Version 4
SHEET 1 2144 1496
WIRE -144 112 -224 112
WIRE -144 112 -144 -16
WIRE -144 160 -144 112
WIRE -144 272 -144 224
WIRE -144 496 -144 464
WIRE -144 608 -144 576
WIRE -16 464 -144 464
WIRE 32 -16 -144 -16
WIRE 48 112 -144 112
WIRE 272 464 64 464
WIRE 272 560 272 464
WIRE 304 -16 112 -16
WIRE 304 112 112 112
WIRE 304 112 304 -16
WIRE 304 160 304 112
WIRE 304 272 304 224
WIRE 320 560 272 560
WIRE 352 464 272 464
WIRE 496 112 304 112
WIRE 496 176 496 112
WIRE 496 464 432 464
WIRE 496 464 496 256
WIRE 560 112 496 112
WIRE 560 464 496 464
FLAG -144 272 0
FLAG 304 272 0
FLAG -224 112 in
IOPIN -224 112 Out
FLAG 560 112 out1
IOPIN 560 112 Out
FLAG -144 608 0
FLAG 560 464 out
IOPIN 560 464 Out
FLAG 320 560 out0
IOPIN 320 560 Out
SYMBOL Misc\\xtal 112 96 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
WINDOW 39 65 14 VTop 0
SYMATTR InstName Q1
SYMATTR Value {C1}
SYMATTR SpiceLine Rser=50 Lser={L1} Cpar=5p
SYMATTR SpiceLine2 ic=0
SYMBOL cap 288 160 R0
SYMATTR InstName C9
SYMATTR Value 20p
SYMBOL cap -160 160 R0
SYMATTR InstName C10
SYMATTR Value 20p
SYMBOL res 16 0 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R1
SYMATTR Value 10MEG
SYMBOL bv -144 480 R0
SYMATTR InstName B1
SYMATTR Value V=-atan(2*V(in))
SYMBOL res 480 160 R0
SYMATTR InstName R2
SYMATTR Value 200
SYMBOL res 336 480 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R3
SYMATTR Value 100
SYMBOL voltage 80 464 R90
WINDOW 0 -32 56 VBottom 0
WINDOW 3 32 56 VTop 0
WINDOW 123 60 56 VTop 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 0
SYMATTR Value2 AC 1
TEXT -808 480 Left 0 ;.ac lin 10000 20MEG 20.01MEG
TEXT -816 120 Left 0 !.options plotwinsize=0
TEXT -824 -856 Left 0 ;FS=1/(2*pi*sqrt(L1*C1)) = 20MHz\n \nFP-FS=C1/C0/2\n \nRS=2*pi*FS*L1/Q\n \ndf = Fs*C1/(CL+C0)/2 \n  \nQ1 is the quartz crystal\nmoeled with the capacitor \nmodel of LTspice.\nC0 is Cpar of Q1\nC1 is C of Q1\nL1 is Lser of Q1\nRS is Rser of Q1
TEXT -824 -304 Left 0 !.param FS=20e6\n.param Q=25000\n.param RS=50\n.param L1=RS*Q/(2*pi*FS)\n.param C1=1/(2*pi*FS*2*pi*FS*L1)
TEXT -848 304 Left 0 !.save V(out0)
TEXT -816 -72 Left 0 ;.options nomarch
TEXT -816 -40 Left 0 !.options numdgt=12
TEXT -816 -8 Left 0 !.options measdgt=12
TEXT -816 88 Left 0 !.options trtol=1
TEXT -848 272 Left 0 ;.tran 0 2m 0 0.5n uic      ; Q=250 or 2500
TEXT -816 24 Left 0 !.options reltol=1e-3
TEXT -816 56 Left 0 !.options method=trap
TEXT 112 440 Left 0 ;virtual internal node
TEXT -112 616 Left 0 ;V=-tanh(2*V(in))
TEXT 40 392 Left 0 ;V1 is only for .AC simulation
TEXT -200 360 Left 0 ;"Ideal nonlinear" Inverter
TEXT -808 616 Left 0 ;.save V(in)
TEXT -808 656 Left 0 ;V(in) is needed in .AC to find \nthe freqency where phi(Vin)=0
TEXT -280 -304 Left 0 !.measure tran t1 FIND time WHEN V(out0)=0.1 TD=50u RISE=1\n.measure tran t2 FIND time WHEN V(out0)=0.1 TD=50u RISE=10001\n.measure tran f0 PARAM 10000/(t2-t1)
TEXT 104 80 Left 0 ;Crystal
TEXT 544 -792 Left 0 ;Some results\n--------------------\nBe aware that one .tran simulation may require 1 hour.\n \n\n.tran without minimum timestep: about 20.117 MegHz\n \n*19.4ms to 19.9ms, 0.05n trtol 1 reltol 1e-3 gain=2\nf0: 10000/(t2-t1)=20004771.1534\n \n*19.4ms to 19.9ms, 0.05n trtol 1 reltol 1e-3 gain=5\nf0: 10000/(t2-t1)=20004828.3131\n \n*19.4ms to 19.9ms, 0.05n trtol=1 reltol=1e-4 gain=2 238ppm\nf0: 10000/(t2-t1)=20004771.1612\n \n*19.4ms to 19.9ms, 0.05n trtol 1 reltol 1e-4 gain=5  241ppm\nf0: 10000/(t2-t1)=20004828.2828\n \n  \n19.4m to 19.9m, 0.05n reltol=1e-4, trtol=1 gain=2 tanh()\nf0: 10000/(t2-t1)=20004771.4519
