Reverse bias behavior: Visual walkthrough — Hardware deep dive

We will assume you know only this: a PN junction is a crystal where one side has extra holes (missing electrons, we draw them as "+" mobile dots) and the other side has extra free electrons ("−" mobile dots). Everything else we build here.

Step 1 — The junction before we touch it

WHAT. Put the two sides together. Near the boundary, electrons from the N-side wander into the P-side and holes wander the other way — they meet and cancel. This leaves behind fixed, charged atoms that cannot move: negative ones on the P-side, positive ones on the N-side. That stripped-bare band is the depletion region.

WHY. Those fixed charges create an internal electric field, and that field creates a voltage hill called the built-in potential $V_{bi}$ . Nothing external is connected yet, but the junction already has a barrier. We must draw this first because reverse bias is nothing but making this barrier bigger.

PICTURE. The middle strip has no mobile dots — only fixed circled charges. The little arrow $E_{bi}$ inside it points from the N-side (+) to the P-side (−).

Step 2 — Connect the battery the reverse way

WHAT. Attach the battery's + terminal to the N-side and − terminal to the P-side. That is the definition of reverse bias. In the diode equation we then write $V < 0$ .

WHY. The + terminal pulls the N-side electrons away from the junction; the − terminal pulls the P-side holes away from the junction. Both crowds of mobile carriers retreat. Where they retreat from becomes newly bare — so the depletion region grows and exposes more fixed charge.

PICTURE. Watch the two black arrows: electrons slide right toward "+", holes slide left toward "−". The grey depletion strip visibly widens compared to Step 1.

Step 3 — Turn the field into a voltage hill

WHAT. Draw the potential energy an electron feels as it crosses from N to P. It is a hill. The built-in field made a hill of height $q V_{bi}$ ; the battery adds its own push, raising the top by $q ∣ V ∣$ , so the total climb is proportional to $V_{bi} - V$ (with $V$ negative, $- V$ is positive, so this is bigger).

WHY. Majority carriers can only cross by rolling over this hill (diffusion). A taller hill means far fewer of them make it — this is the whole reason the "forward-style" current is choked off. We need this hill picture before we can talk about who does get across.

PICTURE. Two curves: the dashed lower hill (no bias, height $\propto V_{bi}$ ) and the solid taller hill (reverse bias, height $\propto V_{bi} - V$ ). The extra height is labelled $q ∣ V ∣$ .

$Total barrier \propto built-in hill V_{bi} - battery push (V < 0, so this adds) V$

Step 4 — Who still crosses? The two carrier populations

WHAT. There are two kinds of traffic at the junction:

Majority carriers climbing up the hill (diffusion) — now nearly stopped.
Minority carriers: the few stray electrons on the P-side and stray holes on the N-side. For them the hill is a downhill slide — the field helps them across (drift).

WHY. Reverse bias kills the uphill diffusion but aids the downhill drift. So whatever tiny current survives comes entirely from minority carriers sliding down. This is the seed of the saturation current $I_{S}$ .

PICTURE. On the hill from Step 3, a big crossed-out arrow marks the blocked uphill majority flow; a small green arrow shows a minority carrier sliding downhill across the junction.

Step 5 — Why the current saturates (the bottleneck)

WHAT. Those minority carriers are created by heat: thermal energy occasionally snaps a bond and frees a pair. That happens at some fixed rate for a given temperature. The field's only job is to sweep the created ones across.

WHY (the key insight). Once the field is strong enough to catch every thermally created minority carrier before it disappears, making the field stronger cannot summon more carriers — the supply, not the field, is the limit. That is why the current flattens to a constant no matter how negative $V$ goes. It is called the saturation current $I_{S}$ .

PICTURE. A conveyor belt of "rain drops" (heat-generated carriers) drips at a fixed rate; two different battery sizes both catch every drop — same throughput. Bigger battery, same current.

Step 6 — Put it in the Shockley equation

WHAT. The general junction current (from Shockley diode equation) is:

$I = I_{S} (e^{V / (n V_{T})} - 1)$

Term by term, right where each lives:

$I = saturation current I_{S} (e^{V < 0 now / (ideality n k T / q V_{T})} - 1)$

WHY. We derived physically (Steps 4–5) that only the minority-drift trickle survives. The equation must reproduce that. Plug in a reverse voltage with $∣ V ∣ ≫ V_{T}$ : the exponent is a large negative number, so $e^{(big negative)} \to 0$ .

$I \approx I_{S} (0 - 1) = - I_{S}$

PICTURE. The $I$ – $V$ curve: a flat floor sitting at $- I_{S}$ for all $V < 0$ (the saturation shelf), rising steeply for $V > 0$ (forward). The reverse shelf is essentially flat — matching Step 5's bottleneck.

