In 1988, I received my Ph.D. from Purdue University, and as part of that thesis, this code was written to simulate, in two dimensions, and time, the internal state of the device as it was under bias. The details of the code are really best described in the Thesis, which is still available from University Microfilms, which is interesting to me today... but hey, it's not a lot of storage, so why not?
Shark is based on an input deck like so many codebases that were meant to run on supercomputers without even so much as terminal access. One such input file is:
*
* This will be a comparative test of Sutherland's work for a Silicon
* MOSFET with SiO2, biased into weak inversion, triode mode
*
source thick=0.35
drain thick=0.35
bulk doping=2.0e16
oxide type=SiO2 thick=1000
channel type=MODFET length=3.0 material=Si
* setting dDo=0 will make D constant
DvsE Do=18.13 dDo=0.0
* setting A=C=inf. will make vsat=inf. and then v=u*E
VvsE mobility=700.0 A=2.0e38 C=2.0e38
* bias the device in the triode region
bias vsource=0.00 vdrain=0.35 eavg=0.0 vg1=2.5
* do the steady state operation: delt -> infinity
time tout=10.0 tstop=100.0 delt=1.0e38
solution type=linpk
start with=DC maxit=300 debug=sortof guess=MOS
and simulates a simple MISFET transistor in weak-inversion, triode operation. This was a similar test case given in Sutherland's work, and was used to check that Shark was matching his work for correctness.
When run with something like:
$ shark < wito > wito.outyou will get something like:
*******************************************************************************************************************
* *
* SSSSSSSSS h k 2222222 000000000 *
* S h aaaaaa k 2 2 0 0 *
* S h a r rrrrr k kkk 2 0 0 *
* SSSSSSSSS h hhhhh aaaaa a rr r k kkk v v 2222222 0 0 *
* S hh h a aa r kkk v v 2 0 0 *
* S h h a a r k kkk v v 2 .. 0 0 *
* SSSSSSSSS h h aaaaaaa r k kkk v 222222222 .. 000000000 *
* *
* *
* T W O D I M E N S I O N A L T R A N S I E N T G a A s F E T S I M U L A T O R *
* *
* Written by Robert E. Beaty as a part of a Ph.D. in Electrical Engineering at *
* Purdue University, West Lafayette, Indiana 47906 June 1988 *
* *
*******************************************************************************************************************
==== Shark v2.0 REB 1988 input deck ====
*
* This will be a comparative test of Sutherland's work for a Silicon
* MOSFET with SiO2, biased into weak inversion, triode mode
*
source thick=0.35
drain thick=0.35
bulk doping=2.0e16
oxide type=SiO2 thick=1000
channel type=MODFET length=3.0 material=Si
* setting dDo=0 will make D constant
DvsE Do=18.13 dDo=0.0
* setting A=C=inf. will make vsat=inf. and then v=u*E
VvsE mobility=700.0 A=2.0e38 C=2.0e38
* bias the device in the triode region
bias vsource=0.00 vdrain=0.35 eavg=0.0 vg1=2.5
* do the steady state operation: delt -> infinity
time tout=10.0 tstop=100.0 delt=1.0e38
solution type=linpk
start with=DC maxit=300 debug=sortof guess=MOS
==== Shark v2.0 REB 1988 run parameters ====
The length of the simulated channel is 3.00 (um).
The thickness of the simulated device is 5.10 (um).
The doping density of the substrate, given as (Na-Nd), is 0.20E+17 (/cm**3).
The insulator is SiO2 with a thickness of 1000.0 (Angst.).
The dielectric constants for the materials are:
semiconductor = 11.70
insulator = 3.90
The channel is based on the MODFET channel.
The channel material type is Si .
The biasing on the terminals of the device is:
source voltage = 0.000 (V).
drain voltage = 0.350 (V).
source end of the gate = 2.500 (V).
drain end of the gate = 2.500 (V).
field in the gate = 0.0 (V/cm).
The parameters for the velocity versus field equation are:
low-field mobility = 700.0 (cm**2/V-s).
