Forward bias behavior: Worked examples — Hardware deep dive

Before we start, plain-word reminders so no symbol is unearned:

$V$ = the voltage you apply, measured P-side minus N-side. Positive $V$ = forward push, negative $V$ = reverse pull.
$I_{S}$ = reverse saturation current, the tiny leak set by minority carriers (see Reverse bias behavior). Think of it as "the drip that flows when the hill is un-climbable."
$k$ = Boltzmann constant $= 1.38 \times 1 0^{- 23}$ J/K — the conversion factor between temperature and energy; it tells you how much thermal jiggle energy one degree of temperature is worth (see Boltzmann distribution).
$q$ = elementary charge $= 1.6 \times 1 0^{- 19}$ C — the charge on one electron (or one hole).
$T$ = absolute temperature in kelvin (K).
$V_{T} = k T / q$ = the thermal voltage (see [[Thermal voltage $V_{T}$ ]]): take the thermal energy $k T$ and divide by the charge $q$ to turn "energy per carrier" into "volts." At 300 K it is about $25.9$ mV.
$n$ = the ideality factor, a plain number between 1 and 2 that says how closely a real diode follows the ideal law. $n = 1$ is the textbook ideal diode (pure diffusion). $n = 2$ describes a junction dominated by recombination inside the depletion region. Bigger $n$ = a lazier, less-steep exponential. Unless we say otherwise we take $n = 1$ ; Ex 10 shows what $n > 1$ changes.

The scenario matrix

Every question about a diode falls into one of these boxes. We will fill each one.

#	Case class	What is special	Example
A	Forward, ordinary ( $V ≫ V_{T}$ )	The $- 1$ is negligible, current is large	Ex 1
B	Invert: find $V$ for a target $I$	Use $ln$ , drop the $+ 1$	Ex 2
C	Zero bias ( $V = 0$ )	Degenerate point: $I$ must be exactly 0	Ex 3
D	Reverse bias ( $V < 0$ )	Exponential collapses, $I \to - I_{S}$	Ex 3
E	Small forward ( $V \sim V_{T}$ )	The $- 1$ is not negligible — must keep it	Ex 4
F	Ratio / decade rule	Voltage difference, $I_{S}$ cancels	Ex 5
G	Temperature change	$V_{T}$ and $I_{S}$ both move	Ex 6
H	Real circuit (diode + resistor)	Two equations, solve together	Ex 7
I	Two diodes in series	Voltages add for same current	Ex 8
J	Exam twist (limiting behaviour + breakdown)	$V \to \pm \infty$ , and the avalanche caveat	Ex 9
K	Small-signal / dynamic resistance	Linearize around an operating point	Ex 10a
L	Non-ideal diode ( $n > 1$ )	Ideality factor stretches the exponential	Ex 10b

Unless said otherwise: silicon diode, ideal ( $n = 1$ ), $I_{S} = 1 0^{- 14}$ A, $T = 300$ K so $V_{T} = 0.0259$ V.

Ex 1 — Case A: ordinary forward current

Worked example Statement

$V = 0.6$ V. Find $I$ .

Forecast: Guess the order of magnitude. Is it microamps? Milliamps? Amps? Write your guess before reading on.

Step 1. Compute the exponent $V / V_{T} = 0.6/0.0259 = 23.17$ . Why this step? The exponent is the "how many thermal-voltage-steps up the hill are we un-climbing." Everything hinges on it.

Step 2. $e^{23.17} \approx 1.16 \times 1 0^{10}$ . The $- 1$ next to it is nothing — subtracting one from ten billion. Why this step? This is exactly why Case A lets us drop the $- 1$ : $e^{V / V_{T}} ≫ 1$ .

Step 3. $I = I_{S} e^{V / V_{T}} = 1 0^{- 14} \times 1.16 \times 1 0^{10} \approx 1.16 \times 1 0^{- 4}$ A $= 0.12$ mA. Why this step? We multiply the tiny scale current $I_{S}$ by the huge Boltzmann factor to land on the actual current — $I_{S}$ sets the vertical scale, the exponential sets how far up the curve we are.

Verify: A silicon diode at $0.6$ V should carry a fraction of a milliamp — right on the ramp before the "0.7 V knee." ✔ Units: A × (dimensionless) = A. ✔

Ex 2 — Case B: what voltage for a target current?

Worked example Statement

Same diode. You want $I = 1$ mA. What $V$ do you apply?

Forecast: Will it be more or less than the $0.6$ V that gave $0.12$ mA? By how much?

