Dear Linh,<br><br>I am trying to understand what you just explain.<br>If I keep the bottom and middle layer fixed (BN layers) and I only translate the third layer on the top of them (graphene) by an amount corresponding to the interlayer increase d. For example delta d =0.1 Ang.<br>
Originally, the interlayer distance is for example d=3.3 Ang is increased to 3.4 Ang by moving top layer up by 0.1 Ang and increasing the supercell size in the z direction by 0.1 Ang. I mean adding 0.1 Ang more vacuum on the top layer to keep a vacuum distances on the top graphene and bottom BN layer of 8 Ang for each side.<br>
Is it the right way?<br><br>Thank you very much for your help.<br><br><div class="gmail_quote">On Tue, Jan 26, 2010 at 1:00 PM, Ngoc Linh Nguyen <span dir="ltr"><<a href="mailto:nnlinh@sissa.it">nnlinh@sissa.it</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;"><div class="im">mohamed sabri majdoub wrote:<br>
> Dear Linh,<br>
><br>
> Thank you for your response.<br>
> So, do you suggest that I should increase the supercell size (by the<br>
> same amount of d increase) each time I increase the value of d?<br>
</div>You should do that if d values change by moving 1st and 3rd layer and<br>
keep middle layer.<br>
Somehow, let try to keep away the interactions of the layers between<br>
neighbor supercells.<br>
<div class="im">> Thanks,<br>
><br>
> On Tue, Jan 26, 2010 at 11:02 AM, Ngoc Linh Nguyen <<a href="mailto:nnlinh@sissa.it">nnlinh@sissa.it</a><br>
</div><div><div></div><div class="h5">> <mailto:<a href="mailto:nnlinh@sissa.it">nnlinh@sissa.it</a>>> wrote:<br>
><br>
> > When I plot the graph, it looks like ecutwfc =60 is optimal. Is it<br>
> right? Is it the right way to do > it? Is the final energy (obtained<br>
> from last relaxation step) the right quantity?<br>
> I think what you did for checking E vs. ecutwfc is right. However,<br>
> with<br>
> the big model it could be quite expensive. You can do like that: you<br>
> choose an ecutwfc value firstly, for example ecutwfc = 40 Ry, and<br>
> relax<br>
> your model with that value, then increase the value ecutwcf and<br>
> run scf<br>
> only, you can obtain a relation E vs. ecutwfc in similar.<br>
><br>
> > For Final energy (obtained from last relaxation step) vs d<br>
> interlayer<br>
> distance at cutoff ecutwfc = 50.0:<br>
> > .... ecutwfc = 60.0<br>
> I am confusing about increasing of d values in your models.<br>
> If you keep the sizes of supercell unchanged, when increasing the<br>
> value of d, the 3rd layer of the above supercell and 1st layer of the<br>
> below supercell should be approached too close together, i.e. ,<br>
> let try<br>
> visualizing by Xcryden at different values of d. And it can reduce an<br>
> increase of energy of your models as d is lager.<br>
><br>
> Good luck,<br>
> Linh<br>
> mohamed sabri majdoub wrote:<br>
> > Dear all,<br>
> ><br>
> > I run calculations of 1 layer of graphene on bilyaer boron nitride.<br>
> > I am trying to study the Energy vs cutoff and Energy vs interlayer<br>
> > distance d between the graphene layer and top of BN layer.<br>
> > I got the following results after several runs. Is it the right<br>
> way to<br>
> > do it?<br>
> > Here is the input file I am using:<br>
> ><br>
> ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------<br>
> > &control<br>
> > title = 'GphBN'<br>
> > calculation = 'relax'<br>
> > outdir = '/pwscf/pwscftemp'<br>
