Calculation of thermal noise
At a room temperature of 300K, calculate the thermal noise generated by two resistors of 10KΩ and 20 KΩ when the bandwidth is 10 KHz.
Option:
4.071 * 10-6 V, 5.757 * 10-6 V
6.08 * 10-6 V, 15.77 * 10-6 V
16.66 * 10-6 V, 2.356 * 10-6 V
1.66 * 10-6 V, 0.23 * 10-6 V
Correct Answer: 4.071 * 10-6 V, 5.757 * 10-6 V
Explanation:
Noise voltage Vn= √(4R KTB)
Where, K = 1.381×10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
T noise temperature
R is the resistance
Noise voltage by individual resistors
Vn1= √(4R1 KTB)
= √(4*10*103 *1.381*10-23 *3000* 10*103)
= √16.572*10-12
= 4.071 * 10-6 V
Vn2= √(4R2 KTB)
= √(4*20*103 *1.381*10-23 *3000* 10*103)
= √33.144*10-12
= 5.757 * 10-6 V
