Chapter 5: Problem 9
Calculate the rms thermal noise associated with a \(1.0-\mathrm{M} \Omega\) load resistor operated at room temperature if an oscilloscope with a 1-MHz bandwidth is used. If the bandwidth is reduced to \(100 \mathrm{~Hz}\), by what factor will the noise be reduced?
Short Answer
Expert verified
The noise is reduced by a factor of 315.
Step by step solution
01
Understanding Thermal Noise Formula
The thermal noise voltage for a resistor can be calculated using the formula: \[ V_n = \sqrt{4 k T R B} \] where \( k \) is Boltzmann's constant \( (1.38 \times 10^{-23} \, \text{J/K}) \), \( T \) is the temperature in Kelvin, \( R \) is the resistance, and \( B \) is the bandwidth in Hertz.
02
Determine Parameters for Room Temperature
To proceed with the calculations at room temperature, which is approximately \( 300 \, \text{K} \), we will use \( R = 1.0 \times 10^6 \, \Omega \) and initial \( B = 1,000,000 \, \text{Hz} \).
03
Calculating RMS Thermal Noise at Initial Bandwidth
Substitute known values into the formula: \[ V_n = \sqrt{4 \times 1.38 \times 10^{-23} \times 300 \times 1.0 \times 10^6 \times 1,000,000} \] Simplifying the expression leads to: \[ V_n \approx \sqrt{1.656 \times 10^{-14}} \] Therefore, \( V_n \approx 4.07 \times 10^{-7} \, \text{V} \) or \( 407 \, \text{nV} \).
04
Calculating RMS Thermal Noise at Reduced Bandwidth
Now calculate for \( B = 100 \, \text{Hz} \): \[ V_n = \sqrt{4 \times 1.38 \times 10^{-23} \times 300 \times 1.0 \times 10^6 \times 100} \] Simplifying gives: \[ V_n \approx \sqrt{1.656 \times 10^{-18}} \] Therefore, \( V_n \approx 1.29 \times 10^{-9} \, \text{V} \) or \( 1.29 \, \text{nV} \).
05
Factor of Noise Reduction Calculation
To find out the factor of reduction, divide the initial noise by the reduced bandwidth noise: \[ \text{Reduction Factor} = \frac{407 \, \text{nV}}{1.29 \, \text{nV}} \approx 315 \] Hence, the noise is reduced by a factor of 315.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Thermal Noise Formula
Thermal noise, sometimes referred to as Johnson-Nyquist noise, is an intrinsic noise produced by the thermal agitation of electrons within a resistance. It is ever-present in electronic devices, contributing to the baseline level of noise. The thermal noise voltage, denoted as \(V_n\), is calculated using a specific formula:\[ V_n = \sqrt{4 k T R B} \]Where:- \(k\) is Boltzmann's constant (\(1.38 \times 10^{-23} \, \text{J/K}\)),- \(T\) is the absolute temperature in Kelvin,- \(R\) represents resistance in ohms,- \(B\) is the bandwidth over which the noise is measured (in Hertz).This equation shows that thermal noise is dependent on four key factors: the physical constants, the temperature of the system, the resistance, and the bandwidth.
Resistance and Bandwidth
Resistance and bandwidth play crucial roles in determining the level of thermal noise within a system. Higher resistance values result in increased noise production, which is evident from the thermal noise formula \( V_n = \sqrt{4 k T R B} \).
- Resistance (R): Measured in ohms, resistance denotes how much a resistor impedes the flow of electric current. In our exercise, a \(1.0-\text{M} \Omega\) load resistor was used for calculations.
- Bandwidth (B): This refers to the range of frequencies that the oscilloscope can accommodate, measured in Hertz. Initially specified at \(1 \, \text{MHz}\) for our calculation, and later reduced to \(100 \, \text{Hz}\).
Room Temperature Calculations
Conducting calculations at room temperature (approximately \(300 \, \text{K}\)) is significant because it provides a standard condition under which the system's thermal noise can be assessed.To calculate the RMS thermal noise for a \(1.0-\text{M} \Omega\) resistor with a bandwidth of \(1 \, \text{MHz}\), we substitute these values into the thermal noise formula. The result is an estimated noise voltage of \(407 \, \text{nV}\). This calculation reflects typical operational conditions in many electronic circuits and is crucial in designing systems that effectively manage thermal noise.
Noise Reduction Factor
Reducing the bandwidth has a significant effect on the thermal noise voltage. This relationship is evident in the formula \( V_n = \sqrt{4 k T R B} \), where a decrease in bandwidth \(B\) leads to a decrease in thermal noise.When the bandwidth was reduced from \(1 \, \text{MHz}\) to \(100 \, \text{Hz}\) in the exercise, the thermal noise voltage decreased from \(407 \, \text{nV}\) to \(1.29 \, \text{nV}\). The noise reduction factor, calculated as the initial noise divided by the reduced noise, amounts to 315.
- This means that by narrowing the bandwidth, the noise level was reduced significantly, making systems more efficient in environments where noise considerations are critical.