t0.00ms
V-65.0mV
m0.053
h0.596
n0.318
gNa0.00mS/cm²
gK0.00mS/cm²
peak V—mV
Membrane potential V(t) — mV
V
time (ms)
Gating variables — fraction open (0–1)
m
h
n
time (ms)
m activates fast; h inactivates Na⁺; n activates K⁺ more slowly — that timing gap is what makes a spike a spike.
Conductances gNa, gK — mS/cm²
gNa
gK
time (ms)
gNa switches on and back off quickly; gK turns on slower and stays on longer — this asymmetry shapes the spike.
m∞, h∞, n∞ vs V
membrane potential V (mV)
τm, τh, τn vs V (ms)
membrane potential V (mV)
The dashed vertical line tracks the neuron's current V as the sweep plays. Note: blocking a channel zeroes its maximal conductance (ḡ) — it does not change these kinetics.
The equations
$$C_m\frac{dV}{dt} = I - \bar g_{Na}\,m^3 h\,(V-E_{Na}) - \bar g_K\,n^4\,(V-E_K) - g_L\,(V-E_L)$$
$$\frac{dx}{dt} = \alpha_x(V)\,(1-x) - \beta_x(V)\,x \qquad x \in \{m,\,h,\,n\}$$
Cm=1, ḡNa=120, ḡK=36, gL=0.3 mS/cm²; ENa=50, EK=−77, EL=−54.387 mV. Forward Euler, dt = 0.01 ms.
$$\alpha_m = \dfrac{0.1(V+40)}{1-e^{-(V+40)/10}}$$
$$\beta_m = 4e^{-(V+65)/18}$$
$$\alpha_h = 0.07\,e^{-(V+65)/20}$$
$$\beta_h = \dfrac{1}{1+e^{-(V+35)/10}}$$
$$\alpha_n = \dfrac{0.01(V+55)}{1-e^{-(V+55)/10}}$$
$$\beta_n = 0.125\,e^{-(V+65)/80}$$
Both \(\alpha_m\) and \(\alpha_n\) have a removable 0/0 singularity (at V=−40 and V=−55 mV). This demo uses the analytic limit there: \(\alpha_m(-40)=1.0\), \(\alpha_n(-55)=0.1\), matching the companion Python script.