Hello Sir Thank you for this great video I have concern regarding k value calculation.... Is the V in volt or Mili volt? And same for the current ampere or milliampere?
Yes. When I first started. I also tried to calculate at every potential. The results show different K1 and K2 value, which might due to different intensity of current response at different potential. Thus, it is important to use the maximum redox peak. Then you can fix the K1 and K2 value for the rest of potential. You just need to mention that in your paper the analyzation is based on fix potential at certain voltage.
Thank you for this video. Great work. When we plot the step 7 graph, what are the values taken from the step 6 table(specifically which current value)? Kindly please explain.
the value at step 6 you can get by obtaining the current response at fixed potential for both oxidation and reduction peak. For example, at 0 V what is the current response at both oxidation and reduction. Follow by 0.05, 0.1, 0.15, until desired PW. In this video until 0.5 V. To plot the step 7, used the current response after multiply with the percentage contribution, the you can successfully plot diffusion contribution or diffusion limiting contribution. Hope You understand.
Thanks for the explanation! But I have a small question, why we need to fix a certain potential rather than choose the highest point which is shifted with scan rate. If it's possible, please provide me the reference. Thanks!!
The ideal case scenario is peak the voltage that produce the highest current. But at every scan rate the voltage must be the same. For reference you can visit my google scholar. Just type Muhammad Norhaffis Electrochromic Supercapacitor.
Thank you, it is really helpful. I have a question, regarding the log current vs log scan rate, is it peak current or it is the current at the voltage of 0.35 V? Please inform.
The current at voltage 0.35V. If your peak current exists at a different voltage you can choose diff voltage. For example 0.4 V. It is necessary to pick voltage at a peak current. The answer is no.
@@DrHaffeast I have another question regarding the R^2 fit (procedure 9). Most of the papers use anodic peak and cathodic peaks subtracting from the base line. But you used current at a constant voltage i.e. 0.35 V. Would you please explain that? If possible with a reference. Thank you in advance. :)
Just applicable for battery type material. Normally battery grade material was used as positive electrode. For negative potential window please used TRASATTI method.
sir, I have a question when we get cathodic peaks we also get negative values of current, as you know log of negative values is possible so we can neglect the negative value of cathodic current?
Thanks brother for nice explanation and procedure. But I don't understand why we should measure the current only at fixed potential for determination of b value? I thought we should take the peak current in each scan rate since after reaching the peak current values the diffusion/mass control portion at the electrode interface take place? Can you clear that part please. Thanks again.
Based on my understanding, increasing the scan rate, the diffusion control mechanism will decrease compared to the capacitive control mechanism. This is because at a higher scan rate, there is insufficient time for the electrolytes to interact with the bulk of the materials. As a result the prominent redox peak in CV became less obvious. If you take the peak current in each scan rate without fixing the voltage, then there is no point to see the behaviour of your electrode-electrolyte interaction. We need a constant parameter for good comparison. Above all, my knowledge might be little, thus I recommended you to seek others advice. Thank you.
Run GCD for 10000 cycles. Calculate the specific capacitance for the first gcd cycle and every 500 cycles. Use formula (specific capacitance at each 500 cycles) divided by (specific capacitance at first cycle) times hundred.
Sir for the determination of b value, which peak , oxidation or reduction peak will take . in this explanation oxidation peak (0.35 v ) taken. can we take reduction peak?
Thank you so much I am getting EDLC behaviour like CV,I have calculated capacitance and diffusion using the dunn method is it correct or ? Can you explain about trasatti method?
@@DrHaffeast Why Dunn method can't be used in EDLC behaviour? I have seen few papers in which edlc based electrodes are studied using dunn method. Please explain?
After you get the percentage contribution. Focus on the capacitive contribution. For example your percentage capacitive contribution is 50%. After that using origin and find the oxidation and reduction current response at several potential (0V, 0.05V, 0.1V............0.5V) Then use mathematic formula Contribution*current response at specific potential to get the new Ocurrent response and new Rcurrent response.
With this formula, I need to find contribution rate for each scan rate? But we get k1 and K2 values from the sqrt of scan rate vs I/sqrt of scan rate graph. So, the k1 and K2 values is common for all scan rates. Then how will I calculate it for individual scan rate sir?
@@kavipriyahnagarajan571 you can use same K1 and K2 value. Sorry some clarifications for formula. It should be K1 * scan rate/(K1* scan rate + K2* sqrt scan rate) for contribution ratio and vice versa. Different scan rate, you can just change the scan rate value. The K value remains the same.
Unit does not matter. If the value from the potentiostat is Ampere, then you can use Ampere. If you want to change to mA also can. Just times with 1000.
Fix a voltage, then read the current response from multiple CV at different scan rate (1 to 5 mV/s). I have uploaded another video on how to obtain the current response.
