MOSFET as a switch: Exercises — Hardware deep dive

The only formulas you need, all earned in the parent note:

Definition The three regions — one place to look them up

Every exercise below classifies the MOSFET into exactly one of three operating regions. Here are the complete, collated criteria (N-channel enhancement) so you never have to guess from context:

Cut-off (switch OPEN): $V_{GS} < V_{T H}$ . No channel exists, so $I_{D} \approx 0$ regardless of $V_{D S}$ .
Triode / ohmic (switch CLOSED): $V_{GS} > V_{T H}$ and $V_{D S} < V_{o v}$ (where $V_{o v} = V_{GS} - V_{T H}$ ). Channel exists and is continuous drain-to-source; the device behaves like the resistor $R_{D S (o n)}$ . This is where we drive a switch.
Saturation / active (constant current): $V_{GS} > V_{T H}$ and $V_{D S} \geq V_{o v}$ . Channel is pinched near the drain; $I_{D}$ is nearly independent of $V_{D S}$ . Good for amplifiers, bad for a switch (drops too much voltage).

Quick test order: (1) Is $V_{GS} > V_{T H}$ ? No ⇒ cut-off. Yes ⇒ (2) compare $V_{D S}$ to $V_{o v}$ : below ⇒ triode, at-or-above ⇒ saturation.

Level 1 — Recognition

Exercise 1.1

A device has $V_{T H} = 2.5 V$ . The gate is driven to $V_{GS} = 1.8 V$ . Source is grounded. Is the switch OPEN or CLOSED, and which region is it in?

Recall Solution 1.1

What decides it? Only $V_{GS}$ versus $V_{T H}$ — never the drain voltage. Why this test and not, say, the drain current? The channel is created by the vertical field from gate to source; that field is set by $V_{GS}$ . Until it exceeds $V_{T H}$ there is literally no conducting layer for current to use. $V_{GS} = 1.8 V < V_{T H} = 2.5 V$ , so no channel forms. ⇒ Cut-off region → switch OPEN. The overdrive is $V_{o v} = 1.8 - 2.5 = - 0.7 V$ (negative ⇒ below threshold ⇒ confirms OFF).

Exercise 1.2

Same device, $V_{T H} = 2.5 V$ , now $V_{GS} = 6 V$ , and the drain drops only $0.08 V$ across itself when carrying current. State OPEN/CLOSED and region.

Recall Solution 1.2

$V_{GS} = 6 V ≫ V_{T H} = 2.5 V$ ⇒ channel exists ⇒ switch CLOSED. Why "triode" and not "saturation"? By the region table above, triode needs $V_{D S} < V_{o v}$ . Here $V_{D S} = 0.08 V$ is far below $V_{o v} = V_{GS} - V_{T H} = 3.5 V$ , so the channel is uniform and resistor-like — exactly where we want a closed switch. Overdrive $= 6 - 2.5 = 3.5 V$ .

Exercise 1.3

The same MOSFET is OFF ( $V_{GS} = 0$ ). A circuit now forces the drain $1.5 V$ below the source (reverse polarity). Is the drain-to-source path truly open, or does current flow anyway? By what path?

Recall Solution 1.3

Why the channel test is not the whole story here. With $V_{GS} = 0$ there is no channel, so the conductive channel is indeed open. But the intrinsic body diode points source→drain (cathode at drain). "Drain $1.5 V$ below source" forward-biases that diode by $1.5 V > 0.7 V$ . ⇒ Current flows backward through the body diode, not through a channel. The MOSFET is not an ideal open switch for reverse voltage. This is why H-bridges and buck converters rely on (or must tame) the body diode — see Low-side vs High-side switching.

Level 2 — Application

Use the low-side topology in the schematic below for Exercises 2.2–2.3: $V_{D D}$ feeds the load $R_{L}$ , whose bottom connects to the MOSFET drain; the source is grounded; the drain node is the output.

