Visual walkthrough — Body effect and substrate bias

Step 0 — The cast of characters (read this first)

Before any math, meet the pieces. A MOSFET is a sandwich: a metal gate on top, a thin insulating oxide below it, then the silicon body (also called substrate or bulk).

Look at the figure: the gate is the top bar, the red slab is the oxide, and below it the silicon. The dots are the two kinds of charge we will keep track of: fixed negative ions (they cannot move) and mobile holes (they can). Everything that follows is just bookkeeping of these two.

Step 1 — The gate's job is to balance charge

WHAT. Put a positive voltage on the gate. It stores positive charge $Q_G$ per unit area on the metal. That field reaches through the oxide and must be answered by an equal-and-opposite negative charge in the silicon.

WHY. An insulator (the oxide) lets no charge cross, so it behaves like a capacitor: whatever charge sits on one plate is mirrored on the other. So $Q_G$ is pinned to the silicon's response.

WHAT IT LOOKS LIKE. Two things can supply that negative charge on the silicon side:

The picture shows the gate charge on top (red) being balanced from below by two layers: a wall of fixed ions ($Q_{dep}$), and — only once the gate is strong enough — a thin sheet of electrons ($Q_{inv}$). The key insight: the gate must pay for the ions first before it gets any channel.

Step 2 — Why depletion appears (holes run away)

WHAT. Raise the gate voltage from zero. The positive gate first pushes the nearby positive holes downward, away from the surface.

WHY. Like charges repel. The gate is positive; holes are positive; they flee. What is left behind where the holes used to be? The fixed negative acceptor ions — they cannot move. That stripped-bare zone is the depletion region.

WHAT IT LOOKS LIKE. A growing dark band under the oxide, empty of mobile carriers, full of stuck negative ions.

Step 3 — Poisson's equation turns width into voltage

WHAT. We need a link between the depletion width $W$ and the voltage (band bending) $\psi$ it holds off. That link is Poisson's equation.

WHY THIS TOOL, not another? We have a region of constant charge density (uniform doping $-q{N}_A$) and we want the potential across it. Poisson's equation is exactly the rule that converts "how much charge sits where" into "what voltage results". No other tool does that job.

WHAT IT LOOKS LIKE. A constant second derivative means the potential is a parabola in $x$. Integrate twice from the depletion edge (where $\psi =0$, field $=0$) to the surface:

$\psi =\frac{q{N}_A{W}^2}{2{\epsilon }_{si}}$

Read it: the voltage the band bends by grows with the square of the width. Now flip it around to solve for the width — because we control the voltage and want to know the resulting width:

$W=\sqrt{\frac{2{\epsilon }_{si}\psi }{q{N}_A}}$

The figure plots the parabolic potential across the depletion band. Notice the flat start (zero slope at the edge) and the steepening curve — that is the signature of constant charge density.

Step 4 — Depletion charge as a square-root law

WHAT. Multiply the width by the charge per volume to get charge per area: $Q_{dep}=q{N}_{A}W.$

WHY. $Q_{dep}$ is what the gate has to answer for (Step 1). We want it in terms of the voltage $\psi$, so substitute the $W$ we just found.

WHAT IT LOOKS LIKE. The two $W$'s combine and the square root survives:

The red curve in the figure is $\sqrt{\psi }$: steep at first, then flattening. This shape is the fingerprint of the whole body effect. Whenever someone draws a straight line here, they are wrong.

Step 5 — Threshold: what voltage just makes the channel?

WHAT. At threshold two things must be paid for by the gate:

1. bend the surface enough to invert it — that costs a fixed band bending of $2{\varphi }_F$, and
2. support the depletion charge underneath — that costs $Q_{dep}/C_{ox}$ volts.

WHY $2{\varphi }_F$? ${\varphi }_F$ is how far the bulk Fermi level sits from mid-gap; to flip the surface from p-type to n-type (strong inversion) you must bend by twice that. It is the price of building any channel at all, before body effect enters. See MOSFET threshold voltage.

WHY $Q_{dep}/C_{ox}$? The oxide capacitance $C_{ox}={\epsilon }_{ox}/t_{ox}$ converts a charge per area into the gate voltage needed to hold it (voltage = charge / capacitance).

WHAT IT LOOKS LIKE. A stack of voltage "costs" the gate must clear before the channel forms.

Step 6 — Back-bias adds to the band bending

WHAT. Now lift the source above the body by $V_{SB}>0$. This reverse-biases the channel-to-body junction and makes it hold off extra voltage.

WHY. The depletion region must now sustain not just $2{\varphi }_F$ but $2{\varphi }_F+V_{SB}$. So we replace $\psi \to 2{\varphi }_F+V_{SB}$ in the depletion charge:

The figure overlays two depletion bands: thin (no bias) and thick (back-biased). The extra ions in the wider band are the new charge the gate must now pay for.

Step 7 — Subtract the baseline (or you double-count)

WHAT. Insert $Q_{dep}(V_{SB})$ into $V_T$. But $V_{T0}$ — the threshold at $V_{SB}=0$ — already contains the depletion cost for $2{\varphi }_F$. So we only add the difference.

WHY. If we wrote the full new root without subtracting the old one, we would charge the gate twice for the same baseline band bending. Subtract $\sqrt{2{\varphi }_F}$ to keep only the extra.

Step 8 — Every case, including the degenerate ones

Cover the whole range so you never hit an unshown scenario:

The figure shows $V_T$ vs $V_{SB}$: it starts exactly at $V_{T0}$, climbs steeply, then bends over — the square root made visible. This same rising $V_T$ on stacked devices is what slows a CMOS NAND gate, and it sits alongside Channel-length modulation as a second-order effect designers must respect.

Worked example, checked against the walkthrough

The one-picture summary

This final figure chains the whole story left-to-right: holes flee → ions exposed → Poisson turns width into a parabolic voltage → charge is a square root of that voltage → threshold stacks the costs → back-bias adds inside the root → subtract baseline → the rising $V_T$ curve.

Active recall