Thank you so much for the valuable vid eo. When I plot v1/2 vs I/v(1/2) the plot shows linear points in the decreasing order not in the increasing order as u have shown in the video. Why this 8:398:39 happends. Did I do any mistake
Sir, how ahould be the relation between capacitive contribution and diffusion controlled contribution? For eg If capacitive contribution is 60% then diffusion controlled should be 40%, if we use the vice versa formula. But I'm getting 4% if i use the vice versa formula.
lama dah cari penerangan macam ni, akhirnya jumpa juga. terima kasih untuk ilmu yang disampaikan.
Sama2
Thank you for adding invaluable content..👍
Welcome 👍
Useful information..Great..Thank you...
Welcome 😁
Most awaited video, thank you sir
Welcome👍. Sorry for the delayed.
@@DrHaffeast Our workstation no separate oxidation and oxidation current response is there?
reduction
@@jithujoseph1102 Just use origin and you can manually obtained the oxidation and reduction current response.
In origin linear fit , how can i get the slope and intercept values?
great one... thank you sir..
welcome 👍
Hello Sir
Thank you for this great video
I have concern regarding k value calculation....
Is the V in volt or Mili volt?
And same for the current ampere or milliampere?
Volt and ampere
Hello Sir.
Very nice work.
What do you do when each voltage point has a different percentage contribution due to the different K1 and K2 values?
Yes. When I first started. I also tried to calculate at every potential. The results show different K1 and K2 value, which might due to different intensity of current response at different potential. Thus, it is important to use the maximum redox peak. Then you can fix the K1 and K2 value for the rest of potential. You just need to mention that in your paper the analyzation is based on fix potential at certain voltage.
thank you so much,,,, bb
Welcome
Thank you for this video. Great work.
When we plot the step 7 graph, what are the values taken from the step 6 table(specifically which current value)? Kindly please explain.
the value at step 6 you can get by obtaining the current response at fixed potential for both oxidation and reduction peak. For example, at 0 V what is the current response at both oxidation and reduction. Follow by 0.05, 0.1, 0.15, until desired PW. In this video until 0.5 V.
To plot the step 7, used the current response after multiply with the percentage contribution, the you can successfully plot diffusion contribution or diffusion limiting contribution.
Hope You understand.
@@DrHaffeast Thank you so much.
Thanks for the explanation! But I have a small question, why we need to fix a certain potential rather than choose the highest point which is shifted with scan rate. If it's possible, please provide me the reference. Thanks!!
The ideal case scenario is peak the voltage that produce the highest current. But at every scan rate the voltage must be the same. For reference you can visit my google scholar. Just type Muhammad Norhaffis Electrochromic Supercapacitor.
@@DrHaffeast OK THanks~
Thank you, it is really helpful. I have a question, regarding the log current vs log scan rate, is it peak current or it is the current at the voltage of 0.35 V? Please inform.
The current at voltage 0.35V. If your peak current exists at a different voltage you can choose diff voltage. For example 0.4 V. It is necessary to pick voltage at a peak current. The answer is no.
@@DrHaffeast Thanks
@@DrHaffeast I have another question regarding the R^2 fit (procedure 9). Most of the papers use anodic peak and cathodic peaks subtracting from the base line. But you used current at a constant voltage i.e. 0.35 V. Would you please explain that? If possible with a reference. Thank you in advance. :)
Good explaination. Just a doubt, is it applicable for negative potential window?
Just applicable for battery type material. Normally battery grade material was used as positive electrode. For negative potential window please used TRASATTI method.
@@DrHaffeast thanks a lot Dr.
@@gayathryganesh561 Welcome 😁
sir, I have a question when we get cathodic peaks we also get negative values of current, as you know log of negative values is possible so we can neglect the negative value of cathodic current?
Thanks brother for nice explanation and procedure. But I don't understand why we should measure the current only at fixed potential for determination of b value? I thought we should take the peak current in each scan rate since after reaching the peak current values the diffusion/mass control portion at the electrode interface take place? Can you clear that part please. Thanks again.
Based on my understanding, increasing the scan rate, the diffusion control mechanism will decrease compared to the capacitive control mechanism. This is because at a higher scan rate, there is insufficient time for the electrolytes to interact with the bulk of the materials. As a result the prominent redox peak in CV became less obvious.
If you take the peak current in each scan rate without fixing the voltage, then there is no point to see the behaviour of your electrode-electrolyte interaction. We need a constant parameter for good comparison.
Above all, my knowledge might be little, thus I recommended you to seek others advice. Thank you.
Thank you sir
Welcome.
Could you please tell me how you are calculation rafter, oafter. I didn’t get that.
let say your oxidation current is 0.1A. o after is the current multiply the percentage contribution. if 60% . Thus oafter is 0.06A.
Sir, How to find Capacitance retention from GCD and how to make its graph in origin?
