How Do You Calculate Vout in an Instrumentation Amplifier Circuit?

In summary, the conversation involved a question about calculating Vout using a given equation. The attempt at a solution involved using the values for V2IN and VIN1, which were both 0.1Vpk, but this did not give the correct answer. It was suggested to use real-world datasheet specs for the LM324, and to connect any unused opamps in a benign configuration. The correct method for calculating differential gain was shown, and the conversation also touched on calculating CMRR using the equation CMRR = 20log(Ad/|Acm|), but there was some confusion about the value of Acm.
  • #1
nothing909
170
4

Homework Statement


I'm trying to calculate Vout.

Homework Equations


[/B]
VDIFF = VIN2 - VIN1 = VDIFF1 + VDIFF2

VOUT=(VIN2–VIN1)(1+2R1/RG)(R6/R5)

The Attempt at a Solution


[/B]
VOUT=(VIN2–VIN1)(1+2R1/RG)(R6/R5)
VOUT=(V2IN-VIN1)(1+2(10x10^3)/2x10^3)(10x10^3/10x10^3)

What are the values of V2IN and VIN1? I was just looking at the circuit and it's 0.1Vpk for both so 0.1-0.1=0 but obviously that's not right. what am i missing?
 

Attachments

  • INA.png
    INA.png
    17.3 KB · Views: 349
Physics news on Phys.org
  • #2
You would need to use the real-world datasheet specs for the LM324 to calculate Vout, I would think. If you treat them as ideal opamps, of course you will get infinite CM rejection.

Also, it's good to get in the habit of connecting up any unused opamps (like U1C) in a benign configuration, to keep them from oscillating and interfering with the operation of the circuit (same thing goes for unused gates in a multi-gate logic part). In this case you could just show it connected as a unity-gain voltage follower with ground at its + input.
 
  • #3
It asks me to calculate the differential gain (which is 11 in the answers). The equation is Ad = VOUT/VDIFF

If I do VOUT=(VDIFF1+VDIFF2)(1+2R1/RG)(R6/R5) = 2.2

Ad = 2.2/0.2 = 11

This corresponds with the answer. Is this the correct way I've done it?
 
  • #4
Also if you look at this equation CMRR = 20log(Ad/|Acm|)

if my differential gain is 11 and my common mode gain is 20μ, is it CMRR= 20log(11/20x10^-6) = 114.8

what does the lines mean |Acm|, does this change the of 20μ in the equation?
 
  • #5
nothing909 said:
If I do VOUT=(VDIFF1+VDIFF2)(1+2R1/RG)(R6/R5) = 2.2

Ad = 2.2/0.2 = 11

This corresponds with the answer. Is this the correct way I've done it?
I believe so.
nothing909 said:
if my differential gain is 11 and my common mode gain is 20μ, is it CMRR= 20log(11/20x10^-6) = 114.8
Where did the Acm = 20uV/V come from? Sorry if I'm missing something obvious.
 
  • #6
you're not missing something. I'm doing transient analysis in multisim and i get the Vout for the cm off the graph and then just Acm = Vout/Vcm
 
  • Like
Likes berkeman

Related to How Do You Calculate Vout in an Instrumentation Amplifier Circuit?

1. What is an instrumentation amplifier?

An instrumentation amplifier is a type of electronic amplifier that is designed to amplify small signals accurately, while rejecting any noise or interference. It is commonly used in scientific and medical instruments, as well as in industrial and communication equipment.

2. How does an instrumentation amplifier work?

An instrumentation amplifier typically consists of three operational amplifiers (op amps) connected in a specific configuration. It amplifies the difference between two input signals, while rejecting any common-mode signals (noise) that are present in both inputs. The gain of an instrumentation amplifier can also be adjusted by changing the values of its external resistors.

3. What are the advantages of using an instrumentation amplifier?

Instrumentation amplifiers offer several advantages over traditional amplifiers. They have high input impedance, low output impedance, and high common-mode rejection ratio (CMRR), making them suitable for amplifying small signals accurately. They also have a wide bandwidth, low noise, and high gain accuracy.

4. What are some common applications of instrumentation amplifiers?

Instrumentation amplifiers are used in a variety of applications, including medical equipment (such as ECG and EEG machines), sensors and transducers, data acquisition systems, strain gauges, and audio equipment. They are also commonly used in precision measurement and control systems, where accurate amplification of small signals is crucial.

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

The most important factors to consider when choosing an instrumentation amplifier are the required gain, bandwidth, input impedance, and CMRR. You should also consider the power supply requirements, noise specifications, and any other specific features that may be necessary for your application. It is best to consult with a knowledgeable engineer or refer to the manufacturer's datasheet for guidance in selecting the appropriate instrumentation amplifier for your needs.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
1
Views
450
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
2K
  • Engineering and Comp Sci Homework Help
2
Replies
38
Views
8K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
12
Views
2K
  • Electrical Engineering
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
772
  • Engineering and Comp Sci Homework Help
Replies
10
Views
3K
Back
Top