L2 · Chapter 3

The action potential from one equation, all or none

Sandbox Physics · 2026-10-31

The last chapter looked at the sodium and potassium currents separately under voltage clamp. Now release the clamp and let the voltage move. Opening sodium channels depolarises the membrane, and depolarisation opens more sodium channels; this positive feedback is the rising edge of the spike. The central result of Hodgkin and Huxley in 1952 is that equations fitted only to voltage-clamp data predict a spike they never fitted, with the right shape, threshold and ion movements.

Sodium feedback and two brakes

Put the three gating equations together with the membrane equation and you have four coupled ordinary differential equations. A small depolarisation lets the fast sodium activation catch up, sodium current flows in, and the voltage rises faster. Nothing stops this step, and the voltage heads for the sodium reversal potential.

The brakes come from two slower processes. The sodium inactivation variable $h$ shuts the sodium channels within about a millisecond, while the potassium activation $n$ opens the potassium channels. The voltage falls, overshoots rest, stays for a while about 11 mV below it, and slowly returns.

Hodgkin and Huxley integrated these equations step by step with Hartree’s method (p. 523 of the paper), with steps of 0.02 ms on the rising phase. We integrate with a fourth-order Runge–Kutta scheme; halving the step changes the peak by less than $0.002\ \mathrm{mV}$ .

C_M\frac{du}{dt}=-g_{\mathrm{Na}}\,m^{3}h\,(u-E_{\mathrm{Na}})-g_{\mathrm{K}}\,n^{4}\,(u-E_{\mathrm{K}})-g_L\,(u-E_L)+I_{\mathrm{ext}}(t)

Eq. (26) of the 1952 paper, rewritten with depolarisation positive.

g_{\mathrm{Na}}=120

g_{\mathrm{K}}=36

and

g_L=0.3\ \mathrm{mS\,cm^{-2}}

, with

C_M=1\ \mu\mathrm{F\,cm^{-2}}

Source: A. L. Hodgkin and A. F. Huxley · W. Gerstner

The threshold lies between 6 and 7 nC/cm²

Hodgkin and Huxley computed a series of initial depolarisations at 6 °C: 6 mV gave no spike and 7 mV did (Figs. 12 and 21 of the paper). Bisection in our model at 6.3 °C puts the threshold at $6.502\ \mathrm{nC\,cm^{-2}}$ ; a second integrator, written independently in the paper’s own signs, gives 6.504.

Above threshold, a larger shock hardly changes the peak. Table 4 lists a 102.1 mV peak for a 7 mV start, 105.4 mV for 15 mV and 108.5 mV for 90 mV. The starts differ by more than ten times and the peaks by 6%. That is what all or none means.

The threshold is not a fixed voltage. Right after a spike most sodium channels are inactivated and the potassium channels are still open, so the same shock again gives no spike; this is the refractory period (Fig. 20). Conversely, hold the membrane 30 mV below rest and release it suddenly: part of the sodium inactivation has been removed and the potassium channels are shut, so the membrane fires on its own. This is anode break (Fig. 22).

Source: A. L. Hodgkin and A. F. Huxley · W. Gerstner

Worked example: one 15 nC/cm² shock

Problem. At 6.3 °C a resting membrane receives an instantaneous shock of 15 nC/cm². Find the peak, the maximum rate of rise and the extra sodium that enters and potassium that leaves, and compare them with Tables 4 and 5 of 1952. How much sodium would it take just to charge the capacitor to the peak, and how does that compare with the sodium that actually enters?

Model. Integrate Eq. (26) from $u_0=15\ \mathrm{mV}$ . As in the paper, integrate the movements up to the third return to the resting level after the peak, then subtract the flux that would have flowed at rest over the same time. That subtraction is our inference from the numbers; the paper does not say so.

Numbers. The peak is $105.41\ \mathrm{mV}$ against 105.4 printed in Table 4; the maximum rate of rise is $311\ \mathrm{V\,s^{-1}}$ against 311; the peak conductance is $37.0\ \mathrm{mS\,cm^{-2}}$ against 37.0. The extra sodium entry is $14.46\ \mathrm{pmol\,cm^{-2}}$ against 14.46 in Table 5, and the extra potassium exit 14.33 against 14.32. Charging $1\ \mu\mathrm{F\,cm^{-2}}$ to 105 mV needs only $105\ \mathrm{nC\,cm^{-2}}/F=1.09\ \mathrm{pmol\,cm^{-2}}$ .

Explanation. Thirteen times more sodium enters than charging requires. During the second half of the rise the potassium channels are already open, so most of the charge sodium brings in is cancelled by potassium leaving at the same time, and only a small part stays on the capacitor. A spike is cheap in charge and expensive in ions, and pumps must later move all that sodium and potassium back.

Check. Repeat the calculation at 18.5 °C: the model gives 96.9 mV and 563 V/s against 96.8 and 564 in Table 4, and extra sodium and potassium of 4.01 and 4.10 pmol/cm² against 3.99 and 4.07 in Table 5. The largest difference is the potassium of the anode-break row, 16.59 in the model against 16.77 printed, 1.1%.