Figure s02 — low-side N-channel switch: $V_{D D}$ through load $R_{L}$ into the drain; the drain node is $V_{o u t}$ ; source grounded.

Exercise 2.1

$k = 1.5 A/V^{2}$ , $V_{T H} = 2 V$ , gate driven to $V_{GS} = 5 V$ . Compute $R_{D S (o n)}$ .

Recall Solution 2.1

Why this formula applies. With the gate driven well above threshold and $V_{D S}$ small, the quadratic term in the triode current vanishes and $I_{D} \approx k V_{o v} V_{D S}$ — a straight line through the origin, i.e. a resistor. Its slope's inverse is $R_{D S (o n)} = \frac{1}{k ( V _{GS} - V _{T H} )}$ . Overdrive $= 5 - 2 = 3 V$ . $R_{D S (o n)} = \frac{1}{1.5 \times 3} = \frac{1}{4.5} \approx 0.2222 Ω.$

Exercise 2.2

Using the $R_{D S (o n)}$ from 2.1, the switch carries $I_{D} = 2.5 A$ . Find the conduction power loss and the voltage dropped across the closed switch.

Recall Solution 2.2

Why $P = I_{D}^{2} R_{D S (o n)}$ here. A closed switch is just a resistor (Exercise 2.1). Any current through a resistor dissipates Joule heat $P = I^{2} R$ ; there is no other loss mechanism while conducting steadily, so this single term is the conduction loss. $P_{co n d} = I_{D}^{2} R_{D S (o n)} = 2. 5^{2} \times 0.2222 = 6.25 \times 0.2222 \approx 1.389 W .$ Why Ohm's law gives $V_{D S}$ . Same reason — the closed switch is a resistor, so the voltage it drops is $I \times R$ : $V_{D S} = I_{D} R_{D S (o n)} = 2.5 \times 0.2222 \approx 0.556 V .$ Note $V_{D S} = 0.556 V$ is indeed $≪ V_{o v} = 3 V$ , so the triode (resistor) assumption we used is self-consistent — had $V_{D S}$ come out comparable to $V_{o v}$ , we would have had to revisit the region.

Exercise 2.3

Low-side switch: $V_{D D} = 9 V$ , $R_{L} = 220 Ω$ , $R_{D S (o n)} = 0.15 Ω$ . Find $V_{o u t}$ when ON and when OFF.

Recall Solution 2.3

Why a voltage divider. When ON, $R_{L}$ and $R_{D S (o n)}$ sit in series across $V_{D D}$ (look at the schematic: $V_{D D} \to R_{L} \to$ drain node $\to R_{D S (o n)} \to$ ground). The drain node is the junction of two series resistors, and series resistors split the supply in proportion to their values — that is exactly the divider formula. $V_{o u t} = 9 \cdot \frac{0.15}{0.15 + 220} = 9 \cdot \frac{0.15}{220.15} \approx 0.00613 V \approx 0 (LOW) .$ OFF (channel gone, $R_{D S (o n)} \to \infty$ ): with an open channel no current flows, and no current through $R_{L}$ means no voltage drop across it, so the drain node floats up to the supply: $V_{o u t} \approx V_{D D} = 9 V (HIGH) .$ The topology inverts the logic — see Logic gates from MOSFETs (CMOS).

Level 3 — Analysis

Exercise 3.1

Same $k = 2 A/V^{2}$ , $V_{T H} = 2 V$ device. Compare $R_{D S (o n)}$ at $V_{GS} = 3 V$ versus $V_{GS} = 10 V$ . By what factor does the resistance change?

The figure below plots $R_{D S (o n)}$ (vertical axis, ohms) against the gate voltage $V_{GS}$ (horizontal axis, volts) for this device. The red curve is $R_{D S (o n)} = 1/ [k (V_{GS} - V_{T H})]$ ; the dashed blue line marks $V_{T H} = 2 V$ (left of it the switch is OPEN). The yellow dot is the "barely on" case ( $V_{GS} = 3 V$ ), the green dot the "fully driven" case ( $V_{GS} = 10 V$ ).