> > prefix = 'GphBN'<br>
> > pseudo_dir = '/pseudopot-C-B-N'<br>
> > tprnfor = .t.<br>
> > restart_mode = 'from_scratch'<br>
> > wf_collect = .true.<br>
> > disk_io = 'low'<br>
> > /<br>
> > &system<br>
> > ibrav = 0,<br>
> > celldm(1) = 1.8897261<br>
> > nat = 72,<br>
> > ntyp = 3,<br>
> > ecutwfc = 60.0<br>
> > occupations = 'smearing'<br>
> > smearing = 'gaussian'<br>
> > degauss = 0.003675<br>
> ><br>
> > /<br>
> > &electrons<br>
> > mixing_mode = 'local-TF'<br>
> > mixing_beta = 0.05<br>
> > diagonalization = 'david'<br>
> > conv_thr = 1.D-5<br>
> > /<br>
> > &ions<br>
> > trust_radius_ini = 0.10<br>
> ><br>
> > /<br>
> > &cell<br>
> ><br>
> > ATOMIC_SPECIES<br>
> > B 10.81100 B.pz-vbc.UPF<br>
> > C 12.01070 C.pz-vbc.UPF<br>
> > N 14.00674 N.pz-vbc.UPF<br>
> > ATOMIC_POSITIONS angstrom<br>
> > B 0.710000 8.607000 11.306000<br>
> > B 0.710000 11.066000 11.306000<br>
> > B 2.840000 7.377000 11.306000<br>
> > B 0.710000 6.148000 11.306000<br>
> > B 6.390000 6.148000 8.000000<br>
> > B 6.390000 8.607000 8.000000<br>
> > B 6.390000 11.066000 8.000000<br>
> > B 2.840000 9.836000 11.306000<br>
> > B 7.100000 7.377000 11.306000<br>
> > B 7.100000 9.836000 11.306000<br>
> > B 7.100000 12.295000 11.306000<br>
> > B 4.970000 11.066000 11.306000<br>
> > B 2.840000 12.295000 11.306000<br>
> > B 4.970000 6.148000 11.306000<br>
> > B 4.970000 8.607000 11.306000<br>
> > B 4.260000 7.377000 8.000000<br>
> > B 2.130000 11.066000 8.000000<br>
> > B 2.130000 8.607000 8.000000<br>
> > B 2.130000 6.148000 8.000000<br>
> > B 4.260000 9.836000 8.000000<br>
> > B 0.000000 7.377000 8.000000<br>
> > B 0.000000 9.836000 8.000000<br>
> > B 4.260000 12.295000 8.000000<br>
> > B 0.000000 12.295000 8.000000<br>
> > C 2.130000 6.148000 14.106000<br>
> > C 2.130000 8.607000 14.106000<br>
> > C 2.130000 11.066000 14.106000<br>
> > C 0.000000 12.295000 14.106000<br>
> > C 0.000000 9.836000 14.106000<br>
> > C 0.000000 7.377000 14.106000<br>
> > C 0.710000 11.066000 14.106000<br>
> > C 0.710000 8.607000 14.106000<br>
> > C 0.710000 6.148000 14.106000<br>
> > C 6.390000 8.607000 14.106000<br>
> > C 6.390000 6.148000 14.106000<br>
> > C 4.970000 11.066000 14.106000<br>
> > C 6.390000 11.066000 14.106000<br>
> > C 7.100000 12.295000 14.106000<br>
> > C 7.100000 9.836000 14.106000<br>
> > C 7.100000 7.377000 14.106000<br>
> > C 4.970000 8.607000 14.106000<br>
> > C 2.840000 12.295000 14.106000<br>
> > C 2.840000 9.836000 14.106000<br>
> > C 2.840000 7.377000 14.106000<br>
> > C 4.260000 7.377000 14.106000<br>
> > C 4.970000 6.148000 14.106000<br>
> > C 4.260000 12.295000 14.106000<br>
> > C 4.260000 9.836000 14.106000<br>
> > N 0.000000 12.295000 11.306000<br>
> > N 0.000000 9.836000 11.306000<br>
> > N 0.000000 7.377000 11.306000<br>
> > N 2.840000 7.377000 8.000000<br>
> > N 2.840000 9.836000 8.000000<br>
> > N 2.840000 12.295000 8.000000<br>
> > N 4.970000 11.066000 8.000000<br>
> > N 4.970000 8.607000 8.000000<br>
> > N 4.970000 6.148000 8.000000<br>
> > N 7.100000 12.295000 8.000000<br>
> > N 7.100000 9.836000 8.000000<br>
> > N 7.100000 7.377000 8.000000<br>
> > N 0.710000 6.148000 8.000000<br>
> > N 0.710000 8.607000 8.000000<br>
> > N 0.710000 11.066000 8.000000<br>
> > N 6.390000 11.066000 11.306000<br>
> > N 6.390000 8.607000 11.306000<br>
> > N 6.390000 6.148000 11.306000<br>
> > N 2.130000 11.066000 11.306000<br>
> > N 2.130000 8.607000 11.306000<br>
