Body effect and substrate bias: Question bank — Hardware deep dive

The one equation everything here orbits: $V_{T} = V_{T 0} + γ (2 ϕ_{F} + V_{S B} - 2 ϕ_{F}), γ = \frac{2 q ε _{s i} N _{A}}{C _{o x}}$

Here $V_{S B}$ is the source-to-body voltage, $γ$ the body-effect coefficient, $2 ϕ_{F}$ the surface potential needed for strong inversion, and $Q_{d e p}$ the fixed depletion charge under the channel. If any of these is fuzzy, revisit MOSFET threshold voltage and Depletion region and Poisson's equation first.

True or false — justify

The body effect only matters at high frequencies.

False. It is a DC shift of

V_{T}

set purely by

V_{S B}

; it changes drive current and gate delay at every speed, and appears even in a static, held-open transistor.

V_{S B} = 0

the body effect term vanishes entirely.

True. The bracket becomes

2 ϕ_{F} - 2 ϕ_{F} = 0

, so

V_{T} = V_{T 0}

— by construction

V_{T 0}

is defined as the zero-back-bias threshold.

V_{T}

rises linearly with

V_{S B}

False. It follows a square-root law because

Q_{d e p} \propto 2 ϕ_{F} + V_{S B}

from integrating Poisson's equation; the rise is steep early and flattens out.

For an nMOS, the body effect requires the body to be at a higher potential than the source.

False. It's the opposite: the source must sit above the body (

V_{S B} > 0

) so the source-body junction is reverse-biased and the depletion layer widens.

A larger oxide capacitance

C_{o x}

weakens the body effect.

True.

γ = 2 q ε_{s i} N_{A} / C_{o x}

, so bigger

C_{o x}

(thinner oxide) means each unit of extra depletion charge costs fewer gate volts, shrinking

γ

Body effect can be eliminated by tying every transistor's source to its own body.

True in principle. If source and body share a node then

V_{S B} = 0

always, killing the term — but in a shared-well CMOS process you usually cannot give every device its own isolated body.

The body-effect coefficient

γ

has units of volts.

False. Since it multiplies a difference of square-roots-of-volts to yield volts,

γ

carries units of

V

Increasing channel doping

N_{A}

makes the transistor less sensitive to substrate bias.

False.

γ \propto N_{A}

, so heavier doping means more fixed ions to expose per volt of bending — the device becomes more sensitive.

Body effect makes a MOSFET easier to turn on.

False. It raises

V_{T}

, so for a fixed

V_{GS}

the overdrive

V_{GS} - V_{T}

shrinks, making the device harder to drive into strong conduction.

Spot the error

"We plug

V_{B S}

(a negative number for nMOS) into the root, so the term is negative."

The correct quantity is the positive

V_{S B} = - V_{B S}

. Putting a negative argument under the square root would give an imaginary (nonsensical) result; the physical reverse-bias magnitude is what widens depletion.

"Since

Q_{d e p} = q N_{A} W

and

W

grows with bias,

Q_{d e p}

grows linearly with

V_{S B}

W

itself grows as

ψ

, not linearly, because Poisson's equation gives

ψ \propto W^{2}

. So

Q_{d e p} \propto 2 ϕ_{F} + V_{S B}

— sub-linear, not linear.

V_{T} = V_{T 0} + γ 2 ϕ_{F} + V_{S B}

— clean single root."

The baseline

- γ 2 ϕ_{F}

is missing.

V_{T 0}

already accounts for the depletion charge at bending

2 ϕ_{F}

; dropping the subtraction double-counts and inflates

V_{T}

even at

V_{S B} = 0

"Body effect is a property of BJTs where the base back-biases the emitter."

No — it is a MOSFET phenomenon involving gate/oxide/depletion charge balance; there is no oxide-capacitance

C_{o x}

or field-controlled inversion layer in a BJT.

"Since the body is a 'second gate', it can turn the channel fully on by itself."

