Worked examples — DRAM refresh and charge leakage

Before we start, one promise: every symbol we use was earned in the parent note. Quick roll-call so you never scroll back —

The scenario matrix

Every question this topic can ask lands in one of these cells. The worked examples below each carry a [Cell X] tag so you can see the whole grid get filled.

Two decay pictures dominate everything — know when each applies:

Example 1 — Retention with a half-threshold [Cell A]

Forecast: guess — will it survive milliseconds, or seconds? (Jot a number before reading.)

1. Compute $\tau =R_{leak}C$. $\tau =(2\times 10^{12})(30\times 10^{-15})=60\times 10^{-3}\text{s}=60\text{ms}$ Why this step? $\tau$ is the decay's natural clock; every retention answer is $\tau$ times some pure number. We need it first. Why does $\Omega \times \text{F}$ come out in seconds? An ohm is volts-per-amp $=\text{V}/(\text{C/s})$, and a farad is coulombs-per-volt $=\text{C}/\text{V}$. Multiply: $\frac{\text{V}\cdot \text{s}}{\text{C}}\times \frac{\text{C}}{\text{V}}=\text{s}$. The volts and coulombs cancel, leaving pure seconds — so $\tau$ is genuinely a time.

2. Recognise the threshold is exactly half. $V_{th}=0.6=V_0/2$. Why this step? When $V_{th}=V_0/2$, the log collapses: $\ln (V_0/V_{th})=\ln 2\approx 0.693$. No calculator gymnastics.

3. Apply the retention formula (the boxed $t_{ret}$ above). $t_{ret}=\tau \ln \frac{V_0}{V_{th}}=60\text{ms}\times \ln 2\approx 60\times 0.693=41.59\text{ms}$ Why this step? This is the exact moment $V(t)$ crosses the red decision line in the left panel of the figure (the curved resistor case).

Verify: Plug back into $V(t)=V_0{e}^{-t/\tau }$: $V(41.59\text{ms})=1.2e^{-41.59/60}=1.2\times e^{-0.693}=1.2\times 0.5=0.6\text{V}$. ✓ Exactly the threshold. Units: ms in, ms out. Since $41.59\text{ms}<64\text{ms}$ standard period, this cell is a borderline weak cell — refresh saves it.

Example 2 — Constant-current drain, half threshold [Cell B]

Forecast: will this be shorter or longer than the 41.6 ms of Ex 1? Guess.

1. Charge to remove. We must lose $\Delta V=V_0-V_{th}=0.6\text{V}$ worth of charge: $\Delta Q=C\Delta V=30\times 10^{-15}\times 0.6=1.8\times 10^{-14}\text{C}$ Why this step? $Q=CV$ converts a voltage drop into the charge the current must carry away.

2. Constant current ⇒ straight-line drain. Time = charge ÷ current: $t=\frac{\Delta Q}{I_{leak}}=\frac{C\Delta V}{I_{leak}}=\frac{1.8\times 10^{-14}}{2\times 10^{-15}}=9\text{s}$ Why this step? With current fixed, $\Delta Q=I\Delta t$ — this is the straight ramp in the figure's right panel, not a curve.

Verify: Units: $\text{C}/\text{A}=\text{C}/(\text{C/s})=\text{s}$. ✓ And note it's way longer than Ex 1's 41.6 ms — because a real resistor drains fast while the voltage is high, whereas constant current is gentle throughout. Same cell, two models, two answers: always check which model the problem states.

Example 3 — A threshold that is NOT half (exponential) [Cell C]

Forecast: because the threshold is higher (closer to $V_0$), do we get more or less retention than the half case?

1. Cannot use $\ln 2$ — the ratio isn't 2. Compute the true ratio: $\frac{V_0}{V_{th}}=\frac{1.0}{0.8}=1.25$ Why this step? The formula $t_{ret}=\tau \ln (V_0/V_{th})$ is general; only the special case $V_{th}=V_0/2$ simplifies to $\ln 2$. Here we must take a real log.

2. Apply the general formula. $t_{ret}=50\text{ms}\times \ln (1.25)=50\times 0.2231=11.16\text{ms}$ Why this step? $\ln 1.25\approx 0.223$, far less than $\ln 2=0.693$.

Verify: Plug into $V(t)=V_0{e}^{-t/\tau }$: $V(11.16\text{ms})=1.0e^{-11.16/50}=e^{-0.2231}=0.8\text{V}$. ✓ And the lesson lands: a stricter (higher) threshold gives you LESS time — you demand the bit stay strong, so it fails sooner. Confirms the forecast: higher $V_{th}$ ⇒ shorter retention.

