I am trying to calculate LOD and Sensitivity. In calibration plot, my X-axis is Concentration (mM) and Y axis is Current (microampere).
Slope is 3.3417 and standard deviation of blank is 0.2057.
Linear detection range is 1-9mM.
Now how to calculate LOD? I tried 3sigma method (3*Standard Deviation)/Slope, which gives some value, but I am confused in units.
Will it be mM (milimolar)?
People have published similar work, their linear detection range is also in mM, but their LOD comes out to be in micromolar or nanomolar, how it is done?
Also they write "based on S/N method", how to use this method?
The simple and wery objective method is the determination through the repeatability of small concentrated sample measurement (for example the lowest calibration standard). LOD = 3*SD and LOQ = 10*SD.
Dear Jiří Kokšal,
My main problems is units.
Please read above . .and reply. What will be the unit of LOD(Milimolar or MIcromolar)?
I think that Milimolar - it is the same like the measured unit ;-) It depends on the expression. If you will have a LOD = 0,001 mM you can use 1 uM. But in my recomanded method is "unit" simple because you calculate the limits directly from SD of measured concentration - it is the same unit.
ok thanks.. but in papers they have done like me . .their calibration plot is also in milimolar but LOD they showed is in micomolar.
In this paper the autors have probably a wery good S/N ratio - it could be due the filtration of signal (current). LOD is 3time biger than nois - principle of S/N algoritm - and this method is no objective because it depends on signal filtration methods. If you will have very low level of nois than you will have a very low level of LOD - 3 times biger than nois ;-) It could be better the measure of low concentration sample repeatability and than the SD you will use for LOD and LOQ quantification.
Detection limit can be calculates using the following formula;
LOD= (k * sb)/m
Where k is chosen to be 2 or 3; k value 2 corresponds to a confident level of 92.1%.. k value 3 corresponds to a confident level of 98.3%. sb is the standard deviation of the blank. and m is the calibration sensitivity that is slope of the linear plot between concentration vs current..
how to calculate standard deviation of blank?
What I understand is, I need to take CV of blank solution 10-20 times and then calculate S.D.
Am I right?
According to literature,singnal/noise shoud be equal to 3.
the equation: y=mx+b
Add the lower concent to the equation and got lower signal Then the lower concentration should be converted to current (signal) acording the linear regtetion you got. Finally just do s/n=3. The baseline is the noise. I have confirmed calculating from anothers papers
i have done this..
its very easy...
please refer to my articles..
Article Development of Highly Sensitive Polysilicon Nanogap with APT...
Article A Point-of-Care Immunosensor for Human Chorionic Gonadotropi...
Usually, I use the following equation:
LOD=3*SD/m
where SD=magnitude of error bar at blank, m=slope of the calibration curve at blank
Sensitivity= slope of the calibration curve at a certain target concentration
José M. Campiña
how can I calculate the SD of blanks when they do not show any response, I mean there is no current observable without analyte at all??
First, we should know that limit of detection is described as the concentration of analyte giving a signal equal to the blank signal, Sb, plus three times the standard deviation of blank, σb:
Limit of detection= Sb+3σb
In the aforementioned formula, 3 is a statistical number that guaranteed that the probability of each of two types of statistical error (error type one and error type two) occurring is less than 5%.
If we obtain a signal, for example, peak current in voltammetry, which is equal to LOD, we can claim that there is more than 95% chance to detect analyte if there is exist.
If we want to use LOD= k *s/m we should analysis at least 5 blank solutions and calculate the standard deviation of these measurements, s. Afterward, the calibration curve should be obtained for different concentrations of analyte and the corresponding slope should be calculated, m. then, according to k *s/m LOD can be calculated.
In the LOD= k *s/m, acquired LOD has a dimension of Concentration which we use for plotting calibration curve because the dimension of the numerator is Signal, for example, µA in voltammetry, and the dimension of the denominator is Signal/Concentration, e.g. µA/molar.
I would add a caution that the formula LOD = (3 * SD)/m only works if the 3*SD value actually intersects the linear part of the calibration curve having the slope m. For most calibration curves, there is a part between the linear range and the background where the curve is not linear and has a slope of less than m. If the 3*SD value intersects this portion of the calibration curve, then one must use the calibration curve to graphically determine the LOD intersection point.
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