The body is a weak, reverse-biased back-gate: it modulates

V_{T}

by widening depletion, but it cannot inject the inversion charge the front gate creates — it only makes inversion harder or easier.

"Because

γ

contains

ε_{s i}

, a higher silicon permittivity lowers

γ

Wrong direction.

γ \propto ε_{s i}

appears in the numerator, so higher

ε_{s i}

raises

γ

— it lets the depletion region store more charge per unit bending.

Why questions

Why does exposing fixed ions (not mobile carriers) raise the threshold?

The gate must supply charge to balance both inversion and depletion charge; the newly exposed immobile acceptor ions add to

Q_{d e p}

, so more gate voltage is "spent" before any inversion begins. See Depletion region and Poisson's equation.

Why does the

V_{T}

increase saturate at large

V_{S B}

instead of running away?

The square-root law means each extra volt of

V_{S B}

widens depletion by an ever-smaller amount (

W \propto ψ

), so the marginal

V_{T}

gain diminishes — diminishing returns baked into the physics.

Why does the top transistor in a stacked NAND suffer body effect but the bottom one often doesn't?

The bottom device's source sits at ground (body potential), so

V_{S B} \approx 0

; the top device's source is lifted by the devices below it, giving

V_{S B} > 0

and a raised

V_{T}

. This is exactly why CMOS NAND gate delay grows with stack height.

Why do we use

2 ϕ_{F}

(twice the Fermi potential) rather than

ϕ_{F}

as the inversion condition?

Strong inversion is defined as bending the surface until the electron concentration equals the bulk hole concentration, which requires the band to bend by the full

2 ϕ_{F}

— one

ϕ_{F}

to reach intrinsic, another to build the mirror-image inversion layer.

Why does the flat-band voltage

V_{F B}

not appear in the body-effect increment even though it's in

V_{T}

V_{F B}

is fixed by materials and work-function difference (see Flat-band voltage and work function); it doesn't depend on

V_{S B}

, so it cancels when we subtract the baseline and only the bias-dependent depletion term survives.

Why is body effect a threshold shift and not a channel-length effect like Channel-length modulation?

Body effect changes the vertical charge balance under the channel (how much

V_{T}

you need to invert), while channel-length modulation changes the effective length of an already-inverted channel in saturation — different axes, different mechanisms.

Edge cases

What happens to

V_{T}

V_{S B}

is made negative (source below body) on an nMOS?

This forward-biases the source-body junction, which is not a normal operating condition and can inject current; the depletion narrows and

V_{T}

would drop, but the diode conduction makes the simple square-root model invalid.

In the limit

N_{A} \to 0

(intrinsic body), what happens to the body effect?

γ = 2 q ε_{s i} N_{A} / C_{o x} \to 0

, so there are essentially no fixed ions to expose and

V_{T}

becomes independent of

V_{S B}

— the body effect disappears.

In the limit of very thin oxide (

t_{o x} \to 0

, so

C_{o x} \to \infty

), what happens to

γ

γ \to 0

: the huge gate capacitance means depletion charge is neutralised by a negligible gate voltage, so back-bias barely shifts

V_{T}

What does the model predict at exactly

V_{S B} = 0

for the slope

d V_{T} / d V_{S B}

The slope is

γ / (2 2 ϕ_{F})

— finite and at its maximum there. The sensitivity is largest at zero bias and monotonically decreases as

V_{S B}

grows.

If a pMOS is used instead of an nMOS, does the same story hold?

Yes, with mirrored signs: the relevant bias is the source-to-body voltage that reverse-biases the (now n-type) body junction, doping becomes donor

N_{D}

, and

∣ V_{T} ∣

increases in magnitude as the reverse bias grows.

Recall One-sentence self-test

If someone claims " $V_{T}$ doubles when $V_{S B}$ doubles," what single word refutes them? ::: "Square-root" — doubling $V_{S B}$ raises the root argument, which raises $V_{T}$ by less than double, never proportionally.