Calculate the thermal noise power available from any resistor at a temperature of 290 K for a bandwidth of 1 MHz. Calculate also the corresponding noise voltage if the resistance R = 100 $\Omega$

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written 7.8 years ago by

teamques10 ★ 65k

modified 2.2 years ago by

sagarkolekar ★ 11k

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2.1kviews

written 7.8 years ago by

teamques10 ★ 65k

Given Data:

Temperature, T = 290 K

Bandwidth, B = 1 MHz

R = 100 Ω

To find:

Thermal noise power, $P_N$=?

Noise voltage, $V_N$=?

Solution:

Thermal noise power is given as:

$P_N$ = k T B

$= 1.38 \times 10^{-23} \times 290 \times 1 \times 10^6 \\ = 4.003 …

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