﻿ Empower 3 - calibration curve problem - Chromatography Forum

## Empower 3 - calibration curve problem

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### Empower 3 - calibration curve problem

Help.

Standard calibration curve not working .. only 2 standards at same level but intercept is much larger than the peak areas
ie zero amount has a peak response in calibration curve but not a calibration point
Yvalue - area , xvalue- amount, fit - linear- weighting - none

Y= -2.11E+003x + 3.61e+005 ?

X values
23.69
24.964
y value
311000
3308764

much appreciated

### Re: Empower 3 - calibration curve problem

If you have only two calibration points at the same concentration, constructing a calibration curve by linear regression doesn't make much sense...
I'd use one-point calibration in this case.

### Re: Empower 3 - calibration curve problem

Hi ,

Not exactly what I was asking but OK
What happens if method states mean 2 standards and use these to quantify by calibration curve

### Re: Empower 3 - calibration curve problem

Not sure how or why, but once or twice in my experience, I changed from "Linear" to "Linear through Zero" in the processing method to get this sort of thing sorted. The original cal. curve looked like it had a negative slope(!).
Thanks,
DR

### Re: Empower 3 - calibration curve problem

ydna1977 wrote:
Hi ,

Not exactly what I was asking but OK
What happens if method states mean 2 standards and use these to quantify by calibration curve

If you have only two standards at the same concentration, constructing a calibration line by linear regression will give you nonsense. I don't work with Empower, but I'd suppose if you select "linear" as calibration model it will use linear regression.
If you calibrate at one concentration only (no matter how many replicates) you're doing a one-point calibration. This is indeed a linear calibration where the second calibration point is the origin 0/0. Either select "one-point calibration" (or whatever it's called in Empower), this will of course also be a calibration with a calibration line. If you still want to use linear regression (which mathematically doesn't make much sense in this case, but of course it's possible to do)you need to force the origin through zero.

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