Body effect and substrate bias: Exercises — Hardware deep dive

This page is your self-test range for the body-effect topic. Work each problem before opening its solution. Every constant, symbol, and step you need was built in the parent note — but we re-state each formula the first time it appears here so you never have to guess.

Definition Constants and symbols used on this page

$q = 1.602 \times 1 0^{- 19} C$ (electron charge), $ε_{0} = 8.854 \times 1 0^{- 12} F/m$ , $ε_{s i} = 11.7 ε_{0}$ (silicon), $ε_{o x} = 3.9 ε_{0}$ (SiO $_{2}$ ). $N_{A}$ is the substrate acceptor doping concentration — the number of fixed acceptor (negative-when-ionised) dopant atoms per unit volume in the p-type body. It appears in $γ$ and in $W$ because those very ions are the depletion charge the gate must fight. Datasheets quote it in $cm^{- 3}$ ; every formula on this page needs it in SI $m^{- 3}$ (multiply by $1 0^{6}$ ). $ψ$ (psi) is the band bending — how much the energy landscape at the surface has been pushed, measured in volts. $V_{S B}$ is the source-to-body voltage (source held above the bulk for an nMOS). $V_{GS}$ is the gate-to-source voltage — the voltage you actually apply to the gate to try to turn the transistor on; the device conducts once $V_{GS}$ exceeds the threshold $V_{T}$ . $V_{T 0}$ is the threshold voltage at zero substrate bias — the value of $V_{T}$ measured with $V_{S B} = 0$ . It is the baseline the body effect adds to; see MOSFET threshold voltage. $ϕ_{F}$ (phi-F) is the Fermi potential of the p-type body — a measure (in volts) of how far the doping has pushed the material's Fermi level below the mid-gap. Strong inversion, the moment the channel truly forms, requires the surface to bend by $2 ϕ_{F}$ (twice this potential), which is why the combination $2 ϕ_{F}$ appears everywhere below. $V_{F B}$ is the flat-band voltage — the gate voltage at which the semiconductor bands are flat (no bending at all); it is the fixed offset that sits underneath $V_{T 0}$ . See Flat-band voltage and work function.

Intuition Sign convention and the two edge cases (read before Level 1)

For an nMOS the reverse junction bias that widens depletion is source above bulk, i.e. $V_{S B} > 0$ — the only case where the square root is real and $V_{T}$ rises.

Forward body bias ( $V_{S B} < 0$ ): the source sits below the bulk, the junction is forward-ish, the depletion region narrows, so $Q_{d e p}$ shrinks and $V_{T}$ falls below $V_{T 0}$ . The same master equation still holds — you just plug a negative $V_{S B}$ (kept small so $2 ϕ_{F} + V_{S B} > 0$ , otherwise the root goes imaginary and the model breaks). Low-power chips exploit this to speed up devices on demand.
pMOS: everything mirrors with signs flipped. The controlling variable is $V_{B S}$ (body-to-source $> 0$ ), the doping is donor $N_{D}$ , and $γ$ carries a minus sign so that a more negative $V_{T}$ becomes more negative (harder to turn on). Magnitude behaviour is identical — a square-root rise in $∣ V_{T} ∣$ .

Level 1 — Recognition

L1.1

Which way does $V_{T}$ move when $V_{S B}$ increases from $0$ to $+ 1 V$ (nMOS)? And which underlying charge grows to cause it?

Recall Solution

$V_{S B} ↑$ pulls mobile electrons out from under the channel, uncovering more fixed ionised acceptor ions. That is the depletion charge $Q_{d e p}$ , and it grows. Since a chunk of gate voltage $Q_{d e p} / C_{o x}$ is "spent" supporting that charge, the gate needs a higher voltage to invert — so $V_{T}$ increases. Answer: $V_{T}$ goes up; $Q_{d e p}$ is the charge that grows.

L1.2

Read off the units of $γ$ directly from the master equation.

Recall Solution

Inside the bracket we have $2 ϕ_{F} + V_{S B}$ , which carries units of $V$ . The whole term $γ \cdot V$ must come out in volts (it adds to $V_{T 0}$ ). So $γ = \frac{V}{V} = V$ . Answer: $γ$ has units of $V$ .

L1.3

At $V_{S B} = 0$ , what does the master equation collapse to? Explain the disappearing term.

Recall Solution

Set $V_{S B} = 0$ : the bracket becomes $2 ϕ_{F} - 2 ϕ_{F} = 0$ , so $V_{T} = V_{T 0}$ . The subtraction of $2 ϕ_{F}$ exists precisely so that $V_{T 0}$ (the zero-bias threshold, which already contains the depletion charge for band bending $2 ϕ_{F}$ ) is recovered exactly. Answer: $V_{T} = V_{T 0}$ .

