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 9.4 years ago by

teamques10 ★ 70k

modified 3.7 years ago by

sagarkolekar ★ 11k

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2.4kviews

written 9.4 years ago by

teamques10 ★ 70k

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

$= 1.38 \times 10^{-23} \times 290 \times 1 \times 10^6 \\ = 4.003 \times 10^{-15} W$

Noise voltage is given as:

$V_N = \sqrt{4RkTB} $ $= \sqrt{4×100 ×1.38 ×10^{-23} ×290 ×1 × 10^6} = 1.2655 μV$

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