Example 3b — Constant current AND a non-half threshold [Cell B×C]

Forecast: the linear formula only cares about how many volts we drop, not the ratio — so will the "non-half" twist matter as much as it did in Example 3?

1. Write the general linear-drain law. For constant current the drain is a straight line, so time is simply the charge lost over the current: $t=\frac{C(V_0-V_{th})}{I_{leak}}$ Why this step? Unlike the exponential (which uses the ratio $V_0/V_{th}$ inside a log), the linear case uses the difference $V_0-V_{th}$. Constant current removes charge at a fixed rate, so only the total charge removed matters — and that charge is $C(V_0-V_{th})$ by $Q=CV$.

2. Plug the numbers. The drop is $V_0-V_{th}=1.0-0.8=0.2\text{V}$: $t=\frac{40\times 10^{-15}\times 0.2}{3\times 10^{-15}}=\frac{8\times 10^{-15}}{3\times 10^{-15}}=2.667\text{s}$ Why this step? We're evaluating the straight-ramp crossing (right panel of the figure) for a shallow $0.2\text{V}$ drop instead of a half drop.

Verify: Units: $\frac{\text{F}\cdot \text{V}}{\text{A}}=\frac{\text{C}}{\text{C/s}}=\text{s}$. ✓ Sanity: a smaller voltage window ($0.2\text{V}$ vs a half-drop of $0.5\text{V}$) means less charge to lose, so a shorter time — consistent. And the forecast lands: in the linear case the non-half threshold enters as a simple difference $V_0-V_{th}$, far tamer than the logarithm of a ratio it became in the exponential Example 3. Same twist, different model, different math.

Example 4 — Refresh interval from row count [Cell D]

Forecast: bigger $N$ than the parent's 8192 — will the interval be longer or shorter?

1. Divide the budget evenly. $t_{RI}=\frac{t_{REF}}{N}=\frac{64\text{ms}}{16384}=\frac{64\times 10^{-3}}{16384}\text{s}$ Why this step? Distributed refresh spreads $N$ REFs uniformly over $t_{REF}$; one every $t_{REF}/N$ by definition.

2. Evaluate. $t_{RI}=3.906\times 10^{-6}\text{s}\approx 3.9\mu s$ Why this step? Twice the rows as the parent's example ⇒ half the interval ($7.8\mu s\to 3.9\mu s$).

Verify: Multiply back: $3.906\mu s\times 16384=64000\mu s=64\text{ms}$. ✓ Units: ms ÷ (dimensionless count) = ms. Forecast confirmed: more rows ⇒ shorter interval (REFs fire faster).

Example 5 — Overhead scaling, a real-world twist [Cell E]

Forecast: true or false? Guess before computing.

1. Overhead = busy time ÷ interval. $\text{overhead}=\frac{t_{RFC}}{t_{RI}}=\frac{550\text{ns}}{3906\text{ns}}$ Why this step? Every $t_{RI}$ the bank is unusable for $t_{RFC}$; the fraction of time lost is that ratio.

2. Evaluate. $=0.1408\approx 14.1\%$ Why this step? Both in ns, so ratio is pure. $14\%\gg 5\%$ — the vendor's claim is false for this large chip.

Verify: Sanity: 550 out of every 3906 ns busy → clearly more than a tenth, i.e. >10%. ✓ This is exactly the scaling wall the parent warned about: as chips grow, $t_{RFC}$ rises while $t_{REF}$ stays $64\text{ms}$, so overhead climbs toward 20%. This is why self-refresh and profiling schemes exist.

Example 6 — Hot chip corner [Cell F]

Forecast: if $\tau$ halves, does retention halve too? (Yes/no — think about the formula.)

1. Retention is proportional to $\tau$. $t_{ret}=\tau \ln (V_0/V_{th})$; the log factor is unchanged, only $\tau$ scales. Why this step? Temperature changes leakage, hence $R_{leak}$ and $\tau$ — but not the voltage ratio. So retention scales linearly with $\tau$.

2. Halve it. $t_{ret}^{hot}=\frac{1}{2}\times 80\text{ms}=40\text{ms}$ Why this step? $\tau$ halved ⇒ retention halved. Direct proportionality.

