Worked examples — Emerging memories (MRAM, ReRAM, PCM)

This page is the calculator's dojo for the parent topic on emerging memories. The parent explained why MRAM, ReRAM and PCM work. Here we grind through every kind of number these devices can hand you — every sign of TMR, every resistance ratio, the degenerate cases where a formula almost breaks, the limits, and a couple of exam-style traps.

Before line one: three little tools reappear on every page. If any of these feels shaky, open the linked note first.

---

The scenario matrix

Every numerical question this topic can throw at you falls into one of these case classes. The whole point of this page is that after reading it, you will have seen each cell solved at least once.

#	Case class	What makes it tricky	Covered by
A	TMR — normal, both resistances given	plug-and-chug baseline	Ex 1
B	TMR — degenerate: $R_{AP}=R_P$	window collapses, TMR $=0$	Ex 2
C	TMR — inverse / sign flip: $R_{AP}<R_P$	TMR goes negative, is that legal?	Ex 2
D	TMR — reverse solve: given TMR, find a resistance	algebra, not just substitution	Ex 3
E	Read divider — pick $R_L$ to maximise the gap	optimisation / limiting behaviour	Ex 4
F	Read divider — limit cases $R_{cell}\to 0$ and $R_{cell}\to \infty$	what does the amp see at extremes	Ex 4
G	PCM RESET — solve for current from energy	square-root, unit gymnastics	Ex 5
H	PCM SET vs RESET — compare two operations	which is hotter, by how much	Ex 6
I	Real-world word problem — array read power	scale one cell up to millions	Ex 7
J	Exam twist — the "which is the 1?" trap + endurance budget	reading past the convention	Ex 8

---

Worked examples

Ex 1 — Cell A: the baseline TMR

1. Write the definition. $\text{TMR}=\frac{R_{AP}-R_P}{R_P}$ Why this step? TMR is defined as the relative change, normalised by the low state. We never invent a new formula; we use the one the parent earned.
2. Substitute (keep units matched — both in kΩ, so they cancel). $\text{TMR}=\frac{4.5-1.5}{1.5}=\frac{3.0}{1.5}=2.0$ Why this step? The subtraction gives the gap; dividing by $R_P$ turns raw ohms into a dimensionless ratio a designer can compare across devices.
3. Convert to percent. $2.0=200\%$ Why this step? "$\times 100$" is just the convention for reporting a dimensionless ratio as a percent.

Verify: $R_{AP}=R_P(1+\text{TMR})=1.5\times (1+2.0)=4.5\text{ k}\Omega$. ✓ The trap in the forecast: $3\times$ the resistance is $200\%$, not $300\%$ — because TMR measures the extra on top of $R_P$, not the total.

---

Ex 2 — Cells B & C: the degenerate and the sign-flip

1. Device B — substitute. ${\text{TMR}}_B=\frac{3-3}{3}=\frac{0}{3}=0=0\%$ Why this step? When the two magnetic states have identical resistance, the numerator (the gap) is exactly zero. This is the degenerate case: the memory window has collapsed. A sense amp cannot tell "0" from "1" — the device is dead as a memory.
2. Device C — substitute (mind the sign of the numerator). ${\text{TMR}}_C=\frac{2-3}{3}=\frac{-1}{3}=-0.333=-33.3\%$ Why this step? The subtraction is honest: if $R_{AP}<R_P$ the numerator is negative, so TMR is negative. This is called inverse TMR and it is physically real (it appears in certain junction stacks). The formula does not break — a negative sign just means "the state we labelled AP happens to be the low-resistance one."
3. Interpret the sign. Why this step? Sign carries meaning: positive TMR ⇒ AP is the high state (the usual case). Negative TMR ⇒ AP is the low state. Zero ⇒ no window at all.

Verify: Rebuild each. B: $R_P(1+0)=3\text{ k}\Omega =R_{AP}$ ✓. C: $R_P(1+(-1/3))=3\times \frac{2}{3}=2\text{ k}\Omega =R_{AP}$ ✓. The magnitude $|-33.3\%|$ is a real (if small) read margin; the sign just tells the amp which way round the states sit.

---

Ex 3 — Cell D: reverse-solve for a resistance

1. Start from the definition and rearrange for $R_{AP}$. $\text{TMR}=\frac{R_{AP}-R_P}{R_P}\Rightarrow R_{AP}=R_P(1+\text{TMR})$ Why this step? Multiply both sides by $R_P$, then add $R_P$. We're inverting the formula because the unknown moved from the output slot to inside.
2. Convert the percent to a plain ratio first. $150\%=1.5$ Why this step? Percent must become a pure number before it enters the arithmetic, or you'll be off by $100\times$.
3. Substitute. $R_{AP}=2\times (1+1.5)=2\times 2.5=5\text{ k}\Omega$

Verify: plug back into Ex 1's definition: $\frac{5-2}{2}=\frac{3}{2}=1.5=150\%$ ✓. Sanity: bigger than $R_P$, as any positive TMR demands.

