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I am trying to implement a small python script to detect peaks and calculate the corresponding area. The y-values are in mAU and x-values in minutes but there are some unit conversions in most HPLC softwares. For example in this image (found here):

HPLC example

The peak at 7.87 min has a height of 329451 but only approximately 20 V in the chromatogram. How can I calculate this height?

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    $\begingroup$ I suggest finding something much more substantive than what is at the link. Astonishingly, it actually says “What Is a Chromatogram? Now that you have understood everything about chromatography, let’s learn about chromatography and chromatographic analysis.” $\endgroup$
    – Ed V
    Commented Jan 31, 2021 at 13:58
  • $\begingroup$ Just took this example from their article because I got the same problem with our data. It is just about the unit conversion factors on the y-axis. $\endgroup$
    – SLx64
    Commented Jan 31, 2021 at 14:02
  • $\begingroup$ The table reads «329451» as area, e.g. after baseline correction (if there was one, but e.g., in powder diffraction crystallography [PXRD] one would look to perform a suitable one), followed by an area integration. Thus, first digitize the plot (e.g., automeris.io/WebPlotDigitizer). For the subsequent processing, there already are solutions published which may serve as inspiration for you, e.g., fityk.nieto.pl, PyMS, HappyTools. Area = half-peak width x peak height is one way. $\endgroup$
    – Buttonwood
    Commented Jan 31, 2021 at 15:35
  • $\begingroup$ @Buttonwood, the area calculation by "half-peak width x peak height" gives really bad results in HPLC. He should rather use rectangular, Simpson's or trapezoidal method for Gaussian or exponentially modified Gaussians to get anywhere close to the true area. $\endgroup$
    – ACR
    Commented Jan 31, 2021 at 18:13
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    $\begingroup$ @Buttonwood, Interesting...I have done paper cutting once or twice...at that time I did not have a peak fitting software and it works exceptionally well. Modern students may not heard of the fact one could use a non-computer method to determine peak area :-) $\endgroup$
    – ACR
    Commented Feb 2, 2021 at 20:31

2 Answers 2

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Height in HPLC simply refers to the height of the peak (= maximum value of the peak) above the baseline. The point by z1273 is right in the sense that this voltage to height conversion factor is instrument specific but this height cannot be in milliabsorbance units- it is too large for absorbance. This height is apparently "intensity". Anyway, you do not need to worry about the "true height" for peak detection. It will be fine to use the height of 20 V for the first peak at 7.867 min. In older instruments, the detector output was displayed as Volts or mV and the manufacturer provided a conversion factor from volts to light absorbance or sometimes as intensity in the manual. See this discussion here (https://www.chromforum.org/viewtopic.php?t=6442). You should really look at modern HPLC data. The one you are showing here is from an ancient instrument.

I am assuming you are using derivatives to detect peaks? For any chromatography program, I would suggest that you do a simulation first, because all variables will be in your hand. Make a chromatogram as sum of large and small exponentially modified Gaussians or simple Gaussians of various heights and areas and add some noise to test your program.

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The Height ($h$) in mAU is a function of the Signal A measured in volts at 254nm. The function should look like $h=16700\cdot A+1500$ or so, in this case. The calibration equation would be specific to a given installation.

I suppose the Height may be back-calculated from the area, but it looks circular reasoning to me.

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    $\begingroup$ Maurce, he used the term "circular reasoning" not circular height. Circular reasoning refers to the chicken and egg type problem. Who came first? $\endgroup$
    – ACR
    Commented Jan 31, 2021 at 17:11
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    $\begingroup$ @Maurice Please take a moment to review the OP where mAU, Signal A at 254 nm and the context of the numbers are specified. As for the units, I suppose mAU/Volt and mAU may be appropriate for 16700 and 1500, respectively. What I meant by 'circular reasoning' is that deriving the height from an area that is itself calculated using the height appeared to me as a circular reasoning. To be sure, the link in the OP provides relevant specifics as to how they view the calculation of the area. Hopefully it is noted that I focused my answer to the OP's question 'How can I calculate this height?' $\endgroup$
    – z1273
    Commented Jan 31, 2021 at 17:27

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