Figure s01 — $R_{D S (o n)}$ falls off a cliff just above threshold, then flattens as you drive the gate hard.

Recall Solution 3.1

At $V_{GS} = 3 V$ : $V_{o v} = 1 V$ , $R = \frac{1}{2 \times 1} = 0.5 Ω$ . At $V_{GS} = 10 V$ : $V_{o v} = 8 V$ , $R = \frac{1}{2 \times 8} = 0.0625 Ω$ . Ratio $= \frac{0.5}{0.0625} = 8$ . Why exactly 8×? $R_{D S (o n)} \propto 1/ V_{o v}$ , and the overdrive went from $1 V$ to $8 V$ — an 8× overdrive gives an 8× lower resistance. On the figure: near the dashed threshold the red curve rockets up (a cliff of huge resistance), then flattens as you drive the gate hard — the yellow dot sits high on the cliff, the green dot low on the flat.

Exercise 3.2

A load draws $I_{D} = 4 A$ . Compare the conduction loss for a "barely on" MOSFET ( $R_{D S (o n)} = 0.5 Ω$ ) versus a "fully driven" one ( $R_{D S (o n)} = 0.0625 Ω$ ). What is the ratio of powers?

Recall Solution 3.2

Why the current is the same in both cases. We fix the load current at $4 A$ and vary only the switch, so we compare heat at equal current — the fair way to judge two switches. Barely on: $P = 4^{2} \times 0.5 = 16 \times 0.5 = 8 W$ . Fully driven: $P = 4^{2} \times 0.0625 = 16 \times 0.0625 = 1 W$ . Ratio $= 8 : 1$ . Insight: because $P \propto R_{D S (o n)}$ at fixed current, and $R \propto 1/ V_{o v}$ , driving the gate 8× harder cuts the heat 8×. "Drive it hard, drop stays small."

Exercise 3.3

A logic-level MOSFET must switch $I_{D} = 2 A$ and you require $V_{D S} \leq 0.1 V$ when ON. Given $k = 3 A/V^{2}$ , $V_{T H} = 1.5 V$ , find the minimum $V_{GS}$ you must apply.

Recall Solution 3.3

Step 1 — required $R_{D S (o n)}$ . Why start from the voltage spec? Because a closed switch is a resistor, so $V_{D S} = I_{D} R_{D S (o n)}$ ties the allowed drop directly to the allowed resistance: $R_{D S (o n)} \leq \frac{V _{D S}}{I _{D}} = \frac{0.1}{2} = 0.05 Ω.$ Step 2 — required overdrive. From $R_{D S (o n)} = \frac{1}{k V _{o v}}$ , invert to get the overdrive that achieves it: $V_{o v} \geq \frac{1}{3 \times 0.05} = \frac{1}{0.15} \approx 6.667 V .$ Step 3 — required gate voltage. Overdrive is measured above threshold, so add $V_{T H}$ back: $V_{GS} = V_{o v} + V_{T H} \geq 6.667 + 1.5 = 8.167 V .$ So you must drive the gate to at least $\approx 8.17 V$ .

Level 4 — Synthesis

Exercise 4.1

Design a low-side switch. $V_{D D} = 12 V$ drives an LED string modelled as $R_{L} = 60 Ω$ . You have a MOSFET with $k = 2.5 A/V^{2}$ , $V_{T H} = 2 V$ , driven at $V_{GS} = 10 V$ . (a) Find $R_{D S (o n)}$ . (b) Find the load current (treat the divider: current through $R_{L} + R_{D S (o n)}$ ). (c) Find $V_{o u t}$ (the drain node) when ON. (d) Find the conduction loss in the MOSFET.

