Forward bias behavior: Visual walkthrough — Hardware deep dive

We only assume you can read a graph and that you believe a battery has a $+$ side and a $-$ side. Everything else — barriers, "Boltzmann", the letter $e$ , thermal voltage — is built from scratch below.

Step 1 — The junction before we touch it

WHAT: We draw the junction with no battery attached. See PN Junction at Equilibrium for why this strip forms.

WHY: You cannot understand "forward bias narrows the strip" until you can see the strip in its resting state.

PICTURE: In the figure, the P-side (left) is packed with orange holes, the N-side (right) with blue electrons. The gray band in the middle is empty of carriers but holds fixed ions: negative acceptor ions on the P-edge, positive donor ions on the N-edge.

Those exposed ions set up an internal field (an invisible push on charges) pointing from the $+$ donor ions toward the $-$ acceptor ions — that is, N → P. This field is the traffic cop that stops carriers from pouring across.

Step 2 — The barrier: turn the field into a hill

WHAT: We replace the field arrows with an energy hill. Its height is $q V_{bi}$ .

Term-by-term, right where it lives: $barrier energy = charge of one carrier q \cdot built-in voltage (volts) V_{bi}$

$q$ = the charge carried by a single hole or electron, $q \approx 1.6 \times 1 0^{- 19}$ coulomb. Multiplying a voltage by a charge gives an energy — that is what a volt is (energy per charge).
$V_{bi}$ = the Built-in potential, the voltage the junction builds on its own.

WHY a hill and not the field itself? Because the next step asks "what fraction of carriers can get across?" — and "getting across" is much easier to picture as "who can climb this hill?" than as "who can fight this field?".

PICTURE: A smooth hill. On the left, orange holes sit at the low ground of the P-side, wanting to reach the N-side but blocked by the climb. The hilltop is labelled $q V_{bi}$ .

Step 3 — Who can climb the hill? (Boltzmann enters, and why)

WHY this exact tool? We need a rule for "what fraction of jiggling particles have at least energy $E$ ?" The Boltzmann distribution answers precisely that question — that is the tool's whole job — and its answer is: $fraction over the hill = e^{- E / k T}$

Term-by-term:

$E$ = the energy needed to cross. Here $E = q V_{B}$ , the hill height.
$k$ = Boltzmann's constant, $1.38 \times 1 0^{- 23}$ J/K — it just converts temperature into energy.
$T$ = temperature in kelvin. Bigger $T$ = more jiggle = easier climb.
$e^{\cdot}$ = the exponential function. Why $e$ and not just a fraction? Because doubling the hill doesn't halve the crowd — it squares down the crowd. Each extra rung of hill cuts the survivors by the same multiplying factor, and "same multiplier per equal step" is the exact fingerprint of the exponential.

PICTURE: The same hill, now with a scatter of carriers at different heights (their jiggle energy). A dashed line marks the hilltop $q V_{B}$ ; only the tall-arrow carriers poke above it. A tiny inset curve $e^{- E / k T}$ shows the crowd thinning fast as we go up.

So the current that diffuses over the hill is proportional to that fraction: $I_{diffusion} \propto e^{- q V_{B} / k T}$

Step 4 — Attach the battery: forward bias lowers the hill

WHAT: Two opposing fields partly cancel. Weaker net field ⟹ narrower depletion strip ⟹ shorter hill. The new hill height is $V_{B} = V_{bi} - V$ .

$V_{B} = original hill V_{bi} - applied forward volts V$

$V$ = the battery voltage. Every volt you apply shaves one volt off the hill.
The minus sign is the whole story of forward bias: you are subtracting from the barrier.

WHY subtract, not add? Because the battery's field opposes the built-in one. (Flip the battery and you'd add — that's Reverse bias behavior, the opposite page.)

PICTURE: Two panels side by side. Left: tall hill, height $q V_{bi}$ , few carriers cross. Right: after forward bias, the hill is chopped down to $q (V_{bi} - V)$ , the depletion band is visibly thinner, and a flood of orange holes streams over the top.

Step 5 — Put the shorter hill into Boltzmann

WHAT: Substitute $V_{B} = V_{bi} - V$ into the Boltzmann fraction from Step 3.

$I_{diffusion} \propto e^{- q (V_{bi} - V) / k T} = fixed number e^{- q V_{bi} / k T} \cdot grows with V e^{q V / k T}$

WHY split it into two factors? Because the algebra rule $e^{a + b} = e^{a} e^{b}$ lets us peel off the part that never changes (the original tall-hill term, a constant) from the part that responds to your battery ( $e^{q V / k T}$ ). The constant we'll pin down in the next step; the growing factor is where all the action lives.

