Forward bias behavior: Exercises — Hardware deep dive

Before the toolkit, one symbol you must own — it appears in every formula below:

The two tools you will reuse constantly:

Prerequisite links if a step feels shaky: Thermal voltage $V_T$, Diode I-V characteristics, Boltzmann distribution, and the parent Forward bias behavior.

Level 1 — Recognition

Can you spot the right tool and plug in?

L1.1 — Which polarity is forward?

A battery's (+) terminal is wired to the N-side of a silicon diode, and (−) to the P-side. Is the diode forward or reverse biased? What happens to the depletion region?

Recall Solution

Forward bias requires (+) → P and (−) → N. Here it is the opposite: (+) → N. So the diode is reverse biased. The external field now adds to the built-in field (both point N→P), so more ions get exposed and the depletion region widens. Almost no current flows (only $- I_{S}$ ). See Reverse bias behavior.

L1.2 — Read the thermal voltage

Compute $V_{T}$ at $T = 350$ K.

Recall Solution

$V_{T} = \frac{k T}{q} = \frac{( 1.38 \times 1 0 ^{- 23} ) ( 350 )}{1.6 \times 1 0 ^{- 19}} = 0.0302 V \approx 30.2 mV$ Why it grew: $V_{T} \propto T$ . Hotter carriers jiggle harder, so the natural energy-per-charge scale is larger.

L1.3 — Straight substitution

Silicon diode: $I_{S} = 1 0^{- 15}$ A, $V = 0.65$ V, $T = 300$ K. Find $I$ .

Recall Solution

$\frac{V}{V _{T}} = \frac{0.65}{0.0259} = 25.10$ $I \approx I_{S} e^{V / V_{T}} = 1 0^{- 15} e^{25.10} = 1 0^{- 15} \times 8.03 \times 1 0^{10} \approx 8.03 \times 1 0^{- 5} A \approx 0.080 mA$ The $- 1$ was dropped because $e^{25.10} \approx 8 \times 1 0^{10} ≫ 1$ .

Level 2 — Application

Invert, rearrange, chain the tools together.

L2.1 — Voltage for a target current

Diode with $I_{S} = 2 \times 1 0^{- 14}$ A. What forward voltage $V$ gives $I = 5$ mA at 300 K?

Recall Solution

Invert Shockley (drop the $+ 1$ since $I ≫ I_{S}$ ): $V = V_{T} ln (\frac{I}{I _{S}}) = 0.0259 ln (\frac{5 \times 1 0 ^{- 3}}{2 \times 1 0 ^{- 14}})$ $\frac{I}{I _{S}} = 2.5 \times 1 0^{11}, ln (2.5 \times 1 0^{11}) = 26.25$ $V = 0.0259 \times 26.25 \approx 0.680 V$ Feynman check: a silicon diode turns on near 0.6–0.7 V. ✔

L2.2 — Ratio of two currents (no $I_{S}$ needed)

The same diode is driven at $V_{1} = 0.60$ V then at $V_{2} = 0.72$ V. By what factor does the current grow?

Recall Solution

Because $I \approx I_{S} e^{V / V_{T}}$ , the unknown $I_{S}$ cancels in a ratio: $\frac{I _{2}}{I _{1}} = e^{(V_{2} - V_{1}) / V_{T}} = e^{(0.72 - 0.60) /0.0259} = e^{0.12/0.0259} = e^{4.633} \approx 102.8$ The current grows about 100× — consistent with the 60 mV/decade rule: $0.12$ V is $2$ decades, i.e. $1 0^{2}$ .

L2.3 — Find $I_{S}$ from a measurement

A diode carries $I = 1$ mA at $V = 0.62$ V, $T = 300$ K. Find its reverse saturation current $I_{S}$ .

