<div dir="ltr"><div>Dear Users,</div><div>I am doing Au slab with ZnO Slab. My calculation contain 6 layers of Au and 12 Double Layer of ZnO. I run the "relax" calculation and it gives the optimized atomic positions. Using that position when i run the "scf" run by applying the dipole correction, then that calculation is not converging. Following is the input of the "scf"</div><div><br></div><div>&CONTROL </div><div> calculation = 'scf' , </div><div> restart_mode = 'from_scratch' , </div><div> outdir = './OUT',</div><div> pseudo_dir = './',</div><div> prefix = 'Au-ZnO-NW',</div><div> etot_conv_thr = 2.0d-5 , </div><div> forc_conv_thr = 1.0d-3 , </div><div> tstress = .true. , </div><div> tprnfor = .true. ,</div><div> wf_collect = .true.,</div><div> tefield=.true.,</div><div> dipfield=.true.,</div><div> / </div><div> &SYSTEM </div><div> ibrav = 4,</div><div> celldm(1) = 6.1413449010,</div><div> celldm(3) = 15.8461538461,</div><div> nat = 30,</div><div> ntyp = 3,</div><div> ecutwfc = 56, </div><div> ecutrho = 448,</div><div> occupations ='smearing',</div><div> smearing ='m-v',</div><div> degauss =0.03,</div><div> lda_plus_u = .true.,</div><div> Hubbard_U(1) =8.0, Hubbard_U(2)=0.0,</div><div> Hubbard_U(3) =0.0,</div><div> edir=3,</div><div> emaxpos=0.55,</div><div> eopreg=0.06,</div><div> eamp=0,</div><div>/ </div><div> &ELECTRONS </div><div> mixing_beta = 0.4 , </div><div> diagonalization = 'david' , </div><div> diago_thr_init = 1.D-2 , </div><div> diago_david_ndim = 10,</div><div> conv_thr = 1.0d-6, </div><div><span class="" style="white-space:pre"> </span> electron_maxstep = 140,</div><div> / </div><div> &IONS </div><div> ion_dynamics ='bfgs',</div><div> bfgs_ndim = 3, </div><div> / </div><div> &CELL</div><div> cell_dynamics = 'bfgs' ,</div><div> press = 0.00,</div><div> wmass = 0.010</div><div>/</div><div>ATOMIC_SPECIES </div><div>Zn 65.39 Zn.pbe-van.UPF</div><div> O 15.9994 O.pbe-van_ak.UPF</div><div>Au 196.97 Au.pbe-van_ak.UPF</div><div>ATOMIC_POSITIONS (Angstrom)</div><div>Zn 7.499742896 5.752629371 3.230876064</div><div>Zn 5.874757485 6.690733893 5.921325325</div><div>Zn 7.499651212 5.752560151 8.611774586</div><div>Zn 5.874685478 6.690665275 11.302223847</div><div>Zn 7.499814131 5.752654773 13.985634015</div><div>Zn 5.874963177 6.690829207 16.671576079</div><div>Zn 7.499874447 5.752675885 19.356809802</div><div>Zn 5.874856769 6.690780480 22.041789006</div><div>Zn 7.499724155 5.752620086 24.729390916</div><div>Zn 5.874792904 6.690719489 27.417963638</div><div>Zn 7.499686920 5.752565718 30.107473986</div><div>Zn 5.874692982 6.690668726 32.896386195</div><div>O 5.874814089 6.690782677 2.514551440 </div><div>O 7.499685927 5.752583792 5.205000701 </div><div>O 5.874718327 6.690713410 7.895449962 </div><div>O 7.499618111 5.752518096 10.585899223 </div><div>O 5.874878823 6.690804905 13.310596326</div><div>O 7.499890469 5.752674934 15.997050233</div><div>O 5.874943952 6.690827750 18.682493540</div><div>O 7.499787626 5.752628781 21.367966551</div><div>O 5.874797778 6.690771718 24.053807177</div><div>O 7.499721925 5.752568423 26.742414673</div><div>O 5.874754539 6.690718084 29.433188626</div><div>O 7.499626837 5.752521379 32.126332530</div><div>Au 7.499703856 5.752469031 35.061709956</div><div>Au 9.124636220 6.690621085 37.151452057</div><div>Au 7.499709255 7.628772145 39.211201531</div><div>Au 7.499712252 5.752462483 41.258545283</div><div>Au 9.124644992 6.690614771 43.331292891</div><div>Au 7.499717550 7.628769078 45.387107036</div><div>K_POINTS automatic</div><div> 6 6 1 1 1 1</div><div><br></div><div>Following is the few lines of the output file</div><div><br></div><div><div> <font color="#cc0000"> total cpu time spent up to now is 48274.1 secs</font></div><div><font color="#cc0000"><br></font></div><div><font color="#cc0000"> total energy = -459.60706118 Ry</font></div><div><font color="#cc0000"> Harris-Foulkes estimate = -9029.54246705 Ry</font></div><div><font color="#cc0000"> estimated scf accuracy < 645684.43872681 Ry</font></div><div><font color="#cc0000"><br></font></div><div><font color="#cc0000"> iteration</font> <font color="#cc0000">#105 ecut= 56.00 Ry beta=0.40</font></div><div><font color="#cc0000"> Davidson diagonalization with overlap</font></div><div><font color="#cc0000"> ethr = 1.00E-02, avg # of iterations = 14.7</font></div><div><font color="#cc0000"><br></font></div><div><font color="#cc0000"> negative rho (up, down): 1.422E-04 0.000E+00</font></div><div><font color="#cc0000"><br></font></div><div><font color="#cc0000"> Adding external electric field</font></div><div><font color="#cc0000"><br></font></div><div><font color="#cc0000"> Computed dipole along edir(3) :</font></div><div><font color="#cc0000"> Dipole 12.1613 Ry au, 30.9110 Debye</font></div><div><font color="#cc0000"> Dipole field 0.0481 Ry au</font></div><div><font color="#cc0000"><br></font></div><div><font color="#cc0000"> Potential amp. -8.7961 Ry</font></div><div><font color="#cc0000"> Total length 91.4777 bohr</font></div><div><font color="#cc0000"><br></font></div><div><font color="#cc0000"><br></font></div><div><font color="#cc0000"> total cpu time spent up to now is 48690.9 secs</font></div><div><font color="#cc0000"><br></font></div><div><font color="#cc0000"> total energy = -1539.58421411 Ry</font></div><div><font color="#cc0000"> Harris-Foulkes estimate = -6830.90427333 Ry</font></div><div><font color="#cc0000"> estimated scf accuracy < 340940.96928352 Ry</font></div></div><div><font color="#cc0000"><br></font></div><div><font color="#000000">Please help me in this regard</font></div><div><br></div></div><div class="gmail_extra"><br><div class="gmail_quote">On Mon, Mar 30, 2015 at 4:27 PM, Giovanni Cantele <span dir="ltr"><<a href="mailto:giovanni.cantele@spin.cnr.it" target="_blank">giovanni.cantele@spin.cnr.it</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div style="word-wrap:break-word">Provided that you have a sufficient number of layers to guarantee in both sides a “bulk-like” region, you can choose the innermost layers.<div><br></div><div>But, because you have an interface, what is the quantity are you interested in? The interface barrier (in your case the Schottky barrier, because you have a metal-semiconductor</div><div>junction) or the work function? Consider that, in the latter case, because the two sides of the slab are different:</div><div>i) the vacuum level is different at the two sides, so you just choose the interplanar distance of the material you are interested in (either ZnO or Au)</div><div>ii) because the vacuum level is different at the two sides, you must consider to add a correction for a “spurious” dipole introduced by boundary conditions, using tefield=.true.<br> dipfield=.true., and the related variables (see PW/DOC/INPUT_PW.txt for explanation) edit, eamp=0.D0, eopreg and examples</div><div><br></div><div>In the case you are instead interested in the interface barrier, you easily figure out that by choosing, as the window for the macroscopic average either the ZnO or the Au interplanar distance you get rid of microscopic oscillations only on one side. In this case, you can compute the averages with both windows, and plot them on the same plot. If needed, you might want to have a look at Fig. 6 of N.R. D’amico, et al, J. Phys.: Condens. Matter 27 (2015) 015006 </div><div><br></div><div>Hope this helps,</div><div><br></div><div> Giovanni</div><div><br></div><div>PS Warning: the ZnO orientation you are studying is polar, so the macroscopic average does not give you a constant value of the potential in the innermost part of the ZnO slab,</div><div>but a linearly varying one. There are examples in the literature on how the handle the electrostatic potential in polar slabs, see references in the above mentioned paper.<div><div class="h5"><br><div><br></div><div><br><div><blockquote type="cite"><div>On 30 Mar 2015, at 11:47, Bipul Rakshit <<a href="mailto:bipulrr@gmail.com" target="_blank">bipulrr@gmail.com</a>> wrote:</div><br><div><div dir="ltr">Really thanks for your suggestion Giovanni,<div>I just have few more doubts that if i want to find the work function of Au-ZnO Slab. I have created Au(111) and ZnO(0001) Slab. So it has an interface of Zn-Au and a vacuum of 10 Angstrom. I took 6 layers of Au and 4 double layers of ZnO. Then how to proceed.</div><div>Means</div><div>1) Which two consecutive atomic planes I have to choose Au-Au, Au-Zn or the Zn-O planes. for the microscopic average calculation?</div><div><br></div><div>2) Also in-order to choose the inner part of the slab, so that part is inside the Au-Slab or the ZnO Slab?