Term-by-term on the right piece:

$q V / k T$ = (applied energy $q V$ ) ÷ (jiggle energy $k T$ ). It measures your battery in units of "how many jiggles."
As $V$ grows, this exponent grows, so the whole current multiplies upward — the exponential rise.

PICTURE: The exponent $- q (V_{bi} - V) / k T$ visibly factored: the left brace collapses the $V_{bi}$ term into a small fixed dot on the axis, while the right term sweeps a growing curve as $V$ increases.

Step 6 — Pin the constant using the $V = 0$ balance

WHAT: At $V = 0$ the diffusion current must exactly equal the backward drift current. Call that fixed backward trickle $I_{S}$ , the reverse saturation current — it's tiny and set entirely by rare minority carriers.

Setting diffusion at bias $V$ equal to $I_{S}$ scaled by the growing factor: $I_{diffusion} = I_{S} e^{q V / k T}, I_{drift} = - I_{S}$

Add them for the net current: $I = I_{S} e^{q V / k T} - I_{S} = I_{S} (e^{q V / k T} - 1)$

The first term = flood over the shrinking hill (grows exponentially with $V$ ).
The $- 1$ = the constant backward drift, now bundled in.

WHY the $- 1$ matters: at $V = 0$ , $e^{0} = 1$ , so $I = I_{S} (1 - 1) = 0$ . The equation automatically gives zero current at zero bias — proof we pinned the constant correctly.

PICTURE: Two arrows over the hill — a fat orange "diffusion" arrow going over, a thin gray " $- I_{S}$ drift" arrow coming back — and a balance scale showing them equal at $V = 0$ .

Step 7 — All the cases (never leave a scenario unshown)

Case A — strong forward ( $V ≫ V_{T}$ ): $e^{V / V_{T}}$ is huge, the $- 1$ is a rounding error, so $I \approx I_{S} e^{V / V_{T}}$ . Current explodes upward. (right half of the figure)

Case B — zero bias ( $V = 0$ ): $I = I_{S} (1 - 1) = 0$ . The curve passes exactly through the origin.

Case C — reverse bias ( $V < 0$ ): now the exponent is negative, so $e^{V / V_{T}} \to 0$ , leaving $I \to I_{S} (0 - 1) = - I_{S}$ . A tiny, flat, backward current — the saturation floor. This is the domain of Reverse bias behavior.

Degenerate check — $T \to 0$ (frozen crystal): $V_{T} = k T / q \to 0$ , so $e^{V / V_{T}}$ becomes an infinitely sharp switch: nothing below $V_{bi}$ , a cliff above it. Real diodes at low temperature really do get "harder" turn-ons — the maths predicts it.

PICTURE: The full I–V curve on one axis: flat $- I_{S}$ floor for $V < 0$ , passing through the origin, then the steep exponential blast-off for $V > 0$ . Each region is labelled A/B/C.

This is the curve you'll meet in full at Diode I-V characteristics.

The one-picture summary

Everything above, compressed: the hill shrinks by $V$ , Boltzmann converts hill-height into a crowd size, the crowd size is the current, and pinning the $V = 0$ balance stamps the $- 1$ onto the formula.

Recall Retell the whole walkthrough in plain words (Feynman)

Two towns are separated by a hill. Left town is full of "holes," right town full of electrons, and normally the hill is too tall for almost anyone to climb — so barely anyone crosses, and the few who wander over get swept right back. That's the resting junction. Now I bring a bulldozer (the battery, $+$ on the hole-town side) and shave the hill down by an amount $V$ . Heat makes everyone jiggle, and the rule for "how many jigglers can clear a hill of a given height" is the Boltzmann rule — an exponential, meaning every equal chunk of hill I remove multiplies the crowd by the same factor. So a little shaving lets ten times as many pour across. That flood of people is the current. When I've shaved nothing ( $V = 0$ ), the crowd crossing exactly matches the trickle being swept back, so nothing net moves — and that balance is what forces the " $- 1$ " into the equation. Flip the bulldozer around (reverse bias) and I make the hill taller; then only the tiny backward trickle $I_{S}$ remains. One picture, one equation: $I = I_{S} (e^{V / V_{T}} - 1)$ .

Connections

Forward bias behavior — the parent note this page derives in pictures.
PN Junction at Equilibrium — where the resting hill of Step 1–2 comes from.
Built-in potential — the hill height $V_{bi}$ we shave down.
Boltzmann distribution — the exponential of Step 3.
Thermal voltage $V_T$ — the $k T / q$ scale set in Step 6.
Reverse bias behavior — Case C, the hill made taller.
Diode I-V characteristics — the full curve of Step 7.