Level 2 — Application

L2.1

Given $V_{T 0} = 0.45 V$ , $γ = 0.5 V$ , $2 ϕ_{F} = 0.8 V$ , and $V_{S B} = 2.5 V$ . Find $V_{T}$ .

Recall Solution

Substitute straight into the master equation: $V_{T} = 0.45 + 0.5 (0.8 + 2.5 - 0.8)$ $3.3 = 1.8166$ , $0.8 = 0.8944$ , difference $= 0.9222$ . $V_{T} = 0.45 + 0.5 (0.9222) = 0.45 + 0.4611 = 0.911 V$ Answer: $V_{T} \approx 0.911 V$ (a $0.46 V$ jump).

L2.2

Compute $γ$ from device parameters: substrate doping $N_{A} = 1 \times 1 0^{17} cm^{- 3}$ , oxide thickness $t_{o x} = 4 nm$ . Then, using $2 ϕ_{F} = 0.8 V$ , find $V_{T}$ at $V_{S B} = 1 V$ given $V_{T 0} = 0.4 V$ .

Recall Solution

Step 1 — Get $C_{o x}$ . (Oxide capacitance Cox.)

= 8.633\times10^{-3}\ \text{F/m}^2 $$ **Step 2 — Convert doping to SI.** $N_A=10^{17}\ \text{cm}^{-3}=10^{23}\ \text{m}^{-3}$. **Step 3 — Build the square-root numerator.** $$ \sqrt{2q\varepsilon_{si}N_A}=\sqrt{2(1.602\times10^{-19})(11.7\times8.854\times10^{-12})(10^{23})} $$ $2q\varepsilon_{si}N_A = 2(1.602\times10^{-19})(1.0359\times10^{-10})(10^{23})=3.319\times10^{-6}$, so its root $=1.8218\times10^{-3}\ \text{C}\,\text{V}^{-1/2}\text{m}^{-2}$. **Step 4 — Divide.** $$ \gamma=\frac{1.8218\times10^{-3}}{8.633\times10^{-3}}=0.2111\ \sqrt{\text{V}} $$ **Step 5 — Threshold at $V_{SB}=1$.** $$ V_T = 0.4 + 0.2111(\sqrt{1.8}-\sqrt{0.8}) = 0.4 + 0.2111(1.3416-0.8944) $$ $$ = 0.4 + 0.2111(0.4472) = 0.4 + 0.0944 = 0.494\ \text{V} $$ **Answer:** $\gamma\approx 0.211\ \sqrt{\text{V}}$, $\;V_T\approx 0.494\ \text{V}$.

L2.3

Two lab measurements: $V_{T} = 0.40 V$ at $V_{S B} = 0$ , and $V_{T} = 0.72 V$ at $V_{S B} = 2 V$ , with $2 ϕ_{F} = 0.75 V$ . Extract $γ$ .

Recall Solution

Subtract the two master equations — this cancels $V_{T 0}$ , the whole reason we take two measurements: $0.72 - 0.40 = γ (0.75 + 2 - 0.75)$ $2.75 = 1.6583$ , $0.75 = 0.8660$ , difference $= 0.7923$ . $0.32 = γ (0.7923) \Rightarrow γ = 0.4039 V$ Answer: $γ \approx 0.404 V$ .

Level 3 — Analysis

L3.1 (geometric)

For $N_{A} = 1 0^{23} m^{- 3}$ , compute the depletion width $W$ at band bending $ψ = 0.8 V$ (i.e. $V_{S B} = 0$ ) and at $ψ = 0.8 + 2.5 = 3.3 V$ (i.e. $V_{S B} = 2.5$ ). By what factor does $W$ grow?

Figure 1 plots exactly this: the horizontal axis is the band bending $ψ = 2 ϕ_{F} + V_{S B}$ in volts, the vertical axis is the depletion width $W$ in nanometres, and the teal curve is $W = 2 ε_{s i} ψ / (q N_{A})$ . The two dashed drop-lines mark our two operating points ( $V_{S B} = 0$ in orange, $V_{S B} = 2.5 V$ in plum). Notice the curve rises steeply near the origin then bends over — that visible flattening is the square-root's diminishing return, the whole moral of this exercise.

