Dear Palla Gopal,

Logarithmic transformation of both concentration (C) and peak current intensity (IP) may be a convenient tool to linearize your data but it will not preserve the error structure of your (C, IP) data and as such it will lead to bias in the calculation of LOD. Consider, for example, that the random error terms will have a mean of zero in the log units but not in the original arithmetic units. If you wish to report the LOD, best defined as 3 times the standard deviation of the blank value, it is better not to log-transform your data.

If possible, place yourself under experimental conditions where your calibration line (or standard addition line) is linear, for example using low-concentration standard solutions (or standard additions) and short deposition times. A least square linear regression of your (C, IP) data will produce an equation of the form IP = aC + b where b is the intercept and a is the slope (a = dIP/dC). Your LOD will be: LOD = 3 s(b)/a where s(b) is the standard deviation of the intercept.

If the relationship between C and IP deviates from linearity, even at low concentration values, then you will have to perform a nonlinear regression to fit an appropriate function to your calibration data. The expression of LOD will be the same as above, except that you will have to express a and b as a function or the regression parameters. In this case, a will be the slope measured at C = 0 and 3 s(b) will be the 99.5% confidence interval of the b parameter.

Note: as you approach the LOD you should be in a region of the calibration line (or standard addition line) that is linear. Nonlinearity in that region may be an indicator of a problem, I think.

Sir, I have investigated the differential pulse voltammetric current response with respect to Rutin concentration and we have not received linear increase in the current response. But our other results (scan rate) suggesting that Rutin has adsorption property towards the electrode surface and I feel that the adsorbed Rutin might have catalytic activity and causes increase in the current response as shown in figure attached here. and we have found linear equations for log of currents and log of concentrations in the following attachament. please suggest me the right way to calculate LOD.

Dear Palla Gopal,

Thank you for sending me this additional information and also for attaching your calibration curve for Rutin. Based on this information, I would like to suggest the following procedure (which is actually the procedure I use in all my voltammetric work) for calculating LOD.

1. Record a IP versus E scan for a blank solution. If you cannot produce a blank solution, then use a low-Rutin sample and apply a 0 second accumulation period prior to the scan.

2. Calculate the standard deviation of IP based on all the IP values recorded in the potential range EP +/- 100 mV, where EP is the peak potential of Rutin that you use for your analyses.

3. Multiply this standard deviation by 3 and then use the calibration curve to determine the concentration that produces a current equal to 3 x sd(IP). To do that, you can either fit your calibration data to an equation such as I = I0 + a exp(bC) or you can take the log of  3 x sd(IP) and use your log-log relationship to calculate log(LOD).

Please get back in touch if anything is unclear.

Thank you very much for your valuable information. Sir, can suggest me any reference for the calculation of log(LOD)  

Sorry, 'calculate' was perhaps not the right word. Also, I should have designated the current as I (not IP) in my previous explanation since the background current does not produce any analytical peak. To summarise:

1. Record a I vs. E scan for a blank solution. If you cannot produce a blank solution then use a low-Rutin sample a apply a 0 second accumulation period before the scan.

2. Fit a short section of this I vs. E scan (+/- 100 mV or less, either side of where your peak potential is expected to be) to a straight line: this segment represents the averaged background current in that region.

3. Consider the set of data made up of all the differences between measured and predicted (from the linear fit) values of I within that potential range. For this set of data, the standard deviation is sd. Multiplying this value by 3 then gives you 3 x sd = the minimum value that a deviation from the baseline must have to be interpreted as a Rutin signal.

4. From your plot of I vs. C (or log(I) vs. log(C)), the value of C corresponding to a current value of 3 x sd  is, by definition, the LOD.

Hope this is more helpful!

Hey people,

I have found R^2 and R equal 0,999 for a delta-I from DPV analysis VS log(concentration/mol L-1).... How should I calculate the LOD? When I use 3 x SD/intercept base on Origin program, I cannot find a reasonable number.... Should I apply "log" at this number?

Hello Mariana,

I am not familiar with the Origin software but I am pretty confident that you can still calculate the LOD from your data. It has to be done in two steps though:

1. Consider the I (or delta-I) vs. E voltammogram resulting from a blank. You can then EITHER calculate SD(Iblank) or in your case SD(delta-Iblank) from the noise in the voltammogram---as I had proposed earlier---OR you can run several blanks and then calculate SD of the blank signals. After that, multiply by 3 to obtain 3 SD(delta-Iblank) .

2. Now you can turn to your delta-I vs. log(C) curve to extract the value of LOD. The LOD is simply the (log of) concentration that produces a signal equal to 3 SD(delta-Iblank).

I hope this helps.