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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 $\Omega$
1 Answer
written 7.8 years ago by |
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} $ $= \sqrt{4×100 ×1.38 ×10^{-23} ×290 ×1 × 10^6} = 1.2655 μV$