---

Ex 4 — Cells E & F: choosing $R_L$, and the two limits

The read divider is Voltage Divider: $V_{sense}=V_{read}\cdot \frac{R_L}{R_L+R_{cell}}$. See the figure — the sensed node sits between the two resistors, and the amp watches that node.

1. LRS reading. $V_{sense}^{LRS}=0.3\cdot \frac{10}{10+1}=0.3\cdot \frac{10}{11}=0.2727\text{ V}$ Why this step? Low cell resistance ⇒ most of the read voltage lands across $R_L$ ⇒ high sensed voltage.
2. HRS reading. $V_{sense}^{HRS}=0.3\cdot \frac{10}{10+100}=0.3\cdot \frac{10}{110}=0.02727\text{ V}$ Why this step? High cell resistance hogs the voltage, leaving little for $R_L$ ⇒ low sensed voltage.
3. The gap the amp must resolve. $\Delta V=0.2727-0.02727=0.2455\text{ V}$ Why this step? The gap is the read margin. A quarter-volt separation is enormous — trivially sensed. This is why $R_L$ is chosen between the two states, so neither reading is pinned near $0$ or near $V_{read}$.
4. Limit F1 — a perfect short, $R_{cell}\to 0$. $V_{sense}\to 0.3\cdot \frac{10}{10+0}=0.3\text{ V}=V_{read}$ Why this step? If the cell had zero resistance, the whole read voltage falls across $R_L$. The sensed node saturates at $V_{read}$ — the absolute ceiling.
5. Limit F2 — a perfect open, $R_{cell}\to \infty$. $V_{sense}\to 0.3\cdot \frac{10}{10+\infty }\to 0\text{ V}$ Why this step? An infinite cell resistance blocks all current, so $R_L$ carries no voltage. The absolute floor is $0$. Every real reading lives strictly between these two rails — which is exactly what the figure shows.

Verify: LRS $0.2727$ V and HRS $0.02727$ V both lie inside $(0,0.3)$ V ✓. Gap $0.2455$ V $>0$ ✓. The higher reading is LRS (bit "1"), matching intuition: low resistance → high sensed voltage.

---

Ex 5 — Cell G: PCM RESET, solve for current

1. Energy needed to heat the volume. $E=C_{th}\Delta T=1.2\text{ fJ/K}\times 600\text{ K}=720\text{ fJ}$ Why this step? Temperature rise = energy in ÷ heat capacity, so energy in $=C_{th}\Delta T$. The kelvins cancel, leaving joules (here, femtojoules).
2. Set electrical energy equal to that heat. $I^{2}Rt=E\Rightarrow I^2=\frac{E}{Rt}$ Why this step? Joule Heating says the electrical energy dumped in the pulse is $I^{2}Rt$. All of it (idealised) goes into the tiny volume, so we equate.
3. Substitute in SI units — this is where care pays off. $I^2=\frac{720\times 10^{-15}\text{ J}}{(8\times 10^3\Omega )(40\times 10^{-9}\text{ s})}=\frac{720\times 10^{-15}}{3.2\times 10^{-4}}=2.25\times 10^{-9}{\text{ A}}^2$ Why this step? fJ $\to 10^{-15}$ J, kΩ $\to 10^3\Omega$, ns $\to 10^{-9}$ s. Mixing prefixes here is the #1 error; go fully to SI.
4. Take the square root. $I=\sqrt{2.25\times 10^{-9}}=4.74\times 10^{-5}\text{ A}\approx 47.4\mu \text{A}$

Verify: feed it back — $I^{2}Rt=(4.74\times 10^{-5}{)}^2\times 8000\times 40\times 10^{-9}=2.25\times 10^{-9}\times 3.2\times 10^{-4}=7.2\times 10^{-13}\text{ J}=720$ fJ ✓, exactly the energy we demanded. Units: ${\text{A}}^2\cdot \Omega \cdot \text{s}={\text{A}}^2\cdot \frac{\text{V}}{\text{A}}\cdot \text{s}=\text{V⋅A⋅s}=\text{J}$ ✓. Answer is microamps, not milliamps — the forecast trap.

---

Ex 6 — Cell H: SET vs RESET, side by side

1. Express each current from the same energy balance. $I=\sqrt{\frac{C_{th}\Delta T}{Rt}}\Rightarrow I\propto \sqrt{\Delta T}$ Why this step? With $C_{th},R,t$ fixed, only $\Delta T$ changes. The current sits under a square root, so it scales as $\sqrt{\Delta T}$, not linearly.
2. Take the ratio — everything fixed cancels. $\frac{I_{RESET}}{I_{SET}}=\sqrt{\frac{\Delta T_{RESET}}{\Delta T_{SET}}}=\sqrt{\frac{600}{250}}=\sqrt{2.4}=1.549$ Why this step? Ratios kill the shared constants, so we never even need $C_{th}$. The temperature ratio was $2.4$; the current ratio is its square root.

