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: Understanding the Concept:
The thermal voltage (\( V_T \)) is a parameter that appears in the equations governing semiconductors. It represents the average thermal energy per unit charge and is directly proportional to the absolute temperature.
Step 2: Key Formula or Approach:
The formula for thermal voltage is:
\[ V_T = \frac{kT}{q} \] where:
\( k \) = Boltzmann's constant (\( 1.38 \times 10^{-23} \) J/K)
\( T \) = Absolute temperature in Kelvin (K)
\( q \) = Elementary charge (\( 1.602 \times 10^{-19} \) C)
Step 3: Detailed Explanation:
1. Convert Temperature to Kelvin:
The given temperature is 47°C. To convert to Kelvin, we add 273.15 (or often just 273 for simplicity).
\[ T(K) = 47 + 273.15 = 320.15 \, K \] 2. Substitute the 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. Convert to millivolts (mV):
To convert from Volts to millivolts, we multiply by 1000.
\[ V_T \approx 2.758 \times 10^{-2} \times 1000 \, \text{mV} \] \[ V_T \approx 27.58 \, \text{mV} \] Step 4: Final Answer:
The calculated value is approximately 27.6 mV, which matches option (A).