Step 7 — Degenerate case: $V = 0$ and small $∣ V ∣$

WHAT. Check the corners so the reader never hits an unshown case.

$V = 0$ : exponent $= 0$ , so $e^{0} = 1$ and $I = I_{S} (1 - 1) = 0$ . No battery, no net current. ✔
Small reverse, e.g. $V = - V_{T}$ : exponent $= - 1$ , $e^{- 1} \approx 0.368$ , so $I = I_{S} (0.368 - 1) = - 0.63 I_{S}$ . Not yet fully saturated — the shelf is still bending.
$∣ V ∣ ≫ V_{T}$ : fully on the shelf, $I = - I_{S}$ .

WHY. The " $- I_{S}$ " answer is only the limit. Between $0$ and a few $V_{T}$ the current is still climbing toward the shelf. Showing this stops the mistake "reverse current is always exactly $- I_{S}$ ."

PICTURE. Zoom on the near-origin part of the $I$ – $V$ curve: mark $V = 0$ (current 0), $V = - V_{T}$ (current $- 0.63 I_{S}$ ), and the approach to the flat $- I_{S}$ shelf.

Step 8 — The depletion width grows as $V_{bi} - V$

WHAT. Now derive how wide the bare strip gets. Poisson's equation says: integrate the fixed charge once to get the field $E$ , integrate again to get the voltage. So voltage supported $\propto W^{2}$ , while the exposed charge $\propto W$ .

WHY. More reverse voltage must be balanced by more uncovered charge (Step 2). Because voltage grows like $W^{2}$ , doubling the voltage does not double the width — the width goes as the square root:

$W = \frac{2 ε}{q} how easily each side depletes \frac{1}{N _{A}} + \frac{1}{N _{D}} total voltage across strip (V_{bi} - V)$

The doping levels $N_{A}, N_{D}$ come from Doping concentration N_A N_D.

And since the strip is a capacitor with plate gap $W$ :

$C_{j} = \frac{ε A}{W} = \frac{C _{j 0}}{1 - V / V _{bi}}$

so a wider gap means less capacitance — the principle behind Varactor diodes.

PICTURE. A plot of $W$ versus reverse voltage: a square-root curve, rising fast at first then flattening. Beside it, a shrinking parallel-plate capacitor as $V$ goes more negative.

Step 9 — Edge case at the far end: breakdown

WHAT. Keep pushing $V$ negative and eventually the field inside the thin strip is so violent that current suddenly explodes — at $V_{B R}$ . Two mechanisms:

Zener (thin, heavily doped, $≲ 5 V$ ): the field literally tears electrons out of their bonds by tunneling. $V_{B R}$ drops as $T$ rises.
Avalanche (wide, lightly doped, $≳ 6 V$ ): one fast carrier knocks loose more carriers, a chain reaction. $V_{B R}$ rises as $T$ rises.

WHY. Our saturation story (Steps 5–6) assumed the field only sweeps carriers, never creates them. Past $V_{B R}$ the field itself starts creating carriers — the bottleneck is gone, so current is no longer capped. This is the one place the flat shelf ends. See Zener diodes.

PICTURE. Extend the $I$ – $V$ curve leftward: flat shelf at $- I_{S}$ , then a sudden near-vertical plunge at $V_{B R}$ .

The one-picture summary

Everything at once: the bare junction (Step 1) widened by the battery (Step 2), the tall barrier hill (Step 3), the minority downhill trickle (Step 4) that saturates because heat sets the supply (Step 5), giving the flat $- I_{S}$ shelf in the $I$ – $V$ curve (Step 6–7), the $V_{bi} - V$ width growth (Step 8), and the breakdown cliff (Step 9).

Recall Feynman retelling of the whole walkthrough

A junction already has a little hill at its middle. Hook a battery the "wrong" way and it drags the crowds away from the hill — so the empty zone gets wider and the hill gets taller. Almost nobody can climb the taller hill, so the ordinary current stops. But a few charges are being made by heat right at the hill, and for them the hill is a downhill ride — the battery gladly sweeps them across. That trickle is all you get, and it stays the same size no matter how hard you push, because heat (not the push) decides how many charges exist. Hotter → more charges → bigger trickle. The empty zone widens like a square root of the push, which makes the junction a smaller and smaller capacitor. Push absurdly hard and the field starts ripping charges free itself — the trickle becomes a flood: breakdown.

Connections

Parent: Reverse bias behavior
Shockley diode equation — the equation Step 6 specializes
Forward bias behavior — the mirror image (hill lowers, current floods)
Depletion region & built-in potential — Steps 1, 3, 8
Doping concentration N_A N_D — sets $N_{A}, N_{D}$ in Step 8
Varactor diodes — uses the shrinking $C_{j}$ of Step 8
Zener diodes — the breakdown of Step 9