Verify: using Ex 5's $I_{RESET}=47.4\mu$A, predict $I_{SET}=47.4/1.549=30.6\mu$A. Check directly: $I_{SET}=\sqrt{\frac{1.2\text{e-}15\times 250}{8000\times 40\text{e-}9}}=\sqrt{\frac{3.0\text{e-}13}{3.2\text{e-}4}}=\sqrt{9.375\text{e-}10}=3.06\times 10^{-5}$ A ✓. So RESET draws only $\approx 1.55\times$ the SET current — not $2.4\times$ — because energy grows with the square of current. This is the quantitative version of the parent's "RESET is the hot, power-hungry one."

---

Ex 7 — Cell I: real-world array read power

1. Current through one LRS cell (ignore the load for a worst-case cell current). $I_{cell}=\frac{V_{read}}{R_{LRS}}=\frac{0.2}{2000}=1.0\times 10^{-4}\text{ A}=100\mu \text{A}$ Why this step? Straight Ohms Law: current = voltage ÷ resistance for one cell.
2. Scale to the whole row (currents in parallel add). $I_{row}=N\cdot I_{cell}=1024\times 100\mu \text{A}=1.024\times 10^{-1}\text{ A}\approx 102\text{ mA}$ Why this step? Parallel branches each draw their own current from the same driver; total current is the sum. This is why "sneak" and total-current limits dominate crossbar design.
3. Total read power. $P_{row}=V_{read}\cdot I_{row}=0.2\times 0.1024=0.02048\text{ W}\approx 20.5\text{ mW}$ Why this step? Power $=VI$. Notice: one cell is only $20\mu$W, but $1024$ of them add to $20$ mW — the array is where the budget bites, a key argument for In-Memory Computing where you want to read many cells at once but must respect this current.

Verify: $P_{row}=N\cdot V_{read}^2/R=1024\times 0.04/2000=1024\times 2\times 10^{-5}=0.02048$ W ✓ (two routes agree). Answer is ~100 mA / ~20 mW — hundreds of mA, not amps.

---

Ex 8 — Cell J: the exam twist (convention + endurance budget)

1. (a) Read the convention, don't assume. Bit "1" = high resistance ⇒ $R_B=9\text{ k}\Omega$ is bit "1", $R_A$ is bit "0." Why this step? The parent's mistake note warns: high-R $=$ "1" is not universal — it's whatever the designer declares. Here they declared it, so we obey the statement, not habit.
2. (b) Window as a resistance ratio (window $=R_{high}/R_{low}$). $\text{window}=\frac{R_B}{R_A}=\frac{9}{1.2}=7.5$ Why this step? The memory window is the factor separating the two states; $7.5\times$ is comfortable for sensing. (As a TMR-style figure it would be $\frac{9-1.2}{1.2}=6.5=650\%$.)
3. (c) Endurance lifetime — divide writes budget by write rate. $t_{sec}=\frac{10^9\text{ writes}}{50\text{ writes/s}}=2\times 10^7\text{ s}$ Why this step? Endurance is a total-writes budget. Lifetime = budget ÷ how fast you spend it.
4. Convert seconds to years. $t_{yr}=\frac{2\times 10^7}{3.1536\times 10^7\text{ s/yr}}=0.634\text{ yr}\approx 7.6\text{ months}$ Why this step? One year $=3.1536\times 10^7$ s. The scary result: hammered at $50$ writes/s, a $10^9$-endurance cell dies in under a year — which is exactly why endurance and wear-levelling matter, and why MRAM's $\sim 10^{15}$ is prized.

Verify: $10^9/50=2\times 10^7$ s ✓; $2\times 10^7/3.1536\times 10^7=0.6342$ yr ✓. Ratio check (b): $R_A\times 7.5=1.2\times 7.5=9=R_B$ ✓. The two traps: "1" is $R_B$ (not $R_A$ by any default), and the lifetime is months, not centuries.

---

Recall Rapid self-test (reveal after guessing)

$R_P=2\text{ k}\Omega ,R_{AP}=6\text{ k}\Omega$: TMR? ::: $\frac{6-2}{2}=200\%$ $R_{AP}=R_P$: TMR? ::: $0\%$ — dead cell, window collapsed $R_{AP}<R_P$: is TMR legal? ::: yes, it's negative (inverse TMR), still a real margin As $R_{cell}\to \infty$, $V_{sense}\to$? ::: $0$ V (the floor) As $R_{cell}\to 0$, $V_{sense}\to$? ::: $V_{read}$ (the ceiling) PCM current scales with $\Delta T$ how? ::: as $\sqrt{\Delta T}$ (energy $\propto I^2$) Which is hotter, SET or RESET? ::: RESET (melt), by $\sqrt{600/250}\approx 1.55\times$ the current here Is high-R always bit "1"? ::: no — it's a designer's convention