Shockley Diode Equation
I = Is × (e^(V/(n×Vt)) − 1)
Quick answer
This calculator solves I = Is × (e^(V/(n×Vt)) − 1), Vt ≈ 25.85 mV at 25°C for I. With Diode Voltage 0.6 V, Saturation Current 10 nA, Ideality Factor 0.5, the result is 1.45 × 10¹².
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Formula
Description
The Shockley diode equation describes the exponential I-V characteristic of a pn junction diode. Is is the reverse saturation current (typically pA to µA), n is the ideality factor (1 for ideal diffusion, 2 for recombination-dominated), and Vt is the thermal voltage (kT/q ≈ 25.85 mV at 25°C). At forward biases above about 100 mV, the −1 term becomes negligible and the current is purely exponential. This equation is the foundation of semiconductor device physics and is used in SPICE simulation for all diode and BJT models.
Variables
- V — Forward voltage across the diode (V)
- Is — Reverse saturation current (A)
- n — Ideality factor (1-2, dimensionless)
Practical Notes
The result is the diode current in amperes. Is roughly doubles for every 10°C temperature increase. Typical Is values: small-signal silicon diode 1-10 nA, Schottky diode 1-100 µA, LED 1-100 pA. The exponential is clamped at e^500 to prevent numerical overflow. For reverse bias, the current approaches −Is (very small). Real diodes deviate from this equation at high currents (series resistance) and high reverse voltage (breakdown).
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| V | Diode Voltage | mV | pV, nV, µV, mV, V, kV |
| Is | Saturation Current | nA | pA, nA, µA, mA, A, kA |
| n | Ideality Factor | — | — |
Step by step
- 1. What you know: Diode Voltage 0.6 V, Saturation Current 10 nA, Ideality Factor 0.5
- 2. The relation: I = Is × (e^(V/(n×Vt)) − 1), Vt ≈ 25.85 mV at 25°C
- 3. Substitute: V = 0.6 V, Is = 10 nA, n = 0.5
- 4. Result: I = 1.45 × 10¹²
Worked examples
Minimum range
Given: Diode Voltage 10 mV, Saturation Current 100 pA, Ideality Factor 0.025
Result: I = 5.25 × 10⁻⁴
Low range
Given: Diode Voltage 100 mV, Saturation Current 1 nA, Ideality Factor 0.25
Result: I = 0.005251
Typical
Given: Diode Voltage 0.6 V, Saturation Current 10 nA, Ideality Factor 0.5
Result: I = 1.45 × 10¹²
High range
Given: Diode Voltage 0.9 V, Saturation Current 100 nA, Ideality Factor 0.8
Result: I = 7.95 × 10¹¹
Maximum range
Given: Diode Voltage 9 V, Saturation Current 1 µA, Ideality Factor 8
Result: I = 7.95 × 10¹²
How I changes with Diode Voltage
| Diode Voltage (input) | I (result) |
|---|---|
| 60 mV | 1.03 × 10⁻⁶ |
| 150 mV | 0.001097 |
| 0.3 V | 120.3 |
| 0.45 V | 1.32 × 10⁷ |
| 0.6 V | 1.45 × 10¹² |
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<p><a href="https://vbatpower.com/formulas/shockley-diode/">Shockley Diode Equation</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Shockley Diode Equation?
Use I = Is × (e^(V/(n×Vt)) − 1), Vt ≈ 25.85 mV at 25°C. With Diode Voltage 0.6 V, Saturation Current 10 nA, Ideality Factor 0.5, I works out to 1.45 × 10¹².
What is the formula for Shockley Diode Equation?
The formula is I = Is × (e^(V/(n×Vt)) − 1), Vt ≈ 25.85 mV at 25°C. The Shockley diode equation describes the exponential I-V characteristic of a pn junction diode.
Related Concepts
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