A = 0.2000E+39
B = 0.1000E+39
C = 0.2000E+39
D = 1.000
F = 1.000
The parameters for the diffusivity versus field equation are:
low-field diffusivity = 18.13 (cm**2/sec).
Do-to-peak value = 0.00 (cm**2/sec).
A = 8.130
B = 0.5960
Starting the channel with the steady state electron concentration in the channel.
The Newton iterations will halt if the largest change in the voltage is less than 0.10E-05 kT/q,
or 300 iterations have been completed.
The systems of equations will be solved by the small LINPACK algorithm.
The starting time of the simulation is 0.0 (psec).
The stopping time of the simulation is 100.0 (psec).
Output will be generated every 10.0 (psec) of the simulation.
==== Shark v2.0 REB 1988 mesh selection ====
The simple MOS electrostatics for the channel region are:
Vgmax = 2.500 V
Vg+Qss/Cox = 2.500 V
Cox = 0.345E-07 F/cm**2
Vo = 2.7840 V
V1 = 0.0000 V
phis = 0.629 V
xd = 0.202 um
Because Shark is selecting the mesh, Shark is setting the thickness of the
substrate to 4.413 um
automatically selecting the y mesh for a MODFET channel.
placing y nodes uniformly in tox...
placing y nodes quadratically in xd...
placing y nodes uniformly past xd to tsub...
placing x nodes non-uniformly in the channel...
setting xnodes = 32
setting ynodes = 78
maximum hx ratio = 1.111 from:
hx max. = 0.1962 (um)
hx min. = 0.0404 (um)
maximum hy ratio = 3.252 from:
hy max. = 0.0181 (um)
hy min. = 0.0051 (um)
==== Shark v2.0 REB 1988 setting up device ====
setting up the oxide...
setting up the channel...
setting the substrate...
==== Shark v2.0 REB 1988 applying biases ====
==== Shark v2.0 REB 1988 initial guess ====
Computing the initial guess based on MOS electrostatics for the
center section device simulation.
Using Pao-Sah to calculate the initial quasi-fermi levels...
Solving the electrostatics for the left boundary...
Iter. # 1 took 0.06 msec. with norm = 0.800 at V( 1, 6).
Iter. # 2 took 0.07 msec. with norm = 0.203 at V( 1, 21).
Iter. # 3 took 0.06 msec. with norm = 0.106E-01 at V( 1, 21).
Iter. # 4 took 0.06 msec. with norm = 0.284E-04 at V( 1, 21).
Iter. # 5 took 0.06 msec. with norm = 0.215E-09 at V( 1, 21).
The Newton solution converged in 5 iterations which took 0.00 sec.
Solving for the exact n and V for the left boundary...
Iter. # 1 took 0.4 msec. with norm = 0.177E-05 at n( 1, 53).
Iter. # 2 took 0.2 msec. with norm = 0.582E-09 at n( 1, 38).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatics for the right boundary...
Iter. # 1 took 0.05 msec. with norm = 0.810 at V( 32, 6).
Iter. # 2 took 0.06 msec. with norm = 0.205 at V( 32, 21).
Iter. # 3 took 0.07 msec. with norm = 0.108E-01 at V( 32, 21).
Iter. # 4 took 0.07 msec. with norm = 0.295E-04 at V( 32, 21).
Iter. # 5 took 0.06 msec. with norm = 0.233E-09 at V( 32, 21).
The Newton solution converged in 5 iterations which took 0.00 sec.
Solving for the exact n and V for the right boundary...
Iter. # 1 took 0.2 msec. with norm = 0.141E-05 at n( 32, 19).
Iter. # 2 took 0.2 msec. with norm = 0.118E-11 at n( 32, 21).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 2.
Iter. # 1 took 0.05 msec. with norm = 0.594E-09 at V( 2, 8).
The Newton solution converged in 1 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 2.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 2, 66).
Iter. # 2 took 0.2 msec. with norm = 0.581E-09 at n( 2, 38).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 3.
Iter. # 1 took 0.04 msec. with norm = 0.594E-09 at V( 3, 8).
