Visual walkthrough — Neuromorphic computing

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Step 1 — A membrane is two things at once: a bucket and a leak

WHAT. Picture the thin fatty wall of a neuron. On one side, positive charge piles up; the wall keeps the two sides apart. That is exactly what a capacitor does — it stores separated charge. But the wall is not perfect: tiny channels let charge trickle back across. Something that lets charge trickle through in proportion to how hard you push is a resistor. So the membrane is a resistor and a capacitor, side by side — an RC circuit.

WHY this picture and not a plain wire. A plain wire has no memory: push current, charge flows, nothing accumulates. But a neuron remembers recent input for a short while and then forgets. Only a store-and-leak object does that. The RC circuit is the simplest object with both a memory (the capacitor) and a fade (the resistor).

PICTURE. The tank on the left fills with input current $I$; the mint hole on the right leaks water out. The water level is the voltage $V$.

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Step 2 — Write down the balance: what comes in must go somewhere

WHAT. Every bit of current that arrives has only two possible fates: it either piles onto the capacitor (raising the level) or escapes through the leak. Nothing vanishes. That single sentence is the whole equation: $I(t)=\underset{\text{fills the tank}}{\underbrace{C\frac{dV}{dt}}}+\underset{\text{leaks out}}{\underbrace{\frac{V}{R}}}$

WHY a derivative appears here. The tool $\frac{dV}{dt}$ is brand new, so we justify it. It reads "how fast the level $V$ is changing right now." We need it because the current charging a capacitor is not about how full it is — it is about how fast it is filling. The question "how fast is this changing?" is exactly the question a derivative answers, so that is the tool the physics forces on us.

Term by term:

• $C\frac{dV}{dt}$ — charge going into storage. Wider tank ($C$ big) or faster rise ($\frac{dV}{dt}$ big) means more current is being swallowed.
• $\frac{V}{R}$ — the leak. Higher level $V$ pushes harder through the drain; narrower hole (big $R$) leaks less.

PICTURE. The incoming arrow splits into two: one into the tank, one out the leak. Their sizes must add up to the input.

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Step 3 — Clean it up, and meet the "forgetting time" ${\tau }_m$

WHAT. Multiply every term by $R$ and gather: ${\tau }_m\frac{dV}{dt}=-V+RI(t),{\tau }_m\equiv RC$

WHY define ${\tau }_m=RC$. We bundle $R$ and $C$ into one symbol because together they set the only timescale in the system. $R$ (ohms) times $C$ (farads) comes out in seconds — it is literally a duration. We name it ${\tau }_m$, the membrane time constant: the neuron's "attention span."

Why the minus sign in front of $V$. Turn off the input ($I=0$). The equation becomes ${\tau }_m\dot{V}=-V$: the fuller the tank, the faster it empties. That minus sign is the leak — it always drags $V$ back toward zero. This is the "leaky" in Leaky Integrate-and-Fire: old input is continuously forgotten.

PICTURE. With the tap off, the level slides down a smooth curve; after one ${\tau }_m$ it has fallen to about $37\%$ of where it started (that $0.37$ is $1/e$, a number we meet properly next step).

Cloze check:

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Step 4 — Turn on a steady tap: the charging curve

WHAT. Hold the input at a constant $I$, start empty ($V(0)=0$). Solving ${\tau }_m\dot{V}=-V+RI$ gives: $V(t)=RI\left(1-e^{-t/{\tau }_m}\right)$

WHY the number $e$ shows up. We did not choose the exponential — the leak did. Whenever a thing's rate of change is proportional to itself ("the more there is, the faster it moves"), the only function that satisfies that is $e^{(\dots )}$. That is $e$'s job description: it is the function that is its own rate of change. The leak makes $\dot{V}$ depend on $V$, so $e$ is unavoidable.

Term by term:

• $RI$ — the ceiling. As $t\to \infty$, the bracket $\to 1$, so $V\to RI$. This is the level where "in" exactly balances "leak," so filling stops. Nail this value; Step 6 hangs on it.
• $e^{-t/{\tau }_m}$ — the approach. At $t=0$ it is $1$ (bracket $=0$, empty). It shrinks as time passes, so the bracket climbs toward $1$.
• After $t={\tau }_m$: $e^{-1}\approx 0.37$, so $V\approx 0.63RI$ — 63% of the way up in one time constant.

PICTURE. A rising curve that starts steep and flattens toward the dashed ceiling $RI$. The tick at ${\tau }_m$ marks the 63% point.

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Step 5 — Add the trigger: threshold, fire, reset

WHAT. The smooth curve of Step 4 never actually spikes — it just glides up to $RI$. A neuron must produce a sharp event. So we bolt on a rule by hand: $\text{if }V(t)\ge V_{th}\Rightarrow \text{emit a spike, then set }V\leftarrow V_{reset}$

WHY this is separate from the ODE. The differential equation is analog and smooth; the spike is digital and abrupt. No smooth equation makes an all-or-nothing event, so the threshold is an extra ingredient. This bolt-on is precisely what turns an RC filter into a spiking neuron — the heart of any SNN.

PICTURE. The charging curve climbs, kisses the coral line $V_{th}$, a vertical spike shoots up, and $V$ drops back to $V_{reset}$ to start charging again.

