Question

A receiver has a 400 Ω input resistance, a bandwidth of 6 MHz and a temperature of 29°C. The input thermal voltage is _______. Given Boltzmann's constant is 1.38 × 10^-23 J/K.
1. 2.24 μV
2. 5.32 μV
3. 6.32 μV
4. 4.48 μV

Answer 1

Correct Answer - Option 3 : 6.32 μV

Thermal Noise:

• Thermal Noise is random and often referred to as White Noise or Johnson Noise
• Thermal noise is generally observed in the resistor or the sensitive resistive components of a complex impedance due to the random and rapid movement of molecules or atoms or electrons

For most cases, the resistive component of the impedance will remain constant over the required bandwidth. It therefore possible to simplify the thermal noise voltage as:

\(V = \sqrt {4 k T B R}\)

Given:

k = 1.38 × 10-23 J/K

B = 6MHz

R = 400 Ω

T = 302 K

\(V=\sqrt{4\times 1.38\times10^{-23}\times302\times6\times10^6\times400}\)

V = 6.32 μV