Q. 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.- Published on 15 Oct 15a. 4.071 * 10-6 V, 5.757 * 10-6 V
b. 6.08 * 10-6 V, 15.77 * 10-6 V
c. 16.66 * 10-6 V, 2.356 * 10-6 V
d. 1.66 * 10-6 V, 0.23 * 10-6 V
ANSWER: 4.071 * 10-6 V, 5.757 * 10-6 V
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