Recall Solution

Use $W = 2 ε_{s i} ψ / (q N_{A})$ with $ε_{s i} = 1.0359 \times 1 0^{- 10} F/m$ . At $ψ = 0.8$ :

=\sqrt{1.0344\times10^{-14}}=1.017\times10^{-7}\ \text{m}=101.7\ \text{nm} $$ **At $\psi=3.3$:** the denominator is $qN_A=(1.602\times10^{-19})(10^{23})=1.602\times10^{4}$, so $$ W=\sqrt{\frac{2(1.0359\times10^{-10})(3.3)}{qN_A}} =\sqrt{\frac{6.837\times10^{-10}}{1.602\times10^{4}}} =\sqrt{4.267\times10^{-14}}=2.066\times10^{-7}\ \text{m}=206.6\ \text{nm} $$ **Ratio:** $206.6/101.7 = 2.03 = \sqrt{3.3/0.8}=\sqrt{4.125}=2.031$. ✓ Because $W\propto\sqrt{\psi}$, quadrupling-ish the bending only *doubles* the width — the square-root's diminishing returns, exactly the flattening seen in Figure 1. **Answer:** $W\approx 101.7\ \text{nm} \to 206.6\ \text{nm}$, factor $\approx 2.03$.

L3.2

Show that the slope $d V_{T} / d V_{S B}$ shrinks as $V_{S B}$ grows, and evaluate it at $V_{S B} = 0$ and $V_{S B} = 3 V$ for $γ = 0.4$ , $2 ϕ_{F} = 0.7 V$ .

Figure 2 makes this slope visible: the orange curve is $V_{T}$ (volts) versus $V_{S B}$ (volts), and the two dashed straight lines are its tangents at $V_{S B} = 0$ (teal) and $V_{S B} = 3 V$ (plum). The tangent tilts down as we move right — that shrinking tilt is precisely the shrinking derivative you are about to compute.

Recall Solution

Reminder of the tool — the chain rule for a square root. For any function $u (x)$ , $\frac{d}{d x} u = \frac{1}{2 u} \cdot \frac{d u}{d x}$ Here $u = 2 ϕ_{F} + V_{S B}$ , a constant plus $V_{S B}$ , so $\frac{d u}{d V _{S B}} = 1$ , leaving $\frac{d}{d V _{S B}} 2 ϕ_{F} + V_{S B} = \frac{1}{2 2 ϕ _{F} + V _{S B}}$ . Differentiate the master equation w.r.t. $V_{S B}$ (only the first root depends on it, and the constant $- 2 ϕ_{F}$ and $V_{T 0}$ vanish): $\frac{d V _{T}}{d V _{S B}} = γ \cdot \frac{1}{2 2 ϕ _{F} + V _{S B}}$ Why a derivative? The question asks how fast $V_{T}$ climbs — that is exactly what a derivative measures: rate of change of $V_{T}$ per volt of back-bias. Since $2 ϕ_{F} + V_{S B}$ sits in the denominator and grows with $V_{S B}$ , the slope falls — diminishing returns, confirming the square-root shape. Units check: $γ$ is $V$ and $2 ϕ_{F} + V_{S B}$ is also $V$ , so the slope is $V / V =$ dimensionless (V/V) — as it must be, being volts-of- $V_{T}$ per volt-of- $V_{S B}$ . At $V_{S B} = 0$ : $\frac{0.4}{2 0.7} = \frac{0.4}{1.6733} = 0.2390$ (V/V). At $V_{S B} = 3$ : $\frac{0.4}{2 3.7} = \frac{0.4}{3.8471} = 0.1040$ (V/V). Answer: slope $\approx 0.239 \to 0.104$ (both dimensionless); it more than halves, matching the flattening tangents in Figure 2.

L3.3

A NAND-stack top transistor (CMOS NAND gate delay) sees $V_{S B} = 0.6 V$ from the device below. With $V_{T 0} = 0.35$ , $γ = 0.45$ , $2 ϕ_{F} = 0.8$ , and gate drive $V_{GS} = 1.2 V$ , find the loss of overdrive $Δ (V_{GS} - V_{T})$ caused by body effect.

Recall Solution

Recall $V_{GS}$ is the gate-to-source voltage we apply, and the overdrive $V_{GS} - V_{T}$ is how far above threshold we are driving — the quantity that sets on-current. Body effect raises $V_{T}$ , so the overdrive drops by exactly the $V_{T}$ increase: $Δ V_{T} = γ (2 ϕ_{F} + V_{S B} - 2 ϕ_{F}) = 0.45 (1.4 - 0.8)$ $1.4 = 1.1832$ , $0.8 = 0.8944$ , difference $= 0.2888$ . $Δ V_{T} = 0.45 (0.2888) = 0.1300 V$ Overdrive without body effect $= 1.2 - 0.35 = 0.85 V$ ; with body effect $= 0.85 - 0.130 = 0.720 V$ . Answer: overdrive drops by $\approx 0.130 V$ (from $0.85$ to $0.720 V$ ), which cuts current (roughly $\propto$ overdrive $^{2}$ in saturation) and slows the gate.

