1
Equilibrium potential (6pt)
At equilibrium, the total current crossing the cell membrane is zero, such
that:
ge (Veq − Ee ) + gi (Veq Ei ) = 0
Veq (ge + gi ) = ge Ee + gi Ei
ge Ee + gi Ei
Veq =
(2pt)
ge + gi
(1)
(2)
(3)
For ge → 0, Veq → Ei (1pt)
For ge → +∞, Veq → Ee (1pt)
And Veq = 0 for ge = −gi Ei /Ee ≈ 1.4mS (1pt)
Correct plot: (1pt)
2
Integrate-and-fire neuron (7pt + bonus)
2.1
The voltage tends exponentially to the equilibrium voltage, which is such
that:
τ
dV
= 0 = El − Veq + RI0
dt
Veq = El + RI0
(4)
(5)
The neuron only spikes if Veq > Vth , that is, if
I0 >
Vth − El
(2pt)
R
(6)
2.2
At t = t0 , the voltage is at El , it then tends towards the equilibrium potential
with time constant τ :
V (t) = Veq + (V (t0 ) − Veq ) exp(−(t − t0 )/τ )
= El + RI0 + (El − El − RI0 ) exp(−(t − t0 )/τ )
= El + RI0 (1 − exp(−(t − t0 )/τ ))
1
(7)
(8)
(9)
Therefore, the neuron only spikes if:
El + RI0 (1 − exp(−T /τ )) > Vth
Vth − El
I0 >
(3pt)
R(1 − exp(−T /τ ))
(10)
(11)
Plot:
For T << τ , exp(−T /τ ) ≈ 1 − T /τ, I0 ≈ Tτ VthR−El (1pt)
For T >> τ , exp(−T /τ ) ≈ 0, I0 ≈ VthR−El (1pt)
2.3
Advanced
Supposing that the neuron hasn’t fired during the first current step, during
the delay, the voltage decays exponentially back to El :
V (t) = Veq + (V (t0 + T ) − Veq ) exp(−(t − (t0 + T ))/τ )
= El + (El + RI0 (1 − exp(−T /τ )) − El ) exp(−(t − (t0 + T ))/τ )
= El + RI0 (1 − exp(−T /τ )) exp(−(t − (t0 + T ))/τ )
(12)
(13)
(14)
So that when the second step starts, the voltage is now:
V (t0 + T + ∆t) = El + RI0 (1 − exp(−T /τ )) exp(−∆t/τ )
(15)
And during the next current step, the voltage increases exponentially to
El + RI0
V (t) = Veq + (V (t0 + T + ∆t) − Veq ) exp(−(t − (t0 + T + ∆t)/τ )(16)
= El + RI0 + (El + RI0 (1 − exp(−T /τ )) exp(−∆t/τ ) − El − RI0 ) exp(−(t − (t0 + T + ∆t)/τ )(17)
El + RI0 (1 + ((1 − exp(−T /τ )) exp(−∆t/τ ) − 1) exp(−(t − (t0 + T + ∆t)/τ ))(18)
Such that by the end of the second step:
V (t0 + 2T + ∆t) = El + RI0 (1 + ((1 − exp(−T /τ )) exp(−∆t/τ ) − 1) exp(−T /τ ))(19)
= El + RI0 (1 − exp(−T /τ )) + exp(−∆t/τ )RI0 (exp(−T /τ ) − exp(−2T /τ )(20)
The neuron fires during the second step if:
Vth < El + RI0 (1 − exp(−T /τ )) + exp(−∆t/τ )RI0 (exp(−T /τ ) − exp(−2T /τ )(21)
Vth − El − RI0 (1 − exp(−T /τ )) < exp(−∆t/τ )RI0 (exp(−T /τ ) − exp(−2T /τ )(22)
2
Vth − El − RI0 (1 − exp(−T /τ ))
(23)
RI0 (exp(−T /τ ) − exp(−2T /τ )
Vth − El − RI0 (1 − exp(−T /τ ))
)(24)
−∆t/τ > log(
RI0 (exp(−T /τ ) − exp(−2T /τ )
RI0 (exp(−T /τ ) − exp(−2T /τ )
∆t < τ log(
)(25)
Vth − El − RI0 (1 − exp(−T /τ ))
exp(−∆t/τ ) >
2.4
2.4.1
Integrate-and-fire with refractory period (7pt)
fI curve without refractory period
We may consider that the neuron’s membrane potential is clamped to El
during a refractory period tr . Supposing that the neuron was reset to El after
a spike occuring at t = 0, without refractory period, the voltage dynamics
are given by:
V (t) = El + RI0 (1 − exp(−t/τ ))
(26)
Such that the neuron spiked when:
Vth = El + RI0 (1 − exp(−T /τ ))(1pt)
El + RI0 − Vth
exp(−T /τ ) =
RI0
RI0
T = τ log(
)(1pt)
El + RI0 − Vth
(27)
(28)
(29)
And the firing rate is given by:
F =
1
RI0
)
τ log( El +RI
0 −Vth
(30)
Which goes to +∞ when I0 goes to +∞. (1pt)
2.4.2
fI curve with refractory period
We may consider that the neuron’s membrane potential is clamped to El
during a refractory period tr . (1pt)
Supposing that the neuron was reset to El after a spike occuring at t = 0,
with refractory period, the voltage dynamics are given by:
V (t) = El + RI0 (1 − exp(−(t − tr )/τ )), t > tr
3
(31)
And the neuron spikes when:
T = tr + τ log(
RI0
)(1pt)
El + RI0 − Vth
(32)
And the firing rate is given by:
F =
1
(1pt)
RI0
)
tr + τ log( El +RI
0 −Vth
(33)
Now, when I0 goes to +∞, the firing rate only goes to 1/tr , since the
neuron cannot fire during the refractory period. (1pt)
4