Recall Solution

Start from the exact Shockley equation $I = I_{S} (e^{V / V_{T}} - 1)$ . Here $V / V_{T} = 0.62/0.0259 = 23.94$ , so $e^{V / V_{T}} \approx 2.5 \times 1 0^{10}$ , which is vastly bigger than $1$ . Subtracting $1$ from $2.5 \times 1 0^{10}$ changes it by less than $1 0^{- 8} %$ — utterly negligible. So we may safely use $I \approx I_{S} e^{V / V_{T}}$ and solve for $I_{S}$ : $I_{S} = I e^{- V / V_{T}} = 1 0^{- 3} e^{- 0.62/0.0259} = 1 0^{- 3} e^{- 23.94}$ $e^{- 23.94} = 4.00 \times 1 0^{- 11} \Rightarrow I_{S} \approx 4.00 \times 1 0^{- 14} A$ A tiny value — exactly as expected, since $I_{S}$ is set by minority carriers.

Level 3 — Analysis

Reason about slopes, decades, and behaviour of the curve.

L3.1 — The 60 mV/decade slope

Starting from $I \approx I_{S} e^{V / V_{T}}$ , show that the voltage increase needed to raise current by a factor of 10 is $V_{T} ln 10$ , and evaluate it at 300 K.

Recall Solution

Take two points with $I_{2} / I_{1} = 10$ : $\frac{I _{2}}{I _{1}} = e^{(V_{2} - V_{1}) / V_{T}} = 10 \Rightarrow \frac{Δ V}{V _{T}} = ln 10 \Rightarrow Δ V = V_{T} ln 10$ $Δ V = 0.0259 \times 2.3026 = 0.05964 V \approx 59.6 mV$ This is the famous "~60 mV per decade".

Read figure s01. The horizontal axis is diode voltage $V$ in volts; the vertical axis is current $I$ in amps drawn on a logarithmic scale (each gridline up is ×10). The cyan curve is $I = I_{S} e^{V / V_{T}}$ — on a log axis it becomes a perfectly straight line. The two amber dots sit one decade apart (a factor of 10 in current); the amber dashed steps between them show the horizontal gap is exactly $Δ V \approx 60$ mV. That constant horizontal step per decade is the slope $1/ (V_{T} ln 10)$ decades per volt.

L3.2 — Temperature shift at fixed current

A diode holds $I = 1$ mA. Its required $V$ at 300 K is $0.65$ V. Assuming $I_{S}$ unchanged, does the required voltage rise or fall when heated to 320 K? Compute the new $V$ (ignore $I_{S}$ 's own temperature dependence for this idealised part).

Recall Solution

With $I_{S}$ frozen, $V = V_{T} ln (I / I_{S})$ , and $V_{T} \propto T$ . First back out $ln (I / I_{S})$ at 300 K: $ln (I / I_{S}) = V / V_{T} = 0.65/0.0259 = 25.10$ . At 320 K: $V_{T}^{'} = k T^{'} / q = (1.38 \times 1 0^{- 23}) (320) / (1.6 \times 1 0^{- 19}) = 0.0276$ V. $V^{'} = V_{T}^{'} \times 25.10 = 0.0276 \times 25.10 \approx 0.693 V$ So with $I_{S}$ frozen the required voltage rises. Real-world caveat: In an actual diode $I_{S}$ climbs steeply with $T$ (roughly doubling every ~10 K). That effect dominates and makes the real required $V$ fall by about $- 2$ mV/K. This part isolated only the $V_{T}$ term to show it alone pushes the other way.

L3.3 — Where is the $- 1$ actually needed?

For a diode with $I_{S} = 1 0^{- 14}$ A at 300 K, at what forward voltage does the true current $I_{S} (e^{V / V_{T}} - 1)$ differ from the approximation $I_{S} e^{V / V_{T}}$ by exactly $1%$ ?

Recall Solution

Relative error of dropping the $- 1$ : $error = \frac{I _{S} e ^{V / V_{T}} - I _{S} ( e ^{V / V_{T}} - 1 )}{I _{S} ( e ^{V / V_{T}} - 1 )} = \frac{1}{e ^{V / V_{T}} - 1}$ Set this $= 0.01$ : $e^{V / V_{T}} - 1 = 100 \Rightarrow e^{V / V_{T}} = 101 \Rightarrow \frac{V}{V _{T}} = ln 101 = 4.615$ $V = 0.0259 \times 4.615 \approx 0.1196 V \approx 0.120 V$ Reading: above ~0.12 V the $- 1$ matters less than 1%. That is why we always drop it in the forward-on region (0.6–0.7 V) — there the error is astronomically tiny. This is the formal justification for the shortcut used back in L2.3.

