- #1

Master1022

- 611

- 117

- Homework Statement
- Context:

A certain instrument measures a signal with rms voltage 10 mV and a significant spectrum between 0 and 250 Hz. The signal is obtained from two probes and there is a coherent interference signal of 1.5 V rms at 50 Hz. We have resistors that have an accuracy of 0.01% and the instrument can operate with an output signal with at most 1% rms 50 Hz interference.

The main bit:

Calculate the differential gain required of instrumentation amplifier. Ignore the effects of tolerances on the resistors in the input stage.

- Relevant Equations
- ## CMRR = \frac{A_{diff}}{A_{cm}} ##

Hi,

I have a question regarding a practical CMRR calculation.

- Signal input: 10 mV rms

- Interference input: 1.5 V rms

- In an earlier part of the question, we found that for the second stage of the instrumentation amplifier (the summation amplifier) that:

$$ \frac{V_{out}}{V_{in}} < |\frac{2 \delta}{100} | $$

where ## \delta ## is the % change in the resistor tolerance value, ## V_{out} ## is the output of the summation amplifier, and ## V_{in} ## is the value of BOTH inputs to the summation amplifier

We can take the common mode gain of the input stage to be unity.

We only want 1% of the current interference amplitude present at the output, and therefore we need the common mode gain to be ## 1/100 ## which will turn the 1.5 V rms at the input into 15 mV rms at the output

I know that ## CMRR = \frac{A_{diff}}{A_{cm}} ##, but I have two unknowns and I don't really know what to do at this point. I also know that I can use the above expression to calculate what the current common mode gain is based on the tolerance values:

$$ \frac{V_{out}}{V_{in}} = \frac{2 \cdot 0.01}{100} = 2 \cdot 10^{-4} $$

Any help would be greatly appreciated.

I have a question regarding a practical CMRR calculation.

**Main information:**- Signal input: 10 mV rms

- Interference input: 1.5 V rms

- In an earlier part of the question, we found that for the second stage of the instrumentation amplifier (the summation amplifier) that:

$$ \frac{V_{out}}{V_{in}} < |\frac{2 \delta}{100} | $$

where ## \delta ## is the % change in the resistor tolerance value, ## V_{out} ## is the output of the summation amplifier, and ## V_{in} ## is the value of BOTH inputs to the summation amplifier

**My attempt:**We can take the common mode gain of the input stage to be unity.

We only want 1% of the current interference amplitude present at the output, and therefore we need the common mode gain to be ## 1/100 ## which will turn the 1.5 V rms at the input into 15 mV rms at the output

I know that ## CMRR = \frac{A_{diff}}{A_{cm}} ##, but I have two unknowns and I don't really know what to do at this point. I also know that I can use the above expression to calculate what the current common mode gain is based on the tolerance values:

$$ \frac{V_{out}}{V_{in}} = \frac{2 \cdot 0.01}{100} = 2 \cdot 10^{-4} $$

Any help would be greatly appreciated.