= Solution
Assume constant current, $g_L>0$, $C_m>0$, and reset below threshold. With $g_E=0$, $V_\infty=V_L+I_e/g_L$. A spike occurs in finite time after reset precisely when $V_\infty>V_\theta$, so the critical boundary is
$$
\boxed{I_c=g_L(V_\theta-V_L)}.
$$
At equality the voltage approaches threshold asymptotically, so the finite-time firing condition is the strict inequality $I_e>I_c$.
Solve the exponential voltage trajectory from reset to threshold. The interspike interval of this <constant-input firing rate of a leaky integrate-and-fire neuron> is
$$
T=\frac{C_m}{g_L}\log\frac{V_\infty-V_0}{V_\infty-V_\theta}=\frac{C_m}{g_L}\log\left(1+\frac{g_L(V_\theta-V_0)}{I_e-I_c}\right).
$$
Thus
$$
\boxed{r(I_e)=0\ (I_e\leq I_c),\qquad r(I_e)=T^{-1}\ (I_e>I_c)}.
$$
The logarithm decreases strictly with current, so $r$ increases, starting from zero as $I_e\downarrow I_c$. At large current, its argument increment is small and
$$
T\sim\frac{C_m(V_\theta-V_0)}{I_e-I_c},\qquad\boxed{r(I_e)\sim\frac{I_e}{C_m(V_\theta-V_0)}}.
$$
The second expression states the leading linear growth, not an exact intercept formula. If an explicit refractory time $t_{\rm ref}$ is added, the rate instead becomes $1/(T+t_{\rm ref})$ and approaches $1/t_{\rm ref}$ at very high current. The printed model contains no such extra interval, which is why its unrestricted high-current rate is linear.
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