Level 4 — Synthesis

Combine the diode with an external circuit — the equation no longer solves by itself.

L4.1 — Diode + series resistor (load line)

A $5$ V source drives a diode in series with $R = 1 k Ω$ . The diode has $I_{S} = 1 0^{- 14}$ A at 300 K. Estimate the diode voltage $V_{D}$ and current $I$ .

Recall Solution

Two equations, two unknowns ( $V_{D}$ , $I$ ):

Circuit (KVL / load line): $5 = I R + V_{D} \Rightarrow I = \frac{5 - V _{D}}{1000}$ .
Device: $I = I_{S} e^{V_{D} / V_{T}}$ .

These meet where the load line crosses the exponential (figure s02). Solve by iteration:

Guess $V_{D} = 0.7$ V $\Rightarrow I_{circuit} = (5 - 0.7) /1000 = 4.30$ mA. Diode needs $V_{D} = V_{T} ln (I / I_{S}) = 0.0259 ln (4.30 \times 1 0^{- 3} /1 0^{- 14}) = 0.0259 \times 26.79 = 0.694$ V.
Update $V_{D} = 0.694$ V $\Rightarrow I = 4.306$ mA, diode voltage $= 0.0259 ln (4.306 \times 1 0^{- 3} /1 0^{- 14}) = 0.694$ V. Converged.

$V_{D} \approx 0.694 V, I \approx 4.31 mA$ The resistor "absorbs" the rest: $I R = 4.31 \times 1000 \times 1 0^{- 3} = 4.31$ V, and $4.31 + 0.69 = 5.00$ V. ✔

Read figure s02. The horizontal axis is diode voltage $V$ in volts; the vertical axis is current $I$ in milliamps (linear this time). The cyan curve is the diode's own $I = I_{S} e^{V / V_{T}}$ , shooting up near 0.7 V. The amber straight line is the load line $I = (5 - V) /1000$ , sloping down from 5 mA at $V = 0$ to zero at $V = 5$ V. The single white dot where they cross is the only $(V, I)$ pair that satisfies both the diode and the circuit — that is the operating point $V_{D} \approx 0.694$ V, $I \approx 4.31$ mA.

L4.2 — Two diodes in series

Two identical diodes ( $I_{S} = 1 0^{- 14}$ A each) are stacked in series and forced to carry $I = 2$ mA at 300 K. What total voltage appears across the pair?

Recall Solution

Same current flows through both (series). Each drops: $V_{1} = V_{T} ln (\frac{I}{I _{S}}) = 0.0259 ln (\frac{2 \times 1 0 ^{- 3}}{1 0 ^{- 14}}) = 0.0259 \times 26.02 = 0.6740 V$ Two identical drops in series: $V_{total} = 2 V_{1} \approx 1.348 V$ Insight: doubling diodes doubles the voltage (each still exponential), whereas doubling the current only adds $V_{T} ln 2 \approx 18$ mV per diode.

Level 5 — Mastery

Design and derive — no template to copy.

L5.1 — Design a current source point

You want a silicon diode to sit at exactly $I = 3$ mA. You measure that at $V = 0.60$ V it carries $I_{0} = 0.5$ mA (300 K). What voltage delivers the $3$ mA, and what is this diode's $I_{S}$ ?

Recall Solution

Step 1 — voltage, using the ratio trick (no $I_{S}$ yet): $V = V_{0} + V_{T} ln (\frac{I}{I _{0}}) = 0.60 + 0.0259 ln (\frac{3}{0.5}) = 0.60 + 0.0259 \times 1.7918$ $V = 0.60 + 0.0464 = 0.6464 V \approx 0.646 V$ Step 2 — extract $I_{S}$ from the known point: $I_{S} = I_{0} e^{- V_{0} / V_{T}} = 0.5 \times 1 0^{- 3} e^{- 0.60/0.0259} = 0.5 \times 1 0^{- 3} e^{- 23.166}$ $e^{- 23.166} = 8.68 \times 1 0^{- 11} \Rightarrow I_{S} \approx 4.34 \times 1 0^{- 14} A$ Both answers are self-consistent: plug $I_{S}$ and $V = 0.646$ V back into Shockley and you recover $3$ mA.

