Common Mode Rejection Ratio: Instrumentation Amplifier Calculation

[Master1022]

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.

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.

 

[KataKoniK]

Hi there,

Thank you for your question. To calculate the common mode rejection ratio (CMRR) in this scenario, you will need to know the differential gain (A_diff) and the common mode gain (A_cm) of your instrumentation amplifier.

Since you have already determined that the common mode gain of the input stage is unity, you can focus on finding the differential gain. This can be calculated by dividing the output voltage by the input voltage, as you have already done in your attempt.

To find the output voltage, we can use the equation for the summation amplifier given in the question:
$$V_{out} = \frac{R_3}{R_1}(V_1-V_2)$$
where V_1 and V_2 are the input voltages and R_3 and R_1 are the resistances in the summation amplifier.

Since we know that the signal input is 10 mV rms and the interference input is 1.5 V rms, we can plug those values into the equation to find the output voltage:
$$V_{out} = \frac{R_3}{R_1}(10\times10^{-3}-1.5)$$
Simplifying this equation, we get:
$$V_{out} = \frac{R_3}{R_1}(-1.49)$$

Now, to find the differential gain, we need to divide this output voltage by the input voltage. Since both inputs are being amplified in the summation amplifier, we can use the combined input voltage of 1.51 V rms (10 mV rms + 1.5 V rms). Plugging this into the equation, we get:
$$A_{diff} = \frac{V_{out}}{V_{in}} = \frac{\frac{R_3}{R_1}(-1.49)}{1.51}$$
Simplifying, we get:
$$A_{diff} = \frac{-1.49R_3}{1.51R_1}$$

Now that we have the differential gain, we can plug this into the CMRR equation:
$$CMRR = \frac{A_{diff}}{A_{cm}} = \frac{\frac{-1.49R_3}{1.51R_1}}{1}$$
Simplifying, we get: