= Passive spatial and conductance integration
The spatial panel uses two compartments with <membrane potentials> $V_s,V_d$, one representing the <soma> and one a <dendrite>. Set $v_s=V_s-E_L$, $v_d=V_d-E_L$, with leak <reversal potential> $E_L=-65\,\mathrm{mV}$, each <membrane capacitance> $C=100\,\mathrm{pF}$, each leak <electrical conductance> $g_L=5\,\mathrm{nS}$ and axial <electrical conductance> $g_c=10\,\mathrm{nS}$. <Electric charge conservation> gives
$$
C\dot v_s=-(g_L+g_c)v_s+g_cv_d+I_s,\qquad C\dot v_d=g_cv_s-(g_L+g_c)v_d+I_d.
$$
This chosen passive model has two decay <time constants>, $C/g_L=20\,\mathrm{ms}$ for the equal-voltage mode and $C/(g_L+2g_c)=4\,\mathrm{ms}$ for the voltage-difference mode. Starting from rest, a $100\,\mathrm{pA}$ step into the <soma> gives, with $u$ the time since onset,
$$
v_s(u)=10(1-e^{-u/20})+2(1-e^{-u/4})\quad\mathrm{mV},
$$
while the same step into the <dendrite> gives
$$
v_s(u)=10(1-e^{-u/20})-2(1-e^{-u/4})\quad\mathrm{mV}.
$$
$u$ is expressed in $\mathrm{ms}$. Their long-time rises are $12$ and $8\,\mathrm{mV}$ respectively. Applying both inputs gives $20(1-e^{-u/20})\,\mathrm{mV}$ exactly: spatial attenuation coexists with linear summation of prescribed <electric currents>. After the steps end, both modes decay with their respective <time constants>. This deliberately coarse model is not a reconstructed <dendrite> or a fit to the experiment above.
The conductance panels use the existing one-compartment <synaptic potential> equation, with $C=200\,\mathrm{pF}$, $g_L=10\,\mathrm{nS}$, $E_L=-65\,\mathrm{mV}$ and excitatory <reversal potential> $E_e=0\,\mathrm{mV}$. For a constant excitatory <synaptic conductance> $g_e$, the <steady state> voltage rise and <membrane time constant> are
$$
\Delta V_\infty=\frac{g_e(E_e-E_L)}{g_L+g_e},\qquad \tau_{\mathrm{eff}}=\frac{C}{g_L+g_e}.
$$
Each isolated $2\,\mathrm{nS}$ input has a limiting rise $130/12\approx10.833\,\mathrm{mV}$. Adding those isolated rises predicts $21.667\,\mathrm{mV}$, but opening both together gives $260/14\approx18.571\,\mathrm{mV}$. The two conductances share a changing driving force. This is sublinear input summation even though the voltage <ordinary differential equation> is linear for each prescribed conductance waveform.
Add $g_i=20\,\mathrm{nS}$ with <reversal potential> $E_i=E_L$ to the joint $4\,\mathrm{nS}$ excitation. The limiting rise becomes $260/34\approx7.647\,\mathrm{mV}$ and the <membrane time constant> becomes $200/34\approx5.882\,\mathrm{ms}$. The shunt alone leaves rest unchanged. At the joint-excitation-only <steady state> $V\approx-46.429\,\mathrm{mV}$, however, its inward <electric current> would be $20(-65+46.429)\approx-371.429\,\mathrm{pA}$: it opposes that excursion. The finite $50\,\mathrm{ms}$ traces in the diagram approach, but need not reach, these limiting voltages. See <shunting inhibition> for the constant-<electric current> comparison.
Back to article page