> > N 2.130000 6.148000 11.306000<br>
> > N 4.260000 12.295000 11.306000<br>
> > N 4.260000 9.836000 11.306000<br>
> > N 4.260000 7.377000 11.306000<br>
> > K_POINTS automatic<br>
> > 10 10 1 1 1 0<br>
> > CELL_PARAMETERS<br>
> > 8.51980 0.00000 0.00000<br>
> > 0.00000 7.37600 0.00000<br>
> > 0.00000 0.00000 22.6120<br>
> ><br>
> -------------------------------------------------------------------------------------------------------------------------------------------------------------------------<br>
> > The results are :<br>
> ><br>
> > 1- Final energy (obtained from last relaxation step) vs cutoff<br>
> (ecutwfc)<br>
> ><br>
> > cut20: Final energy = -857.6291985882 Ry<br>
> > cut30: Final energy = -876.5012327526 Ry<br>
> > cut40: Final energy = -885.4765196610 Ry<br>
> > cut50: Final energy = -890.0493283821 Ry<br>
> > cut60: Final energy = -892.0114549785 Ry<br>
> > cut70: Final energy = -892.8706506804 Ry<br>
> > cut80: Final energy = -893.2520149989 Ry<br>
> ><br>
> > When I plot the graph, it looks like ecutwfc =60 is optimal. Is it<br>
> > right? Is it the right way to do it? Is the final energy (obtained<br>
> > from last relaxation step) the right quantity?<br>
> ><br>
> ><br>
> ------------------------------------------------------------------------------------------------------------------------------------------------------------<br>
> > 2- For Final energy (obtained from last relaxation step) vs d<br>
> > interlayer distance at cutoff ecutwfc = 60.0:<br>
> ><br>
> > d=2.5= Final energy = -892.0122348924 Ry<br>
> > d=3.0= Final energy = -892.0119194861 Ry<br>
> > d=3.1= Final energy = -892.0089055437 Ry<br>
> > d=3.2= Final energy = -892.0104083862 Ry<br>
> > d=3.3= Final energy = -892.0115677649 Ry<br>
> > d=3.4= Final energy = -892.0113018510 Ry<br>
> > d=3.5= Final energy = -892.0121118001 Ry<br>
> > d=3.6= Final energy = -892.0113544429 Ry<br>
> > d=3.7= Final energy = -892.0096184164 Ry<br>
> > d=3.9= Final energy = -892.0041914533 Ry<br>
> ><br>
> > There are fluctuations in the data? Is it normal?<br>
> ><br>
> -----------------------------------------------------------------------------------------------------------------------------------------------------<br>
> ><br>
> > 3- For Final energy (obtained from last relaxation step) vs d<br>
> > interlayer distance at cutoff ecutwfc = 50.0:<br>
> ><br>
> > d=2.5= Final energy = -890.0504214490 Ry<br>
> > d=2.8= Final energy = -890.0503533642 Ry<br>
> > d=2.9= Final energy = -890.0503471460 Ry<br>
> > d=3.0= Final energy = -890.0486572147 Ry<br>
> > d=3.1= Final energy = -890.0491375298 Ry<br>
> > d=3.2= Final energy = -890.0480863744 Ry<br>
> > d=3.3= Final energy = -890.0492074574 Ry<br>
> > d=3.4= Final energy = -890.0501303488 Ry<br>
> > d=3.6= Final energy = -890.0488317041 Ry<br>
> > d=3.5= Final energy = -890.0492797825 Ry<br>
> > d=3.7= Final energy = -890.0470527582 Ry<br>
> > d=3.8= Final energy = -890.0443250729 Ry<br>
> > d=3.9= Final energy = -890.0412177248 Ry<br>
> ><br>
> > Same thing for these data.<br>
> > Is there any thing wrong?<br>
> > Your suggestions are welcomed.<br>
> ><br>
> > Thank you in advance for your help!<br>
> ><br>
> > Regards,<br>
> ><br>
> > ME<br>
> > University of Houston<br>
> ><br>
> ><br>
> ><br>
> ><br>
> ><br>
> ------------------------------------------------------------------------<br>
> ><br>
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