Refractory period · Lobeworks/17
The refractory period is the stretch after each action potential during which a neuron cannot fire again (the absolute period) or needs a stronger push to do so (the relative period), and it is what caps how fast a neuron can fire and keeps a spike travelling one way.
Refractory period. The refractory period is the stretch after each action potential during which a neuron cannot fire again (the absolute period) or needs a stronger push to do so (the relative period), and it is what caps how fast a neuron can fire and keeps a spike travelling one way.
Its cause is the inactivation gate of the voltage-gated sodium channel. During the spike that gate, written hhh in the Hodgkin-Huxley equations, closes most of the channels, and no voltage can open a channel until its gate has reopened. Reopening is a relaxation, never a switch: once the membrane is back near rest, the fraction of available channels recovers as
h(t)=h∞−(h∞−h0) e−t/τhh(t) = h_\infty - (h_\infty - h_0)\, e^{-t/\tau_h}h(t)=h∞−(h∞−h0)e−t/τh
so availability climbs back exponentially with a time constant τh\tau_hτh of a few milliseconds. The absolute period lasts until enough channels are back for the sodium current to beat the potassium current at any input. The relative period is the tail of that curve, made longer by the potassium channels the spike opened (their gate nnn is still high), so the threshold starts high and falls back as both recover. A ceiling follows: an absolute period tabst_{\text{abs}}tabs allows at most fmax≈1/tabsf_{\max} \approx 1/t_{\text{abs}}fmax≈1/tabs, about 1000 spikes per second for 1 ms, and real neurons stay well under it, the fastest at a few hundred.
Temperature changes the clock. The gates are proteins changing shape, and their rates grow by a factor Q10Q_{10}Q10 of about 3 for every 10 °C, so τ(T)=τ(T0) Q10−(T−T0)/10\tau(T) = \tau(T_0), Q_{10}^{-(T-T_0)/10}τ(T)=τ(T0)Q10−(T−T0)/10. Hodgkin and Huxley measured at 6.3 °C; at body temperature the same gates run almost thirty times faster (33.073^{3.07}33.07), which is why a mammal's spikes and refractory periods are a fraction of the squid's.
Potassium outside moves the starting line. The availability at rest is h∞(V)=1/(1+e(V−V1/2)/k)h_\infty(V) = 1/\big(1 + e^{(V - V_{1/2})/k}\big)h∞(V)=1/(1+e(V−V1/2)/k), a sigmoid that falls as the membrane depolarises. More potassium outside raises the potassium equilibrium potential EK=RTFln[K+]o[K+]iE_K = \frac{RT}{F}\ln\frac{[K^+]_o}{[K^+]_i}EK=FRTln[K+]i[K+]o towards zero and drags the resting membrane potential up with it, so fewer channels are available and they recover more slowly: a modest rise makes the neuron easier to fire, a large one leaves so many channels inactivated that it cannot fire at all, which is depolarisation block.
Fatigue is a slower refractoriness. Firing for long adds processes with time constants from hundreds of milliseconds to seconds: calcium entering with each spike opens calcium-activated potassium channels whose afterhyperpolarisation lengthens the next interval, some sodium channels fall into a slow inactivated state that takes seconds to leave, and the sodium-potassium pump, working harder, adds an outward current. Under a constant push the firing rate therefore falls over time, which is spike-frequency adaptation.
It gives the spike a direction. The patch of axon just behind a travelling spike is refractory, so the spike can only excite the patch ahead of it.
The refractory period is the recovery curve of one gate, read as a clock.
Time, temperature, potassium and fatigue each change either where the curve starts, where it ends or how fast it runs, and the neuron's top speed follows.
Questions: How fast do inactivated sodium channels come back, and what shape does their recovery take? Exponentially. Once the membrane is back near rest, the fraction of channels whose inactivation gate has reopened follows h(t)=h∞−(h∞−h0) e−t/τhh(t) = h_\infty - (h_\infty - h_0),e^{-t/\tau_h}h(t)=h∞−(h∞−h0)e−t/τh, so it climbs fast at first and then ever more slowly towards the resting availability h∞h_\inftyh∞. The time constant τh\tau_hτh is a few milliseconds and depends on voltage and temperature: a more depolarised membrane or a colder one recovers more slowly. The threshold for a second spike falls along the same curve, which is why the relative refractory period has no sharp end. How can you check that a sorted unit is really one neuron? Look for impossible intervals. One neuron cannot fire twice within its refractory period, about a millisecond or two, so a unit with many spikes closer than that is contaminated by another cell. Units are graded as single units or multi-unit activity, and drift of the brain against the shank, which moves every footprint, is corrected before sorting. What sets the highest rate at which a neuron can fire? The absolute refractory period. No second spike can start until enough sodium channels have recovered from inactivation, so a period tabst_{\text{abs}}tabs allows at most fmax≈1/tabsf_{\max} \approx 1/t_{\text{abs}}fmax≈1/tabs spikes per second: about 1000 for 1 ms. Real neurons stay well under that ceiling, the fastest at a few hundred per second, because the relative period and the potassium currents that follow each spike stretch every interval further. What is the time between two spikes called, and how do you turn it into a frequency? The inter-spike interval, ISIn=tn+1−tn\mathrm{ISI}_n = t_{n+1} - t_nISIn=tn+1−tn, and its inverse is the instantaneous frequency, fn=1/ISInf_n = 1/\mathrm{ISI}_nfn=1/ISIn. Spikes 5 ms apart mean 1/0.005=2001/0.005 = 2001/0.005=200 Hz; 20 ms apart, 50 Hz. The interval can never be shorter than the absolute refractory period, about 1 ms, which caps any neuron near 1000 Hz. Why does a neuron fire more slowly after seconds of steady input? Slow processes pile up behind the fast refractory period. Calcium entering with each spike opens calcium-activated potassium channels, some sodium channels drop into a slow inactivated state that takes seconds to leave, and the sodium-potassium pump, pumping harder to restore the gradients, adds an outward current of its own with a time constant of up to a few seconds. Each pushes the next spike later, so under a constant input the rate falls over time, which is spike-frequency adaptation. Why does a large rise of potassium outside silence neurons that a small rise excites? Because the same depolarisation that brings the membrane nearer threshold also inactivates sodium channels. More potassium outside raises EK=RTFln[K+]o[K+]iE_K = \frac{RT}{F}\ln\frac{[K^+]_o}{[K^+]_i}EK=FRTln[K+]i[K+]o towards zero and the resting potential follows it up. A small shift puts the cell closer to threshold while most channels are still available, so it fires more easily. A large one slides far down the availability curve h∞(V)h_\infty(V)h∞(V), leaving too few channels to make a spike at all, and the neuron falls silent in depolarisation block.