Recall Solution 4.1

(a) $V_{o v} = 10 - 2 = 8 V$ . $R_{D S (o n)} = \frac{1}{2.5 \times 8} = \frac{1}{20} = 0.05 Ω.$ (b) Why divide by the series sum? $R_{L}$ and $R_{D S (o n)}$ are in series across $V_{D D}$ , so the same current flows through both: $I_{D} = \frac{V _{D D}}{R _{L} + R _{D S (o n)}} = \frac{12}{60 + 0.05} = \frac{12}{60.05} \approx 0.1998 A .$ (c) $V_{o u t} = V_{D D} \frac{R _{D S (o n)}}{R _{L} + R _{D S (o n)}} = 12 \cdot \frac{0.05}{60.05} \approx 0.00999 V \approx 10 mV (LOW) .$ (d) Why $P = I_{D}^{2} R_{D S (o n)}$ here? The ON MOSFET is a resistor carrying the load current $I_{D}$ , and the only steady loss in a resistor is Joule heating $I^{2} R$ — so this term is the whole conduction loss: $P_{co n d} = I_{D}^{2} R_{D S (o n)} = 0.199 8^{2} \times 0.05 \approx 0.001997 W \approx 2 mW .$ Reading: the MOSFET wastes ~ $2 mW$ while the LEDs get essentially all $12 V$ minus $10 mV$ — an excellent switch.

Exercise 4.2

Compare two design choices for switching $I_{D} = 5 A$ : MOSFET-A with $R_{D S (o n)} = 0.02 Ω$ , and MOSFET-B with $R_{D S (o n)} = 0.12 Ω$ . Compute each conduction loss and the extra watts B wastes. If both sit on a board that can shed only $2 W$ passively, which survives without a heatsink?

Recall Solution 4.2

A: $P_{A} = 5^{2} \times 0.02 = 25 \times 0.02 = 0.5 W .$ B: $P_{B} = 5^{2} \times 0.12 = 25 \times 0.12 = 3 W .$ Extra: $P_{B} - P_{A} = 3 - 0.5 = 2.5 W$ . A survives ( $0.5 W < 2 W$ ). B needs a heatsink ( $3 W > 2 W$ ). The whole reason we chase low $R_{D S (o n)}$ is this thermal budget — see R_DS(on) and switching losses.

Level 5 — Mastery

The high-side topology for Exercises 5.1 and 5.3 is drawn below: drain to $V_{D D}$ , source to the load, load to ground. The source is the output and floats up when the switch conducts.

Figure s03 — high-side N-channel: the source floats up to ~ $V_{D D}$ when ON, collapsing $V_{GS}$ toward zero.

Exercise 5.1

An engineer tries to use a plain N-channel MOSFET as a high-side switch: drain to $V_{D D} = 12 V$ , source to the load (load returns to ground), gate tied straight to $V_{D D} = 12 V$ . $V_{T H} = 2 V$ . When the load pulls the source up toward $V_{D D}$ , explain quantitatively why the switch fails to stay on, and compute the gate voltage needed to hold $V_{GS} = 8 V$ once the source sits at $11.5 V$ .

Recall Solution 5.1

The failure — quantitatively. For a high-side N-channel, the source is the output and rises toward $V_{D D}$ as the load conducts. What matters is $V_{GS} = V_{G} - V_{S}$ , not $V_{G}$ alone. Start: gate and drain both at $V_{G} = 12 V$ , source near ground, so $V_{GS} = 12 - 0 = 12 V$ — the channel opens and current flows. But as current flows, the source is pulled up toward $V_{D D}$ . Suppose it reaches $V_{S} = 11.5 V$ while the gate is stuck at $V_{G} = 12 V$ : $V_{GS} = V_{G} - V_{S} = 12 - 11.5 = 0.5 V .$ Since $V_{GS} = 0.5 V < V_{T H} = 2 V$ , the channel collapses ⇒ switch turns OFF. The device self-chokes: the very act of conducting raises the source and kills its own gate drive, settling near $V_{S} \approx V_{D D} - V_{T H}$ where it barely conducts. This is the fundamental reason a gate at $V_{D D}$ cannot hold a high-side N-channel switch on. The fix — how high must the gate go? To hold a healthy $V_{GS} = 8 V$ with the source at $11.5 V$ , solve $V_{GS} = V_{G} - V_{S}$ for $V_{G}$ : $V_{G} = V_{GS} + V_{S} = 8 + 11.5 = 19.5 V .$ You need a gate voltage $19.5 V$ — above $V_{D D}$ — supplied by a bootstrap capacitor or charge pump, or you switch to a P-channel device. See Low-side vs High-side switching.

