Intuition The one-sentence ideaA MOSFET behaves like a voltage-controlled switch : put enough voltage on the gate and a conducting channel forms between drain and source (switch CLOSED); remove it and the channel vanishes (switch OPEN). Because the gate is insulated by an oxide, controlling the switch costs almost zero steady current .
For an N-channel enhancement MOSFET (the most common switch), three terminals matter:
Gate (G) — the control knob.Drain (D) — usually the load/high side.Source (S) — the reference (tied to ground in a low-side switch ).
Definition Threshold voltageV T H V_{TH} V T H V G S V_{GS} V GS
The switch state is decided entirely by V G S V_{GS} V GS , not by the drain voltage:
V G S < V T H V_{GS} < V_{TH} V GS < V T H cut-off → switch OPEN.V G S ≫ V T H V_{GS} \gg V_{TH} V GS ≫ V T H triode / ohmic region → switch CLOSED, acts like a tiny resistor R D S ( o n ) R_{DS(on)} R D S ( o n )
Intuition Why we WANT the triode region, not saturationAs an amplifier we bias in saturation (constant-current). As a switch we want the transistor either fully OFF or driven so hard it sits deep in triode , where V D S V_{DS} V D S
Why start here? Because "closed switch" = deep triode, and we need to know how resistive it is.
From the gradual-channel model, the triode-region current is:
I D = k [ ( V G S − V T H ) V D S − V D S 2 2 ] , k = μ n C o x W L I_D = k\left[(V_{GS}-V_{TH})V_{DS} - \frac{V_{DS}^2}{2}\right], \qquad k=\mu_n C_{ox}\frac{W}{L} I D = k [ ( V GS − V T H ) V D S − 2 V D S 2 ] , k = μ n C o x L W
Why this step? μ n C o x \mu_n C_{ox} μ n C o x W / L W/L W / L W / L W/L W / L
When the switch is closed, V D S V_{DS} V D S V D S 2 / 2 V_{DS}^2/2 V D S 2 /2
I D ≈ k ( V G S − V T H ) V D S I_D \approx k(V_{GS}-V_{TH})\,V_{DS} I D ≈ k ( V GS − V T H ) V D S
Why this step? A switch carrying real current should drop only millivolts across itself; the quadratic term is second order.
Rearrange V D S / I D V_{DS}/I_D V D S / I D
R D S ( o n ) = V D S I D = 1 k ( V G S − V T H ) \boxed{R_{DS(on)} = \frac{V_{DS}}{I_D} = \frac{1}{k(V_{GS}-V_{TH})}} R D S ( o n ) = I D V D S = k ( V GS − V T H ) 1
When ON and carrying load current I D I_D I D
P c o n d = I D 2 R D S ( o n ) P_{cond} = I_D^2\, R_{DS(on)} P co n d = I D 2 R D S ( o n )
Why this step? It's just Joule heating in the on-resistance — the fundamental reason low R D S ( o n ) R_{DS(on)} R D S ( o n )
With load resistor R L R_L R L V D D V_{DD} V D D
V o u t = V D D R D S ( o n ) R D S ( o n ) + R L V_{out}=V_{DD}\,\frac{R_{DS(on)}}{R_{DS(on)}+R_L} V o u t = V D D R D S ( o n ) + R L R D S ( o n )
OFF: R D S ( o n ) → ∞ ⇒ V o u t ≈ V D D R_{DS(on)}\to\infty \Rightarrow V_{out}\approx V_{DD} R D S ( o n ) → ∞ ⇒ V o u t ≈ V D D
ON: R D S ( o n ) ≪ R L ⇒ V o u t ≈ 0 R_{DS(on)}\ll R_L \Rightarrow V_{out}\approx 0 R D S ( o n ) ≪ R L ⇒ V o u t ≈ 0
Why this step? Shows the switch inverts logic in this topology and why we want R D S ( o n ) ≪ R L R_{DS(on)}\ll R_L R D S ( o n ) ≪ R L
Worked example Example 1 — Is the switch ON?V T H = 2 V V_{TH}=2\text{ V} V T H = 2 V V G S = 5 V V_{GS}=5\text{ V} V GS = 5 V Step 1: Compare V G S V_{GS} V GS V T H V_{TH} V T H Why? State is set by V G S V_{GS} V GS 5 > 2 5 > 2 5 > 2 switch ON . Overdrive = 5 − 2 = 3 V =5-2=3\text{ V} = 5 − 2 = 3 V
Worked example Example 2 — Conduction lossR D S ( o n ) = 0.05 Ω R_{DS(on)}=0.05\ \Omega R D S ( o n ) = 0.05 Ω I D = 3 A I_D=3\text{ A} I D = 3 A Step 1: P = I D 2 R D S ( o n ) P=I_D^2R_{DS(on)} P = I D 2 R D S ( o n ) Why? Joule heating in on-resistance.
