Body effect and substrate bias: Worked examples — Hardware deep dive

This page is the drill hall for the parent note. We take the master equation $V_{T} = V_{T 0} + γ (2 ϕ_{F} + V_{S B} - 2 ϕ_{F})$ and push it into every corner: positive bias, zero bias, "wrong-sign" bias, extracting $γ$ , extreme values, a real circuit, and an exam trap. Before we compute anything, we lay out the map so you can see which cases exist and which example covers each one.

The scenario matrix

Every problem this topic can ask lives in one of these cells. Each row is a distinct behaviour of the formula, not just different numbers.

#	Case class	What is special about it	Covered by
A	$V_{S B} = 0$ (zero input)	Both roots cancel → $V_{T} = V_{T 0}$ exactly	Ex 1
B	$V_{S B} > 0$ , moderate	The "normal" body effect, $V_{T}$ rises	Ex 2
C	Two-point measurement	Solve for $γ$ (inverse problem)	Ex 3
D	$V_{S B} < 0$ (forward / "wrong sign")	Root of a smaller number → $V_{T}$ drops; watch the danger zone	Ex 4
E	Large- $V_{S B}$ limit	Diminishing returns — the slope $d V_{T} / d V_{S B} \to 0$	Ex 5
F	Degenerate $γ \to 0$ or thin oxide	Body effect vanishes; what makes it vanish	Ex 6
G	Real-world word problem	NAND stack — lifted source raises $V_{T}$	Ex 7
H	Exam twist — solve for $V_{S B}$	Invert the square root to hit a target $V_{T}$	Ex 8

The figure above plots the whole $V_{T}$ – $V_{S B}$ curve once so you can see where each cell lives: the flat crossing at zero (A), the steep early climb (B), the mirror-image forward-bias dip on the left (D), and the flattening tail on the right (E).

Example 1 — Case A: the zero-input sanity check

Example 2 — Case B: the normal body effect

Example 3 — Case C: extract $γ$ from two measurements

Example 4 — Case D: the "wrong-sign" forward bias

Worked example

$V_{S B} < 0$ — threshold drops, and the danger zone Device from Ex 1 ( $V_{T 0} = 0.5$ , $γ = 0.4$ , $2 ϕ_{F} = 0.7$ ). What is $V_{T}$ at $V_{S B} = - 0.3 V$ ? Then ask: what $V_{S B}$ makes the root break?

Forecast: if lifting the source raises $V_{T}$ , what should lowering it do?

Inner root: $0.7 + (- 0.3) = 0.4 = 0.632$ .
- Why this step? A forward-biased source–body junction narrows depletion, so the argument shrinks below the baseline.
$V_{T} = 0.5 + 0.4 (0.632 - 0.837) = 0.5 + 0.4 (- 0.205) = 0.5 - 0.082 = 0.418 V$ .
- Why this step? The correction is now negative — $V_{T}$ falls, mirroring case B.
Danger zone: the root demands $2 ϕ_{F} + V_{S B} \geq 0$ , i.e. $V_{S B} \geq - 0.7 V$ . Below that the formula returns an imaginary number and the model has broken (the junction is heavily forward-biased and injecting — the depletion picture no longer holds).
- Why this step? This is exactly the parent's warning about plugging $V_{B S}$ (negative) blindly and getting $negative$ .

Verify: at $V_{S B} = - 0.3$ the shift $- 0.082$ V is negative and smaller in size than case B's $+ 0.322$ (the curve is steeper near zero on the positive side but a small forward step is close to the tangent). Argument $0.4 > 0$ , so no imaginary result. ✓

Example 5 — Case E: the large-bias limit (diminishing returns)

Worked example Push

$V_{S B}$ high — how much extra do you buy? Same device. Compare the $V_{T}$ gain from $0 \to 1 V$ against the gain from $9 \to 10 V$ .

Forecast: the second step is also 1 V of bias — same gain, or less?

Gain $0 \to 1$ : $0.4 (1.7 - 0.7) = 0.4 (1.304 - 0.837) = 0.4 (0.467) = 0.187 V$ .
Gain $9 \to 10$ : $0.4 (10.7 - 9.7) = 0.4 (3.271 - 3.114) = 0.4 (0.157) = 0.063 V$ .
- Why this step? Same $Δ V_{S B} = 1$ V, but far up the square-root curve the slope is gentle.
So identical bias steps give less and less threshold shift — the square-root's flattening.
- Why this step? The slope is $\frac{d V _{T}}{d V _{S B}} = \frac{γ}{2 2 ϕ _{F} + V _{S B}}$ , which $\to 0$ as $V_{S B} \to \infty$ .

Verify: $0.063 < 0.187$ , ratio $\approx 0.34$ , confirming diminishing returns exactly as $1/ V_{S B}$ predicts (this is the parent's "climbs fast, then flattens" mistake-fix). ✓

Example 6 — Case F: degenerate device, $γ \to 0$

Worked example When body effect nearly disappears

Two processes share $N_{A}$ but process X has oxide thickness $t_{o x} = 10 nm$ and process Y has $t_{o x} = 1 nm$ . If process X has $γ_{X} = 0.5 V$ , what is $γ_{Y}$ ?