Please make a video
Run GCD for 10000 cycles. Calculate the specific capacitance for the first gcd cycle and every 500 cycles. Use formula (specific capacitance at each 500 cycles) divided by (specific capacitance at first cycle) times hundred.
Sir for the determination of b value, which peak , oxidation or reduction peak will take . in this explanation oxidation peak (0.35 v ) taken. can we take reduction peak?
I used oxidation peak.
Both also can.
Thank you so much
I am getting EDLC behaviour like CV,I have calculated capacitance and diffusion using the dunn method is it correct or ?
Can you explain about trasatti method?
Normally people use TRASATTI method in your case. The video on trasatti method already been explained in other video.
@@DrHaffeast Why Dunn method can't be used in EDLC behaviour? I have seen few papers in which edlc based electrodes are studied using dunn method. Please explain?
Sir In step 6 how you plot this graph
Which current you are using and also which values you are using on x axis
Fix a voltage on x axis, for example 0.35 V and read the current value ( Y- axis). FYI, you can choose the prominent redox to select the x axis.
Hi sir,
In my CV curves there is no any redox peak..So how to do the transactive method .
th-cam.com/video/0gGgYxac6Wk/w-d-xo.htmlsi=qSzlZk3uu9Vs-QxL
I have made some video about trasatti method. Please refer to video above
@@DrHaffeast Thank you very much sir 🙏
How you calculated R after and O after is unclear
Please explain.
After you get the percentage contribution. Focus on the capacitive contribution. For example your percentage capacitive contribution is 50%.
After that using origin and find the oxidation and reduction current response at several potential (0V, 0.05V, 0.1V............0.5V)
Then use mathematic formula Contribution*current response at specific potential to get the new Ocurrent response and new Rcurrent response.
@@DrHaffeast you chose any potential for drawing graphs of log ip vs Log V. Can we choose the peak current of each scan rate instead.
@@maheshnaikwade5694 Yes. What you suggested is the best practices.
@@DrHaffeast If there are two anodic peak currents then do we have to consider whether two or one anodic peak current is sufficient. Please explain
@@maheshnaikwade5694 one is sufficient.
Hello sir, Can you explain how to calculate contribution rate after finding the k1 and k2 values?
K1/(K1+K2) *100% = capacitive contribution. K2/(K1+K2) *100% = diffusion controlled.
Thank you sir.
@@kavipriyahnagarajan571 Welcome
With this formula, I need to find contribution rate for each scan rate? But we get k1 and K2 values from the sqrt of scan rate vs I/sqrt of scan rate graph. So, the k1 and K2 values is common for all scan rates. Then how will I calculate it for individual scan rate sir?
@@kavipriyahnagarajan571 you can use same K1 and K2 value. Sorry some clarifications for formula. It should be K1 * scan rate/(K1* scan rate + K2* sqrt scan rate) for contribution ratio and vice versa.
Different scan rate, you can just change the scan rate value. The K value remains the same.
When we plot graph of i//V Vs /V what will be the unit of current A or Milli ampere
Unit does not matter. If the value from the potentiostat is Ampere, then you can use Ampere. If you want to change to mA also can. Just times with 1000.
@@DrHaffeast okey. but changing unit can be change the y-intercept thus asked
@@maheshnaikwade5694 it is recommended to use Ampere
hello sir,
sir, how we can take the log of current in cathodic peaks for the b slope
?
Fix a voltage, then read the current response from multiple CV at different scan rate (1 to 5 mV/s). I have uploaded another video on how to obtain the current response.
@@DrHaffeast Sir can I contact with via email?
@@DrHaffeast hello dear, but how can we take log of negative cathodic values.. please help me
@@jabirshahbaz-v2l for b value just focus on positive anodic current response
@@DrHaffeast sir in some papers reported, also taking cathodic b values in graph.
thank you so much
Thank you so much for the valuable vid eo. When I plot v1/2 vs I/v(1/2) the plot shows linear points in the decreasing order not in the increasing order as u have shown in the video. Why this 8:39 8:39 happends. Did I do any mistake
Depends on your results. Both trends also applicable. try to get the k1 and K2 value by gradient and y intercept, respectively.
@@DrHaffeast thank you so much. But for this case k1 is -0. 933 and k2 is 14.66 is it ok?
@@shobanamohanasundaram6215 So far I ignored the negative value. So K1 will necame 0.933
Thanks a lot🤝
My diffusion contribution is 98.02% the capacitive contribution is 16.75% but the total exceed 100.is it correct
Sir, how ahould be the relation between capacitive contribution and diffusion controlled contribution? For eg If capacitive contribution is 60% then diffusion controlled should be 40%, if we use the vice versa formula. But I'm getting 4% if i use the vice versa formula.
Maybe you used the wrong formula. Please revise the formula.
how did you get the R after O after values, I couldn't get it properly
The value of R and O times percentage contribution.
@@DrHaffeast Ok, Thank you Sir