Dear Linh,<br><br>Thank you for your response.<br>So, do you suggest that I should increase the supercell size (by the same amount of d increase) each time I increase the value of d?<br><br>Thanks,<br><br><div class="gmail_quote">
On Tue, Jan 26, 2010 at 11:02 AM, 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"> > 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>
</div>I think what you did for checking E vs. ecutwfc is right. However, 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 relax<br>
your model with that value, then increase the value ecutwcf and run scf<br>
only, you can obtain a relation E vs. ecutwfc in similar.<br>
<div class="im"><br>
> For Final energy (obtained from last relaxation step) vs d interlayer<br>
distance at cutoff ecutwfc = 50.0:<br>
</div> > .... 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. , 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>
<font color="#888888">Linh<br>
</font><div><div></div><div class="h5">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 way to<br>
> do it?<br>
> Here is the input file I am using:<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>
> The results are :<br>
><br>
> 1- Final energy (obtained from last relaxation step) vs cutoff (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>
> 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>
> 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>
</div></div>> ------------------------------------------------------------------------<br>
<div><div></div><div class="h5">><br>
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