Diode I-V characteristic curve — Hardware notes

WHAT is the I-V curve?

The whole curve is captured by the Shockley diode equation:

$I = I_S\left(e^{V/(nV_T)} - 1\right)$

$I_S$ = reverse saturation current (tiny, ~ $10^{-12}$ A for Si)
$n$ = ideality factor (1–2, ≈1 ideal)
$V_T = kT/q$ = thermal voltage (≈ 25.85 mV at 300 K)

WHY does it look like this? (Derivation from scratch)

The PN junction has a built-in depletion region with a potential barrier. Carriers must climb this barrier to cross. The number of carriers with enough thermal energy to climb an energy barrier $\Delta E$ follows Boltzmann statistics:

$n(\text{can cross}) \propto e^{-\Delta E / kT}$

Step 1 — barrier at equilibrium. With no bias, diffusion current (carriers climbing) exactly cancels drift current (carriers rolling back). Call the balanced drift current $I_S$ .

Why this step? At equilibrium net current must be zero, so the two flows are equal and opposite — this pins the reference current to $I_S$ .

Step 2 — apply forward voltage $V$ . Forward bias lowers the barrier by energy $qV$ . New diffusion current:

$I_{\text{diff}} = I_S\, e^{qV/kT}$

Why this step? Lowering barrier energy by $qV$ multiplies the Boltzmann factor by $e^{qV/kT}$ , so exponentially more carriers cross.

Step 3 — drift current unchanged. Drift depends on minority carriers swept by the field, whose supply is limited, not on the barrier height. So it stays at $I_S$ (flowing the other way):

$I = I_{\text{diff}} - I_{\text{drift}} = I_S e^{qV/kT} - I_S = I_S\left(e^{qV/kT}-1\right)$

Step 4 — define $V_T=kT/q$ and insert ideality $n$ for real (non-ideal) recombination:

$\boxed{I = I_S\left(e^{V/(nV_T)}-1\right)}$

The knee (turn-on) voltage

There is no single true threshold — the exponential is smooth — but practically the current becomes "noticeable" around:

Silicon: $V_\gamma \approx 0.7$ V
Germanium: $\approx 0.3$ V
LED (varies): 1.8–3.3 V

Worked Examples

Common Mistakes

Flashcards

Shockley diode equation

I = I_S\left(e^{V/(nV_T)} - 1\right)

What is

V_T

and its value at 300 K?

Thermal voltage

kT/q \approx 25.85

Why is the forward I-V exponential?

Barrier-crossing carriers follow Boltzmann

e^{qV/kT}

; voltage lowers the barrier exponentially

Reverse-bias current value

\approx -I_S

(reverse saturation / leakage), tiny and roughly constant

What does

n

(ideality factor) represent?

Non-ideality from recombination; 1 (ideal) to 2

Voltage change per decade of current (60 mV rule)

V_T\ln 10 \approx 59.6

Invert Shockley for V

V = nV_T\ln(I/I_S + 1)

Why can we drop the "−1" in forward bias?

Because

e^{V/nV_T}\gg 1

once

V\gg nV_T

Silicon vs Germanium knee voltage

Si ≈ 0.7 V, Ge ≈ 0.3 V

What happens beyond reverse breakdown

V_{BR}

Large reverse current flows (avalanche/Zener); diode conducts backwards

Effect of temperature on

I_S

Increases (roughly doubles per ~10 °C), raising leakage

Recall Feynman: explain to a 12-year-old

Imagine a gate at the top of a hill and a crowd of kids trying to cross. Normally the hill is too tall — only a few super-energetic kids make it (that's the tiny leakage). If you push the gate to make the hill shorter (forward voltage), suddenly WAY more kids can climb over — and every little bit shorter you make it, ten times as many rush through. Push the wrong way and you make the hill taller, so almost nobody crosses. That "almost nobody, then suddenly a flood" is exactly the diode curve.

Connections

PN Junction — the barrier whose lowering is the exponential
Depletion Region — where the potential barrier lives
Boltzmann Distribution — source of the $e^{qV/kT}$ term
Thermal Voltage — $V_T=kT/q$
Zener Diode — engineered breakdown region
Doping — sets $I_S$ and barrier height
Diode Circuit Models — the 0.7 V approximation of this curve

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, diode ka I-V curve basically ek one-way valve ka fingerprint hai. Agar tum forward bias doge (anode positive), toh current pehle kaafi der tak lagbhag zero rehta hai, aur phir ~0.7 V ke aas-paas achanak exponentially phatt se badhne lagta hai. Isko "knee" ya turn-on voltage bolte hain. Reverse bias me current lagbhag zero (bas thoda sa leakage $-I_S$ ) rehta hai, jab tak breakdown na aa jaye.

Yeh exponential kyun? Kyunki PN junction me ek energy barrier hota hai, aur sirf wahi carriers cross kar paate hain jinke paas kaafi thermal energy ho — yeh Boltzmann $e^{-\Delta E/kT}$ follow karta hai. Jab tum forward voltage lagate ho, barrier neeche aata hai, aur har chhoti si voltage badhne pe carriers ki sankhya exponentially badhti hai. Poora rishta ek hi equation me: $I = I_S(e^{V/nV_T}-1)$ , jahan $V_T=kT/q \approx 26$ mV room temperature pe.

Ek fatafat trick yaad rakho: current ko 10 guna karne ke liye sirf ~60 mV extra voltage chahiye ( $V_T \ln 10$ ). Isiliye curve itna steep dikhta hai. Aur ek common galti se bacho — 0.7 V koi "brick wall" nahi hai; uske neeche bhi current behta hai, bas microamps me, isliye nazar nahi aata. Reverse me bhi "bilkul zero" nahi, thoda leakage hamesha rehta hai jo temperature ke saath badhta hai.

Exam aur circuits dono ke liye yeh curve base hai — rectifier, clipper, LED, Zener sab isi exponential nature se samajh aate hain. Formula ko derive karke yaad rakho, ratna mat.

Test yourself — Doping & PN Junctions

Connections