Your Equation

Library / Electrical / ELE-43

Electrical

Thermal voltage

Voltage that sets the scale of semiconductor junction behaviour at absolute temperature T: about 25.85 mV at 300 K.

Ref
ELE-43
Rev
v1
Field
Electrical
Result
VT [mV]

The thermal voltage VT = kBT/q is the natural voltage scale of a semiconductor junction, with kB = 1.381 × 10−23 J/K, q = 1.602 × 10−19 C and T in kelvin. It is about 25.9 mV at 300 K and rises by 86 µV for every kelvin. It sets how fast diode and transistor currents rise with voltage, and also appears in thermal-noise and ion-transport calculations.

Voltage that sets the scale of semiconductor junction behaviour at absolute temperature T: about 25.85 mV at 300 K. It appears in the diode and transistor equations.

Calculator

Solve for
mV
K
200500
  • Rate of change with temperature, kB/q86.17 µV/K
  • Voltage rise for ten times the current in an ideal diode, VT ln 1059.53 mV
  • Thermal voltage at 25 °C (298.15 K)25.69 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.

Thermal voltage (mV) current value 25.85 mV
TAbsolute temperatureKey driver
17.243.1
  • At 300 K the thermal voltage is 25.85 mV. An ideal diode needs 59.5 mV more to pass ten times the current, close to the 60 mV per decade that is quoted.
  • At 25 °C it is 25.7 mV and at 100 °C it is 32.2 mV: a diode’s voltage–current curve stretches as the temperature rises, which is why a transistor’s gain and bias drift with temperature.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (mV)Local stepChange in result
T Absolute temperature200 to 500 K17.2343.0925.85+10%+10 %

Sweep of Absolute temperature

Absolute temperature (K)Thermal voltage (mV)
20017.23
237.520.47
27523.7
312.526.93
35030.16
387.533.39
42536.62
462.539.86
50043.09

Sweep one variable

Thermal voltage (mV)Hover, touch, or use the arrow keys on the chart
202530354045 200250300350400450500 Absolute temperature (K) now 25.85

Variables and how each one contributes

VT

Thermal voltage

Result · mV

What the equation gives you. Everything else feeds it.

T

Absolute temperature

K · typical 200 to 500

The absolute temperature, in kelvin. Raises the thermal voltage in direct proportion. Raising it by 10 K (to 310 K) moves the result by +3.33 %, from 25.85 to 26.71 mV.

Worked example with the values above

  1. Constants: kB/q = 1.380649×10−23 / 1.602177×10−19 = 8.617×10−5 V/K.
  2. Multiply by the temperature: 8.617×10−5 × 300 K.
  3. VT = 25.85 mV.

Assumptions and limits

  • Absolute temperature in kelvin and the standard values of kB and q.
  • A junction in thermal equilibrium at a single temperature.

Common mistakes

  • Using degrees Celsius for T.
  • Remembering it as 26 mV regardless of temperature, which is only right near room temperature.

Statistics and uncertainty

The thermal voltage is exact for a given temperature; the uncertainty is in the temperature, and the junction temperature may be well above the ambient value due to self-heating.

Where it is used

  • The scale factor in diode, bipolar-transistor and thermistor characteristics.
  • Estimating the thermal-noise voltage scale and ion transport voltages.

Maths you need

  • Boltzmann constant: links temperature to energy
  • Elementary charge: the charge of an electron

Related formulas

Practise this formula