calibration of the voltage

[gauravmcgill]

What is happening is i am using labview interface. But i am using measurement computings 1008 minilab DAQ board.
Now the minilab says that analog input in single ended configuration is 11 bits resolution.
I am using minilab to acquire from the LVDT. The resolution of LVDT is practically infinite. Now i need to know what is the smallest voltage that i can get from the LVDT.
The LVDT is a 5volt DC type made by packard and i have no further info(i could need some though).
Importnat points the minilab has a range of +10 to -10 Volts in single ended and -20 to +20V in differential so what i want to know is if i want to conver the bits reading that i get from the A/D converter how should i convert it back to the voltage regarding. I would need
a back of the envelope calculation.

[CoastalMaineBird]

11 bits resolution (an unusual number, by the way) is equal to 2^11 = 2048 discrete values that a single reading can take.
If your card has a range of +/- 10 V, then that's a 20-volt spread, divided into 2048 steps (I'll assume -1024 to +1023).
That means you have a resolution of 0.009766 Volts. (20 V / 2048 steps)

If you could set your card to be UNIPOLAR, it would be better, since your signal is unipolar (not negative voltage).

By using a 0 to +5V signal into a -10 to +10 Volt range, you are using only one fourth of the range (wasting 2 bits).

If the range is -1024 to +1023 counts, your signal is only going to range from 0 to +511.

So if you had a reading of 256, for example, the voltage would be 256 * 0.009766 = 2.5000 Volts.

Hope th
at helps.

[gauravmcgill]

thanks a lot i have few more questions, how am i wasting 2 bits.
secondly i was thinking since the full scale voltage reading is 0 to 5 volts then should it not be 5/2048 resolution.
although i may be wasting 1/4 of the scale but is it having an effect on the calculation as the calculations look the same even when the signal goes from 0 to 5.
thanks

[Dennis_Knutson]

Measurement resolution is calculated from the range of the daq board and not your measured signal. You had stated that it's range was +/-10 volts so CoastalMaineBird's 20/2048 is absolutely correct. To get more resolution, you have to either add bits or change the range.

[CoastalMaineBird]

how am i wasting 2 bits.
By using only one fourth of the input range, you are in effect throwing away two bits (2^2 = 4)

since the full scale voltage reading is 0 to 5 volts then should it not be 5/2048 resolution.
But you said the resolution of the CARD was +/- 10 Volts. The card doesn't know that you're only connecting a 0-5V signal to it. If it's configured to measure +/- 10V, then that's what it will do. Half of your measurement range is wasted because your signal can't be negative. Half of the remainder is wasted because your signal can't be above 5 Volts.

although i may be wasting 1/4 of the scale
You are wasting THREE fourths.

but is it having an effect on the calculation as the calculations look
the same even when the signal goes from 0 to 5.

No - that's my point. The calculations will be the same, regardless of whether your signal goes from 0 to 5V or 0 to 10 V or -10 to 10 V.
It's the range of the CARD divided by the number of different values that the CARD produces.
In your case this is 20 V divided by 2^11 = 0.009766 V.

If that's not clear, keep asking.

[gauravmcgill]

I understand thanks a lt. Since i am using an LVDT which measures the displacement and gives out the voltage signal for the corresponding displacement. Is it possible to find out the resolution in terms of the displacement. What would i need to do this.
thanks

[CoastalMaineBird]

Well, we know that the VOLTAGE resolution is 20 V divided by 2^11 steps = 0.009766 Volts per step.
Suppose your LVDT produces 0-5 V for a displacement of 0-10 inches.
That's a factor of 10 inches divided by 5 Volts = 2 inches per volt.

Multiply 2 inches per Volt times 0.009766 Volts per step, and you get 0.019532 inches per Step - that's your displacement resolution.

If you can possibly configure the DAQ card to use a 0-5 V range instead of +/- 10V, you're better off doing that. That will get you 5-mil (.004883 inches) resolution instead of 20-mils (.019532).

In any case just remember:

Inches Volts Inches
------- x ------ = -------
Volt Step Step