P = 3 2 × 0.05 = 9 × 0.05 = 0.45 W P = 3^2\times0.05 = 9\times0.05 = 0.45\text{ W} P = 3 2 × 0.05 = 9 × 0.05 = 0.45 W Interpretation: manageable without heatsink; if R D S ( o n ) R_{DS(on)} R D S ( o n ) 0.5 Ω 0.5\ \Omega 0.5 Ω 4.5 4.5 4.5
Worked example Example 3 — Low-side switch outputV D D = 12 V V_{DD}=12\text{ V} V D D = 12 V R L = 100 Ω R_L=100\ \Omega R L = 100 Ω R D S ( o n ) = 0.1 Ω R_{DS(on)}=0.1\ \Omega R D S ( o n ) = 0.1 Ω ON: V o u t = 12 ⋅ 0.1 100.1 ≈ 0.012 V V_{out}=12\cdot\frac{0.1}{100.1}\approx 0.012\text{ V} V o u t = 12 ⋅ 100.1 0.1 ≈ 0.012 V Why? R D S ( o n ) ≪ R L R_{DS(on)}\ll R_L R D S ( o n ) ≪ R L OFF: V o u t ≈ 12 V V_{out}\approx 12\text{ V} V o u t ≈ 12 V Why? channel open, no current, no drop on R L R_L R L
Worked example Example 4 — Overdrive lowersR D S ( o n ) R_{DS(on)} R D S ( o n ) k = 2 A/V 2 k=2\ \text{A/V}^2 k = 2 A/V 2 V T H = 2 V_{TH}=2 V T H = 2 V G S = 3 V_{GS}=3 V GS = 3 R = 1 2 ( 3 − 2 ) = 0.5 Ω R=\frac1{2(3-2)}=0.5\ \Omega R = 2 ( 3 − 2 ) 1 = 0.5 Ω V G S = 7 V_{GS}=7 V GS = 7 R = 1 2 ( 7 − 2 ) = 0.1 Ω R=\frac1{2(7-2)}=0.1\ \Omega R = 2 ( 7 − 2 ) 1 = 0.1 Ω Why? More overdrive = denser channel charge = lower resistance.
V D S V_{DS} V D S Why it feels right: In many circuits we apply voltage across a device to make current flow. The fix: The MOSFET is gate -controlled. V G S V_{GS} V GS V D S V_{DS} V D S
Common mistake "A closed MOSFET switch is a perfect short (0 Ω)."Why it feels right: We idealize switches as 0 or ∞. The fix: Closed = a real resistor R D S ( o n ) R_{DS(on)} R D S ( o n ) I 2 R D S ( o n ) I^2R_{DS(on)} I 2 R D S ( o n ) R D S ( o n ) R_{DS(on)} R D S ( o n )
Common mistake "Just barely exceedV T H V_{TH} V T H Why it feels right: Above threshold = ON, so any margin should do. The fix: Just above V T H V_{TH} V T H huge R D S ( o n ) R_{DS(on)} R D S ( o n ) Fully enhance the gate (e.g. V G S = 10 V_{GS}=10 V GS = 10 R D S ( o n ) R_{DS(on)} R D S ( o n )
Common mistake "For a high-side N-channel switch, tie the gate toV D D V_{DD} V D D Why it feels right: V D D V_{DD} V D D The fix: For high-side N-channel, the source floats near V D D V_{DD} V D D V G S = V G − V S → 0 V_{GS}=V_G-V_S\to 0 V GS = V G − V S → 0 gate voltage above V D D V_{DD} V D D (bootstrap/charge pump) — or use a P-channel MOSFET.