Step 1. Start from $I = I_{S} (e^{V / V_{T}} - 1)$ and solve for $V$ . Since $I ≫ I_{S}$ we drop the $+ 1$ after inverting: $V = V_{T} ln (\frac{I}{I _{S}} + 1) \approx V_{T} ln (\frac{I}{I _{S}}) .$ Why this step? $ln$ is the tool that undoes $e^{\cdot}$ . We use it (and not, say, a square root) precisely because the current is an exponential of voltage.

Step 2. $\frac{I}{I _{S}} = \frac{1 0 ^{- 3}}{1 0 ^{- 14}} = 1 0^{11}$ , so $ln (1 0^{11}) = 11 ln 10 = 25.33$ . Why this step? $ln 1 0^{11} = 11 \times 2.3026$ .

Step 3. $V = 0.0259 \times 25.33 \approx 0.656$ V. Why this step? Multiplying the dimensionless log by $V_{T}$ restores volts — $V_{T}$ is the "volts per e-fold," so the number of e-folds times $V_{T}$ is the required voltage.

Verify: $1$ mA is about 8× the $0.12$ mA current, and $0.656 - 0.6 = 0.056$ V $\approx 60$ mV — one decade would be 60 mV, and 8× is just under a decade. ✔ Consistent with Ex 1.

Ex 3 — Cases C & D: zero bias and reverse bias (the degenerate + negative cases)

The figure below is the master map for this whole page. Read it like this: the horizontal axis is the voltage you apply, the vertical axis is the resulting current in milliamps. The magenta curve is the diode equation. Notice three features the text keeps returning to: (1) on the far left the curve lies flat at a tiny negative value — that is the reverse floor $- I_{S}$ ; (2) it crosses exactly through the origin (orange dot) — zero volts, zero current; (3) on the right it shoots up almost vertically — the forward explosion. Every example on this page is just "where am I sitting on this one curve?"

Figure 1 — The complete diode I–V curve. Left: flat reverse floor at $I = - I_{S}$ . Centre (orange dot): passes through the origin because of the " $- 1$ ." Right: exponential forward rise. Ex 3 lives on the left, Ex 1–2 on the right, the origin is Ex 3(i).

Worked example Statement

Find $I$ at (i) $V = 0$ , and (ii) $V = - 0.5$ V (reverse).

Forecast: At exactly zero volts, is the current zero, tiny-positive, or tiny-negative? And deep in reverse, does it keep falling forever?

Step 1 — (i) $V = 0$ . $e^{0/ V_{T}} = e^{0} = 1$ , so $I = I_{S} (1 - 1) = 0.$ Why this step? This is the whole reason the $- 1$ exists: at zero bias, diffusion current exactly equals drift current, net zero. On Figure 1 this is the orange dot sitting exactly on the origin.

Step 2 — (ii) $V = - 0.5$ V. Exponent $= - 0.5/0.0259 = - 19.3$ , so $e^{- 19.3} \approx 4.1 \times 1 0^{- 9}$ , essentially 0 next to 1. $I = I_{S} (4.1 \times 1 0^{- 9} - 1) \approx I_{S} (- 1) = - 1 0^{- 14} A .$ Why this step? A negative exponent makes $e^{V / V_{T}}$ collapse toward zero, so the bracket becomes $- 1$ . The current saturates at $- I_{S}$ — the flat floor on the far left of Figure 1.

Verify: As $V \to - \infty$ , $e^{V / V_{T}} \to 0$ and $I \to - I_{S}$ — it does not keep falling (in the ideal model); it flattens. That flat leak is exactly the reverse saturation current from Reverse bias behavior. ✔

[!mistake] Real caveat — breakdown The ideal equation says the reverse current stays a flat $- I_{S}$ forever. A real diode does not: past a certain large reverse voltage $V_{B R}$ (the breakdown voltage), carriers gain enough energy to knock loose more carriers — an avalanche — and the reverse current shoots up (nearly vertically downward on Figure 1). Zener diodes are built to run there on purpose. The Shockley equation simply does not model this region; treat $V < - V_{B R}$ as outside its domain.

Ex 4 — Case E: tiny forward bias where the "−1" matters

Worked example Statement

$V = 5$ mV $= 0.005$ V. Find $I$ . Then show why dropping the $- 1$ here would be wrong.

Forecast: Is the current close to $I_{S} e^{V / V_{T}}$ , or wildly different?

Step 1. Exponent $= 0.005/0.0259 = 0.1931$ , and $e^{0.1931} = 1.2130$ . Why this step? When $V$ is comparable to $V_{T}$ , the exponent is small and $e^{small}$ is close to 1 — not a huge number.

