Forward bias behavior: Question bank — Hardware deep dive

The traps below attack the exact places forward bias fools people: the shape of the curve, the meaning of "turn-on", what the depletion region does, and the boundary cases at $V = 0$ and beyond.

Definition Symbols used on this page (define before you use)

$V$ — the applied voltage across the diode; positive means forward (battery + on the P-side).
$q$ — the charge of one electron, $q \approx 1.6 \times 1 0^{- 19}$ C. It converts an energy into a voltage.
$k$ — Boltzmann's constant, $k \approx 1.38 \times 1 0^{- 23}$ J/K; it links temperature to energy.
$T$ — absolute temperature in kelvin (room temperature $\approx 300$ K).
$V_{T} = k T / q$ — the thermal voltage, the voltage that a single thermal jiggle of energy $k T$ "is worth" ( $\approx 25.9$ mV at 300 K). It sets how sharp the exponential is. See Thermal voltage $V_T$.
$I_{S}$ — the reverse saturation current: a tiny current (set by minority carriers) that fixes the overall size of the curve.
$V_{bi}$ — the built-in potential: the internal barrier voltage already present at equilibrium, before any battery is connected. See Built-in potential.
The device obeys the Shockley equation $I = I_{S} (e^{V / V_{T}} - 1)$ , which the parent note derives.

True or false — justify

Every answer here must give the reason, never a bare "true/false".

A forward-biased diode obeys Ohm's law once it turns on.

False — the current is

I \approx I_{S} e^{V / V_{T}}

, an exponential, so there is no fixed

I / V

ratio; the "resistance" changes at every point. See Diode I-V characteristics.

Below 0.7 V a silicon diode carries no current.

False — current flows at every

V > 0

; it is just exponentially small below ~0.6 V, so 0.7 V is where it becomes practically large, not where it starts.

Forward bias widens the depletion region.

False — the applied field opposes the built-in field, so fewer exposed ions are needed and the region narrows; widening is what reverse bias does.

At exactly

V = 0

no charge carriers cross the junction.

False — diffusion and drift currents are both large but equal and opposite, so only the net current is zero; carriers are constantly crossing.

Doubling the forward voltage doubles the current.

False — because

I \propto e^{V / V_{T}}

, doubling

V

squares the exponential factor, multiplying

I

by an astronomically larger amount (potentially billions).

The "−1" in

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

matters in strong forward bias.

False — there

e^{V / V_{T}} ≫ 1

so the

- 1

is negligible; it only matters near

V = 0

(to force

I = 0

) and in reverse bias (to give

I \to - I_{S}

Raising temperature always lowers the forward voltage needed for a given current.

True in practice — although

V_{T} = k T / q

rises with

T

I_{S}

rises far faster, so the net effect drops the required

V

by roughly

2 mV /^{\circ} C

The built-in potential

V_{bi}

appears as a usable voltage across the diode's terminals.

False —

V_{bi}

is an internal equilibrium barrier from the space-charge field; you cannot extract it as EMF because the contact potentials around the loop cancel it. See Built-in potential.

Spot the error

Each statement contains one flawed step — name it in the answer.

I \propto e^{q V / k T}

, so if I make

V

negative I get a large negative current."

The exponential only shrinks toward 0 for negative

V

; the reverse current saturates at

- I_{S}

from the

- 1

term, it does not grow large negatively.

"Since forward bias lowers the barrier to

q (V_{bi} - V)

, at

V = V_{bi}

the barrier is zero so current is infinite."

As current grows, voltage is dropped across the resistance of the neutral bulk semiconductor (series resistance), so most of the applied

V

falls on that resistance rather than the junction; the junction voltage self-limits well below

V_{bi}

and the current stays finite. See Diode I-V characteristics.

V_{T} = k T / q

so at 300 K it's about 25 V."

Unit slip — the numbers give

\approx 0.0259

volts = 25.9 mV, not 25 V; see [[Thermal voltage

V_{T}

]].

"Every extra 60 mV doubles the current."

It gives one decade (×10), not ×2, because

V_{T} ln 10 \approx 60

mV; a ×2 change needs only

V_{T} ln 2 \approx 18

mV.

"Forward bias adds the battery voltage to

V_{bi}

, so the total barrier is

V_{bi} + V

The applied field opposes the built-in field, so voltages subtract: the net barrier is

V_{bi} - V

, which is why current turns on.

