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The Shockley equation models current through an idealized pn diode as a function of voltage and temperature.
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Formula
ID and IS in A; VD and VT in V; n dimensionless; T in K
| Symbol | Meaning | Unit |
|---|---|---|
| Diode current, positive forward | A | |
| Reverse saturation current parameter | A | |
| Diode voltage | V | |
| Ideality factor | dimensionless | |
| Thermal voltage kT/q | V |
I_D=I_S\left(e^{V_D/(nV_T)}-1\right),\quad V_T=\frac{kT}{q}The relation models an ideal pn junction at a specified temperature with constant IS and ideality n. At 300 K, VT≈25.85 mV. Real devices depart at high current because of series resistance and high-level injection, at breakdown, at very low current, and as temperature changes. IS is strongly temperature-dependent; this is not a substitute for a device model or datasheet. The exponent must be dimensionless.
Idealized pn diode at specified temperature; breakdown, high injection and series resistance excluded.
The diffusion model gives exponential forward current and a saturation term.
Thermal voltage
Use voltage and temperature consistently in the stated model.
Check the example current at 0.50 V and 300 K.
VT≈25.85 mV, exponent≈19.34 and ID≈0.25 mA for Is=1 pA.
Using a dimensional exponent or Celsius temperature.
The exponent must be dimensionless; VT uses absolute temperature in kelvin.
Formula cluster
The equation explains the exponential trend, not every real-device detail.
At 300 K, n=1, IS=1 pA and VD=0.50 V, VT≈25.85 mV and VD/VT≈19.34. Thus ID≈10−12(e^19.34−1)≈0.25 mA. This idealized estimate can differ from a real diode because of series resistance.
Understand the basic diode current–voltage relation and its temperature dependence.
Here is how to work through a typical Shockley diode equation (ID = IS(exp(VD/(nVT)) − 1)) task step by step:
Task
Check the example current at 0.50 V and 300 K.
Solution path
VT≈25.85 mV, exponent≈19.34 and ID≈0.25 mA for Is=1 pA.
VT=kT/q relates Boltzmann’s constant k, absolute temperature T and elementary charge q. Near 300 K it is about 25.85 mV. It sets the voltage scale for exponential current change. Use temperature in kelvin; inserting Celsius directly is physically invalid even if a spreadsheet accepts the number. Shockley is an idealized pn-junction model at a specified temperature. Breakdown, series resistance and high-level injection are excluded. Application: Understand the basic diode current–voltage relation and its temperature dependence. The numerical example At 300 K, n=1, IS=1 pA and VD=0.50 V, VT≈25.85 mV and VD/VT≈19.34. Thus ID≈10−12(e^19.34−1)≈0.25 mA. This idealized estimate can differ from a real diode because of series resistance anchors the idea in a concrete measurement case.
n scales the exponent and represents departures from ideal junction diffusion. Simple models often use values around one to two, but it depends on device and current range. Do not assume every real diode has n=1; a datasheet model or measurement may be needed for the required accuracy. Shockley is an idealized pn-junction model at a specified temperature. Breakdown, series resistance and high-level injection are excluded. The corresponding units are ID and IS in A; VD and VT in V; n dimensionless; T in K. At 300 K, n=1, IS=1 pA and VD=0.50 V, VT≈25.85 mV and VD/VT≈19.34. Thus ID≈10−12(e^19.34−1)≈0.25 mA. This idealized estimate can differ from a real diode because of series resistance. This checks that the reference quantities are compatible.
The equation predicts about 0.25 mA at 0.50 V for the selected parameters. As current rises, internal resistance drops voltage and injection effects may matter. The ideal junction equation then no longer gives the full characteristic. This number illustrates the exponential trend; it is not a substitute for a device datasheet. Shockley is an idealized pn-junction model at a specified temperature. Breakdown, series resistance and high-level injection are excluded. Scope: The relation models an ideal pn junction at a specified temperature with constant IS and ideality n. At 300 K, VT≈25.85 mV. Real devices depart at high current because of series resistance and high-level injection, at breakdown, at very low current, and as temperature changes. IS is strongly temperature-dependent; this is not a substitute for a device model or datasheet. The exponent must be dimensionless. A practical application is Understand the basic diode current–voltage relation and its temperature dependence.
An exponential argument must be dimensionless. Thermal voltage VT=kT/q has volts, as does VD, so VD/(nVT) is dimensionless because n has no unit. Temperature must be in kelvin, and VD refers to the junction. Series resistance can make this junction voltage differ from the applied terminal voltage. Shockley is an idealized pn-junction model at a specified temperature. Breakdown, series resistance and high-level injection are excluded. Check the calculation with this case: At 300 K, n=1, IS=1 pA and VD=0.50 V, VT≈25.85 mV and VD/VT≈19.34. Thus ID≈10−12(e^19.34−1)≈0.25 mA. This idealized estimate can differ from a real diode because of series resistance. Keep the model limits and units in view.
The full relation ID=IS(exp(VD/(nVT))−1) gives zero net current at VD=0. The subtraction remains relevant for small positive or negative voltages. Only when the exponent is much larger than one may exp(x)−1 be approximated by exp(x). Under strong reverse bias the ideal model approaches −IS until breakdown mechanisms matter. Shockley is an idealized pn-junction model at a specified temperature. Breakdown, series resistance and high-level injection are excluded. The relation describes The Shockley equation models current through an idealized pn diode as a function of voltage and temperature. This helps distinguish its target quantity from neighbouring quantities.
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Which formula describes Shockley diode equation?
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How do you rearrange Shockley diode equation for Thermal voltage?
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