Thermal voltage

The thermal voltage $V_T$ is the voltage scale set by the thermal energy of a charge carrier:

$\boxed{V_T=\frac{kT}{q}}$

with $k$ the Boltzmann constant, $T$ the absolute temperature and $q$ the electron charge. At room temperature it is about $25\text{mV}$, and that one number sets the shape of every exponential in this course.

It is not a voltage anything applies. It is the size of the voltage step that matters to a carrier trying to climb an energy barrier: apply $V_T$ and you change the barrier by exactly one unit of $kT$, so the population able to cross it changes by a factor of $e$.

The number

$k/q=86.2\mu \text{V/K}$, so $V_T$ is proportional to absolute temperature and rises about $86\mu \text{V}$ per kelvin:

$T$	$V_T$
$273\text{K}$ (0\,^\circC)	$23.5\text{mV}$
$290\text{K}$ (17\,^\circC)	$25.0\text{mV}$
$300\text{K}$ (27\,^\circC)	$25.9\text{mV}$
$398\text{K}$ (125\,^\circC)	$34.3\text{mV}$

That table explains the textbook disagreement. Both $25\text{mV}$ and $26\text{mV}$ are quoted as “room temperature”, and they differ only because one is evaluated near $290\text{K}$ and the other near $300\text{K}$. This course uses $25\text{mV}$ throughout, so every worked number here does too. Pick one and stay with it; mixing them shifts small-signal answers by 4%.

Why it turns up everywhere

Every non-linear device law in the course puts $V_T$ in the denominator of an exponent, and every small-signal parameter that comes out of linearising one of those laws is a current divided by $V_T$.

• Diode equation: $i_D=I_S(e^{v_D/V_T}-1)$. $V_T$ is the only thing setting how steep the turn-on is.
• BJT collector current: $i_C=I_S{e}^{v_{BE}/V_T}$, the same exponential on the base-emitter junction.
• Diode small-signal resistance: $r_d=V_T/I_D$, straight from differentiating the diode equation.
• Emitter resistance: $r_e=V_T/I_E$, the same result on a BJT.
• BJT transconductance: $g_m=I_C/V_T$. This is why a BJT beats a MOSFET on transconductance per unit bias current — the BJT divides by $25\text{mV}$ while the MOSFET divides by an overdrive of a few hundred millivolts.
• Built-in voltage: $V_0=V_T\ln (N_A{N}_D/n_i^2)$ across a PN junction.
• Validity of Small-signal analysis: the linearisation holds while the signal swing stays well below $V_T$, so “small signal” on a diode or a BJT literally means a few millivolts.

The two numbers worth memorising

Both follow from the exponential and both are quick sanity checks on any diode or BJT answer.

One $V_T$ multiplies the current by $e$. Raise $v_D$ by $25\text{mV}$ and $e^{v_D/V_T}$ picks up a factor of $e=2.72$.

Sixty millivolts multiplies it by ten. A decade of current needs

$\Delta v=V_T\ln 10=25\text{mV}\times 2.303=57.6\text{mV}\approx 60\text{mV}.$

Turn that around and it is the reason a forward-biased silicon diode is so often treated as a fixed $0.7\text{V}$ drop: moving from $1\mu \text{A}$ to $10\text{mA}$ is four decades, which costs only $230\text{mV}$ of junction voltage. Four orders of magnitude of current, a quarter of a volt of voltage. That near-constancy is the whole justification for the Constant-voltage-drop model.

Temperature drift

Because $V_T\propto T$, everything built on it drifts. $g_m=I_C/V_T$ falls as a part heats up at fixed bias current, so an amplifier’s gain drifts with temperature unless the bias circuit compensates. Working the other way, the proportionality is clean enough to be useful on purpose: a bandgap reference is built by summing a voltage proportional to $V_T$ against a junction voltage that falls with temperature, and trimming the two slopes to cancel.

Note that the diode’s forward voltage moves the opposite way from $V_T$ at fixed current, dropping roughly $2\text{mV/K}$, because the temperature dependence of [[Reverse saturation current|$I_S$]] swamps the $V_T$ in the exponent.