Individual Neurons - Computational Explorations in Cognitive Neuroscience - page 37

in this section. We will see later that this concept will
be used in determining how long to present patterns to
a network, and in the analysis presented in section 2.7
showing the relationship between the biology of a neu-
ron and the mathematics of hypothesis testing.
Within a reasonably short period of time after excita-
tory and/or inhibitory channels are opened (and barring
the effects of any voltage-gated channels that subse-
quently become activated), the membrane potential will
always settle into a new stable value that reflects the
new balance of forces acting on the neuron. At this new
equilibrium membrane potential, the net current ( I net )
will return to zero, even though the individual currents
for particular channels will have non-zero values (they
will all just add up to zero, thereby canceling each other
out). Thus, a net current is present only when changes
are taking place, not when the membrane potential is
steady (as would be expected given that current is the
mathematical derivative of the membrane potential).
It is useful to be able to compute a value of this equi-
librium membrane potential for a given configuration of
steady conductances. Clearly the equation for comput-
ing the value of V m (equation 2.8) is relevant for figur-
ing out what the equilibrium membrane potential should
be. However the V m equation is a recursive equation for
computing the membrane potential — the membrane
potential is a function that depends on itself! Thus, it
provides a recipe for updating the membrane potential,
but it doesn't tell you directly what value the membrane
potential will settle on if you provided a constant input
to the neuron.
Nevertheless, we can easily solve for the equilibrium
membrane potential equation by noting that when I net
is equal to zero, then the value of V m doesn't change
(as should be obvious from equation 2.8). Thus, setting
the equation for I net (equation 2.6) equal to zero and
solving for the value of V m (which appears in several
places in this equation) should give us the equilibrium
value of the membrane potential. When we do this, and
solve for the value of V m (which is no longer a function
of time, because everything is steady-state), we get:

Equilibrium V_m by g_e (g_l = .1)

1.0

0.8

0.6

0.4

0.2

0.0

0.0 0.2 0.4 0.6 0.8 1.0

g_e (excitatory net input)

Figure 2.10: Plot of the equilibrium membrane potential as a
function of level of excitatory input g e , assuming no inhibitory
current and a leak current of .1. For simplicity, E e =1 and

,so V m goes from 0 to approaching 1 as the strength
of the excitatory input increases.

We can now use equation 2.9 to directly solve for the
equilibrium membrane potential for any fixed set of in-
puts, just by plugging in the appropriate conductance
values (and the reversal potential constants). Also, we
will see in section 2.7 that the form of this equation can
be understood in terms of a hypothesis testing analysis
of a detector. Even without the benefit of that analysis,
this equation shows that the membrane potential is basi-
cally just a weighted average based on the magnitude of
the conductances for each type of channel. This can be
made somewhat more obvious by rewriting equation 2.9
as follows:

(2.10)

Thus, the membrane potential moves toward the driv-
ing potential for a given channel c ( E c ) in direct propor-
tion to the fraction that the current for that channel is of
the total current (figure 2.10). For example, let's exam-
ine a simple case where excitation drives the neuron to-
ward a membrane potential (in arbitrary units) of 1 (i.e.,

(2.9)

) and leak and inhibition drive it toward 0 (i.e.,