Recall Active recall — cover the answers
What variable decides ON/OFF? ⇒ ==V G S V_{GS} V GS V T H V_{TH} V T H
Which region = "closed switch"? ⇒ triode/ohmic
Formula for on-resistance? ⇒ R D S ( o n ) = 1 k ( V G S − V T H ) R_{DS(on)}=\frac{1}{k(V_{GS}-V_{TH})} R D S ( o n ) = k ( V GS − V T H ) 1
Conduction power loss? ⇒ P = I D 2 R D S ( o n ) P=I_D^2R_{DS(on)} P = I D 2 R D S ( o n )
How to make a better switch? ⇒ ==increase overdrive V G S − V T H V_{GS}-V_{TH} V GS − V T H W / L W/L W / L
Recall Feynman: explain to a 12-year-oldImagine a garden hose with a squeeze-valve. The gate is your hand. Squeeze hard (high gate voltage) and water rushes through easily — the "pipe" is wide open, switch CLOSED. Let go and the pipe pinches shut, no water — switch OPEN. The clever part: pinching only needs your fingers pressing (voltage), not your muscles pumping (current). That's why a MOSFET switch controls big power with a tiny gentle signal. But even a fully-open pipe has some friction (that's R D S ( o n ) R_{DS(on)} R D S ( o n )
"Gate Votes, Drain Obeys." The Gate's Voltage votes ON/OFF; the Drain just carries whatever current results. And "Drive it hard, drop stays small."
Enhancement vs Depletion MOSFET Triode vs Saturation regions Threshold voltage VTH BJT as a switch (compare: current-controlled vs voltage-controlled)R_DS(on) and switching losses Low-side vs High-side switching Logic gates from MOSFETs (CMOS)
What terminal voltage controls a MOSFET switch's ON/OFF state? V G S V_{GS} V GS relative to
V T H V_{TH} V T H Which operating region corresponds to a CLOSED MOSFET switch? Triode (ohmic) region
Which region corresponds to an OPEN switch? Cut-off (
V G S < V T H V_{GS}<V_{TH} V GS < V T H )
Write the triode-region drain current equation. I D = k [ ( V G S − V T H ) V D S − V D S 2 / 2 ] I_D=k[(V_{GS}-V_{TH})V_{DS}-V_{DS}^2/2] I D = k [( V GS − V T H ) V D S − V D S 2 /2 ] Give the ON-resistance formula from the small-V D S V_{DS} V D S R D S ( o n ) = 1 / [ k ( V G S − V T H ) ] R_{DS(on)}=1/[k(V_{GS}-V_{TH})] R D S ( o n ) = 1/ [ k ( V GS − V T H )] How does increasing V G S V_{GS} V GS R D S ( o n ) R_{DS(on)} R D S ( o n ) It decreases it (larger overdrive → denser channel → lower resistance)
What is the conduction power loss in an ON MOSFET? P = I D 2 R D S ( o n ) P=I_D^2 R_{DS(on)} P = I D 2 R D S ( o n ) Why isn't a closed MOSFET a perfect short? It has finite
R D S ( o n ) R_{DS(on)} R D S ( o n ) that dissipates heat
Why can't a plain N-channel high-side switch turn on with gate at V D D V_{DD} V D D Source rises to ~
V D D V_{DD} V D D , so
V G S → 0 < V T H V_{GS}\to0<V_{TH} V GS → 0 < V T H ; needs gate boosted above
V D D V_{DD} V D D In a low-side switch, what is V o u t V_{out} V o u t ≈
V D D V_{DD} V D D (no current, no drop across load)
Conduction loss I_D squared R
Intuition Hinglish mein samjhoDekho, MOSFET ko ek switch ki tarah samajho jise voltage se control karte hain. Iska sabse important terminal hai Gate . Jab tum gate pe threshold voltage V T H V_{TH} V T H V G S > V T H V_{GS}>V_{TH} V GS > V T H
Switch ke liye hum chahte hain ki transistor ya to poori tarah OFF ho, ya deep triode region mein ho (fully driven). Triode mein V D S V_{DS} V D S R D S ( o n ) R_{DS(on)} R D S ( o n ) R D S ( o n ) = 1 k ( V G S − V T H ) R_{DS(on)}=\frac{1}{k(V_{GS}-V_{TH})} R D S ( o n ) = k ( V GS − V T H ) 1 V G S − V T H V_{GS}-V_{TH} V GS − V T H R D S ( o n ) R_{DS(on)} R D S ( o n )
Power loss ka funda simple hai: ON state mein P = I D 2 R D S ( o n ) P=I_D^2 R_{DS(on)} P = I D 2 R D S ( o n ) R D S ( o n ) R_{DS(on)} R D S ( o n ) V D D V_{DD} V D D V D D V_{DD} V D D V G S V_{GS} V GS Gate votes, Drain obeys.
