Instrumentation Amplifier question

In summary, the conversation discusses obtaining an expression for v0 as a function of v1, v2, and vcm, and how the common mode signal vcm can be canceled from the output by adjusting one of the resistors. The theory of superposition is applied to find v0, and it is explained why R1 is not considered in the closed loop voltage gain of the upper op amp. It is stated that in the cases where V1 = Vcm = 0V and V2 = Vcm = 0V, R1 and R3, R4, and R5 have no effect on the output. The conversation concludes with the understanding that R1 can be considered as an open circuit in these cases.
  • #1
cyeokpeng
69
0
Hi,

Just a little question.
The question wants me to obtain an expression for v0 as a function of v1, v2 and vcm, and show how the commom mode signal vcm can be canceled from the output by a proper adjustment of one of the resistors, R3, R4 or R5.

To find v0, I have to apply the theory of superposition.

Output voltage due to v2 alone (v1=0, vcm=0) is
--> The voltage at the output of the lower op amp is (-R4/R3)v2
--> The voltage at the output of the upper op amp is -(R2/R5)*voltage at
the output of the lower op amp.

My question is: Why is R1 not considered in the closed loop voltage gain of
the upper op amp?

Thanks
 

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  • #2
Because in the case of V1 = Vcm = 0, there is no potential difference across r1, hence no current through it. So whatever the value of r1, no current flows through it, so it has no effect on the output Vo.

Likewise, in the case of V2 = Vcm = 0, there is no potential difference across r3, r4 nor r5, so the output Vo will be independent of r3, r4 and r5.
 
  • #3
Thanks,

Actually, I suspect that this is the answer.
When we consider the case V1 = Vcm = 0V,
the inverting input of the upper op amp is close to 0V due to virtual ground, so there is no pd across R1, so we can consider R1 as an open circuit right?
Similarly, for the case when V2 = Vcm = 0V.

Thanks for your help!
 

Related to Instrumentation Amplifier question

1. What is an instrumentation amplifier?

An instrumentation amplifier is a type of electronic amplifier that is used to amplify small signals with high precision. It is commonly used in measurement and instrumentation systems to accurately amplify and process signals from sensors or other sources.

2. How is an instrumentation amplifier different from a regular amplifier?

An instrumentation amplifier has a high input impedance, low output impedance, and a high common-mode rejection ratio (CMRR) which allows it to accurately amplify small signals while rejecting any common-mode noise. Regular amplifiers typically have a lower input impedance and CMRR, and are not designed for precision applications.

3. What are the main applications of instrumentation amplifiers?

Instrumentation amplifiers are commonly used in systems that require precise measurement and control, such as medical equipment, industrial process control, and scientific research. They are also used in audio and video equipment to amplify and filter signals.

4. How do I choose the right instrumentation amplifier for my application?

The main factors to consider when choosing an instrumentation amplifier are the required gain, input impedance, and CMRR. You should also consider the input voltage range, bandwidth, and power supply requirements. It is important to select an amplifier that is specifically designed for your application to ensure accurate and reliable results.

5. Can I build my own instrumentation amplifier?

While it is possible to build your own instrumentation amplifier using discrete components, it requires a thorough understanding of circuit design and can be challenging to achieve the same level of precision and performance as a commercially available amplifier. It is recommended to use a pre-built instrumentation amplifier for most applications.

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