The Newton solution converged in 1 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 3.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 3, 53).
Iter. # 2 took 0.2 msec. with norm = 0.582E-09 at n( 3, 37).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 4.
Iter. # 1 took 0.05 msec. with norm = 0.238E-04 at V( 4, 7).
Iter. # 2 took 0.06 msec. with norm = 0.402E-11 at V( 4, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 4.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 4, 66).
Iter. # 2 took 0.2 msec. with norm = 0.580E-09 at n( 4, 37).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 5.
Iter. # 1 took 0.05 msec. with norm = 0.305E-03 at V( 5, 7).
Iter. # 2 took 0.06 msec. with norm = 0.666E-09 at V( 5, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 5.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 5, 73).
Iter. # 2 took 0.2 msec. with norm = 0.569E-09 at n( 5, 41).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 6.
Iter. # 1 took 0.05 msec. with norm = 0.332E-03 at V( 6, 7).
Iter. # 2 took 0.06 msec. with norm = 0.795E-09 at V( 6, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 6.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 6, 63).
Iter. # 2 took 0.2 msec. with norm = 0.558E-09 at n( 6, 38).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 7.
Iter. # 1 took 0.04 msec. with norm = 0.376E-03 at V( 7, 7).
Iter. # 2 took 0.06 msec. with norm = 0.103E-08 at V( 7, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 7.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 7, 66).
Iter. # 2 took 0.2 msec. with norm = 0.543E-09 at n( 7, 37).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 8.
Iter. # 1 took 0.05 msec. with norm = 0.413E-03 at V( 8, 7).
Iter. # 2 took 0.06 msec. with norm = 0.125E-08 at V( 8, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 8.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 8, 54).
Iter. # 2 took 0.2 msec. with norm = 0.528E-09 at n( 8, 38).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 9.
Iter. # 1 took 0.05 msec. with norm = 0.460E-03 at V( 9, 7).
Iter. # 2 took 0.06 msec. with norm = 0.156E-08 at V( 9, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 9.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 9, 55).
Iter. # 2 took 0.2 msec. with norm = 0.512E-09 at n( 9, 40).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 10.
Iter. # 1 took 0.04 msec. with norm = 0.514E-03 at V( 10, 7).
Iter. # 2 took 0.06 msec. with norm = 0.197E-08 at V( 10, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 10.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 10, 66).
Iter. # 2 took 0.2 msec. with norm = 0.493E-09 at n( 10, 38).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 11.
Iter. # 1 took 0.05 msec. with norm = 0.569E-03 at V( 11, 7).
Iter. # 2 took 0.06 msec. with norm = 0.245E-08 at V( 11, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 11.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 11, 76).
Iter. # 2 took 0.2 msec. with norm = 0.473E-09 at n( 11, 76).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 12.
Iter. # 1 took 0.05 msec. with norm = 0.631E-03 at V( 12, 7).
Iter. # 2 took 0.06 msec. with norm = 0.305E-08 at V( 12, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 12.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 12, 62).
Iter. # 2 took 0.2 msec. with norm = 0.449E-09 at n( 12, 38).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 13.
Iter. # 1 took 0.05 msec. with norm = 0.705E-03 at V( 13, 7).
Iter. # 2 took 0.06 msec. with norm = 0.385E-08 at V( 13, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 13.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 13, 60).
Iter. # 2 took 0.2 msec. with norm = 0.423E-09 at n( 13, 37).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 14.
Iter. # 1 took 0.05 msec. with norm = 0.781E-03 at V( 14, 7).
Iter. # 2 took 0.06 msec. with norm = 0.480E-08 at V( 14, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 14.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 14, 51).
Iter. # 2 took 0.2 msec. with norm = 0.394E-09 at n( 14, 37).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 15.
Iter. # 1 took 0.05 msec. with norm = 0.870E-03 at V( 15, 7).
Iter. # 2 took 0.06 msec. with norm = 0.605E-08 at V( 15, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 15.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 15, 57).
Iter. # 2 took 0.2 msec. with norm = 0.362E-09 at n( 15, 36).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 16.