L5.2 — Derive the small-signal resistance $r_{d}$

Around an operating current $I$ , a diode looks like a tiny resistor to small wiggles. Derive $r_{d} = d V / d I$ and evaluate it at $I = 1$ mA, 300 K.

Recall Solution

Why a derivative — and why it is the small-signal resistance. Resistance is defined as "voltage change per unit current change," $Δ V /Δ I$ . On a straight I–V line (an ohmic resistor) this ratio is the same everywhere, so one number describes it. A diode's I–V curve is bent (exponential), so the ratio $Δ V /Δ I$ depends on where you sit. If you only wiggle by a tiny amount around an operating point, the curve looks locally straight, and the ratio of a tiny $d V$ to the tiny $d I$ it produces is exactly the slope's reciprocal at that point — that is $d V / d I$ , a derivative. So the small-signal resistance is the local inverse slope of the Shockley curve.

Start from $I \approx I_{S} e^{V / V_{T}}$ and differentiate with respect to $V$ : $\frac{d I}{d V} = I_{S} \cdot \frac{1}{V _{T}} e^{V / V_{T}} = \frac{1}{V _{T}} = I I_{S} e^{V / V_{T}} = \frac{I}{V _{T}}$ (we used $I_{S} e^{V / V_{T}} = I$ ). The small-signal resistance is the reciprocal of this slope: $r_{d} = \frac{d V}{d I} = \frac{V _{T}}{I}$ Evaluate at $I = 1$ mA, 300 K: $r_{d} = \frac{0.0259}{1 0 ^{- 3}} = 25.9 Ω$ Beautiful result: the small-signal resistance depends only on current (through $V_{T} / I$ ), not on $I_{S}$ or the material details. Halve the current, double the resistance.

L5.3 — Power dissipated at an operating point

The diode of L4.1 sits at $V_{D} = 0.694$ V, $I = 4.31$ mA. How much power does the diode dissipate, and how much does the resistor?

Recall Solution

Diode power (its own voltage × its own current): $P_{D} = V_{D} I = 0.694 \times 4.31 \times 1 0^{- 3} = 2.99 \times 1 0^{- 3} W \approx 2.99 mW$ Resistor power: $P_{R} = I^{2} R = (4.31 \times 1 0^{- 3})^{2} \times 1000 = 1.858 \times 1 0^{- 2} W \approx 18.6 mW$ Source delivers $P_{S} = 5 \times 4.31 \times 1 0^{- 3} = 21.55$ mW, and $2.99 + 18.6 = 21.6$ mW ✔ (rounding). Energy conserved.

Recall Checkpoints

Recall What cancels when you take a current

ratio on one diode? $I_{S}$ cancels: $I_{2} / I_{1} = e^{(V_{2} - V_{1}) / V_{T}}$ . You never need $I_{S}$ for ratio/decade questions.

Recall Why does a diode-plus-resistor need iteration?

The equation $V_{src} = I R + V_{T} ln (I / I_{S})$ mixes a linear term and a log term — transcendental, no closed form. Iterate or use the load line.

Recall What is the small-signal resistance of a diode and what does it depend on?

$r_{d} = V_{T} / I$ — depends only on the operating current, not on $I_{S}$ or material.

Ratio of currents from two voltages

I_{2} / I_{1} = e^{(V_{2} - V_{1}) / V_{T}}

Small-signal diode resistance

r_{d} = V_{T} / I

Voltage step for one decade of current

V_{T} ln 10 \approx 60

mV at 300 K

Above what voltage is the "

- 1

" negligible (<1%)?

About

0.12

V for a typical Si diode

Connections

Forward bias behavior — the parent theory these exercises drill.
Diode I-V characteristics — the curve every problem samples.
Thermal voltage $V_T$ — the $k T / q$ scale in every formula.
Boltzmann distribution — the exponential's origin.
Reverse bias behavior — L1.1 lands here.
Built-in potential — barrier the bias fights.
2.2.07 Forward bias behavior (Hinglish) — same content in Hinglish.