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.
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.
@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 OPs question How can I calculate this height?More
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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.
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.
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.
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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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.
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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