Channel length and short-channel effects: Visual walkthrough

Step 0 — The words we will use (so nothing is a surprise)

Before any picture, let us name the actors in plain English. A MOSFET is a tiny electrical switch built on a slab of silicon.

Everything below is about one question:

Prerequisites worth a peek: MOSFET operation and regions, Threshold voltage and body effect, Depletion region physics of pn junctions.

Step 1 — The depletion region: where the fixed charge lives

WHAT. When you apply voltage to the gate, it repels the mobile carriers directly below it, leaving behind a slab of silicon stripped of moving charge but full of fixed, immobile ions. That slab is the depletion region. (See Depletion region physics of pn junctions.)

WHY. The gate can only make current flow once it has paid for this fixed charge — supported it electrically. The amount of fixed charge it must support is what sets the turn-on voltage. So we must first draw exactly where that charge sits and how deep it goes.

PICTURE. Look at the figure: the yellow slab beneath the gate is the depletion region. Its depth into the bulk is $x_{d m}$ — the maximum vertical depletion depth. Every plus sign in that slab is one fixed ion the gate must cope with.

Figure — Channel length and short-channel effects

Step 2 — The long-channel picture: a clean rectangle

WHAT. In a long transistor the depletion region under the gate is a simple rectangle: length $L$ wide, depth $x_{d m}$ deep. All the charge inside belongs to the gate.

WHY. This is our baseline — the "gate is boss" case. We need it drawn cleanly so we can see later exactly what the short device steals from it.

PICTURE. The whole yellow box is the gate's rectangle. Notice the source and drain junctions at the two ends barely poke in — they carve only a whisker off a very long box, so we can ignore them.

Step 3 — The short-channel truth: two triangles get stolen

WHAT. Shrink $L$ . Now the source and drain junctions each carve out a triangular wedge of the depletion region. Those wedges belong to the junctions, not the gate — the junction fields, not the gate, hold that charge in place.

WHY. Because the source and drain are pn-junctions, they push their own depletion into the bulk. Near the corners, whose charge is nearer the junction than the gate? The junction's. So the true "gate-owned" region is not a rectangle — it is a trapezoid: the rectangle minus two corner triangles.

PICTURE. The yellow trapezoid is what the gate still owns. The two red triangles at the bottom corners are the charge the source and drain have quietly taken over. When $L$ was huge these triangles were invisible; now they eat a real fraction of the box.

Step 4 — Measuring the triangle with plain geometry

WHAT. We find the horizontal base $b$ of each stolen triangle using the source/drain junction depth $x_{j}$ and the gate depth $x_{d m}$ .

WHY. To subtract the triangles we must know their size. The only lengths available are $x_{j}$ (how deep the junction islands go) and $x_{d m}$ (how deep the gate depletion goes). The junction depletion spreads as a quarter-circle-ish arc of radius $\sim x_{j}$ ; where it meets the gate's depth $x_{d m}$ sets the base. Pythagoras on the corner triangle gives:

$b = x_{j} (1 + \frac{2 x _{d m}}{x _{j}} - 1)$

PICTURE. The figure isolates one bottom corner. The junction depth $x_{j}$ runs down; the gate depth $x_{d m}$ runs across; the hypotenuse-like arc slices off the base $b$ marked in green. The square-root term is literally the length of that slanted cut.

Step 5 — From area to charge fraction: the trapezoid ratio

WHAT. The two triangles together have total base $2 b$ , but by symmetry we book them as reducing the effective rectangle length. The fraction of charge left to the gate is:

$\frac{Q _{B}^{'}}{Q _{B}} = 1 - \frac{x _{j}}{L} (1 + \frac{2 x _{d m}}{x _{j}} - 1)$

WHY. Charge in the depletion slab is proportional to area (charge density is uniform). The rectangle's "length" is $L$ ; the two triangles remove a length-equivalent of $b$ each. Divide removed by total → the fraction stolen is $b / L$ . Notice the $1/ L$ : a fixed corner is a bigger slice of a shorter box. That single $1/ L$ is the entire origin of roll-off.