3. Compare to the period. $40\text{ms}<64\text{ms}$ ⇒ UNSAFE at the normal rate. Why this step? The refresh rule demands $t_{REF}<t_{ret}$ of the weakest cell. It now fails.

4. JEDEC fix. Halve the period to $32\text{ms}$ (temperature-compensated self-refresh). Now $32\text{ms}<40\text{ms}$ ⇒ safe again. Why this step? This is precisely why hot DRAM refreshes twice as fast — the "robot walks twice as fast on a hot day."

Verify: Check the fix: $32\text{ms}<40\text{ms}$? ✓ Yes, margin restored. Forecast confirmed: retention does halve when $\tau$ halves.

Example 7 — Degenerate limit: zero leakage [Cell G]

Forecast: does a leak-free cell ever need refresh?

1. Take the limit of $\tau$. $\tau =R_{leak}C\stackrel{R_{leak}\to \infty }{\to }\infty$ Why this step? $\tau$ is proportional to $R_{leak}$; send resistance to infinity and the natural clock stops ticking.

2. Retention explodes. $t_{ret}=\tau \ln (V_0/V_{th})\to \infty$. Why this step? Infinite $\tau$ times any finite log factor is infinite — the voltage in the figure never bends downward; it's a flat line forever.

3. Refresh requirement. $t_{REF}$ can be anything — refresh becomes unnecessary. Why this step? If the bit never decays below $V_{th}$, there is nothing to top up.

Verify: This is the idealisation of SRAM — a static latch actively holds its bit and never needs refresh (at the cost of 6 transistors). ✓ The limit $R_{leak}\to \infty$ smoothly turns "dynamic" into "static". Sanity: flat line never crosses threshold ⇒ infinite retention. ✓

Example 8 — Infinite leakage & the weakest-cell exam twist [Cell H]

Forecast: for part (b), will the datasheet quote 200 ms or 50 ms? Why?

1. (a) Restate the retention law for this cell. Since $V_{th}=V_0/2$, the log is $\ln (V_0/(V_0/2))=\ln 2$, so $t_{ret}=\tau \ln 2$ with $\tau =R_{leak}C$. Why this step? We must anchor which formula applies before taking a limit; the half-threshold assumption is what turns the general log into $\ln 2$.

2. (a) Take $R_{leak}\to 0$. $\tau =R_{leak}C\to 0\Rightarrow t_{ret}=\tau \ln 2\to 0$ Why this step? Zero resistance ⇒ instant drain; the exponential $V(t)=V_0{e}^{-t/\tau }$ collapses vertically. The bit is lost immediately — a dead cell.

3. (a) Consequence. No finite $t_{REF}$ can save it; it's a hard defect (mapped out by redundancy/repair, not refresh). Why this step? Refresh assumes the cell can hold charge between visits. If $t_{ret}=0$, refresh cannot keep up.

4. (b) Datasheet uses the WEAKEST cell. The one 50 ms cell dictates the guarantee. A single period is quoted, and it must be strictly less than the worst retention: $t_{REF}<t_{ret}^{min}=50\text{ms}$ Why this step? The refresh rule is a worst-case guarantee: one lost bit corrupts data, so the whole chip runs at the pace of its slowest cell. A single number is published (not a range) — the standard picks the largest JEDEC-defined period that still sits safely under $50\text{ms}$, which is the existing $32\text{ms}$ grade (the next standard step below $64\text{ms}$). Here $32\text{ms}<50\text{ms}$, so it is safe with margin.

5. (b) Over-refresh fraction. Almost every cell could safely wait $200\text{ms}$ but is forced to the $32\text{ms}$ pace: $\frac{999,999,999}{1,000,000,000}=0.999999999\approx 99.9999999\%\approx 100\%$ Why this step? Over-refresh = fraction of cells that could have waited longer than the quoted period but don't. Only the single weak cell truly needs 32 ms; all others are refreshed far more often than physics demands — pure wasted energy. This exact inefficiency is what retention-profiling (RAIDR) and self-refresh optimisations attack.

Verify: Part (a): $\tau \to 0\Rightarrow t_{ret}\to 0$. ✓ Part (b): the quoted period $32\text{ms}<50\text{ms}$ worst-case (safe with margin), and the over-refresh fraction $=1-10^{-9}\approx 0.999999999$ — essentially all one billion cells. ✓ Forecast confirmed: the datasheet follows the 50 ms weakest cell (choosing the standard $32\text{ms}$ grade under it), never the 200 ms majority.

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