Step 2 (keep the $- 1$ ): $I = 1 0^{- 14} (1.2130 - 1) = 1 0^{- 14} \times 0.2130 = 2.13 \times 1 0^{- 15}$ A. Why this step? We must subtract the 1 explicitly because here it is a big fraction of $e^{V / V_{T}}$ , unlike Case A.

Step 3 (the wrong shortcut): if we'd dropped the $- 1$ : $I \approx 1 0^{- 14} \times 1.2130 = 1.21 \times 1 0^{- 14}$ A — nearly 6× too big! Why this step? This is the whole point of Case E: near $V = 0$ the $1$ we subtract is a big chunk of $e^{V / V_{T}}$ , so it cannot be ignored.

Verify: For small $V$ , $e^{V / V_{T}} - 1 \approx V / V_{T}$ (the linear start of the exponential), giving $I \approx I_{S} \cdot V / V_{T} = 1 0^{- 14} \times 0.1931 = 1.93 \times 1 0^{- 15}$ A — same ballpark as our $2.13 \times 1 0^{- 15}$ . ✔ So near zero, a diode looks almost like a resistor — a fact we cash in fully in Ex 10a.

Ex 5 — Case F: the 60 mV-per-decade rule (ratio kills $I_{S}$ )

Ex 6 — Case G: change the temperature

Worked example Statement

The same diode is warmed to $T = 350$ K. (i) What is the new $V_{T}$ ? (ii) At a fixed $V = 0.6$ V, roughly which way does the current move — up or down?

Forecast: Hotter chip — more current or less at the same voltage?

Step 1 — (i). $V_{T} = k T / q = (1.38 \times 1 0^{- 23}) (350) / (1.6 \times 1 0^{- 19}) = 0.0302$ V $= 30.2$ mV. Why this step? $V_{T}$ is directly proportional to $T$ (see Thermal voltage $V_T$); more heat, bigger thermal jiggle voltage.

Step 2 — (ii). Two things change: the exponent $V / V_{T} = 0.6/0.0302 = 19.87$ is smaller than the 23.17 at 300 K (that alone would lower current), but $I_{S}$ rises steeply with temperature. Why this step? You must track both $V_{T}$ and $I_{S}$ — this is what makes Case G a trap.

Step 3 — where the " $I_{S}$ doubles every ~10 K" rule comes from. $I_{S}$ is proportional to the intrinsic carrier density squared, $n_{i}^{2} \propto T^{3} e^{- E_{g} / (k T)}$ , where $E_{g}$ is the bandgap. The exponential $e^{- E_{g} / (k T)}$ dominates. Its fractional change per kelvin is $\frac{1}{I _{S}} \frac{d I _{S}}{d T} \approx \frac{E _{g}}{k T ^{2}} .$ Why this step? To know the direction of the current change we need the growth rate of $I_{S}$ ; differentiating the dominant exponential $e^{- E_{g} / (k T)}$ gives exactly this fractional slope. For silicon $E_{g} \approx 1.12$ eV $= 1.12 \times 1.6 \times 1 0^{- 19}$ J at $T = 300$ K: $\frac{E _{g}}{k T ^{2}} = \frac{1.792 \times 1 0 ^{- 19}}{( 1.38 \times 1 0 ^{- 23} ) ( 300 ) ^{2}} \approx 0.144 per K .$ A doubling means $ln 2 = 0.693$ of fractional growth, so it takes $0.693/0.144 \approx 4.8$ K to double — a few K, of the same order as the popular "~10 K" rule of thumb (the rule is loose because $E_{g}$ , the $T^{3}$ prefactor, and recombination all shift it). The takeaway is rigorous even if the exact number is approximate: $I_{S}$ grows exponentially fast with $T$ .

Verify: $V_{T}$ went $25.9 \to 30.2$ mV, a $+ 16.6%$ rise — matches $350/300 = 1.167$ . ✔ The $I_{S}$ growth swamps the $V_{T}$ change, so at fixed voltage the forward current rises with temperature (equivalently, the turn-on voltage drops by the well-known $\approx - 2$ mV/K). ✔

Ex 7 — Case H: diode in series with a resistor (a real circuit)

The next figure shows how a circuit picks one point on the diode curve. The magenta curve is the same diode law as Figure 1 (zoomed into the forward region). The dashed violet line is the load line: for a battery $V_{B} = 5$ V through resistor $R = 1 k Ω$ , Kirchhoff forces $I = (V_{B} - V_{D}) / R$ , a straight line sloping down. The diode can only sit where both are satisfied at once — the orange dot where the two lines cross. That crossing is the answer.