PICTURE. Two panels side by side: a long box (thin red slivers, big yellow trapezoid) and a short box (fat red triangles, shrunken yellow trapezoid). Same triangles, wildly different fractions.

Step 6 — Assembling $Δ V_{T}$ : less charge, less voltage

WHAT. The gate now supports only $Q_{B}^{'}$ instead of $Q_{B}$ . The threshold shift is the voltage cost of the missing charge $Q_{B} - Q_{B}^{'}$ :

$Δ V_{T} = \frac{Q _{B}^{'} - Q _{B}}{C _{o x}} = - \frac{Q _{B}}{C _{o x}} \cdot \frac{x _{j}}{L} (1 + \frac{2 x _{d m}}{x _{j}} - 1)$

WHY. Threshold voltage's charge term is $Q / C_{o x}$ (Step 2). Replace $Q_{B}$ by the smaller $Q_{B}^{'}$ : the difference, converted to voltage by dividing by $C_{o x}$ , is the roll-off. The minus sign says the threshold drops — less charge to support means an easier turn-on.

PICTURE. The bar chart: full $Q_{B}$ bar for the long device, a shorter $Q_{B}^{'}$ bar for the small device, and the red gap between them labelled " $Δ V_{T} \times C_{o x}$ " — the stolen charge, made visible.

Step 7 — Every case: long, short, and the degenerate limits

WHAT. Let us push the formula to its extremes to be sure it never surprises us.

WHY. A formula you trust must behave sensibly at its edges. We check four scenarios.

PICTURE. A single plot of $∣Δ V_{T} ∣$ versus $L$ — flat and near-zero for large $L$ , exploding upward as $L \to$ the corner size.

The one-picture summary

This single frame stacks the whole story: a long device (clean rectangle, gate owns everything, high $V_{T}$ ) beside a short device (fat corner triangles stolen by source/drain, small trapezoid, dropped $V_{T}$ ), with the arrow showing $Δ V_{T} \propto 1/ L$ .

Recall Feynman retelling — say it back in plain words

A transistor turns on when its gate can shove aside a fixed pile of charge sitting in the silicon. In a long transistor the gate owns that whole pile — a neat rectangular block of it — so it takes a certain voltage $V_{T 0}$ to turn on. Now make the transistor short. The source and drain are little charged islands with their own pull, and near the corners they grab two triangular chunks of the pile for themselves. The gate no longer has to pay for those chunks, so it turns on at a lower voltage. The triangles are a fixed size, but in a short channel they are a huge fraction of the whole pile — that's the $1/ L$ . Push it too far and the two triangles touch in the middle: the gate loses control entirely and the switch fails. That whole cartoon — steal two triangles, pay less voltage, fraction grows as $1/ L$ — is exactly the boxed formula.

Recall Check yourself

Why does $Δ V_{T}$ scale as $1/ L$ and not, say, $1/ L^{2}$ ? ::: The stolen charge is a fixed length $b$ removed from a rectangle of length $L$ ; the fraction removed is $b / L$ , giving one power of $1/ L$ . Which depth sets the triangle height, $x_{d m}$ or $x_{d S}$ ? ::: $x_{d m}$ , the vertical gate depletion depth. $x_{d S}$ is a lateral width that sets short-channel onset, not the triangle geometry. What happens to the formula when $x_{d m} \to 0$ ? ::: $1 + 0 - 1 = 0$ , so $Δ V_{T} = 0$ — no depletion means no stolen triangles means no roll-off. Physically, what does the minus sign in $Δ V_{T}$ mean? ::: The threshold drops: with less fixed charge to support, the gate turns the device on at a lower voltage.

See also: Scaling theory (Dennard scaling), Drain-Induced Barrier Lowering (DIBL), Subthreshold conduction and leakage, Velocity saturation and carrier transport, Surface scattering and effective mobility.