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Step 6 — Solve for the time-to-spike (and meet the logarithm)

WHAT. The neuron fires the first instant the curve reaches $V_{th}$. Call that instant $T$. Set $V(T)=V_{th}$ in the Step-4 curve and unravel it: $V_{th}=RI\left(1-e^{-T/{\tau }_m}\right)\Rightarrow e^{-T/{\tau }_m}=1-\frac{V_{th}}{RI}$ $\boxed{T={\tau }_m\ln \frac{RI}{RI-V_{th}}}$

WHY a logarithm now. We have an equation where the unknown $T$ is trapped inside an exponent: $e^{-T/{\tau }_m}$. To free it we need the tool that undoes an exponential — that is exactly what $\ln$ (the natural logarithm) is: it answers "$e$ to what power gives this number?" Applying $\ln$ to both sides pulls $T$ down out of the exponent. It is the only tool that does that job cleanly, which is why it appears.

Term by term inside the box:

• ${\tau }_m$ out front — scales everything: a sluggish membrane (big ${\tau }_m$) takes proportionally longer to fire.
• $\frac{RI}{RI-V_{th}}$ — the ratio of "ceiling" to "gap remaining." The closer the ceiling $RI$ sits to the threshold $V_{th}$, the smaller the denominator, the bigger the log, the longer it takes.

PICTURE. Same curve, now with $T$ marked where it crosses $V_{th}$, and a bracket showing the shrinking gap $RI-V_{th}$.

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Step 7 — The degenerate case: when the neuron NEVER fires

WHAT. Look at the denominator $RI-V_{th}$. What if the ceiling $RI$ is below the threshold?

• If $RI<V_{th}$: the fraction $\frac{RI}{RI-V_{th}}$ has a negative denominator → the whole argument is negative → $\ln$ of a negative number does not exist. Mathematically the formula breaks; physically, the curve flattens out below the coral line and never touches it. No spike, ever.
• If $RI=V_{th}$: the denominator is $0$, the fraction blows up, $\ln (\infty )=\infty$ → $T\to \infty$. The neuron only just reaches threshold after infinite time — still effectively silent.

WHY this matters — this is the whole energy story. A neuron whose steady level sits under threshold burns essentially no power: it produces no spikes, so no downstream circuits wake up. That is the sparsity that lets brains (and neuromorphic chips) run on a light-bulb's worth of power, unlike a von Neumann machine clocking every cell whether or not it has work.

PICTURE. Two curves side by side: one ceiling above $V_{th}$ (fires), one below (flat, silent forever).

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Step 8 — Put a number in: does this neuron fire, and how fast?

WHAT. Take $R=10\text{M}\Omega$, $C=1\text{nF}$, $V_{th}=15\text{mV}$, $I=2\text{nA}$.

WHY do the steps in this order. Ceiling first (does it even fire?), then timescale, then the time itself.

1. Ceiling $RI=(10\times 10^6)(2\times 10^{-9})=20\text{mV}$. Since $20>15=V_{th}$ → it fires. ✅ (Step 7 passed.)
2. Timescale ${\tau }_m=RC=(10^7)(10^{-9})=0.01\text{s}=10\text{ms}$.
3. Time-to-spike $T={\tau }_m\ln \frac{RI}{RI-V_{th}}=10\text{ms}\cdot \ln \frac{20}{20-15}=10\text{ms}\cdot \ln 4\approx 13.9\text{ms}$.
4. Rate $f=1/T\approx 72\text{Hz}$.

If we instead drop to $I=1\text{nA}$: $RI=10\text{mV}<15\text{mV}$ → argument negative → never fires (the silent branch of Step 7).

PICTURE. The actual charging curve for these numbers, crossing $15\text{mV}$ at about $13.9\text{ms}$, with the $1\text{nA}$ curve shown staying flat below threshold.

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The one-picture summary

Every idea on this page in a single frame: current in → tank charges along an $e$-curve toward ceiling $RI$ → crosses $V_{th}$ at time $T={\tau }_m\ln \frac{RI}{RI-V_{th}}$ → spike → reset → repeat; and if $RI<V_{th}$, the curve flattens under the line and stays silent forever.

Recall Feynman retelling — say it like a story

A neuron is a leaky bucket. Current is a tap filling it; a hole in the side lets water dribble out. The water level is the voltage. Because of the hole, the level does not rise forever — it climbs fast at first, then eases up to a ceiling where dripping-in exactly matches dripping-out. That ceiling is $RI$. The bucket has a memory span, ${\tau }_m=RC$: turn the tap off and it forgets in a few ${\tau }_m$. Now paint a red line at height $V_{th}$. If the ceiling is above the line, the water crosses it at a moment we can compute — that crossing is a spike, and afterward we dump the bucket and start over. The crossing time uses a logarithm, because we had to un-trap the time from inside an exponent. If the ceiling sits below the red line, the water never touches it: no spike, no energy spent. That silence — a bucket quietly holding still — is the trick that lets a whole ocean of these buckets run on almost no power. See In-memory computing and Hebbian learning for what happens once buckets start talking to buckets.

Recall Quick self-test

Why does an exponential (not a straight line) appear in $V(t)$? ::: Because the leak makes $V$'s rate of change proportional to $V$ itself, and $e$ is the unique function whose rate of change is proportional to itself. Why does the logarithm appear when solving for $T$? ::: The unknown $T$ is stuck inside an exponent; $\ln$ is the tool that undoes $e^{(\cdot )}$ and frees it. When does the neuron never fire? ::: When $RI\le V_{th}$: the ceiling sits at/below threshold, so the curve never crosses it — the $\ln$ argument becomes non-positive. What is ${\tau }_m$ physically? ::: $RC$, the membrane's forgetting time; after one ${\tau }_m$ the voltage has moved 63% of the way to its ceiling.

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See also: GPU vs Neuromorphic accelerators, Memristors and ReRAM (how a synapse physically stores its weight next to the compute).