Developer of the CRYSTAL code!
Hi,
job314 said in SCANMODE io error Read_int_1d:
So it turns out having fort.13 and fort.20 is not optional for restart.
Yes, the lack of these two files was the origin of the I/O error.
job314 said in SCANMODE io error Read_int_1d:
PS. I would also like to ask developers to allow uploading compressed files to this forum
Now also .zip, .tar, .tgz, .tar.gz files can be uploaded.
Cheers,
hello 
about VLAB I'll ask to the developer...
As for the MULTIWALL nanotubes.. there is another procedure you can follow, which use the keyword MULTIWALL. These are the steps, using as a test system a graphene sheet:
1) build a two-wall nanotube ---> see 2wall.jpg
title
SLAB
77
2.47
1
6 0.33333333 0.6666666667 0.000000
NANOMULTI
2
WALL
1
ROLLINGV
6 6
WALL
2
ROLLINGV
12 12
NANOJMOL
ENDWALL
..basis set
title
EXTERNAL
ATOMSUBS
24
1 7
2 7
3 7
4 7
5 7
6 7
7 7
8 7
9 7
10 7
11 7
12 7
13 7
14 7
15 7
16 7
17 7
18 7
19 7
20 7
21 7
22 7
23 7
24 7
END
..basis set..
the ZnO@GaN were done exactly in this way... if you roll up a two-layers slab you end with a double wall nanotube
goodmorning,
ok... now I understand better. This is an example. I chose the space group 73 because the 78 has a mirror plane... I don't know which kind of lattice do you want to simulate ...neither if the two lattice (BeO and MgO) are commensurable ...but the idea is to build a two-layers slab, one layer of MgO and the other layer of BeO and then roll the slab
The lattice parameter is chosen randomly ...
title
SLAB
73
3.47
4
8 0.5 0.0 0.000000
4 0.0 0.0 0.0000
12 0.5 0.0 2.200000
8 0.0 0.0 2.2000
SWCNT
10 10
TESTGEOM
END
END
Hello,
thank you for your question. Could you be more precise? I don't understand exactly what you mean but this is the input file for a double wall nanotube created starting from a graphene layer:
title
SLAB
77
2.47
1
6 0.33333333 0.6666666667 0.000000
NANOMULTI
2
WALL
1
ROLLINGV
6 6
WALL
2
ROLLINGV
12 12
NANOJMOL
ENDWALL
END
END
you can have a look at the Tutorial, on the Crystal web page:
https://tutorials.crystalsolutions.eu/tutorial.html?td=mwnanotube&tf=MW_tutorial
I hope it helps...
Hi,
Before running the actual single-point or geometry optimization calculations on the configurations with RUNCONFS, a list of configurations needs to be generated by use of the CONFRAND option. The list of generated configurations is saved into a file CONFIGURATIONS.DAT that is then read by the next RUNCONFS calculation.
Let us go through this step-by-step. I take your system as an example.
Title
CRYSTAL
0 0 0
194
3.065 17.656
4
22 0 0 0.5
14 0 0 0.75
22 0.666666 0.333333 0.364919
6 0.333333 0.666666 0.427507
SCELCONF
1 0 0
0 1 0
0 0 1
CONFRAND
1
5
2
END
Here I am selecting just one crystallographic site for substitution, specified by atom number 5, which is a Ti atom. Given the symmetry of this system, that atom has a multiplicity of 4 (i.e. there are other 3 Ti atoms symmetry-related to it). This can be inspected from here (in bold the selected atoms, in italic its symmetry-equivalents):
N. ATOM EQUIV AT. N. X Y Z
1 1 1 22 TI 0.00000000000E+00 0.00000000000E+00 -5.00000000000E-01
2 1 2 22 TI 0.00000000000E+00 0.00000000000E+00 0.00000000000E+00
3 2 1 14 SI 0.00000000000E+00 0.00000000000E+00 -2.50000000000E-01
4 2 2 14 SI 0.00000000000E+00 0.00000000000E+00 2.50000000000E-01
**5 3 1 22 TI -3.33334000000E-01 3.33333000000E-01 3.64919000000E-01**
*6 3 2 22 TI 3.33334000000E-01 -3.33333000000E-01 -1.35081000000E-01
7 3 3 22 TI 3.33333000000E-01 -3.33333000000E-01 -3.64919000000E-01
8 3 4 22 TI -3.33333000000E-01 3.33334000000E-01 1.35081000000E-01*
9 4 1 6 C 3.33333000000E-01 -3.33334000000E-01 4.27507000000E-01
10 4 2 6 C -3.33333000000E-01 3.33334000000E-01 -7.24930000000E-02
11 4 3 6 C -3.33333000000E-01 3.33334000000E-01 -4.27507000000E-01
12 4 4 6 C 3.33334000000E-01 -3.33333000000E-01 7.24930000000E-02
The last input parameter of CONFRAND, which I set to 2, determines how many of these 4 Ti atoms will be substituted.
