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Chapter 11: Problem 94

In a certain instrumentation amplifier, the input signal consists of a \(20-\mathrm{mV}-\mathrm{rms}\) differential signal and a 5-V-rms 60-Hz interfering common-mode signal. It is desired that the common-mode contribution to the output signal be at least \(60 \mathrm{~dB}\) lower than the contribution from the differential signal. What is the minimum CMRR allowed for the amplifier in decibels?

Short Answer

Expert verified
The minimum CMRR required is 60 dB.

Step by step solution

01

Understanding the Problem

We need to determine the minimum Common-Mode Rejection Ratio (CMRR) required such that the common-mode signal at the output is at least 60 dB lower than the differential signal contribution.
02

Expressing the CMRR in Decibels

The CMRR is defined as the ratio of the differential gain ( G_d) to the common-mode gain ( G_c). In decibels, it is given by:\[\text{CMRR (dB)} = 20\log_{10}\left(\frac{G_d}{G_c}\right)\]
03

Determining the Required Ratio

Given that the common-mode contribution to the output should be 60 dB lower than the differential signal, the ratio \frac{G_d}{G_c} must be at least:\[\frac{G_d}{G_c} = 10^{\left(\frac{60}{20}\right)} = 10^3\]
04

Calculating Minimum CMRR

Substitute the ratio found into the decibel formula:\[\text{CMRR (dB)} = 20\log_{10}\left(10^3\right) = 20 \times 3 = 60 \text{ dB}\]
05

Finalizing the Solution

The minimum CMRR required for the amplifier is determined to be 60 dB. This ensures the common-mode signal is suppressed adequately compared to the differential signal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common-Mode Rejection Ratio (CMRR)
Common-Mode Rejection Ratio (CMRR) is a key performance metric of instrumentation amplifiers. It measures the ability of an amplifier to reject common-mode signals, like noise, while amplifying the desired differential signals. These common-mode signals could be unwanted noise as seen in the 60 Hz interfering signal in our example.
CMRR is expressed as a ratio of differential gain to common-mode gain, indicating how well the amplifier can separate the actual signal from the noise:
  • High CMRR means better noise rejection.
  • Low CMRR could result in poor performance where noise is as amplified as the signal.
Professionals in the field aim for a CMRR that provides at least 60 dB difference, creating a clear distinction between signal and noise.
Differential Signal
Differential signals are crucial in instrumentation amplifiers for their noise immunity and improved signal-to-noise ratio. These signals consist of two opposite polarity voltages applied to different inputs of the amplifier. The device then amplifies the difference between these voltages, effectively reducing noise which affects both lines equally.
In the exercise context, the differential signal is noted as 20 mV rms. This contrasts with the common-mode signal intended to be minimized. Understanding differential signals involves recognizing benefits such as:
  • Reduced interference from external noise sources.
  • Increased accuracy in signal measurement.
  • Enhanced amplifier efficiency through precise processing and interpretation.
Differential signal handling is a staple in sensitive electronic measurement systems.
Decibel Calculation
In electrical engineering, decibel (dB) calculations are essential to understand and express power relationships logarithmically. They compress a broad range of values into an understandable format, making it simpler to interpret small versus large differences.
Specifically, CMRR in decibels can be expressed as:\[\text{CMRR (dB)} = 20 \log_{10}\left(\frac{G_d}{G_c}\right)\]
Where:
  • \(G_d\) is the differential gain
  • \(G_c\) is the common-mode gain
This formula helps translate the performance of an amplifier system into a measurable and comprehensible number, enhancing both design and diagnostic capabilities.
Electrical Engineering Problem Solving
Solving electrical engineering problems involves understanding theoretical concepts and applying them to practical situations. In the case of instrumentation amplifiers, it requires one to work with many factors such as CMRR, differential gains, and decibels.
The process often involves:
  • Identifying and defining the problem, like unwanted noise in measurements.
  • Translating problem requirements into mathematical expressions, such as determining gain ratios.
  • Applying relevant formulas, like decibel calculations, to solve for desired outcomes.
Each problem-solving step consolidates theoretical understanding and practical skills, leading to effective amplifier design and usage.

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Most popular questions from this chapter

Draw a voltage-amplifier model. Is the gain parameter measured under open- circuit or short-circuit conditions? Repeat for a current amplifier model, a transresistanceamplifier model, and a transconductanceamplifier model.

An amplifier has an open-circuit voltage gain of 1000 , an input resistance of \(20 \mathrm{k} \Omega\), and an output resistance of \(2 \Omega\). A signal source with an internal resistance of \(10 \mathrm{k} \Omega\) is connected to the input terminals of the amplifier. An \(8-\Omega\) load is connected to the output terminals. Find the voltage gains \(A_{v s}=V_{o} / V_{s}\) and \(A_{v}=V_{o} / V_{i}\). Also, find the power gain and current gain.

An amplifier having \(R_{i}=1 \mathrm{M} \Omega, R_{o}=1 \mathrm{k} \Omega\), and \(A_{v o c}=-10^{4}\) is operated with a \(1-\mathrm{k} \Omega\) load. A source having a Th茅venin resistance of \(2 \mathrm{M} \Omega\) and an open- circuit voltage of \(3 \cos (200 \pi t) \mathrm{mV}\) is connected to the input terminals. Determine the output voltage as a function of time and the power gain.

What type of ideal amplifier is needed if we need to sense the short-circuit current of a sensor and drive a proportional current through a variable load? Explain your answer.

An amplifier has an input resistance of \(20 \Omega\), an output resistance of \(10 \Omega\), and a short-circuit current gain of 3000 . The signal source has an internal voltage of \(100 \mathrm{mV}\) rms and an internal impedance of \(200 \Omega\). The load is a \(5-\Omega\) resistance. Find the current gain, voltage gain, and power gain of the amplifier. If the power supply has a voltage of \(12 \mathrm{~V}\) and supplies an average current of \(2 \mathrm{~A}\), find the power dissipated in the amplifier and the efficiency.
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