Visual walkthrough — DRAM refresh and charge leakage

This page is the picture-companion to the parent note. Every step here is one idea and one figure.

Step 1 — The bit is a puddle of charge on a capacitor

WHAT. A DRAM cell is one tiny capacitor (a charge-storage bucket) plus one transistor (a switch). We store a 1 by filling the bucket with charge, and a 0 by leaving it empty.

WHY start here. Every later symbol ($V$, $Q$, $C$) is born on this picture. If we don't anchor them to the bucket now, the equations later will float free.

Let us define the three words we will use forever:

These three are tied together by the most basic capacitor law (see Capacitance Q=CV): $\underset{\text{charge stored}}{\underbrace{Q}}=\underset{\text{bucket fatness}}{\underbrace{C}}\times \underset{\text{level / pressure}}{\underbrace{V}}$

PICTURE. The bucket, its water level $V$, and the little drip hole that will ruin everything.

Step 2 — Name the leak as a resistor

WHAT. The bucket has a slow drip. In electronics, "charge slowly escaping through something" is modelled as a resistor $R_{leak}$ connecting the storage node to ground ($0\text{V}$).

WHY a resistor? Because a resistor is precisely "the component whose current is proportional to the voltage across it." A wider drip hole = a smaller resistance = a faster leak. This single number $R_{leak}$ bundles together all three physical leak paths (subthreshold, junction, dielectric) into one handle we can do algebra with.

The escaping (leakage) current is therefore: $\underset{\text{leak current}}{\underbrace{I}}=\frac{\stackrel{\text{pressure pushing charge out}}{\overbrace{V}}}{\underset{\text{narrowness of hole}}{\underbrace{R_{leak}}}}$

PICTURE. The same bucket, but now the drip hole is drawn as a resistor symbol, with the current $I$ arrow flowing out toward ground.

Step 3 — Bookkeeping: current out = charge lost per second

WHAT. We now write down the rate at which charge is disappearing. Whatever current leaves through the hole is exactly the charge the bucket loses each second.

WHY we need a rate. We want to know how the level $V$ changes over time. "How fast is something changing" is exactly the job of a derivative — that is why calculus enters here and nowhere earlier.

Charge conservation says the current out equals the drop in stored charge: $I=-\frac{dQ}{dt}=-\frac{d(CV)}{dt}=-C\frac{dV}{dt}$

Term by term:

• $-\frac{dQ}{dt}$ — the rate charge is leaving (minus sign: stored charge is going down).
• $Q=CV$, and $C$ is a fixed constant of the bucket, so we can pull it out: the charge rate is $C$ times the level rate.

PICTURE. Two arrows on the bucket: the current $I$ leaving through the hole, and the level $V$ ticking downward — the same event seen two ways.

Step 4 — Glue the two facts into one equation (an ODE)

WHAT. We have written the leak current two different ways. Step 2 said $I=V/R_{leak}$. Step 3 said $I=-CdV/dt$. They are the same current, so set them equal. The result is an ODE — short for ordinary differential equation, meaning simply "an equation that links a quantity to its own rate of change."

WHY. This is the whole trick of physics: describe one thing from two angles, then equate. What pops out is a rule linking the level to its own rate of change.

$\underset{\text{charge-loss view (Step 3)}}{\underbrace{-C\frac{dV}{dt}}}=\underset{\text{Ohm view (Step 2)}}{\underbrace{\frac{V}{R_{leak}}}}$

Divide both sides by $-C$ to isolate the rate: $\boxed{\frac{dV}{dt}=-\frac{V}{R_{leak}C}}$

Read this out loud: "the speed at which the level drops is proportional to how full the bucket still is." A near-full bucket drips fast; a near-empty one barely dribbles. That self-referential shape — rate of change proportional to current amount — is the fingerprint of exponential decay (the same maths as RC discharge and radioactive decay).

PICTURE. Three bucket snapshots — full, half, low — with the drip-rate arrow shrinking as the level drops.

Step 5 — Solve it: the exponential curve appears

WHAT. We solve the ODE from Step 4 for $V$ as an actual function of time — not by guessing, but by separating the variables and integrating.

WHY this method. Our equation mixes $V$ and $t$ together. The trick is to shove everything about $V$ to one side and everything about $t$ to the other, so each side can be integrated on its own. That is why "separation of variables" is the right tool: it untangles the two quantities.