</div><div><br></div><div>regards</div><div><br></div></div><div class="gmail_extra"><br><div class="gmail_quote">On Wed, Mar 18, 2015 at 5:02 PM, Giovanni Cantele <span dir="ltr"><<a href="mailto:giovanni.cantele@spin.cnr.it" target="_blank">giovanni.cantele@spin.cnr.it</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div style="word-wrap:break-word"><div>The average electrostatic potential you are calculating/plotting does show microscopic oscillations, revealing the atomic planes</div><div>(plot the 2d column of reference/Al100.avg.out as a function of the 1st one).</div><div><br></div><div>The calculation of the work function requires a “constant” energy level to compare the bulk and the slab calculation. For this purpose,</div><div>you run a macroscopic average of the x-y averaged electrostatic potential. The window you choose for the macroscopic average</div><div>is just the distance between two consecutive atomic planes (in a.u.).</div><div>Because the input positions are in alat units, you get</div><div><br></div><div>(2.8284271247461898 - 2.1213203435596428) * 5.4235090117 = 3.835</div><div><br></div><div>In this way, microscopic oscillations with period 3.835 are averaged and a constant level (in the inner region of the slab) is obtained.</div><div><br></div><div>Concerning 17.8087, in order to get this constant value, you can choose any position in the inner part of the slab where the</div><div>macroscopic average does not show significant variations.</div><div><br></div><div>Giovanni</div><div><br></div><br><div><blockquote type="cite"><div><div><div>On 18 Mar 2015, at 10:49, Bipul Rakshit <<a href="mailto:bipulrr@gmail.com" target="_blank">bipulrr@gmail.com</a>> wrote:</div><br></div></div><div><div><div><div dir="ltr"><div><div><div>In espresso, there is an example to find the workfunction of Al. In the run_example, the input for the the macroscopic average is the following<br>cat > <a href="http://al100.avg.in/" target="_blank">Al100.avg.in</a> <<EOF<br>1<br>Al100.pot<br>1.D0<br>1440<br>3<br><span style="color:rgb(255,0,0)">3.835000000</span><br>EOF<br><br></div>In this file the quantity "3.835" i saw in average.f90 as "awin ! the size of the window for macroscopic averages"<br><br></div>So my doubt is how we can choose this no. Can we get the information from another file prerun file, like Al100.pot, or something else.<br><br></div>Also in run_example there is another quantity vSlab<br><br>vSlab=`grep "17.8087" Al100.avg.out | cut -d \ -f 10`<br clear="all"><div><div><div><br></div><div>So how the value correspond to "17.8087" is assign as vSlab?<br><br></div><div>Kindly help me in this matter.<br><br></div><div>regards<br></div><div>-- <br><div><div dir="ltr"><div>Dr. Bipul Rakshit<br>Research Associate,<br>Institute of Physics (IOP),<br>Bhubaneswar- <span>751 005</span></div><div>Orissa <br></div><div>India</div></div></div>
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-- <br><br>Giovanni Cantele, PhD<br>CNR-SPIN<br>c/o Dipartimento di Fisica<br>Universita' di Napoli "Federico II"<br>Complesso Universitario M. S. Angelo - Ed. 6<br>Via Cintia, I-80126, Napoli, Italy<br>e-mail: <a href="mailto:giovanni.cantele@spin.cnr.it" target="_blank">giovanni.cantele@spin.cnr.it</a><br>Phone: +39 081 676910<br>Skype contact: giocan74<br><br>ResearcherID: <a href="http://www.researcherid.com/rid/A-1951-2009" target="_blank">http://www.researcherid.com/rid/A-1951-2009</a><br>Web page: <a href="http://people.na.infn.it/~cantele" target="_blank">http://people.na.infn.it/~cantele</a><br>
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-- <br><br>Giovanni Cantele, PhD<br>CNR-SPIN<br>c/o Dipartimento di Fisica<br>Universita' di Napoli "Federico II"<br>Complesso Universitario M. S. Angelo - Ed. 6<br>Via Cintia, I-80126, Napoli, Italy<br>e-mail: <a href="mailto:giovanni.cantele@spin.cnr.it" target="_blank">giovanni.cantele@spin.cnr.it</a><br>Phone: +39 081 676910<br>Skype contact: giocan74<br><br>ResearcherID: <a href="http://www.researcherid.com/rid/A-1951-2009" target="_blank">http://www.researcherid.com/rid/A-1951-2009</a><br>Web page: <a href="http://people.na.infn.it/~cantele" target="_blank">http://people.na.infn.it/~cantele</a><br>
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