TEXT -848 208 Left 0 ;Transient Analysis
TEXT -808 392 Left 0 ;AC Analyisis
TEXT -840 -112 Left 0 ;Important options for .tran
TEXT -848 -352 Left 0 ;The crystal model
TEXT -280 -856 Left 0 ;Transient Analysis And AC Analysis Of A Crystal Oscillator\n \nHelmut Sennewald, 27.06.2005\n \nThe goal was to find out which parameters are necessary to \nachieve an accuray of very few ppm in the .tran analysis \nat a frequency of about 20MHz.\n \nMy understanding is that the correct frequency is where the\nphase of the whole path is at 0 degree in the AC-analysis.\n \nHint, I recommend to start with a low Q value to check \nthat the oscillator works. E.g. Q=250 .  Also change \nthe .tran time  and timestep, e.g.  .tran 0 2m 0 0.5n uic  .\n \nThe results of .measure are always in the Error-Log-file.
TEXT -280 -352 Left 0 ;Measure the frequency with .measure command lines
TEXT -280 -192 Left 0 !.measure ac f0ac FIND freq WHEN im(V(in))=0
TEXT 544 -848 Left 0 ;f0ac: freq=(146.022699265dB,0°) at 20004834.5212
TEXT -1592 -864 Left 0 ;Message 6061:\n \nI have tried to find out the necessary parameters for a ppm-precise\n.TRAN simulation of a crystal oscillator.\nMy simulation setup was a crystal plus an ideal inverter made by\na B-source with atan-transfer function.\nThe Q of the crystal model was 25000.\n \nI used the 0-degree frequency from the .AC simulation as the\nreference to see what simulation comes closest.\n \nHere are my recommendations\n---------------------------\n \n1. Reset all simulator settings in the control panel!\n \nLimit the saved traces. One trace may require 200MByte!\n \n.save V(out)\n \nSwitch marching waveform off.\n \n.options nomarch\n \nIncrease the number of digits.\n \n.options numdgt=12\n.options measdgt=12\n \nSet solver method to trap. Don't use "Alternate".\n \n.options method=trap\n \nSet the minimum timing step to 1/1000*period_time.\nSetting no minimum timestep results in a penalty of 100kHz! offset.\n*** I used "uic" in conjunction with ic=.. for the capacitor in the\n*** crystal model. \nChange: I used "uic" in conjunction with .ic V(out1)=5 .\n \nWith a real gain stage you can apply other methods\nto start the oscillator. The only important thing here is the\nminimum timestep.\n \n.tran 0 20m 19.4m 0.05n uic\n \nSwitch off any data compression for lowest distortion.\n \n.options plotwinsize=0\n \nSet it trtol to 1. I tried with 1e-4, but there was only a very very\nlittle change.\n \n.options trtol=1\n \nI tried with 1e-4 too, but the difference was in the sub-ppm range.\n \n.options reltol=1e-3\n \n \nBy the way, one simulation run may require one hour. You should\nhave a few PCs around to simulate in parallel.
TEXT -808 560 Left 0 ;.ac lin 10000 20MEG 21MEG
TEXT -808 448 Left 0 ;Q=25000
TEXT -808 528 Left 0 ;Q=250
TEXT -280 -80 Left 0 !.ic v(out1)=5
TEXT -848 240 Left 0 !.tran 0 20m 19.4m 0.05n uic  ; Q=25000
LINE Normal -288 528 -288 112 2
LINE Normal -192 528 -288 528 2
LINE Normal -208 512 -192 528 2
LINE Normal -208 544 -192 528 2
LINE Normal -272 112 -288 112 2
RECTANGLE Normal 464 672 -224 336
RECTANGLE Normal -424 704 -872 368
RECTANGLE Normal -424 328 -872 184
RECTANGLE Normal -424 152 -872 -136
RECTANGLE Normal -424 -168 -872 -376
RECTANGLE Normal 424 -168 -312 -376