Level 4 — Synthesis

L4.1

Derive from scratch that $Δ V_{T} = γ (2 ϕ_{F} + V_{S B} - 2 ϕ_{F})$ starting from $V_{T} = V_{F B} + 2 ϕ_{F} + Q_{d e p} / C_{o x}$ and Poisson's result $Q_{d e p} = 2 q ε_{s i} N_{A} ψ$ .

Recall Solution

(Recall $V_{F B}$ , the flat-band voltage, is the fixed gate-voltage offset defined in the symbol block — it will simply cancel, so its exact value never matters here.) Step 1 — What total bending does depletion sustain? With back-bias $V_{S B}$ the depletion region must hold the strong-inversion bending plus the reverse-junction drop: $ψ = 2 ϕ_{F} + V_{S B}$ . Step 2 — Insert into $Q_{d e p}$ : $Q_{d e p} (V_{S B}) = 2 q ε_{s i} N_{A} (2 ϕ_{F} + V_{S B})$ Step 3 — Plug into the threshold expression: $V_{T} = V_{F B} + 2 ϕ_{F} + \frac{2 q ε _{s i} N _{A} ( 2 ϕ _{F} + V _{S B} )}{C _{o x}}$ Step 4 — Write the $V_{S B} = 0$ baseline (this is $V_{T 0}$ , see MOSFET threshold voltage and Flat-band voltage and work function for $V_{F B}$ ): $V_{T 0} = V_{F B} + 2 ϕ_{F} + \frac{2 q ε _{s i} N _{A} ( 2 ϕ _{F} )}{C _{o x}}$ Step 5 — Subtract to cancel $V_{F B} + 2 ϕ_{F}$ and identify $γ = 2 q ε_{s i} N_{A} / C_{o x}$ :

= \gamma\left(\sqrt{2\phi_F+V_{SB}}-\sqrt{2\phi_F}\right) $$ **Done.** The subtraction is what removes the double-counted baseline depletion charge. $\blacksquare$

L4.2

A designer wants $γ \leq 0.30 V$ but must keep $t_{o x} = 4 nm$ . What is the maximum allowed substrate doping $N_{A}$ ? (Use $ε_{s i} = 1.0359 \times 1 0^{- 10}$ , $C_{o x} = 8.633 \times 1 0^{- 3} F/m^{2}$ from L2.2.)

Recall Solution

Invert $γ = 2 q ε_{s i} N_{A} / C_{o x}$ for $N_{A}$ : $N_{A} = \frac{( γ C _{o x} ) ^{2}}{2 q ε _{s i}}$ Why square? $γ$ contains a square root of $N_{A}$ ; solving for $N_{A}$ undoes it by squaring both sides. Step 1 — Numerator. $γ C_{o x} = 0.30 \times 8.633 \times 1 0^{- 3} = 2.590 \times 1 0^{- 3}$ $(γ C_{o x})^{2} = (2.590 \times 1 0^{- 3})^{2} = 6.708 \times 1 0^{- 6}$ Step 2 — Denominator. $2 q ε_{s i} = 2 (1.602 \times 1 0^{- 19}) (1.0359 \times 1 0^{- 10}) = 3.319 \times 1 0^{- 29}$ Step 3 — Divide. $N_{A} = \frac{6.708 \times 1 0 ^{- 6}}{3.319 \times 1 0 ^{- 29}} = 2.021 \times 1 0^{23} m^{- 3}$ Step 4 — Convert back to datasheet units (divide by $1 0^{6}$ ): $N_{A} = 2.021 \times 1 0^{23} m^{- 3} = 2.02 \times 1 0^{17} cm^{- 3}$ Answer: $N_{A} ≲ 2.02 \times 1 0^{17} cm^{- 3}$ (equivalently $2.02 \times 1 0^{23} m^{- 3}$ ).

Level 5 — Mastery

L5.1

For fixed process ( $γ = 0.4$ , $2 ϕ_{F} = 0.7$ , $V_{T 0} = 0.4$ ) a circuit tolerates at most $V_{T} = 0.65 V$ . What is the largest allowed $V_{S B}$ ?