The algebra, step by step. Start from Step 4 and gather $V$-terms left, $t$-terms right: $\frac{dV}{dt}=-\frac{V}{R_{leak}C}⟹\frac{1}{V}dV=-\frac{1}{R_{leak}C}dt$ Now integrate both sides — the left over $V$, the right over $t$: $\int \frac{1}{V}dV=-\frac{1}{R_{leak}C}\int dt⟹\ln V=-\frac{t}{R_{leak}C}+K$ Here $\ln$ appears because the integral of $1/V$ is the natural logarithm, and $K$ is the integration constant fixed by the start. Raise $e$ to both sides to undo the $\ln$: $V=e^{-t/(R_{leak}C)+K}=\underset{\text{a constant}}{\underbrace{e^K}}\cdot e^{-t/(R_{leak}C)}$ At $t=0$ the exponential is $1$, so the leading constant $e^K$ must equal the starting level $V_0$. Substituting gives the boxed result.

Define the combined constant that keeps appearing:

The solution: $V(t)=\underset{\text{starting level}}{\underbrace{V_0}}\underset{\text{decay factor, 1 at }t=0,\text{shrinking}}{\underbrace{e^{-t/\tau }}}$

Sanity checks (always test the ends):

• At $t=0$: $e^0=1$, so $V=V_0$. Correct — the bucket starts full.
• As $t\to \infty$: $e^{-\infty }=0$, so $V\to 0$. Correct — a fully-drained bucket, a lost bit.

PICTURE. The falling exponential curve with $V_0$, the $1/e$ point at $t=\tau$, and the tail approaching zero.

Step 6 — The deadline: retention time and the ½-full line

WHAT. The bit is not lost when the bucket empties — it is lost when the level sinks below the line the reader uses to tell 1 from 0. That line is the sense threshold $V_{th}$, read by the sense amplifier, and it usually sits at half the full level, $V_{th}=V_0/2$.

WHY half? The sense amp compares the cell against a reference sitting midway between a full 1 and an empty 0. Cross that midline and the amp guesses the wrong bit.

Set $V(t)=V_{th}$ and solve for the time — call it the retention time $t_{ret}$: $V_0{e}^{-t_{ret}/\tau }=V_{th}\Rightarrow e^{-t_{ret}/\tau }=\frac{V_{th}}{V_0}$

To pull $t_{ret}$ out of the exponent we need the tool that undoes $e$: the natural logarithm $\ln$. ($\ln$ answers the question "$e$ to what power gives this number?") Taking $\ln$ of both sides and tidying signs: $\boxed{t_{ret}=\tau \ln \frac{V_0}{V_{th}}}$

With the standard $V_{th}=V_0/2$, the ratio is $2$, and $\ln 2\approx 0.693$: $t_{ret}=\tau \ln 2\approx 0.693\tau$

PICTURE. The decay curve again, now with the dashed $V_0/2$ threshold line and the retention deadline $t_{ret}$ marked where the curve crosses it.

Step 7 — The rule that follows: refresh before the deadline

WHAT. Refresh = re-fill the bucket before it crosses $V_{th}$. You must revisit every cell within one refresh period $t_{REF}$, and you must pick $t_{REF}<t_{ret}\text{of the weakest cell on the whole chip}.$

WHY the weakest cell. Different buckets have different-sized drip holes (manufacturing variation). One chip-wide period must satisfy the leakiest one. That is why the JEDEC number is $64\text{ms}$ even though a typical cell could hold for seconds.

Edge case — temperature. Heat widens every drip hole (leakage rises roughly like $e^{-E_a/kT}$), so $R_{leak}$ falls, $\tau$ falls, and $t_{ret}$ falls. Above about $85^{\circ }\text{C}$ the spec halves to $32\text{ms}$ — the robot walks twice as fast (see Low-power DRAM self-refresh). This connects to the DDR timing $t_{REFI}$.

PICTURE. Two decay curves (cool = slow, hot = fast) sharing the $V_0/2$ line, showing the hot cell hits the deadline sooner, forcing the $32\text{ms}$ walk.

The one-picture summary

Everything above, compressed: bucket → resistor leak → the ODE → the exponential → the threshold crossing → the refresh deadline.

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