MOSFET as a switch: Visual walkthrough — Hardware deep dive

We use the N-channel enhancement MOSFET — the everyday switch (see Enhancement vs Depletion MOSFET).

Step 0 — The three knobs and the picture we will keep redrawing

WHAT. Before any formula, meet the object. A MOSFET has three terminals we care about:

Gate (G) — a metal plate sitting on top of a thin glass layer (the oxide). It touches nothing electrically; it just pushes on the charge below.
Source (S) — where electrons enter. We tie it to ground (0 V) in a low-side switch.
Drain (D) — where electrons leave, toward the load.

WHY meet it first. Every equation below is a statement about the sheet of charge under the gate. If you can see that sheet, the algebra is just bookkeeping.

PICTURE.

Step 1 — Why a gate voltage builds a channel (the capacitor idea)

WHAT. The gate + oxide + silicon below form a tiny capacitor: two conductors separated by an insulator. Put a positive voltage $V_{GS}$ on the gate and it pulls negative charge (electrons) up to the surface of the silicon. Those gathered electrons are the channel — a thin conducting bridge from source to drain.

WHY a capacitor, not something exotic? Because a capacitor is exactly "voltage on one plate → charge on the other." That is the only tool that answers our question "how much channel charge does a gate voltage summon?" Nothing simpler will do, nothing more complex is needed.

WHY the threshold. Some of that summoned charge first has to cancel the silicon's natural tendency to repel electrons. Only the part of $V_{GS}$ above $V_{T H}$ actually piles up as useful channel charge. That leftover is so important it gets a name:

PICTURE.

The charge per unit area gathered is proportional to $V_{o v}$ : $Q_{c h} \propto C_{o x} (V_{GS} - V_{T H})$

$C_{o x}$ ::: oxide capacitance per unit area — thinner glass = bigger $C_{o x}$ = more charge per volt.
$(V_{GS} - V_{T H})$ ::: the surviving push $V_{o v}$ ; double it and you double the charge.

Step 2 — Turning "gathered charge" into "current" (why current is charge × speed × width)

WHAT. A channel full of charge does nothing until we pull it sideways. Apply a small $V_{D S}$ : it creates a gentle slope of voltage from drain to source, and electrons drift down that slope. The current is:

$I_{D} = (charge per length) \times (drift speed)$

WHY this decomposition. Current is "how much charge crosses a line each second." That is charge-density times how fast it moves — there is no other honest way to count it. This is the same idea as counting cars on a highway: cars-per-mile times miles-per-hour.

PICTURE.

Two geometry facts enter here, and we name them now so they never appear un-introduced:

$W$ = channel width (how many lanes on the highway). More width → more current.
$L$ = channel length (how long the trip). Longer trip → slower net flow → less current.
$μ_{n}$ = electron mobility — how easily electrons drift per unit of push. It converts the sideways voltage slope into a drift speed.

Bundling the material and shape constants together gives the switch's "gain factor":

Step 3 — Assembling the triode current equation (and reading its two pieces)

WHAT. Combining Step 1 (charge $\propto V_{o v}$ ) and Step 2 (current = charge × drift) — and being careful that the channel is thinner near the drain because the drain end already feels part of $V_{D S}$ — the full triode-region current is:

$I_{D} = k [(V_{GS} - V_{T H}) V_{D S} - \frac{V _{D S}^{2}}{2}]$

WHY the two terms. Read them like two forces:

$"charge summoned \times sideways pull" (V_{GS} - V_{T H}) V_{D S}$ — the honest driving term: more overdrive and more drain pull both raise current.
$"channel thins at the drain end" - \frac{V _{D S}^{2}}{2}$ — a penalty. As $V_{D S}$ grows, the drain end of the channel loses charge (its local $V_{GS}$ shrinks), so the channel narrows there. This term subtracts that loss.

This lives in the triode region — see Triode vs Saturation regions for the neighbouring region we deliberately avoid.

PICTURE.

The figure plots $I_{D}$ against $V_{D S}$ : near the origin it rises almost like a straight line, then bends over as the $- V_{D S}^{2} /2$ penalty bites. A switch lives on that straight starting stretch.

Step 4 — The small- $V_{D S}$ approximation (why we throw one term away)

WHAT. A closed switch should drop only millivolts across itself. So $V_{D S}$ is tiny. Compare the two terms when $V_{D S}$ is, say, 50 mV and $V_{o v} = 3$ V:

linear term $\sim 3 \times 0.05 = 0.15$
penalty term $\sim 0.0 5^{2} /2 = 0.00125$ — over 100× smaller.

So we drop the penalty:

$I_{D} \approx k (V_{GS} - V_{T H}) V_{D S}$

WHY allowed. This is not laziness — it is a definition of what "good switch" means. If $V_{D S}$ were big enough to matter, the switch would be dropping serious voltage and wasting power; we would call that a bad switch. In the regime we care about, the quadratic is genuinely negligible.

WHAT IT LOOKS LIKE. We are zooming into the tiny straight sliver right at the origin of the Step 4 curve.

PICTURE.

Look: $I_{D}$ is now a straight line through the origin versus $V_{D S}$ . A straight line through the origin between voltage and current is exactly Ohm's law $V = I R$ . That is the doorway to the next step.

