Library / Electrical / ELE-42
Electrical
Shockley diode current
Current through an ideal junction diode from the voltage across it, with a saturation current and an ideality factor.
- Ref
- ELE-42
- Rev
- v1
- Field
- Electrical
- Result
- I [mA]
The Shockley equation gives the current through an ideal semiconductor diode: I = Is[eVD/(nVT) − 1], with the thermal voltage VT = kBT/q (about 25.9 mV at 300 K) and the ideality factor n between 1 and 2. The current is almost zero (−Is) for reverse bias and rises exponentially for forward bias, about tenfold for every 60 mV at room temperature. Is is in pA and the current in mA.
Current through an ideal junction diode from the voltage across it, with a saturation current and an ideality factor. The current rises exponentially with voltage. The model leaves out breakdown, series resistance and high-injection effects.
Calculator
- Thermal voltage kBT/q25.85 mV
- Dynamic resistance, nVT/I10.3 Ω
- Voltage rise for ten times the current, nVT ln 1059.53 mV
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- At 0.5 V forward bias a diode with Is = 10 pA and n = 1 carries 2.51 mA at 300 K. Its small-signal resistance is 10.3 Ω.
- Every 59.5 mV more bias multiplies the current by ten: a silicon diode therefore seems to “turn on” at around 0.6 to 0.7 V, though the current rises smoothly from the start.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mA) | Local step | Change in result |
|---|---|---|---|---|---|---|
| VD Voltage across the diode | 0 to 0.8 V | 0 | 275,000 | 275,000 | +10% | +592 % |
| Is Saturation current | 0.1 to 1,000 pA | 0.0251 | 251 | 250.9 | +10% | +10 % |
| T Absolute temperature | 250 to 400 K | 0.01994 | 120.1 | 120.1 | +10% | −82.8 % |
| n Ideality factor | 1 to 2 | 1.584e-4 | 2.51 | 2.51 | +10% | −82.8 % |
Sweep of Voltage across the diode
| Voltage across the diode (V) | Diode current (mA) |
|---|---|
| 0 | 0 |
| 0.1 | 4.685e-7 |
| 0.2 | 2.289e-5 |
| 0.3 | 0.001096 |
| 0.4 | 0.05245 |
| 0.5 | 2.51 |
| 0.6 | 120.1 |
| 0.7 | 5,748 |
| 0.8 | 275,000 |
Sweep one variable
Variables and how each one contributes
Diode current
What the equation gives you. Everything else feeds it.
Saturation current
The saturation (reverse leakage) current of the diode, in pA. It varies strongly with temperature and device. The slider is logarithmic. Raises the current in direct proportion at a given voltage. Raising it by 5 pA (to 15 pA) moves the result by +50 %, from 2.51 to 3.765 mA.
Voltage across the diode
The voltage across the diode, in volts (positive for forward bias). Raises the current exponentially: each extra n × 60 mV or so (at room temperature) multiplies it by ten. Raising it by 0.05 V (to 0.55 V) moves the result by +592 %, from 2.51 to 17.36 mA.
Ideality factor
The ideality factor: 1 for an ideal junction, closer to 2 for many real diodes at low current. Lowers the current at a given voltage: a larger n flattens the exponential. Raising it by 0.1 (to 1.1) moves the result by −82.8 %, from 2.51 to 0.4325 mA.
Absolute temperature
The absolute temperature of the junction, in kelvin. Lowers the current here, because a higher thermal voltage flattens the exponential. In a real diode Is rises much faster with temperature than this effect, so the real current rises with temperature. Raising it by 10 K (to 310 K) moves the result by −46.4 %, from 2.51 to 1.345 mA.
Worked example with the values above
- Thermal voltage: VT = kBT/q = 25.85 mV, so VD/(nVT) = 19.34.
- Exponential: e19.34 − 1 = 2.51×108.
- I = 10 pA × 2.51×108 = 2.51 mA.
Assumptions and limits
- An ideal diode junction with a single exponential characteristic, at steady temperature.
- Low to moderate current: at high current the series resistance of the diode adds an extra voltage drop and high-level injection changes n.
Common mistakes
- Treating the forward voltage as a constant 0.7 V when the current changes a lot.
- Using Celsius for T.
- Ignoring that Is changes with temperature.
Statistics and uncertainty
The current depends on the exponential of VD/(nVT), so a 10 mV error in the voltage changes the current by about 47 % at 300 K. It is the saturation current and the ideality factor that are uncertain; fit them from two or more measured points.
Where it is used
- Modelling a diode, a transistor’s base–emitter junction or a temperature sensor.
- Calculating the bias current of a diode circuit.
Maths you need
- Thermal voltage: kBT/q, about 25.9 mV at 300 K
- Forward and reverse bias: polarity of the voltage across the junction