Forecast: thinner oxide → bigger $C_{o x}$ → is $γ$ bigger or smaller?

Recall $γ = \frac{2 q ε _{s i} N _{A}}{C _{o x}}$ and $C_{o x} = \frac{ε _{o x}}{t _{o x}}$ .
- Why this step? Only $C_{o x}$ changes between X and Y, and it scales as $1/ t_{o x}$ .
So $γ \propto \frac{1}{C _{o x}} \propto t_{o x}$ . Ratio $\frac{γ _{Y}}{γ _{X}} = \frac{t _{o x, Y}}{t _{o x, X}} = \frac{1}{10}$ .
$γ_{Y} = 0.5 \times \frac{1}{10} = 0.05 V$ .
- Why this step? In the limit of very thin oxide the coupling is so strong the gate wins easily — body effect fades toward zero.

Verify: with $γ_{Y} = 0.05$ , a full $V_{S B} = 3 V$ bias shifts $V_{T}$ by only $0.05 (3.7 - 0.7) = 0.05 (1.087) = 0.054 V$ — negligible, matching the intuition that thin-oxide (modern) devices suffer far less body effect. ✓

Example 7 — Case G: the NAND stack (real-world)

Worked example Why the top transistor is slow — with numbers

In a 2-input CMOS NAND pull-down, two nMOS are stacked. The top device's source sits on the drain of the bottom device. Suppose that intermediate node is at $0.8 V$ while the body is at ground, so $V_{S B} = 0.8 V$ for the top transistor. Device: $V_{T 0} = 0.5$ , $γ = 0.4$ , $2 ϕ_{F} = 0.7$ . The gate drives $V_{GS} = 1.2 V$ (referenced to that lifted source). Find the top device's $V_{T}$ and its overdrive $V_{GS} - V_{T}$ , and compare to the bottom device (whose source is grounded, $V_{S B} = 0$ ).

Forecast: the bottom transistor has overdrive $1.2 - 0.5 = 0.7$ V. Will the top one keep most of that?

Top $V_{T}$ : $0.5 + 0.4 (0.7 + 0.8 - 0.7) = 0.5 + 0.4 (1.5 - 0.837)$ .
- Why this step? The lifted source is the $V_{S B}$ ; body effect raises this device's threshold.
$1.5 = 1.225$ , so $V_{T} = 0.5 + 0.4 (1.225 - 0.837) = 0.5 + 0.4 (0.388) = 0.5 + 0.155 = 0.655 V$ .
Top overdrive $= V_{GS} - V_{T} = 1.2 - 0.655 = 0.545 V$ , versus bottom $0.700 V$ .
- Why this step? Drive current scales roughly with overdrive; less overdrive → less current → the stack charges/discharges slower.
Overdrive lost: $0.700 - 0.545 = 0.155 V$ , i.e. about $22%$ of the top device's drive vanishes purely from body effect.

Verify: the drop $0.155$ V equals the top device's body-effect $Δ V_{T}$ from step 2 exactly (overdrive lost = threshold gained, since $V_{GS}$ was fixed). Sign positive, magnitude sub-volt. This is the parent's "stacked transistors slow down" made quantitative. ✓

Example 8 — Case H: exam twist, solve for $V_{S B}$

Worked example Invert the root to hit a target threshold

Given $V_{T 0} = 0.5$ , $γ = 0.4$ , $2 ϕ_{F} = 0.7$ . What source-to-body voltage makes $V_{T} = 0.9 V$ ?

Forecast: we already know from Ex 3 that 3 V gave 0.4 V of shift — so is the answer near 3 V?

Move the baseline: $V_{T} - V_{T 0} = γ (2 ϕ_{F} + V_{S B} - 2 ϕ_{F})$ , so $\frac{0.9 - 0.5}{0.4} = 0.7 + V_{S B} - 0.7$ .
- Why this step? Isolate the unknown root; everything else is known.
$1.0 = 0.7 + V_{S B} - 0.837 \Rightarrow 0.7 + V_{S B} = 1.837$ .
Square both sides: $0.7 + V_{S B} = 1.83 7^{2} = 3.374$ , so $V_{S B} = 2.674 V$ .
- Why this step? Squaring undoes the square root — the inverse operation — because we need the argument, not its root.

Verify: plug $V_{S B} = 2.674$ back: $0.5 + 0.4 (3.374 - 0.7) = 0.5 + 0.4 (1.837 - 0.837) = 0.5 + 0.4 = 0.9 V$ . ✓ And it sits just under 3 V, consistent with the Ex 3 measurement where 3 V gave a slightly larger shift.

Active recall

Recall Which cell? Match the symptom to the matrix

"Same 1 V bias step gives less shift each time" ::: Case E — large-bias limit, slope $\propto 1/ V_{S B}$ . "The formula gave an imaginary number" ::: Case D — you drove $V_{S B}$ below $- 2 ϕ_{F}$ . "Thin-oxide chip barely shows body effect" ::: Case F — $γ \propto t_{o x} \to 0$ . "Top transistor in the stack conducts weakly" ::: Case G — lifted source gives $V_{S B} > 0$ .