Threshold voltage (Vth) — Hardware notes

WHAT is $V_{th}$ ?

The quantity that actually drives current is the overdrive voltage: $V_{ov} = V_{GS} - V_{th}$

WHY does a channel need a threshold at all?

Think about an NMOS on a p-type body (holes are the majority carriers).

To make electrons conduct, you must first push all the holes away from the surface (deplete it). That costs some gate voltage.
Then you must pull enough electrons in so that the surface behaves like n-type. That costs more.

So there is a definite "budget" of gate voltage you must spend before any electron channel appears. That budget is $V_{th}$ . Below it, you're just clearing holes; above it, you're stacking electrons.

HOW to derive $V_{th}$ from first principles

We build $V_{th}$ as a sum of the voltages needed at each physical stage.

Step 1 — Flat-band voltage $V_{FB}$ . Why this step? Even with $V_{GS}=0$ , the different work functions of gate metal and semiconductor plus trapped oxide charge bend the bands. $V_{FB}$ is the voltage that "undoes" this and makes the bands flat: $V_{FB} = \phi_{ms} - \frac{Q_{ox}}{C_{ox}}$ where $\phi_{ms}$ is the metal–semiconductor work-function difference and $Q_{ox}$ is fixed oxide charge.

Step 2 — Band bending to strong inversion, $\phi_s = 2\phi_F$ . Why $2\phi_F$ ? Strong inversion is defined as: surface electron density = bulk hole density. The bulk Fermi level sits $\phi_F$ below intrinsic; to invert, we must bend the surface by $\phi_F$ to reach intrinsic and another $\phi_F$ past it to become as n-type as the bulk was p-type. Hence: $\phi_s = 2\phi_F, \qquad \phi_F = \frac{kT}{q}\ln\!\frac{N_A}{n_i}$

Step 3 — Voltage across the oxide to hold the depletion charge, $V_{ox}$ . Why this step? Pushing holes away exposes a depletion region of ionized acceptors with total charge $Q_{dep}$ . The gate must supply an equal opposite charge across the oxide capacitor $C_{ox}$ : $V_{ox} = \frac{Q_{dep}}{C_{ox}}, \qquad Q_{dep} = \sqrt{2\,q\,\varepsilon_s N_A\,(2\phi_F)}$ $Q_{dep}$ comes from solving Poisson's equation for a fully depleted region of width $W$ : charge $= qN_A W$ and $2\phi_F = \tfrac{qN_A W^2}{2\varepsilon_s}$ → eliminate $W$ .

Step 4 — Add them.

Body effect (why $V_{th}$ can change)

If the source is not tied to the body, a reverse body-source bias $V_{SB}$ widens the depletion region, so more charge must be supported → $V_{th}$ rises: $V_{th} = V_{th0} + \gamma\left(\sqrt{2\phi_F + V_{SB}} - \sqrt{2\phi_F}\right), \qquad \gamma = \frac{\sqrt{2q\varepsilon_s N_A}}{C_{ox}}$ $\gamma$ is the body-effect coefficient. Why the square-root? Because $Q_{dep}\propto\sqrt{\text{band bending}}$ from Step 3.

Worked examples

Common mistakes

80/20 — the essentials

$V_{th}$ = gate voltage to reach strong inversion; $V_{ov}=V_{GS}-V_{th}$ drives current.
Three physical costs: flat-band, $2\phi_F$ band bending, depletion charge over $C_{ox}$ .
Thinner oxide → lower $V_{th}$ , higher doping → higher $V_{th}$ , body bias → higher $V_{th}$ .

Recall Feynman: explain to a 12-year-old

Imagine a garden hose with a kink that blocks the water. The metal gate is your thumb pressing on a plastic cover over the kink. You have to press hard enough to first push the mud out of the way, then open a clean path for water. The exact amount of press where water just starts flowing is the threshold. Press less: nothing (well, a tiny drip). Press more: strong flow.

