 Synaptic potential

A synaptic potential is a postsynaptic change in membrane potential caused by activity at a synapse. Unlike the regenerative action potential, a passive synaptic potential is graded: its size depends on input strength, the initial membrane potential and the receiving cell membrane's properties. Local voltage changes can decay before reaching the soma; their timing and overlap help determine whether an action potential follows. See Ion Channels and the Electrical Properties of Membranes.

Depolarization means that the interior becomes less negative relative to the exterior; hyperpolarization means that it becomes more negative. Excitation and inhibition instead describe whether an input promotes or opposes postsynaptic action potential generation. An inhibitory postsynaptic potential can be depolarizing, and shunting inhibition can occur with no voltage change at rest. The receptor, permeating ions, ionic gradients and reversal potential matter. See Purves and colleagues: excitation, inhibition and depolarizing inhibition.

A worked passive model treats the receiving membrane as one compartment with capacitance $C>0$, leak conductance $g_L>0$, and activated synaptic conductances $g_e,g_i\ge 0$:All electric currents are inward-positive. Each term $g(E-V)$ vanishes at its reversal potential $E$; $\mathrm{nS}\times \mathrm{mV}=\mathrm{pA}$. With constant parameters, the ordinary differential equation hasHere $G$ is total conductance, $1/G$ is effective electrical resistance, $V_{\infty }$ is the steady state and $\tau$ the time constant. Without imposed electric current, $V_{\infty }$ is a conductance-weighted average of the reversal potentials. This approximation omits spatial structure, changing ionic gradients and active voltage-gated ion channels; it calculates passive voltage responses, not actual action potential waveforms. See 3.1 Synapses for the mechanism behind the terms; the inward-positive convention here reverses that source's synaptic-current sign.

Take the same illustrative leak and capacitance as under shunting inhibition, start at $V(0)=-65\mathrm{mV}$ and set $I_{\mathrm{ext}}=0$. Opening $g_e=2\mathrm{nS}$ with $E_e=0\mathrm{mV}$ gives $V_{\infty }=-650/12\simeq -54.167\mathrm{mV}$ and $\tau \simeq 16.667\mathrm{ms}$. Its initial inward electric current is $2(0+65)=130\mathrm{pA}$. Adding $g_i=20\mathrm{nS}$ with $E_i=-65\mathrm{mV}$ changes $V_{\infty }$ to $(-650-1300)/32=-60.9375\mathrm{mV}$ and $\tau =6.25\mathrm{ms}$. This is shunting inhibition of the same excitatory drive.

For comparison, $g_i=20\mathrm{nS}$ with $E_i=-75\mathrm{mV}$ alone gives $V_{\infty }\simeq -71.667\mathrm{mV}$: hyperpolarization. With $E_i=-60\mathrm{mV}$ instead, it alone gives $V_{\infty }\simeq -61.667\mathrm{mV}$: depolarization. Yet adding this depolarizing conductance to the $g_e=2\mathrm{nS}$ excitatory drive lowers that drive's $V_{\infty }$ from $-54.167$ to $(-650-1200)/32=-57.8125\mathrm{mV}$. At an illustrative firing threshold $V=-50\mathrm{mV}$, its electric current is $20(-60+50)=-200\mathrm{pA}$, opposing threshold approach. The combined trace initially rises faster than the excitatory drive alone, before settling below it. Thus the sign of a voltage change at rest alone does not determine excitation or inhibition; a claim about firing must specify input strength, timing and the neuron's state.