N_{\mathrm{Na}}=\frac{1}{F}\int_0^{t_3}\bigl[-I_{\mathrm{Na}}(t)+I_{\mathrm{Na}}^{\mathrm{rest}}\bigr]\,dt,\qquad \frac{Q_{\mathrm{peak}}}{F}=\frac{C_M\,u_{\max}}{F}

Left: the extra sodium entry, with

F

the Faraday constant and

t_3

the third return to the resting level. Right: the charge needed to bring the capacitor to the peak, in moles.

Source: A. L. Hodgkin and A. F. Huxley · W. Gerstner

Try it in the Lab

All or none

Predict first: if the shock goes from 15 to 40 nC/cm², will the peak more than double or hardly change?
Choose the One all-or-none spike preset and read the peak depolarization. Then drag the shock charge to 40, then to about 6.5, and compare.
Explain: above threshold the peak is set by the sodium reversal potential and the speed of inactivation, not by the charge of the shock; near threshold a slightly different start gives a completely different outcome.

Refractory period and anode break

Predict first: a second shock of 20 nC/cm² arrives 8 ms after the first spike. Will the membrane fire again?
Choose the A second shock fails preset and read the height of the second peak and the number of spikes. Then drag the delay of the second shock to 20 ms. Finally choose Release from hyperpolarization fires.
Explain: at 8 ms most sodium channels are still inactivated and the potassium channels are open, so the second shock gives only a ripple; recovery takes longer. Hyperpolarisation does the opposite: it removes inactivation, and on release the membrane fires by itself.

Exercises with explained answers

At the maximum rate of rise, $311\ \mathrm{V\,s^{-1}}$ , what net ionic current crosses the membrane?
Show the answer
All of the voltage change comes from charging the capacitor: $I=C_M\,du/dt=1\ \mu\mathrm{F\,cm^{-2}}\times311\ \mathrm{V\,s^{-1}}=311\ \mu\mathrm{A\,cm^{-2}}$ , a net inward current of 0.31 mA/cm².
How many extra sodium ions enter each square micrometre during one spike? With 60 sodium channels per square micrometre, how many ions pass through each channel on average?
Show the answer
$14.46\times10^{-12}\ \mathrm{mol\,cm^{-2}}\times6.022\times10^{23}/10^{8}=8.7\times10^{4}$ per square micrometre, about 1450 per channel. Chapter 4 uses this density of 60 sodium channels per square micrometre.
Why is the peak lower at 18.5 °C than at 6.3 °C?
Show the answer
Warming multiplies every gating rate by 3.82. Sodium activation was already fast and changes little in effect; sodium inactivation and potassium activation were slow and now step in earlier, pulling the voltage back before it reaches its old maximum. The peak drops from 105 mV to 97 mV.
A steady current of 10 µA/cm² makes the model fire 4 times in 50 ms, about 68 Hz. What happens if the injected current rises very slowly instead?
Show the answer
Page 537 of the paper notes that a slowly enough rising current never excites the model. Sodium inactivation and potassium activation keep up, and the steady ionic current at any depolarisation is outward, cancelling the injected current. This is accommodation.

Sources and what we recomputed

A. L. Hodgkin and A. F. Huxley, A quantitative description of membrane current and its application to conduction and excitation in nerve, Journal of Physiology 117, 500–544 (1952), PMC1392413
We read every page, pp. 500 to 544, from the scanned pages in PubMed Central. Sign convention on pp. 505 and 506: V is the displacement from rest with depolarisation negative, and inward current is positive. Rate constants: Eqs. (12), (13), (20), (21), (23) and (24). Eq. (26) and Table 3 on pp. 518 to 520: the full membrane equation and its constants at 6.3 °C, with the leak potential chosen so that the current is zero at rest. Numerical method on p. 523. Figs. 12, 13 and 21: threshold between 6 and 7 mV. Table 4 on p. 527: spike shape. Table 5 on pp. 532 and 533: sodium and potassium movements. Fig. 20: refractory period. Fig. 22: anode break. P. 541: their warning that the equations are an empirical description. We recomputed Tables 4 and 5 for the membrane action potentials, not the propagated one.
W. Gerstner, W. M. Kistler, R. Naud and L. Paninski, Neuronal Dynamics, Cambridge University Press (2014), online edition, Sections 2.1 and 2.2
Section 2.1: the Boltzmann factor, Eq. (2.1), and the Nernst potential, Eq. (2.2), with the sodium example near +67 mV and the potassium example near −83 mV. Section 2.2 (Ch2.S2.html): charge conservation on the membrane, Eqs. (2.3) and (2.4), the three ionic currents, Eq. (2.5), first-order gating, Eq. (2.6), and the response to a voltage step, Eq. (2.7). This book writes u as the membrane potential with depolarisation positive, the convention our course uses.