Active-recall flashcards

What does

V_{th}

physically mark in a MOSFET?

The

V_{GS}

at which the surface reaches strong inversion, forming a conducting channel.

Define overdrive voltage.

V_{ov}=V_{GS}-V_{th}

; it is what actually controls drain current.

Why is the band bending at threshold equal to

2\phi_F

Because strong inversion means surface electron density equals bulk hole density — need

\phi_F

to reach intrinsic and another

\phi_F

to mirror the bulk.

Write the NMOS

V_{th}

formula.

V_{th}=V_{FB}+2\phi_F+\sqrt{2q\varepsilon_s N_A(2\phi_F)}/C_{ox}

How does thinner oxide affect

V_{th}

C_{ox}

increases, so the depletion-charge term shrinks →

V_{th}

decreases.

How does higher substrate doping

N_A

affect

V_{th}

Increases it (both

2\phi_F

and the

\sqrt{N_A}

term grow).

What is the body effect?

Reverse

V_{SB}

widens depletion → more charge →

V_{th}

rises:

V_{th}=V_{th0}+\gamma(\sqrt{2\phi_F+V_{SB}}-\sqrt{2\phi_F})

Is drain current exactly zero below

V_{th}

No — small exponential subthreshold current still flows.

What is the flat-band voltage?

V_{FB}=\phi_{ms}-Q_{ox}/C_{ox}

; the gate voltage that makes semiconductor bands flat at zero applied field.

Formula for

C_{ox}

per unit area?

C_{ox}=\varepsilon_{ox}/t_{ox}

Connections

MOSFET operating regions — cut-off vs triode vs saturation split at $V_{th}$ .
Overdrive voltage and drain current — $I_D\propto V_{ov}^2$ in saturation.
CMOS scaling — why thin oxides & low $V_{th}$ enabled low-power chips.
Depletion region and Poisson's equation — source of the $\sqrt{N_A}$ term.
Work function and flat-band voltage — origin of $V_{FB}$ .
Subthreshold conduction — leakage below $V_{th}$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, MOSFET basically ek voltage se control hone wala switch hai. Gate ke niche ek insulating oxide hota hai, aur uske niche semiconductor body. Jab tum gate pe voltage badhate ho, to body ke surface se majority carriers (NMOS mein holes) hat jaate hain aur minority carriers (electrons) khinch aate hain. Jis exact gate voltage pe electrons itne ho jaate hain ki ek proper conducting channel ban jaaye — usko hum threshold voltage $V_{th}$ kehte hain. Iske niche transistor OFF, iske upar ON.

$V_{th}$ ka formula hum first principles se banate hain — teen kharche jodke: (1) flat-band voltage $V_{FB}$ jo work-function aur oxide charge ka effect handle karta hai, (2) $2\phi_F$ jitna band bending strong inversion ke liye chahiye, aur (3) $Q_{dep}/C_{ox}$ — yaani depletion charge ko sambhaalne ke liye oxide pe jitna voltage giraana padta hai. Yaad rakho: $2\phi_F$ isliye kyunki surface ko intrinsic tak (ek $\phi_F$ ) aur fir bulk jitna n-type banane tak (aur ek $\phi_F$ ) bend karna padta hai.

Practical baat: thin oxide matlab $C_{ox}$ bada, matlab $V_{th}$ chhota — isiliye modern chips patli oxide use karte hain. Zyada doping $N_A$ matlab $V_{th}$ bada. Aur body effect — agar source aur body alag bias ho ( $V_{SB}$ ), to depletion widen hota hai aur $V_{th}$ badh jaata hai (square-root wala relation).

Ek common galti: log sochte hain $V_{th}$ ke niche current bilkul zero hai. Nahi bhai — subthreshold current exponentially chhota hota hai par exist karta hai (leakage). $V_{th}$ sirf strong inversion ka reference point hai, koi hard zero-current wall nahi. Yeh baat exam aur real low-power design dono mein important hai.