By Anonymous on Tuesday, March 18, 2003 - 01:19 pm:
I am working on a gradient method. As I doubled the flow rate, the peak area decreased in half. Does anyone know the reason? any solution?
Thanks.
By Anonymous on Tuesday, March 18, 2003 - 01:42 pm:
x= amount of analyte passing through detector in time y creates area x*y...double the flow rate and time= y/2.... hence your new area is x*y/2....
If for some reason you need constant area then double the injection volume (if this is feasible) otherwise your response factor will take into consideration the change in area.
By Anonymous on Tuesday, March 18, 2003 - 01:43 pm:
concentration sensitive detector. analyte spends 1/2 the time in front of the detector. Theoretically the peak height will stay the same. You'll find a number of past discussions on the subject.
best of luck
By Anonymous on Tuesday, March 18, 2003 - 03:08 pm:
Note that the sensitivity does not change!
By Anonymous on Friday, March 28, 2003 - 07:43 am:
YOU WOULD HAVE TO PUT A LOWER TIME RESPONSE AND AN ADECUATE TIME CONSTANT: for example ,less than 100ms
By Uwe Neue on Friday, March 28, 2003 - 03:12 pm:
That is a good question:
1. how does peak area change when your peak width approaches the sampling rate?
2. how does the signal to noise ratio change when your peak width approaches the sampling rate?
Not that anybody would want to work there, but it is interesting.