Body effect and substrate bias — Hardware notes

WHAT is the body effect?

The controlling quantity is $V_{SB}$ (for an nMOS, source-to-bulk). The parameter that sets how strongly $V_T$ responds is the body-effect coefficient $\gamma$ (gamma).

WHY does it happen? (first-principles story)

The gate charge $Q_G$ must balance two things inside the semiconductor:

the inversion charge $Q_{inv}$ — the actual channel electrons we want, and
the depletion charge $Q_{dep}$ — the immobile ionised acceptor ions in the depleted region below the channel.

$Q_G = Q_{inv} + Q_{dep}$

To turn on the transistor we need enough gate voltage to (a) create the depletion region and (b) then invert the surface. So threshold is roughly

$V_T = V_{FB} + 2\phi_F + \frac{Q_{dep}}{C_{ox}}$

where $V_{FB}$ is flat-band voltage, $2\phi_F$ is the surface potential needed for strong inversion, and $Q_{dep}/C_{ox}$ is the gate voltage "used up" supporting the depletion charge.

HOW do we derive the depletion charge?

Why this step? Poisson's equation $\dfrac{d^2\psi}{dx^2}=-\dfrac{\rho}{\varepsilon}$ with a constant charge density $\rho=-qN_A$ integrates twice to a parabola; evaluating from the edge to the surface pins $\psi \propto W^2$ . Inverting for $W$ and multiplying by $qN_A$ gives the square-root law — depletion charge grows only as $\sqrt{\psi}$ , not linearly.

Now the band bending needed at threshold. With no back-bias the surface must bend by $2\phi_F$ . With a reverse source-body bias $V_{SB}$ , the total potential the depletion region must sustain is $2\phi_F + V_{SB}$ :

$Q_{dep}(V_{SB}) = \sqrt{2\,q\,\varepsilon_{si}\,N_A\,(2\phi_F+V_{SB})}$

Plug into $V_T$ and split the increase from the $V_{SB}=0$ baseline:

Why the subtraction of $\sqrt{2\phi_F}$ ? Because $V_{T0}$ (threshold at $V_{SB}=0$ ) already includes the depletion charge for band bending $2\phi_F$ . We only add the extra charge caused by the back-bias, so we subtract the baseline term. Otherwise we'd double-count.

Worked examples

Common mistakes

Active recall

Recall Explain to a 12-year-old (Feynman)

Imagine the channel is a tiny water pipe and the gate is a hand pressing to open it. Below the pipe there is soft mud (mobile carriers). If you suck the source higher than the ground (that's the back-bias), you pull the soft mud away and expose hard, dry, stuck rocks (the fixed ions). Now your hand (the gate) has to push harder just to clear the rocks before it can even start opening the pipe. So the "minimum push to turn on" — the threshold — goes up. And because clearing rocks gets easier per extra scoop, the extra push grows like a square root, not a straight line.

Flashcards

What is the body effect?

The dependence of MOSFET threshold voltage

V_T

on source-to-body voltage

V_{SB}

; reverse bias widens depletion, raising

V_T

Give the body-effect threshold equation.

V_T=V_{T0}+\gamma(\sqrt{2\phi_F+V_{SB}}-\sqrt{2\phi_F})

Define the body-effect coefficient

\gamma

\gamma=\sqrt{2q\varepsilon_{si}N_A}/C_{ox}

, units

\sqrt{V}

Why does

V_T

increase with

V_{SB}

Reverse bias widens the depletion region, exposing more fixed acceptor ions, so

Q_{dep}

grows and more gate voltage is needed.

Is the

V_T

–

V_{SB}

relation linear?

No — square-root, because

Q_{dep}\propto\sqrt{2\phi_F+V_{SB}}

from Poisson's equation.

Why does high channel doping

N_A

increase

\gamma

More ionised acceptors per volume means more depletion charge per unit band bending, so

V_T

responds more strongly.

Why does thick oxide increase

\gamma

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

shrinks, so each unit of depletion charge costs more gate voltage.

Why do stacked (series) transistors suffer body effect?

Upper devices have source lifted above ground, giving

V_{SB}>0

, raising their

V_T

and reducing drive current.

Derive

Q_{dep}

vs band bending

\psi

Poisson gives

\psi=qN_AW^2/2\varepsilon_{si}

, so

W=\sqrt{2\varepsilon_{si}\psi/qN_A}

and

Q_{dep}=qN_AW=\sqrt{2q\varepsilon_{si}N_A\psi}

Why subtract

\sqrt{2\phi_F}

in the formula?

Because

V_{T0}

already includes depletion charge for band bending

2\phi_F

; we add only the extra caused by back-bias.

Connections

MOSFET threshold voltage — body effect modifies $V_{T0}$ .
Depletion region and Poisson's equation — source of the √ law.
Oxide capacitance Cox — sets the $\gamma$ scale.
CMOS NAND gate delay — practical impact via series stacks.
Channel-length modulation — another " $V_T$ /current is not constant" effect.
Flat-band voltage and work function — the $V_{FB}$ term in $V_T$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, MOSFET ka threshold voltage $V_T$ koi fixed constant nahi hai. Jab hum source ko body (substrate) ke comparison mein upar utha dete hain — matlab $V_{SB}>0$ (reverse bias) — tab channel ke neeche jo depletion region hoti hai wo wide ho jaati hai. Zyada depletion region ka matlab zyada fixed ionised acceptor ions expose ho gaye. Ab gate ko pehle in ions ko "support" karna padta hai, tabhi channel invert hoga. Isliye transistor ko ON karne ke liye zyada gate voltage chahiye — yaani $V_T$ badh jaata hai. Isi ko body effect kehte hain.

Formula yaad rakho: $V_T = V_{T0} + \gamma(\sqrt{2\phi_F + V_{SB}} - \sqrt{2\phi_F})$ . Yahan $\gamma$ (gamma) body-effect coefficient hai, aur wo depend karta hai channel doping $N_A$ aur oxide capacitance $C_{ox}$ par: $\gamma = \sqrt{2q\varepsilon_{si}N_A}/C_{ox}$ . High doping ya thick oxide ho to $\gamma$ bada — matlab body effect strong. Sabse important baat: yeh relation linear nahi, square-root hai. Poisson equation se depletion charge $\sqrt{\psi}$ ke saath badhta hai, isliye shuru mein $V_T$ tezi se badhta hai phir dheere.

Ye matter kyun karta hai? Real circuits mein, jaise NAND gate ke andar series-connected transistors, upar wale transistor ka source ground se upar chala jaata hai, to uska $V_{SB}>0$ ho jaata hai. Body effect uska $V_T$ badha deta hai, overdrive $(V_{GS}-V_T)$ kam ho jaata hai, current girta hai, gate slow ho jaata hai. Isiliye VLSI designers body effect ko kabhi ignore nahi kar sakte. Bas mantra yaad rakho: "Source up, threshold up — by a root."

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Connections