Question

The voltage equivalent of temperature ($V_T$) in a $p\text{-}n$ junctions is given by}

• A. T/1000 volts
• B. T/300 volts
• C. T/1600 volts
• D. T/11600 volts

Hint:

At standard room temperature ($T = 300\text{ K}$), substituting this approximation gives: $$V_T = \frac{300}{11600} \approx 0.02586\text{ V} = 25.86\text{ mV}$$ This is why we commonly approximate thermal voltage as $26\text{ mV}$ when performing rapid AC small-signal diode or BJT circuit analysis!

Answer 1

The Correct Option is D

Concept: The voltage equivalent of temperature, also referred to as the thermal voltage ($V_T$), is a fundamental scaling parameter that appears in the Shockley diode equation and various semiconductor carrier transport equations. It is mathematically defined as: $$V_T = \frac{k \cdot T}{q}$$ Where:
• $k$ = Boltzmann’s constant $\approx 1.3806 \times 10^{-23} \text{ J/K}$
• $T$ = Absolute temperature measured in Kelvin ($\text{K}$)
• $q$ = Magnitude of the electronic charge $\approx 1.6022 \times 10^{-19} \text{ Coulombs}$ Step-by-step Simplification:
• Let us group the constant scalar values ($\frac{k}{q}$) together to express the formula strictly as a function of temperature $T$: $$V_T = \left( \frac{k}{q} \right) \cdot T$$
• Substitute the values for the constants: $$\frac{k}{q} = \frac{1.3806 \times 10^{-23}}{1.6022 \times 10^{-19}} \approx 8.6173 \times 10^{-5} \text{ V/K}$$
• To convert this multiplier into a fractional form ($\frac{1}{\text{Constant}}$), take the inverse of this numeric result: $$\frac{1}{\text{Constant}} = 8.6173 \times 10^{-5} \quad \Rightarrow \quad \text{Constant} = \frac{1}{8.6173 \times 10^{-5}} \approx 11604.5$$
• Rounding to standard engineering approximation guidelines gives approximately $11600$.
• Therefore, substituting this back into the equation yields: $$V_T \approx \frac{T}{11600} \text{ volts}$$ This derivation matches Option (D).