Step 5 — Reading off $R_{D S (o n)}$ (a slope becomes a resistance)

WHAT. Ohm's law says resistance is voltage divided by current, $R = V / I$ . Divide our approximate equation:

$R_{D S (o n)} = \frac{V _{D S}}{I _{D}} = \frac{V _{D S}}{k ( V _{GS} - V _{T H} ) V _{D S}} = \frac{1}{k ( V _{GS} - V _{T H} )}$

The $V_{D S}$ cancels top and bottom — which is the whole point: a real resistor's value must not depend on the voltage across it, and here it doesn't.

WHY this is the flagship result. It converts a knob you control ( $V_{GS}$ ) into the one number that decides how good the switch is (R_DS(on) and switching losses).

PICTURE.

Two overdrive values give two straight lines of different slope. The steeper line (more overdrive) has the smaller $R_{D S (o n)}$ — because $R$ is $1/ slope$ here.

Step 6 — What the closed switch costs: conduction power

WHAT. A closed switch is a resistor carrying the load current $I_{D}$ . Any resistor carrying current heats up by $P = I^{2} R$ :

$P_{co n d} = I_{D}^{2} R_{D S (o n)}$

$I_{D}^{2}$ ::: the load current, squared — doubling the current quadruples the heat.
$R_{D S (o n)}$ ::: our Step-5 resistance — the smaller, the cooler.

WHY $I^{2} R$ and not $V I$ or $V^{2} / R$ . All three are the same power, but here we know the current (set by the load) and we know $R_{D S (o n)}$ (Step 5), while $V_{D S}$ is a tiny derived quantity. Choosing $I^{2} R$ uses the two things we actually hold in our hands.

PICTURE.

Step 7 — The whole switch in a circuit + every edge case

WHAT. Put a load resistor $R_{L}$ from the supply $V_{D D}$ down to the drain, source grounded — a low-side switch (Low-side vs High-side switching). The drain node is the output. $R_{L}$ and $R_{D S (o n)}$ form a voltage divider:

$V_{o u t} = V_{D D} \frac{R _{D S (o n)}}{R _{D S (o n)} + R _{L}}$

The fraction is "the switch's share of the total resistance."

WHY a divider. Two resistors in series across $V_{D D}$ split the voltage in proportion to their sizes — the most basic circuit law. The output sits at the middle node.

Now walk EVERY case — the reader must never meet a surprise:

Case	$V_{GS}$ vs $V_{T H}$	$R_{D S (o n)}$	$V_{o u t}$	State
OFF (cut-off)	$V_{GS} < V_{T H}$	$\to \infty$	$\approx V_{D D}$ (HIGH)	OPEN
Barely on	$V_{GS} ≳ V_{T H}$	huge, hot	messy, mid-rail	bad
Fully driven	$V_{GS} ≫ V_{T H}$	tiny	$\approx 0$ (LOW)	CLOSED
$V_{GS} = 0$ exactly	equals $- V_{T H} < 0$	$\infty$	$V_{D D}$	OPEN

PICTURE.

The one-picture summary

One flow, left to right: gate voltage → overdrive → channel charge → straight-line current → resistance → heat → logic level. Every arrow is one of the steps above.

Recall Feynman: retell the whole walkthrough in plain words

Think of the gate as a hand pressing a soft hose against the ground. Pressing does nothing until you press past a certain firmness — that firmness is the threshold. Everything you press beyond it is the "overdrive," and it decides how much water (charge) you gather in the pipe. Now tilt the hose slightly (a tiny drain voltage) so water drifts through: the amount flowing is just "how much water is in there" times "how fast it drifts." Because your tilt is tiny, flow rises in a perfectly straight line with tilt — and a straight line between push and flow is a resistance. Divide push by flow and, magically, the tilt cancels: the resistance depends only on how hard you're pressing (overdrive) and the pipe's shape. Press harder, the pipe opens wider, resistance drops. But even a wide-open pipe has a little friction, and pushing current through friction makes heat — that's the cost you pay to keep the switch closed. Finally, wire this pipe below a faucet: closed pipe drains the pressure to zero (output LOW), open pipe leaves full pressure up top (output HIGH). That last flip is why one MOSFET is already half a logic gate.

Recall Active recall — cover the answers

Which piece of $V_{GS}$ actually makes channel? ::: the overdrive $V_{o v} = V_{GS} - V_{T H}$
Why can we drop $V_{D S}^{2} /2$ in a closed switch? ::: $V_{D S}$ is tiny, so it is ~100× smaller than the linear term
How does $R_{D S (o n)}$ come out of the current equation? ::: divide $V_{D S} / I_{D}$ ; the $V_{D S}$ cancels, leaving $1/ [k (V_{GS} - V_{T H})]$
What is the closed-switch heat cost? ::: $P = I_{D}^{2} R_{D S (o n)}$
In a low-side switch, OFF gives what output? ::: $\approx V_{D D}$ (HIGH)
Why can't gate-to- $V_{D D}$ turn on a high-side N-channel? ::: source rises to $V_{D D}$ , so $V_{GS} \to 0 < V_{T H}$

Connections

2.4.14 MOSFET as a switch (Hinglish)
Enhancement vs Depletion MOSFET
Triode vs Saturation regions
Threshold voltage VTH
R_DS(on) and switching losses
Low-side vs High-side switching
Logic gates from MOSFETs (CMOS)
BJT as a switch