 Tunnel-diode load-line stability

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.