*******************************************************************************
SUBSTITUTIONS AT SITES (LABELS) :
5 7 6 8
********************************** COMPOSITION : 2 / 4
********************************** NUMBER OF SIC : 3
*******************************************************************************
---> 1 SIC FOUND AT TRY 1 - CONFIGURATION
0 0 1 1
MULTIPLICITY 2 - RANK 1 - CANONICAL RANK 1
---> 2 SIC FOUND AT TRY 2 - CONFIGURATION
1 0 1 0
MULTIPLICITY 2 - RANK 5 - CANONICAL RANK 2
---> 3 SIC FOUND AT TRY 4 - CONFIGURATION
1 0 0 1
MULTIPLICITY 2 - RANK 3 - CANONICAL RANK 4
*******************************************************************************
3 SIC FOUND AFTER 4 TRIES
*******************************************************************************
that is 3 symmetry-independent configurations (SICs) are found and stored in the external file CONFIGURATIONS.DAT.
Title
CRYSTAL
0 0 0
194
3.065 17.656
4
22 0 0 0.5
14 0 0 0.75
22 0.666666 0.333333 0.364919
6 0.333333 0.666666 0.427507
SCELCONF
1 0 0
0 1 0
0 0 1
RUNCONFS
ATOMSUBS
22 273
END
In this case I ask to substitute Ti with Ta.
Hope this helps,
Dear George,
According to your output, you are using CRYSTAL17 to perform the HFsol-3c calculation.
Please note that HFsol-3c and the other "sol-3c" composite methods are only available in CRYSTAL23.
Best regards,
Bartolomeo
Hi similt,
The convergence difficulties you're facing are likely related to the more complex electronic structure of metallic states. These issues arise when the system is metallic, but they can also appear in insulating systems if the SCF goes through some metallic states, and usually it can be stuck there (as seems to be happening in your case).
The general suggestion is to use the SMEAR keyword , that can help a lot when dealing with metallic states.
Also I would suggest to remove the BROYDEN convergence accelerator; and use the default option DIIS.
For metaGGA functionals such as M06, it’s also a good idea to increase the integration grid size (by defayult it should be raised to XXXLGRID, but you can push it further using HUGEGRID)
Following these points, I tried running your calculation on our cluster, and it converged nicely in 12 cycles, giving a bandgap of 3.29 eV with PBE0.
Indeed, during cycle 1 and 2, the system shows metallic behaviour, but from cycle 3 onward it becomes insulating.
Here there is the input file I used: inputPBE0.d12
Let me know if you need further help.
Hi,
jquertin said in SCF fails spinlock with POB-DZVP-REV2:
In the case of SPINLOCK, the manual clearly explain that the NSPIN value is the difference in number of alpha and beta electrons. For SPINLOC2, the text only refers to the spin while the table gives the same definition for SPIN as for NSPIN (in SPINLOCK).
The argument SPIN of SPINLOC2 still represents a number of electrons, as in SPINLOCK.
jquertin said in SCF fails spinlock with POB-DZVP-REV2:
Furthermore, in the calculation, if using SPINLOC2 with 6 or 6.0 as the spin (as defined in the table), crystal defaults to SPINLOCK.
That's right. SPINLOC2 requires a non integer argument. For integer arguments it reduces to SPINLOCK.
jquertin said in SCF fails spinlock with POB-DZVP-REV2:
In short, if I define SPINLOC2 SPIN as 3 (1/2 * 6) or 3.0, crystal defaults to SPINLOCK with NSPIN 3 which is actually half of what I want.
In both SPINLOCK and SPINLOC2, the argument is meant as a number of electrons. Thus, if you have 6 extra up electrons with respect to down electrons, the input value should be 6, not 3. For integer values, SPINLOC2 is of no use.
Hope this clarifies things a little,
