Week #13: SNR & AD524

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rjagodowski
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Joined: Fri Sep 04, 2015 6:59 pm

Week #13: SNR & AD524

Post by rjagodowski »

Here is the Gain Calculation Problem set with solutions, originally posted and distributed back in Week #10.
Gain HW Problem Solutions_2020-1215.pdf
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We'll start discussion of these during the week of December 6th.

This week we'll finish up our discussion of CMRR and SNR from Week #10.

We discussed the common-mode rejection ratio (CMRR) as the ratio of the differential mode gain divided by the common mode gain. The differential mode gain for a differential amplifier is the ratio of the output voltage divided by the difference between the two differential mode inputs. The common mode gain for a differential amplifier is the ratio of the output voltage divided by the common (same voltage connected to both inputs) input signal. The CMRR is a "figure of merit" of how well the amplifier amplifies the preferred or desired signal in comparison to how much it attenuates (or suppresses) the undesired common-mode signal (usually noise). The Excel spreadsheet example from Week #10 showed how this works for the a problem in the Amplifier Gain Homework assignment.

The Excel file is reposted here for convenience:
EET-210 Differential Amplifier Example v19.1.xlsx
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Stated another way, when expressed in dB's, the CMRR(dB) = ADM(dB) - ACM(dB) (Be careful of the signs, ACM(dB) should be negative.)

The Signal to Noise Ratio (SNR) is the ratio of the desired signal to the noise signal. This is usually expressed in dB's as well, though is usually first calculated using just the voltage ratio of "good" signal to "bad" signal. When expressed in dBs, the sign of the SNR can be indicative of the level of signal to noise. If the SNR is negative, then there is MORE NOISE than there is SIGNAL. If the SNR is positive, then there is MORE SIGNAL than there is NOISE. If the Signal and Noise are equal, then the SNR would be equal to 0 dB. (Just like an amplifier with a ratio gain of 1 has a gain of 0 dB.)

Example: As an example of using dB's to calculate the effects of a high CMRR in reducing noise in a signal.
Example: An input to a differential amplifier has 2 mV of signal with 90 mV of noise present. That is, the NOISE is 45 times stronger than the desired SIGNAL. If the CMRR of the differential amplifier used is rated at +95 dB. Determine the SNR at the input and the output.

Input: SNR(ratio) = 2mV/90mV = 0.0222 which is equal to SNR(dB) = 20* log (0.022222) = -33db

Output: The easy way to determine the output SNR is knowing that:

SNR(dB) output = CMRR(dB) + SNR(dB) input

SNR(dB)output = +95dB + (-33dB) = +62 dB.
If you convert +62dB to a ratio 10^(+62/20) = 10^3.1 = 1,259. So on the output, the desired SIGNAL voltage is 1,259 times stronger than the undesirable noise!!!

Checking: The CMRR(dB) is equal to the SNR(dB)output - SNR(dB)input. CMRR(dB) = +62 - (-33) = +95dB



Here's the link on Instrumentation Amplifiers posted back in Week #10. The IA (Instrumentation Amplifier) is a differential amplifier circuit which provides and easily adjustable gain (using a single resistor) and provides for buffered inputs (high input impedance aka "High Z" inputs). The standard differential amplifier required that two resistors be changed to change the differential mode gain, and considering the resistors should be matched as closely as possible, that makes for a difficult circuit to use IF adjustable gain is a requirement. The IA solves that problem.

We will also discuss an Instrumentation Amplifier on a chip, the Analog Devices AD524. We'll investigate the key specs on the data sheet, the pin-programmable gain options as well as the user-settable gain options.

Sample Programmable Gain circuits are shown on Page 16. On Page 16, note there is a typo on the Gain formula. The formula in the upper right of the page should be: RG=40,000/(G-1) .

We'll also discuss the Sense and the Reference Terminals on page 18.

AD524_short.pdf
This is the shorter version (5 pages) of the AD524 data sheet and is the one distributed in class.
(300.13 KiB) Downloaded 159 times
AD524.pdf
This is the FULL document and is 28 pages long. The shorter 5 page version was distributed in class.
(548.25 KiB) Downloaded 155 times
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