Figure 2 — Graphical circuit solution. Magenta = diode's own $I$ – $V$ curve; dashed violet = the resistor's load line for a 5 V supply and 1 kΩ; orange dot = the unique operating point where they intersect ( $V_{D} \approx 0.69$ V, $I \approx 4.3$ mA).

Worked example Statement

A $5$ V battery drives the same diode through a $1 k Ω$ resistor. Estimate the current $I$ and the diode voltage $V_{D}$ .

Forecast: The diode "wants" ~0.7 V. Guess $I$ before solving.

Step 1 — two equations. Kirchhoff around the loop: $5 = I R + V_{D}, I = I_{S} e^{V_{D} / V_{T}} .$ Why this step? One equation is the straight-line load line (the resistor + battery), the other is the diode's exponential curve. The true operating point is where they cross — the orange dot in Figure 2.

Step 2 — guess-and-refine. Try $V_{D} = 0.7$ V: $I = (5 - 0.7) /1000 = 4.3$ mA. Check the diode: $V_{D} = V_{T} ln (I / I_{S}) = 0.0259 ln (4.3 \times 1 0^{- 3} /1 0^{- 14}) = 0.0259 \times 26.79 = 0.694$ V. Why this step? We plug the current back into the diode law to see if $0.7$ V was consistent. $0.694 \approx 0.70$ — close enough after one pass.

Step 3. Refine once: with $V_{D} = 0.694$ V, $I = (5 - 0.694) /1000 = 4.31$ mA. Stable. Why this step? Feeding the corrected $V_{D}$ back into the load line confirms the answer no longer moves — the two equations are now simultaneously satisfied.

Verify: Answer $I \approx 4.3$ mA, $V_{D} \approx 0.69$ V. Sanity: the resistor drops $5 - 0.69 = 4.31$ V, and $4.31 V /1000 Ω = 4.31$ mA. ✔ Self-consistent. The $0.7$ V "constant-drop" model was an excellent starting guess exactly because the exponential is so steep that $V_{D}$ barely moves.

Ex 8 — Case I: two identical diodes in series

Ex 9 — Case J: the exam twist — limiting behaviour (and breakdown caveat)

Worked example Statement

Without a calculator, describe $I$ in three limits: (i) $V \to + \infty$ , (ii) $V \to - \infty$ , (iii) $V \to 0$ . State the shape of the whole $I$ – $V$ curve.

Forecast: Sketch it in your head first (or glance at Figure 1).

Step 1 — (i) $V \to + \infty$ . $e^{V / V_{T}} \to \infty$ , so $I \to I_{S} e^{V / V_{T}} \to \infty$ : current explodes exponentially with no upper bound (until the diode physically burns). Why this step? The exponential dominates; the $- 1$ is invisible.

Step 2 — (ii) $V \to - \infty$ . In the ideal model $e^{V / V_{T}} \to 0$ , so $I \to I_{S} (0 - 1) = - I_{S}$ : a flat, tiny negative floor. Why this step? Negative exponent → the exponential dies → the $- 1$ is all that's left.

Caveat (real device): this ideal flat floor holds only until the reverse voltage reaches the breakdown value $V_{B R}$ . Beyond that, avalanche multiplication makes the reverse current shoot up sharply (see the [!mistake] callout in Ex 3). So "reaches $- I_{S}$ and stays there forever" is a statement about the equation, not about a real diode.

Step 3 — (iii) $V \to 0$ . $I \to I_{S} (1 - 1) = 0$ , and near zero $I \approx (I_{S} / V_{T}) V$ — a gentle straight-line start. Why this step? This ties Cases C and E together: the curve leaves the origin linearly, then bends into the exponential.

Verify: The three limits describe exactly the diode I–V curve of Diode I-V characteristics: flat $- I_{S}$ floor on the left (until breakdown), through the origin, exploding on the right. ✔ All three cells (C, D, A) reappear as the ends of one smooth curve — the matrix is complete.

Ex 10 — Cases K & L: small-signal resistance and the non-ideal diode

Worked example 10a — Dynamic (small-signal) resistance

A diode carries a DC current $I = 1$ mA. If you add a tiny wiggle of voltage on top, how much does the current wiggle? I.e. find the diode's dynamic resistance $r_{d}$ at this operating point.

Forecast: Will $r_{d}$ be the same everywhere on the curve, or depend on the current?

Step 1 — what "resistance to a small wiggle" means. For a small change, resistance is the slope's reciprocal, $r_{d} = d V / d I$ . We use the derivative here (and not $V / I$ ) because the curve is not a straight line through the origin — only the local slope tells you how a small wiggle responds. Why this step? Ex 4 already hinted the diode looks "almost like a resistor" over a small range; the derivative makes that exact.