Iter. # 1 took 0.05 msec. with norm = 0.966E-03 at V( 16, 7).
Iter. # 2 took 0.06 msec. with norm = 0.759E-08 at V( 16, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 16.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 16, 57).
Iter. # 2 took 0.2 msec. with norm = 0.326E-09 at n( 16, 39).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 17.
Iter. # 1 took 0.05 msec. with norm = 0.107E-02 at V( 17, 7).
Iter. # 2 took 0.05 msec. with norm = 0.958E-08 at V( 17, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 17.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 17, 74).
Iter. # 2 took 0.2 msec. with norm = 0.286E-09 at n( 17, 40).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 18.
Iter. # 1 took 0.05 msec. with norm = 0.968E-03 at V( 18, 7).
Iter. # 2 took 0.09 msec. with norm = 0.791E-08 at V( 18, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 18.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 18, 49).
Iter. # 2 took 0.2 msec. with norm = 0.251E-09 at n( 18, 40).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 19.
Iter. # 1 took 0.05 msec. with norm = 0.872E-03 at V( 19, 7).
Iter. # 2 took 0.06 msec. with norm = 0.653E-08 at V( 19, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 19.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 19, 51).
Iter. # 2 took 0.2 msec. with norm = 0.218E-09 at n( 19, 37).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 20.
Iter. # 1 took 0.05 msec. with norm = 0.786E-03 at V( 20, 7).
Iter. # 2 took 0.06 msec. with norm = 0.537E-08 at V( 20, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 20.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 20, 59).
Iter. # 2 took 0.2 msec. with norm = 0.189E-09 at n( 20, 35).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 21.
Iter. # 1 took 0.05 msec. with norm = 0.707E-03 at V( 21, 7).
Iter. # 2 took 0.06 msec. with norm = 0.440E-08 at V( 21, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 21.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 21, 53).
Iter. # 2 took 0.2 msec. with norm = 0.162E-09 at n( 21, 36).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 22.
Iter. # 1 took 0.05 msec. with norm = 0.636E-03 at V( 22, 7).
Iter. # 2 took 0.06 msec. with norm = 0.360E-08 at V( 22, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 22.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 22, 67).
Iter. # 2 took 0.2 msec. with norm = 0.139E-09 at n( 22, 38).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 23.
Iter. # 1 took 0.05 msec. with norm = 0.573E-03 at V( 23, 7).
Iter. # 2 took 0.06 msec. with norm = 0.295E-08 at V( 23, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 23.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 23, 49).
Iter. # 2 took 0.2 msec. with norm = 0.117E-09 at n( 23, 35).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 24.
Iter. # 1 took 0.04 msec. with norm = 0.516E-03 at V( 24, 7).
Iter. # 2 took 0.06 msec. with norm = 0.241E-08 at V( 24, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 24.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 24, 50).
Iter. # 2 took 0.2 msec. with norm = 0.979E-10 at n( 24, 36).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 25.
Iter. # 1 took 0.05 msec. with norm = 0.465E-03 at V( 25, 7).
Iter. # 2 took 0.06 msec. with norm = 0.197E-08 at V( 25, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 25.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 25, 61).
Iter. # 2 took 0.2 msec. with norm = 0.806E-10 at n( 25, 35).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 26.
Iter. # 1 took 0.05 msec. with norm = 0.418E-03 at V( 26, 7).
Iter. # 2 took 0.06 msec. with norm = 0.161E-08 at V( 26, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 26.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 26, 53).
Iter. # 2 took 0.2 msec. with norm = 0.651E-10 at n( 26, 76).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 27.
Iter. # 1 took 0.04 msec. with norm = 0.377E-03 at V( 27, 7).
Iter. # 2 took 0.06 msec. with norm = 0.131E-08 at V( 27, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 27.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 27, 56).
Iter. # 2 took 0.2 msec. with norm = 0.510E-10 at n( 27, 37).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 28.
Iter. # 1 took 0.04 msec. with norm = 0.339E-03 at V( 28, 7).