Exercise 5.2 (design + cross-check)

Full spec: switch a $3 A$ load with a low-side N-channel MOSFET; you must keep conduction loss $\leq 0.3 W$ and keep the ON-state drain node below $0.12 V$ . Given $k = 4 A/V^{2}$ , $V_{T H} = 1.8 V$ . (a) From the power limit, find the max allowed $R_{D S (o n)}$ . (b) From the voltage limit, find the max allowed $R_{D S (o n)}$ . (c) Which limit binds? Use it to find the minimum $V_{GS}$ .

Recall Solution 5.2

(a) Power limit. $P = I_{D}^{2} R \leq 0.3$ ⇒ $R \leq \frac{0.3}{3 ^{2}} = \frac{0.3}{9} \approx 0.03333 Ω.$ (b) Voltage limit. $V_{D S} = I_{D} R \leq 0.12$ ⇒ $R \leq \frac{0.12}{3} = 0.04 Ω.$ (c) Which binds? The smaller ceiling wins (you must satisfy both, so obey the stricter one): $0.03333 Ω$ (the power limit) is tighter than $0.04 Ω$ . Use $R_{D S (o n)} \leq 0.03333 Ω$ . Overdrive: $V_{o v} = \frac{1}{k R} = \frac{1}{4 \times 0.03333} = \frac{1}{0.13333} = 7.5 V .$ Minimum gate: $V_{GS} = V_{o v} + V_{T H} = 7.5 + 1.8 = 9.3 V .$ Cross-check at $R = 0.03333 Ω$ : $P = 9 \times 0.03333 = 0.30 W$ ✓ (meets power) and $V_{D S} = 3 \times 0.03333 = 0.10 V \leq 0.12 V$ ✓ (comfortably meets voltage). Both constraints satisfied by driving $V_{GS} \geq 9.3 V$ .

Exercise 5.3 (body diode in a bridge)

In an H-bridge, a low-side N-channel MOSFET is switched OFF while an inductive motor winding still forces $2 A$ of current to keep flowing up into its drain (drain negative relative to source). The channel is off. Where does the $2 A$ go, roughly what voltage appears across the device, and why does this matter for "dead-time"?

Recall Solution 5.3

Why the current cannot just stop. An inductor resists sudden change in current; when you open the channel it will drive whatever voltage is needed to keep $2 A$ flowing. Where it goes: the body diode (source→drain) is now forward-biased and carries the full $2 A$ . The device drops roughly the diode forward voltage $\approx 0.7 V$ (not $I R_{D S (o n)}$ — the channel is off). Why it matters: during the brief "dead-time" when both switches of a leg are off, the body diode provides the freewheel path so the inductor current is never interrupted — but it dissipates $\approx 2 A \times 0.7 V = 1.4 W$ , more than a good channel would. This is why designers add fast external diodes or minimise dead-time. See Low-side vs High-side switching.

Recall

Recall One-line takeaways

State test ::: compare $V_{GS}$ to $V_{T H}$ , ignore the drain
Lower $R_{D S (o n)}$ by ::: raising overdrive $V_{GS} - V_{T H}$ (and $W / L$ )
Design ceiling on $R_{D S (o n)}$ ::: min of the power limit and the voltage-drop limit
High-side N-channel needs ::: gate driven above $V_{D D}$ (bootstrap/charge pump)
A MOSFET is NOT a one-way switch because ::: the parasitic body diode conducts reverse current above ~0.7 V

Connections

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