By Anonymous on Monday, September 23, 2002 - 07:08 pm:
We would like to know whether is correct to give units to a peak area or not.
We understand the peak area is an integration of the response obtained with respect to the time,in a limited interval of time. We are not sure if the peak area should be expressed as millivolts/Units of Absorbance, simply express it as a peak area with no units, or alternatively, as millivolts*second/Absorbance Units*second.
Our computer package is Millenium32.
By Anonymous on Tuesday, September 24, 2002 - 05:36 am:
I believe the units for the area of a peak are usually expressed as microvolt-seconds.
By hinsbarlab on Tuesday, September 24, 2002 - 12:39 pm:
Most data analysis packages have provisions for converting the raw signal (mV) into a more meaningful unit (e.g., mAU). That way the absorbance value you see on the chromatogram matches what you see on the detector. Once converted, area would be expressed as (mAU x sec.) Gotta be careful to get the conversion factor right though. Check the specifications in your instrument manual.
By omarikhalid on Wednesday, April 28, 2004 - 06:03 am:
PEAK AREA = h x W½
the peak height (h) and peak width at half-height (W½)
so peak area = unit of signal which is mainly y-axis in the chromatogram X unit of time which is mainly x-axis in the chromatogram
peak area = milli or kilo volts (counts).minute or second (i.e: mV.S)