Calculation of thermal noise when resistors are connected in parallel

Q.  At a room temperature of 300K, calculate the thermal noise generated by two resistors of 10KΩ and 30 KΩ when the bandwidth is 10 KHz and the resistors are connected in parallel.
- Published on 15 Oct 15

a. 30.15 * 10-3
b. 8.23 * 10-23
c. 11.15 * 10-7
d. 26.85 * 10-7

ANSWER: 11.15 * 10-7
 
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 resistors when connected in parallel is
Vn = √{4R KTB}
Here for resistors to be in parallel,
1/R = 1/R1 + 1/R2
= 1/10K + 1/30K
= 0.1333
R = 7.502KΩ
Vn = √{4 * 7.502 * 103 * 1.381×10-23 * 300 * 10 * 103}
= √124.323 * 10-14
= 11.15 * 10-7

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