Step 2 — differentiate the diode law. For $I \approx I_{S} e^{V / V_{T}}$ , $\frac{d I}{d V} = \frac{I _{S}}{V _{T}} e^{V / V_{T}} = \frac{I}{V _{T}} ⟹ r_{d} = \frac{d V}{d I} = \frac{V _{T}}{I} .$ Why this step? The exponential's derivative is itself times $1/ V_{T}$ , so the slope is just the current over $V_{T}$ . Clean and universal.

Step 3 — plug in. $r_{d} = V_{T} / I = 0.0259/1 0^{- 3} = 25.9 Ω$ . Why this step? Substituting the operating current turns the general slope formula into a concrete resistance for this bias point.

Verify: At $I = 1$ mA, $r_{d} = 25.9 Ω$ . Sanity: at ten times the current (10 mA) the diode is ten times stiffer, $r_{d} = 2.59 Ω$ — so $r_{d}$ is not constant; it shrinks as current grows. That matches the curve getting steeper as you climb it in Figure 1. ✔ (This is why Ex 4's "resistor" value $V_{T} / I_{S}$ only applies at the near-zero operating point.)

Worked example 10b — Non-ideal diode (

$n > 1$ ) A real diode has ideality factor $n = 2$ , $I_{S} = 1 0^{- 14}$ A. What voltage gives $I = 1$ mA now, and what is the new decade-slope?

Forecast: Will it need more or less voltage than the ideal $n = 1$ diode (which needed 0.656 V)?

Step 1 — put $n$ into the law. $I = I_{S} (e^{V / (n V_{T})} - 1)$ , so inverting, $V = n V_{T} ln (I / I_{S})$ . Why this step? The ideality factor multiplies $V_{T}$ , stretching the voltage axis of the exponential by a factor $n$ .

Step 2 — plug in. $V = 2 \times 0.0259 \times ln (1 0^{11}) = 2 \times 0.0259 \times 25.33 = 1.312$ V. Why this step? Everything is exactly the ideal Ex 2 answer, doubled, because $n = 2$ doubles the volts-per-e-fold.

Step 3 — new decade slope. $Δ V$ per decade $= n V_{T} ln 10 = 2 \times 0.0596 = 0.119$ V $\approx 120$ mV/decade. Why this step? A lazier exponential needs more volts per 10× of current; scaling Ex 5's result by $n$ gives the new slope.

Verify: $n = 2$ needs $1.312$ V vs $0.656$ V ideal — exactly double ✔, and the slope $120$ mV/decade is double the ideal 60 ✔. So $n > 1$ makes a diode "softer": same current costs more voltage, and the curve is less steep. Real silicon diodes often sit near $n \approx 1$ – $2$ depending on current range.

Recall

Recall Which case forces you to keep the "−1"?

Case E — small forward bias ( $V \sim V_{T}$ ). Here $e^{V / V_{T}}$ is close to 1, so subtracting 1 changes the answer a lot; only for $V ≫ V_{T}$ can you drop it.

Recall In a diode-plus-resistor circuit, what geometric picture finds the current?

The operating point is where the resistor's straight load line crosses the diode's exponential curve (the orange dot in Figure 2).

Recall Why does the 60 mV/decade rule not depend on

$I_{S}$ ? Because it is a ratio of currents: taking $Δ V = V_{T} ln (I_{2} / I_{1})$ cancels $I_{S}$ entirely. (With ideality it becomes $n V_{T} ln 10$ .)

Recall What is the dynamic resistance of a diode carrying current

$I$ ? $r_{d} = V_{T} / I$ — it is not constant; it shrinks as the current grows.

Recall What does the ideality factor

$n$ do to the diode curve? It multiplies $V_{T}$ : bigger $n$ stretches the voltage axis, so the same current costs more voltage and the decade-slope becomes $n V_{T} ln 10$ .

Recall Does the ideal Shockley equation describe reverse breakdown?

No — it predicts a flat $- I_{S}$ forever. Real diodes avalanche past $V_{B R}$ ; that region is outside the equation's domain.

Connections

Forward bias behavior — the parent note these examples drill.
Diode I-V characteristics — the full curve Figure 1 and Ex 9 sketch.
Thermal voltage $V_T$ — the $k T / q$ used in every example, and the star of Ex 6 and Ex 10a.
Reverse bias behavior — the $- I_{S}$ floor and breakdown seen in Ex 3 and Ex 9.
Boltzmann distribution — origin of the exponential and the $k$ that sets $V_{T}$ .
Built-in potential — the barrier the applied $V$ fights.
2.2.07 Forward bias behavior (Hinglish) — same ideas in Hinglish.