}T$. Edge damage, stray coupling to the reference layer, nonuniform domains and Barkhausen jumps cause hysteresis and noise. Johnson noise and low-frequency resistance noise compete with a limited GMR signal; raising sense current is restricted by heating and current-induced disturbance.

One useful design combines a single-domain, gently biased free layer with a flux-compensated synthetic-antiferromagnetic reference. This reduces reference stray fields, suppresses unstable domain switching and holds the free layer near a linear operating point. The active sensing width can remain small while magnetic and electrical design preserve adequate stability and signal. An alternative is a magnetic tunnel junction with an optimized barrier, giving a larger magnetoresistance signal; its resistance-area product, shot noise and barrier reliability still require optimization. Shrinking every magnetic dimension indefinitely is not a sensitivity improvement.

Mean kinetic energy $3k_{B}T/2=12.9$ keV means $T=8.6$ keV, not 12.9 keV. For fully ionized D-T with equal electron and ion temperatures, $p=2n_i{k}_{B}T$ gives $n_i\simeq 4.90\times 10^{19}$ m${}^{-3}$. Thusonly about one third of the stated requirement. The specified operating point therefore does not meet that criterion and requires improved confinement/pressure or continuing external heating. If the criterion's $n$ counts all thermal particles instead of ions, the product doubles but remains below its given threshold. A precise fusion gain or duty cycle cannot be inferred from the triple product alone.

Phase calibration error in an admittance-based capacitance estimate can rotate a large in-phase conductance into a small capacitive quadrature. In the parallel port model $Y=G+i\omega C$ with $G\ge 0$, $C>0$, and the $e^{i\omega t}$ convention, a relative wave phase error $\delta$ gives $Y_{\mathrm{meas}}=Y{e}^{i\delta }$. The exact estimate in input capacitance inferred from measured admittance therefore has relative errorThe ratio $G/(\omega C)$, not the DC resistance alone, determines sensitivity to a small wave phase rotation. The manufacturer's separate differential/common-mode port models motivate specifying the terminals and operating point before fitting an input capacitance.^[ref]

If $|\delta |\le ϵ\le \pi /2$, the worst absolute relative error is bounded byThe negative endpoint $\delta =-ϵ$ attains this bound. For target relative error $r>1-\cos ϵ$, a sufficient frequency condition isThis condition assumes the same $G+i\omega C$ model and the same phase-error bound remain valid over that frequency range.

For an original planning example, take $G=1\mu \mathrm{S}$, $C=50\mathrm{pF}$ and $ϵ=0.5^{\circ }$. The bound is about $2.782\%$ at $1\mathrm{kHz}$ and $0.282\%$ at $10\mathrm{kHz}$. A one-percent target requires approximately $f\ge 2.788\mathrm{kHz}$ in this model. Increasing frequency suppresses the conductance-mixing term, but leaves the $1-\cos ϵ$ term and can expose other unmodelled parasitic effects.

A measurement uses the same voltage/current wave phase reference for the ratio $I/V$, checks that reference with a known port, and retains fixture admittance in the model or subtracts a validated value. Amplitude measurement calibration alone cannot establish this wave phase bound. The numbers above are chosen calculations, not device specifications or a claim that a high-frequency measurement is automatically more accurate.

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.

A solar cell converts incident light into electrical power through the photovoltaic effect. A semiconductor p-n junction separates and collects photogenerated electrons and electron holes. An ideal illuminated junction has output current densityHere $J_{\mathrm{sc}}$ is the short-circuit current density, $J_0$ is the diode saturation current, and $V_{\mathrm{oc}}$ is the open-circuit electric potential difference. The conversion efficiency is the largest delivered $VJ$ divided by the incident power per unit area; the short-circuit and open-circuit operating points themselves deliver zero power.

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.