Thanks
I have tried two ways to use your model
1. Simply copy the model from your QSCH to successfully simulate it
2. Adding. qsym separately to “symbols and IP” will result in the same error, I don’t know why
**3.**Regarding the differences in simulation results, I have reviewed your two models
QK040K7 seems to be Thyristor_TRIAC_Qxx40xx_A_SPICE-Model，This model seems to be closer to the result I want
|. SUBCKT QK040K7 1 2 3\nQnpn1 5 4 3 NoutF OFF\nQpnp1 4 5 7 PoutF OFF\nQnpn2 11 6 7 NoutR OFF\nQpnp2 6 11 3 PoutR OFF\nDfor 4 5 DZ OFF\nDrev 6 11 DZ OFF\nRfor 4 6 12MEG\nRon 1 7 10m\nRhold 7 6 50\nRGP 8 3 10\nRG 2 8 5.8\nRS 8 4 75\nDN 9 2 DIN OFF\nRN 9 3 4\nGNN 6 7 9 3 0.1\nGNP 4 5 9 3 0.1\nDP 2 10 DIP OFF\nRP 10 3 3.56\n.MODEL DIN D (IS=382F)\n.MODEL DIP D (IS=382F N=1.19)\n.MODEL DZ D (IS=382F N=1.5 IBV=50U BV=1000)\n.MODEL PoutF PNP (IS=382F BF=0.45 CJE=380p TF=0.3U)\n.MODEL NoutF NPN (IS=382F BF=3 CJE=380p CJC=76p TF=0.3U)\n.MODEL PoutR PNP (IS=382F BF=2.5 CJE=380p TF=0.3U)\n.MODEL NoutR NPN (IS=382F BF=0.5 CJE=380p CJC=76p TF=0.3U)\n.ENDS
Another Triac_ST, has this been modified by you based on the ST file?Wolud you like to tell me what modifications have been made? It can run now, but the waveform doesn’t seem to be what I want. It’s strange why
|.subckt Triac_ST A K G \nS_S3 A Plip1 positive 0 Smain\nD_DAK1 Plip1 Plip2 Dak\nR_Rlip Plip1 Plip2 1k\nV_Viak Plip2 K DC 0 AC 0\nS_S4 A Plin1 negative 0 Smain\nD_DKA1 Plin2 Plin1 Dak\nR_Rlin Plin1 Plin2 1k\nV_Vika K Plin2 DC 0 AC 0\nR_Rgk G K 10G\nD_DGKi Pg2 G Dgk\nD_DGKd G Pg2 Dgk\nV_Vig Pg2 K DC 0 AC 0\nR_Rlig G Pg2 1k\nR_Rp Controlp positive 2.2\nC_Cp 0 positive 1u\nBE_IF15OR3 Controlp 0 V=IF(((V(CMDIG)>0.5)|(V(CMDILIH)>0.5)|(V(CMDVdrm)>0.5)),400,0)\nR_Rn Controln negative 2.2\nC_Cn 0 negative 1u\nBE_IF14OR3 Controln 0 V=IF(((V(CMDIG)>0.5)|(V(CMDILIHN)>0.5)|(V(CMDVdrm)>0.5)),400,0)\nBE_IF1IG inIG 0 V=IF((ABS(I(V_Vig)))>(Igt-1u),1,0)\nBE_MULT2MULT CMDIG 0 V=V(Q4)V(inIG)\nBE_IF2Quadrant4 Q4 0 V=IF(((I(V_Vig)>(Igt-0.000001))&((V(A)-V(K) )<0)&(Standard==0)),0,1)\nBE_IF10IL inIL 0 V=IF(((I(V_Viak))>(Ih/2)),1,0)\nBE_IF5IH inIH 0 V=IF(((I(V_Viak))>(Ih/3)),1,0)\nBE_IF6DIHIL SDIHIL 0 V=IF((V(inIL)V(inIH)+V(inIH)(1-V(inIL))(V(CMDILIH)))>0.5,1,0)\nC_CIHIL CMDILIH 0 1n\nR_RIHIL SDIHIL CMDILIH 1K\nR_RIHIL2 CMDILIH 0 100Meg\nBE_IF11ILn inILn 0 V=IF(((I(V_Vika))>(Ih/2)),1,0)\nBE_IF3IHn inIHn 0 V=IF(((I(V_Vika))>(Ih/3)),1,0)\nBE_IF4DIHILN SDIHILN 0 V=IF((V(inILn)V(inIHn)+V(inIHn)(1-V(inILn))(V(CMDILIHN)))>0.5,1,0)\nC_CIHILn CMDILIHN 0 1n\nR_RIHILn SDIHILN CMDILIHN 1K\nR_RIHILn2 CMDILIHN 0 100Meg\nBE_IF8Vdrm inVdrm 0 V=IF((ABS(V(A)-V(K))>(Vdrm1.3)),1,0)\nBE_IF9IHVDRM inIhVdrm 0 V=IF((I(V_Viak)>(Vdrm1.3)/1.2meg)|(I(V_Vika)>(Vdrm1.3)/1.2meg), 1,0)\nBE_IF7DVDRM SDVDRM 0 V=IF((V(inVdrm)+(1-V(inVdrm))*V(inIhVdrm)*V(CMDVdrm))>0.5,1,0)\nC_CVdrm CMDVdrm 0 1n\nR_RVdrm SDVDRM CMDVdrm 100\nR_RVdrm2 CMDVdrm 0 100Meg\n.MODEL Smain SW Roff=1.2meg Ron={Rt} Vh=50 Vt=50\n.MODEL Dak D( Is=3E-12 Cjo=5pf)\n.MODEL Dgk D( Is=1E-16 Cjo=50pf Rs=5)\n.ends
