Question

At a temperature of 47°C the thermal voltage is (Given value of Boltzmann's constant = 1.38 x10\(^{-23}\) joule/°K):

• A. 27.6 mV
• B. 27.6 V
• C. 27.6 \(\mu\)V
• D. 2.76 V

Hint:

A useful shortcut for competitive exams is to remember the thermal voltage at room temperature (\(T \approx 300 \, K\) or \(27^\circ C\)). At 300 K, \(V_T \approx 25.85 \, mV \approx 26 \, mV\). You can use this as a benchmark to quickly estimate the answer for other temperatures. Since 47°C is slightly warmer than room temperature, the thermal voltage should be slightly higher than 26 mV.

Answer 1

The Correct Option is A

Step 1: Concept Definition:
Thermal voltage (\( V_T \)) is a semiconductor parameter representing average thermal energy per unit charge, directly proportional to absolute temperature.
Step 2: Formula:
The thermal voltage is calculated using:
\[ V_T = \frac{kT}{q} \]Where:
\( k \) = Boltzmann's constant (\( 1.38 \times 10^{-23} \) J/K)
\( T \) = Absolute temperature (K)
\( q \) = Elementary charge (\( 1.602 \times 10^{-19} \) C)
Step 3: Calculation:
1. Temperature Conversion:
Convert 47°C to Kelvin:
\[ T(K) = 47 + 273.15 = 320.15 \, K \]2. Value Substitution:
Substitute known values into the formula:
\[ V_T = \frac{(1.38 \times 10^{-23} \, \text{J/K}) \times (320.15 \, \text{K})}{1.602 \times 10^{-19} \, \text{C}} \]\[ V_T = \frac{4.41807 \times 10^{-21}}{1.602 \times 10^{-19}} \, V \]\[ V_T \approx 2.758 \times 10^{-2} \, V \]3. Unit Conversion:
Convert Volts to millivolts:
\[ V_T \approx 2.758 \times 10^{-2} \times 1000 \, \text{mV} \]\[ V_T \approx 27.58 \, \text{mV} \]Step 4: Result:
The computed thermal voltage is approximately 27.6 mV.