"The exponential comes from Ohm's law scaled up."

It comes from Boltzmann statistics — the fraction of carriers with thermal energy above a barrier of height

q V_{B}

goes as

e^{- q V_{B} / k T}

; Ohm's law never enters.

I_{S}

is set by the majority carriers because there are so many of them."

I_{S}

is set by minority carriers (drift/generation current); majority carriers drive the diffusion term, not the saturation current.

Why questions

Why does forward current rise exponentially rather than linearly with voltage?

Because only carriers with thermal energy above the barrier cross, and that fraction follows Boltzmann

e^{- q V_{B} / k T}

; lowering the barrier by

V

multiplies the surviving fraction by

e^{q V / k T}

Why is a single equation (

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

) able to describe both forward and reverse bias?

The exponential dominates forward (

≫ 1

, current explodes) while the

- 1

dominates reverse (

e^{V / V_{T}} \to 0

, current

\to - I_{S}

); the sign of

V

automatically selects the regime.

Why does the depletion region get narrower when we forward bias?

The external field cancels part of the built-in field, so fewer exposed ionised dopants are required to sustain the (now smaller) net field, shrinking the space-charge width.

Why does the natural voltage scale of a diode come out to

k T / q

Carriers jiggle with thermal energy

\sim k T

; converting an energy to a voltage means dividing by charge

q

, so

k T / q

is the voltage a thermal jiggle "is worth". See [[Thermal voltage

V_{T}

]].

Why does the "−1" guarantee zero current at zero bias?

V = 0

e^{0} = 1

, so

I_{S} (1 - 1) = 0

exactly — it encodes that diffusion current perfectly cancels drift current at equilibrium.

Why is the built-in barrier what stops diffusion at equilibrium instead of letting carriers keep flowing?

As carriers diffuse they leave charged ions that build an opposing field; this field grows until its drift exactly balances diffusion, freezing the net flow.

Why do we say the diode "turns on" near 0.6–0.7 V if current exists at all voltages?

Because the exponential is so steep that current changes from microscopic to milliamps over a narrow window; "turn-on" is a practical label for where the curve visibly lifts, not a physical threshold.

Edge cases

At exactly

V = 0

, what is the net current and why?

Exactly zero — the diffusion term

I_{S} e^{0} = I_{S}

is cancelled by the drift term

- I_{S}

; carriers still cross both ways in balance.

V \to - \infty

(deep reverse), what does

I

approach?

I \to - I_{S}

, the reverse saturation current, because

e^{V / V_{T}} \to 0

leaves only the

- 1

term times

I_{S}

What happens to the shape of the I–V curve as

T \to 0

V_{T} = k T / q \to 0

makes the exponential infinitely sharp, so under the ideal Shockley equation the current jumps at

V = 0

— any

V > 0

gives a blow-up while any

V < 0

gives

- I_{S}

; the finite thermal voltage at real temperatures is what smears this into a smooth curve.

I_{S}

were somehow zero, what current would flow at any forward voltage?

Zero —

I = 0 \cdot (e^{V / V_{T}} - 1) = 0

;

I_{S}

sets the entire magnitude, and no minority carriers means nothing to inject.

In the limit

V ≫ V_{T}

, is the

- 1

ever safe to keep, and does dropping it change the answer noticeably?

Keeping it is always correct; dropping it is a fine approximation once

e^{V / V_{T}} ≫ 1

, introducing error far below measurement precision.

What is the current at the boundary where applied

V

just equals

V_{bi}

in an ideal Shockley model (no series resistance)?

The ideal formula gives a huge but finite

I = I_{S} (e^{V_{bi} / V_{T}} - 1)

; only real series resistance and high-injection effects then cap it — the ideal model itself never diverges.

Is there a smallest positive voltage below which current is truly zero?

No — for any

V > 0

e^{V / V_{T}} > 1

I > 0

; the current is continuous and smooth through the origin, never a hard cutoff.

Connections

Forward bias behavior — the parent note these traps drill.
PN Junction at Equilibrium — the $V = 0$ balance behind the "−1".
Reverse bias behavior — the polarity these edge cases contrast against.
Built-in potential — the barrier forward bias subtracts from.
Boltzmann distribution — why the curve is exponential.
Diode I-V characteristics — the full curve the traps live on.
Thermal voltage $V_T$ — the $k T / q$ scale two traps attack.