Recall Solution

Rearrange the master equation for $V_{S B}$ : $\frac{V _{T} - V _{T 0}}{γ} + 2 ϕ_{F} = 2 ϕ_{F} + V_{S B}$ $\frac{0.65 - 0.40}{0.4} + 0.7 = 0.625 + 0.8367 = 1.4617$ Square both sides: $2 ϕ_{F} + V_{S B} = 1.461 7^{2} = 2.1366$ . $V_{S B} = 2.1366 - 0.7 = 1.4366 V$ Answer: $V_{S B, m a x} \approx 1.44 V$ .

L5.2

Combine body effect with Channel-length modulation conceptually: a stacked-nMOS device has $V_{S B} = 1 V$ (body effect) and a short channel. Which effect raises $V_{T}$ and which raises output conductance (current slope in saturation)? State each cause in one line, then compute the body-effect $Δ V_{T}$ for $γ = 0.35$ , $2 ϕ_{F} = 0.8$ .

Recall Solution

Body effect raises $V_{T}$ : back-bias widens the depletion layer → more $Q_{d e p}$ → higher threshold. It does not by itself change the saturation slope. Channel-length modulation does not change $V_{T}$ ; it makes the effective channel shorten as $V_{D S}$ rises, so $I_{D}$ keeps creeping up in saturation → nonzero output conductance $g_{d s} = λ I_{D}$ . They are independent mechanisms — one shifts the turn-on point, the other tilts the on-current. Numerically: $Δ V_{T} = 0.35 (0.8 + 1 - 0.8) = 0.35 (1.3416 - 0.8944) = 0.35 (0.4472) = 0.1565 V$ Answer: body effect → $Δ V_{T} \approx 0.157 V$ (raises threshold); channel-length modulation → raises output conductance (tilts $I_{D}$ – $V_{D S}$ ), leaves $V_{T}$ alone.

L5.3

A body-effect curve is measured to have slope $d V_{T} / d V_{S B} = 0.12$ (dimensionless, V/V) at exactly $V_{S B} = 2 V$ , with $2 ϕ_{F} = 0.7 V$ . Recover $γ$ .

Recall Solution

From L3.2, $\frac{d V _{T}}{d V _{S B}} = \frac{γ}{2 2 ϕ _{F} + V _{S B}}$ . Solve for $γ$ : $γ = 2 2 ϕ_{F} + V_{S B} \cdot \frac{d V _{T}}{d V _{S B}} = 2 2.7 (0.12)$ Why multiply by $2 2 ϕ_{F} + V_{S B}$ ? It is the exact inverse of the derivative relation — the slope was $γ$ divided by that factor, so recovering $γ$ multiplies it back. Note the slope is dimensionless, $2.7$ is $V$ , and the factor $2$ is pure number, so $γ$ lands in $V$ — correct. $2.7 = 1.6432$ , so $γ = 2 (1.6432) (0.12) = 0.3944 V$ . Answer: $γ \approx 0.394 V$ .

L5.4

A low-power design applies forward body bias $V_{S B} = - 0.3 V$ to an nMOS with $V_{T 0} = 0.4$ , $γ = 0.4$ , $2 ϕ_{F} = 0.7$ . Find the new $V_{T}$ and explain the sign of the shift.

Recall Solution

The same master equation holds with a negative $V_{S B}$ (still keeping $2 ϕ_{F} + V_{S B} > 0$ ): $V_{T} = 0.4 + 0.4 (0.7 - 0.3 - 0.7) = 0.4 + 0.4 (0.4 - 0.7)$ $0.4 = 0.6325$ , $0.7 = 0.8367$ , difference $= - 0.2042$ . $V_{T} = 0.4 + 0.4 (- 0.2042) = 0.4 - 0.0817 = 0.318 V$ Why it drops: forward body bias narrows the depletion region, uncovering fewer fixed ions, so $Q_{d e p}$ shrinks and the gate needs less voltage to invert. $V_{T}$ falls below $V_{T 0}$ — this is how designers momentarily speed up a transistor. Answer: $V_{T} \approx 0.318 V$ (a $0.082 V$ drop).

Recall One-line self-audit before you leave

Every $V_{T}$ shift is a square-root in $(2 ϕ_{F} + V_{S B})$ , scaled by $γ$ ; the slope $d V_{T} / d V_{S B}$ is dimensionless; forward bias ( $V_{S B} < 0$ ) lowers $V_{T}$ ; to invert for $V_{S B}$ or $N_{A}$ you must square; to convert doping you multiply cm $^{- 3}$ by $1 0^{6}$ ; and the baseline $2 ϕ_{F}$ is always subtracted. Miss any one and the number is wrong.