Part 3 · ~1 ms to ~5 ms
From Channel to Spike: The Neuron as an RC Circuit
The twisted retinal pushes on the surrounding helices, and within about a millisecond a pore opens. Channelrhodopsin-2 is a non-selective cation channel: sodium, potassium, protons and some calcium can pass, and its reversal potential is near 0 mV.
One channel is tiny
The single-channel conductance of ChR2 is only of order 50 fS (an approximate value; see the note on estimates below). At a resting potential of −70 mV, the driving force is about 70 mV, so one open channel carries:
\[ i = g\,(V - E_{\text{rev}}) \approx 50\ \text{fS} \times 70\ \text{mV} \approx 3.5\ \text{fA} \]
The membrane as a circuit
A neuron's membrane behaves like a capacitor C (the lipid bilayer) in parallel with a resistor R (the leak channels). For a typical cortical neuron, take C ≈ 100 pF and R ≈ 150 MΩ, so the time constant is τ = RC ≈ 15 ms. A steady photocurrent I charges the membrane as:
\[ \Delta V(t) = I R \left(1 - e^{-t/\tau}\right) \]
Where the numbers for C and R come from
Capacitance from a parallel-plate model
The lipid bilayer is a thin insulator between two conducting salt solutions, so treat it as a parallel-plate capacitor. Its hydrocarbon core has a relative permittivity εr ≈ 2 and a thickness d ≈ 4 nm:
\[ \frac{C}{A} = \frac{\varepsilon_0\,\varepsilon_r}{d} \approx \frac{(8.85\times10^{-12}\ \text{F/m})(2)}{4\times10^{-9}\ \text{m}} \approx 4\times10^{-3}\ \text{F/m}^2 \approx 0.4\ \mu\text{F/cm}^2 \]
The measured value for nearly all biological membranes is close to 1 µF/cm², the same order of magnitude; the difference comes from the polar head groups and a thinner effective core. A neuron with C ≈ 100 pF therefore has a membrane area of about 100 pF ÷ 1 µF/cm² = 10⁻⁴ cm² = 10,000 µm², counting its cell body and dendrites.
Resistance from leak channels
Leakage scales with area too, so physiologists quote a specific membrane resistance, Rm ≈ 15,000 Ω·cm² for a typical neuron. Dividing by the area gives the input resistance:
\[ R = \frac{R_m}{A} \approx \frac{1.5\times10^{4}\ \Omega\,\text{cm}^2}{10^{-4}\ \text{cm}^2} = 1.5\times10^{8}\ \Omega = 150\ \text{M}\Omega \]
A neat result: τ does not depend on size
Because C grows with area and R shrinks with it, the area cancels in the time constant:
\[ \tau = R\,C = \frac{R_m}{A}\,(C_m A) = R_m C_m \approx (1.5\times10^{4}\ \Omega\,\text{cm}^2)(10^{-6}\ \text{F/cm}^2) = 15\ \text{ms} \]
Small and large neurons integrate input over similar times; what differs is how much current they need, since a small cell has a larger R. This is the point of Exercise 4.
A common sign confusion
By convention, current flowing out of the cell is positive. Cations entering through channelrhodopsin are therefore a negative (inward) current, even though they make the inside more positive. Step 2 of the derivation below sidesteps this by defining I as the inward current.
How much light current fires a spike?
If the threshold sits about 20 mV above rest, the steady current must exceed 20 mV / 150 MΩ ≈ 130 pA. Dividing by 3.5 fA per channel gives roughly 40,000 channels open at once, which is why neurons are engineered to express ChR2 at high density.
How fast?
With a strong 500 pA photocurrent, IR ≈ 75 mV, and threshold is reached after t = −τ ln(1 − 20/75) ≈ 0.31 τ ≈ 4.7 ms. The channel then closes over roughly 10 ms in the dark, which limits how fast it can drive repeated spikes; faster engineered variants push that limit to well over 100 spikes per second.
Derivation: the membrane equation
Step 1 — Kirchhoff's current law
Charge is conserved at the inside node of Figure 3: the capacitive current plus the two ionic currents must sum to zero. Taking outward current as positive:
\[ C\,\frac{dV}{dt} + \frac{V - V_{\text{rest}}}{R} + G_{\text{ChR}}\,(V - E_{\text{rev}}) = 0 \]
Step 2 — Constant-current approximation
Near rest, V − Erev ≈ −70 mV barely changes before threshold, so treat the light-driven current as a constant inward current I = GChR(Erev − Vrest). With u = V − Vrest and τ = RC:
\[ \tau\,\frac{du}{dt} = -u + I R \]
Step 3 — Solve
This is a first-order linear equation. Starting from rest, u(0) = 0:
\[ u(t) = I R\left(1 - e^{-t/\tau}\right) \]
Step 4 — Rheobase and time to threshold
The steady state is u∞ = IR. A spike needs u to reach uth, which is only possible if IR > uth; the minimum current, or rheobase, is Imin = uth / R. Above it, solving u(tth) = uth gives:
\[ t_{\text{th}} = -\tau \ln\!\left(1 - \frac{u_{\text{th}}}{I R}\right) \]
Step 5 — The exact linear solution
Keeping the voltage dependence of the channel current, the two conductances simply add. Define the total conductance G = 1/R + GChR:
\[ \tau_{\text{eff}} = \frac{C}{G}, \qquad V_\infty = \frac{V_{\text{rest}}/R + G_{\text{ChR}}\,E_{\text{rev}}}{G} \]
Two lessons follow. Light makes the neuron faster, since τeff < τ. And V∞ is a conductance-weighted average of −70 mV and 0 mV, so no amount of light can push the membrane past Erev: channelrhodopsin can only depolarise, never hyperpolarise, the neuron.
Step 6 — Back to photons
The channel conductance is set by how many channels are open, GChR = Nch · popen · γ, where γ ≈ 50 fS is the single-channel conductance and popen rises with light intensity. Combining with the estimate above links a light power in mW/mm² to a spike latency in milliseconds — which is exactly what Simulations 3 and 4 compute.
What this model leaves out
These values are order-of-magnitude teaching estimates; real neurons and expression levels vary widely. Real channelrhodopsin has at least four states, including a desensitised one that lowers the current during long pulses, and a real spike is shaped by voltage-gated Na⁺ and K⁺ channels rather than the reset rule used here.