= Phenomenological paired-pulse recovery
{title2=$I_{\mathrm{req}}(\Delta)$}
Choose the following original phenomenological <mathematical model> solely to calculate a paired-pulse readout. It is not experimental data, a fit to the teaching experiment, or a simulation of the <Hodgkin-Huxley model>. The conditioning <action potential> defines time zero. Test pulses have fixed width $0.2\,\mathrm{ms}$; $I$ denotes their inward <electric current> amplitude. Impose an <absolute neuronal refractory period> $t_{\mathrm{abs}}=2\,\mathrm{ms}$. No finite test amplitude generates an event for $0<\Delta<t_{\mathrm{abs}}$. After that hold, stipulate a required amplitude
$$
I_{\mathrm{req}}(\Delta)=I_\infty\left(1+A e^{-(\Delta-t_{\mathrm{abs}})/\tau_r}\right),\qquad \Delta\geq t_{\mathrm{abs}},
$$
with $I_\infty=0.2\,\mathrm{nA}$, dimensionless $A=2$ and recovery <time constant> $\tau_r=3\,\mathrm{ms}$. Emit a test event exactly when $I\geq I_{\mathrm{req}}(\Delta)$, including equality. For this ideal event readout, assign events to pulse onset and assume the conditioning event is successfully evoked. There is no calculated <membrane potential>, spike waveform, response latency or stochastic success <probability>.
\Image[/neuronal-refractory-recovery-curves.png]
{title=Chosen refractory hold and required test-pulse amplitude}
{description=Original <phenomenological paired-pulse recovery> rule, not measured data or a channel simulation. The grey region forbids a second event; after $2\,\mathrm{ms}$, the required <electric current> falls from $0.6$ toward $0.2\,\mathrm{nA}$. Horizontal lines show two chosen test amplitudes; their intersections give different first permitted test delays. The dashed baseline is approached asymptotically.}
The <exponential decay> represents a combined recovery deficit; it is neither the <Hodgkin-Huxley model>'s $h$ nor a measured <potassium> <electrical conductance>. One such curve cannot separately identify <sodium> recovery and persistent <potassium> <electrical conductance>. A constant pulse-width amplitude criterion also does not establish an equivalent rule for other pulse widths or voltage trajectories.
At the hold boundary, $I_{\mathrm{req}}(2\,\mathrm{ms})=0.2(1+2)=0.6\,\mathrm{nA}$. At $\Delta=4\,\mathrm{ms}$,
$$
I_{\mathrm{req}}=0.2(1+2e^{-2/3})\,\mathrm{nA}\approx0.405367\,\mathrm{nA};
$$
at $\Delta=6\,\mathrm{ms}$,
$$
I_{\mathrm{req}}=0.2(1+2e^{-4/3})\,\mathrm{nA}\approx0.305439\,\mathrm{nA}.
$$
Thus a $0.35\,\mathrm{nA}$ test fails at $1\,\mathrm{ms}$ because of the imposed hold, fails at $4\,\mathrm{ms}$ because recovery is insufficient for that amplitude, and succeeds at $6\,\mathrm{ms}$. At $4\,\mathrm{ms}$, a $0.5\,\mathrm{nA}$ pulse would instead succeed: failure of the weaker pulse was relative refractoriness.
\Image[/neuronal-paired-pulse-event-readout.png]
{title=Paired pulses and event readout at three chosen delays}
{description=Each row is a separately reset ideal trial with a $0.65\,\mathrm{nA}$ conditioning pulse assumed to evoke an <action potential> at time zero and a $0.35\,\mathrm{nA}$ test pulse. Rectangles are imposed <electric current> pulses; vertical ticks are abstract events, not voltage recordings. Only the $6\,\mathrm{ms}$ test passes the stipulated recovery rule. The $1\,\mathrm{ms}$ and $4\,\mathrm{ms}$ failures have different causes.}
For $I>I_\infty$, solve $I\geq I_{\mathrm{req}}(\Delta)$ using the <logarithm>. The earliest allowed delay is
$$
\Delta_{\min}(I)=t_{\mathrm{abs}}+\max\left\{0,\tau_r\ln\left(\frac{A}{I/I_\infty-1}\right)\right\}.
$$
For $I=0.35\,\mathrm{nA}$ this is $2+3\ln(8/3)\approx4.942488\,\mathrm{ms}$; for $I=0.5\,\mathrm{nA}$ it is $2+3\ln(4/3)\approx2.863046\,\mathrm{ms}$. If $I\geq0.6\,\mathrm{nA}$, the imposed $2\,\mathrm{ms}$ hold is the only delay restriction. If $I\leq I_\infty$, no finite delay passes the rule: the required amplitude stays strictly above its asymptotic baseline. A finite experimental “fully recovered” time therefore also needs a tolerance and a readout definition.
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