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Signaling feedback returns the effect of a downstream event to an earlier stage of signal transduction, closing a loop analogous to feedback control. Trace the causal arrows: if an earlier relay activates a later relay and the later relay changes the earlier relay's activity, the return path closes the loop. Negative feedback opposes the initiating change; positive feedback reinforces it. Signaling feedback is a possible property of signal transduction, not a requirement of every pathway.

For a schematic example, let relay $A$ activate relay $B$. If $B$ inhibits $A$, increased $A$ eventually produces an opposing influence: negative feedback. If $B$ activates $A$, increased $A$ produces a reinforcing influence: positive feedback. These signs describe effects on activity, rather than the direction in which molecules travel. A closed loop containing two inhibitions can have positive feedback, because inhibiting an inhibitor reinforces the original change. The sign belongs to the whole loop.

Negative feedback can restrain excessive activation and help preserve a usable range of responses. Positive feedback can support a decisive transition: Alberts and colleagues describe an enzyme whose product increases its own activity. Such reinforcement can contribute to switchlike behavior, but a positive feedback arrow alone does not prove bistability. A different structure, an incoherent feedforward loop, sends opposing influences forward from the same input; inhibition there is not a downstream return. ^{[ref] [ref]}

A worked mathematical model shows why negative feedback does not automatically produce complete signaling adaptation. Let $s$ be a fixed normalized input, $r$ a normalized response, $g\ge 0$ a dimensionless feedback gain, and $\tau >0$ a time constant. Suppose response removal contributes $-r$ and a return proportional to the response contributes $-gr$. The ordinary differential equation isSetting the derivative to zero gives the equilibrium point $r_{\ast }=s/(1+g)$. A displacement $\delta r=r-r_{\ast }$ obeys $\tau d(\delta r)/dt=-(1+g)\delta r$, so its decaying exponential function establishes asymptotic stability. For a unit input switched on at $t=0$, initial condition $r(0)=0$, $g=2$ and $\tau =1$ second,The negative feedback reduces the eventual response from $1$ to $1/3$, but this response rises monotonically and never returns to its zero baseline. Attenuation and complete signaling adaptation are different properties. This deliberately simple mathematical model specifies no molecular pathway; integral action provides a contrasting way to remove a lasting offset in a stable feedback control system. ^[ref]

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.