Lecture 5: Integrate-and-fire neurons

A Cell ⟺ An RC circuit

Capacitance effect owing to charge accumulation

At the membrane:

\[Q = C_m V = c_m A V\]

where $c_m = 10 nF/mm^2$

\[\dot{Q} = C_m \dot{V}\]

Resistivity to external currents

\[ΔV = R_m I_e = \frac{r_m}{A} I_e\]

Resistivity to pass through the membrane

⟹ The cell is a RC-circuit:

\[τ_m = R_m C_m\]

Equilibrium potential

Diffusion and repulsion of ions at the membrane (which is an isopotential).

For each charged ion, there an equilibrium potetial $E_i$ at which the migration through the membrane stops.

Ohm’s law:

\[ΔV = R I\]

For an ion $j$:

\[i_j = \frac{ΔV}{r_j} = \frac{V - E_j}{r_j} = \underbrace{g_j}_{\text{conductance}} (V-E_j)\]

Membrane current: \(i_m = \sum\limits_{ j } g_i (V - E_j)\)

\[I_e = \frac{dV}{dt}c_m + I_m\\ ⟺ c_m \frac{dV}{dt} = -i_m + \frac{I_e}{A}\]

Here, we’ll suppose that $i_m$ is a leaky term, i.e. a linear one

Charge conservation

\[\begin{align*} & c_m \frac{dV}{dt} = -i_l + \frac{I_l}{A} \\ ⟺ & c_m \frac{dV}{dt} = -g_l(V - E_l) + \frac{I_l}{A} \\ ⟺ & c_m \frac{dV}{dt} = \frac 1 {r_l} (-V + E_l) + \frac{r_l}{A} I_l \\ ⟺ & \underbrace{r_l c_m}_{≝ τ_m} \frac{dV}{dt} = E_l - V + R_m I_l \\ \end{align*}\]

Integrate-and-fire neuron

Homogeneous equation:

\[τ_m \frac{dV_{hom}}{dt} = - V_{hom}\\ ⟹ V_{hom} = K \exp(-t/τ_{m})\]

Now variation of the constant:

$K = K(t)$ and replace $V$ in the equation:

\[\begin{align*} & τ_m \frac{dV}{dt} = E_l - V + R_m I_l \\ ⟺ & τ_m \left(\frac{dK}{dt} \exp(-t/τ_m) - K/τ_m \exp(-t/τ_m)\right) = E_l - K \exp(-t/τ_m) + R_m I_l \\ ⟺ & τ_m \frac{dK}{dt} \exp(-t/τ_m) = E_l + R_m I_l \\ ⟺ & \frac{dK}{dt} = \frac{E_l + R_m I_l}{τ_m} \, \exp(t/τ_m)\\ ⟺ & K = (E_l + R_m I_l) \, \exp(t/τ_m) + \text{const} \end{align*}\]

So:

\[V(t) = \left[(E_l + R_m I_l) \, \exp(t/τ_m) + \text{const}\right] \exp(-t/τ_m)\\ V(t) = (E_l + R_m I_l) + (V(0) - E_l - R_m I_l) \exp(-t/τ_m)\]

Spiking mechanism:

\[V > V_{threshold} ⟹ V = V_r \text{ (there's a spike)}\]

Output firing rate

• $t=0 \qquad V=V_r$

  • \[V(t) = (E_l + R_m I_l) + (V_r - E_l - R_m I_l) \exp(-t/τ_m)\]

• $t=t^\ast \qquad V=V_{th}$

  • \[V_{th} = (E_l + R_m I_l) + (V_r - E_l - R_m I_l) \exp(-t^\ast/τ_m)\\ ⟺ t^\ast = -τ_m \ln\left(\frac{V_{th} - (E_l + R_m I_l)}{V_r - E_l - R_m I_l}\right) = τ_m \ln\left(\frac{V_r - E_l - R_m I_l}{V_{th} - E_l - R_m I_l}\right)\]

Firing rate $ν$:

\[ν ≝ \left(τ_m \ln\left(\frac{V_r - E_l - R_m I_l}{V_{th} - E_l - R_m I_l}\right) \right)^{-1}\]

Refractory period

Whenever the neuron depolarize until the voltage hits the threshold: it fires, then the voltage is reinitialized at $V_r$ for a time span $τ_{ref}$ (the refractory period), then it depolarizes again.

If we take into account this $τ_{ref}$, the firing rate becomes:

\[ν ≝ \left(τ_m \ln\left(\frac{V_r - E_l - R_m I_l}{V_{th} - E_l - R_m I_l}\right) + τ_{ref}\right)^{-1}\]

Leaky: because $E_l - V$ is linear. More complex model: when it is quadratic.