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DAQ Problem CLAD

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7) How does changing a DAQ board from a 0-10 volt range to a -10 to 10 volt range, affect the minimum voltage change that can be detected?

A   The minimum voltage change that can be detected will be doubled.

B   The minimum voltage change that can be detected will be split in half.

C   The minimum voltage change that can be detected will depend inversely with the resolution of the board.

D   There will be no effect on the minimum voltage change that can be detected


Please explain me about this!!!!!!!!




Prashant Soni
LabVIEW Engineer
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Message 1 of 10

To help you learn, please explain your thinking about this first.


How do you determine the minimum detectable voltage change?  What effect does the full scale range of the DAQ have on resolution?  How do you calculate resolution?  How is resolution related to the minimum detectable voltage change?



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Message 2 of 10

I dont have much knowlledge about it , can u pls explain me bat this or there is any resource by that i can get much knowledge abt this.

Prashant Soni
LabVIEW Engineer
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Message 3 of 10

Answer is A


smallest detectable voltage change is directly proportinal to range

Message 4 of 10

If you have any DAQ device, look at the manuals.  They usually have an explanation of resolution and range.  Also search the Knowledge Base for information.



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Message 5 of 10


With a DAQ board you use an analogue to digital converter (ADC) to read in the analogue voltage to the computer. The greater the resolution of the ADC the smaller the voltage change you can detect.

The resolution of the converter will depend on how many bits are used to cover the expected range of voltage input. For example, if you have 2 bits to represent the analogue number you only have four possible states in the digital number (namely 00, 01, 10, 11) - if these four states are to allow you to measure over a given voltage range, you will only be able to detect changes of one quarter of the total range.

Therefore, if you double the range of voltages you are expecting to measure, you also double the minimum change in voltage that you can detect.


Make sense?

Message 6 of 10
Accepted by Prashant_S1

Hi Prashant,


The minimum voltage change is givein in bits.  For example, let's use minimum voltage change=16bit.  That means you have 16 bits, 65531 of unique digital number, to represent all voltage in a given range (see below concerning how I calculated it).  To find out what is the minimum voltage change, you take the voltage range, if it is -10 to 10V, the range is 20, and divide it by 65531 (20/6531).  As you can see, if the range is direction proportional to the minimum voltage change, if if range double, the other one will double as well.


Answer: A



2^0+2^1+2^2+2^2+2^4+2^5+2^6+2^7+2^8+2^9+2^10+2^11+2^12+2^13+2^14+2^15 = 65531


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Message 7 of 10

Actually 2^16 is 65536.  So divide the range by that.


2^16 means you have 65536 discrete steps that range from 0 to 65535

Message 8 of 10



I hope you get the solution and reading stuff on your board posting, to make it easy for other users navigation reading, mark your solution accepted.



- HS

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Message 9 of 10

I would like to know why a question about hardware is on a Labview exam since Labview is software?  This question has absolutely nothing to do with Labview programming ability.  I know that Labview is used to program DAQ equipment, but believe it or not, there are very many instances where Labview is used without any DAQ eauipment involved at all.  Testing ones knowledge of hardware in a software prdiciency exam has no purpose.  This issue has been brought up time and again, and NI still refuses to acknowledge their mistaken ways.  Their claim that a Labview programmer should have some knowledge of DAQ is just plain wrong.  A person can be a master at Labview and know nothing about DAQ.


- tbob

Inventor of the WORM Global
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Message 10 of 10