Iter. # 2 took 0.06 msec. with norm = 0.107E-08 at V( 28, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 28.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 28, 63).
Iter. # 2 took 0.2 msec. with norm = 0.377E-10 at n( 28, 34).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 29.
Iter. # 1 took 0.05 msec. with norm = 0.305E-03 at V( 29, 7).
Iter. # 2 took 0.06 msec. with norm = 0.870E-09 at V( 29, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 29.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 29, 51).
Iter. # 2 took 0.2 msec. with norm = 0.265E-10 at n( 29, 35).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 30.
Iter. # 1 took 0.04 msec. with norm = 0.275E-03 at V( 30, 7).
Iter. # 2 took 0.06 msec. with norm = 0.709E-09 at V( 30, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 30.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 30, 51).
Iter. # 2 took 0.2 msec. with norm = 0.158E-10 at n( 30, 33).
The Newton solution converged in 2 iterations which took 0.0 sec.
Solving the electrostatic problem for slice i= 31.
Iter. # 1 took 0.05 msec. with norm = 0.247E-03 at V( 31, 7).
Iter. # 2 took 0.06 msec. with norm = 0.577E-09 at V( 31, 20).
The Newton solution converged in 2 iterations which took 0.00 sec.
Solving the exact n and V problem for slice i= 31.
Iter. # 1 took 0.2 msec. with norm = 0.177E-05 at n( 31, 75).
Iter. # 2 took 0.2 msec. with norm = 0.691E-11 at n( 31, 32).
The Newton solution converged in 2 iterations which took 0.0 sec.
+------> 1 x nodes/character maximum voltage = -9.18 kT/q = -0.238 V
! at 32, 14 going v the local dV = 2.32 kT/q = 0.060 V
! elec. conc. = 0.20E+05 nio = 0.30E+15 /cm**3
! 2 y nodes/character minimum voltage = -34.42 kT/q = -0.892 V
! elec. conc. = 0.67E-11 nio = 0.10 /cm**3
V
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s d
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s88888888888888888888888888888888d
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Now attempting to use DC() to solve for the biases:
Vs = 0.000 Vd = 0.350 Vg1 = 2.500 Eavg = 0.0
Solving for t= 0.000 sec. with dt= 0.1000E+33 sec.
Iter. # 1 took 0:00+ 0:00+ 0:00 with norm = 7.77 at n( 32, 19).
Iter. # 2 took 0:00+ 0:00+ 0:00 with norm = 5.11 at n( 32, 23).
Iter. # 3 took 0:00+ 0:00+ 0:00 with norm = 9.71 at n( 32, 24).
Iter. # 4 took 0:00+ 0:00+ 0:00 with norm = 4.45 at n( 2, 23).
Iter. # 5 took 0:00+ 0:00+ 0:00 with norm = 3.08 at V( 32, 23).
Iter. # 6 took 0:00+ 0:00+ 0:00 with norm = 5.09 at n( 27, 7).
Iter. # 7 took 0:00+ 0:00+ 0:00 with norm = 9.43 at n( 27, 16).
Iter. # 8 took 0:00+ 0:00+ 0:00 with norm = 3.00 at n( 16, 67).
Iter. # 9 took 0:00+ 0:00+ 0:00 with norm = 9.98 at n( 6, 19).
Iter. # 10 took 0:00+ 0:00+ 0:00 with norm = 19.2 at n( 5, 18).
Iter. # 11 took 0:00+ 0:00+ 0:00 with norm = 22.6 at n( 4, 18).
Iter. # 12 took 0:00+ 0:00+ 0:00 with norm = 34.9 at n( 30, 7).
Iter. # 13 took 0:00+ 0:00+ 0:00 with norm = 29.6 at n( 30, 12).
Iter. # 14 took 0:00+ 0:00+ 0:00 with norm = 25.5 at n( 3, 18).
Iter. # 15 took 0:00+ 0:00+ 0:00 with norm = 25.3 at n( 3, 19).
Iter. # 16 took 0:00+ 0:00+ 0:00 with norm = 28.5 at n( 30, 13).
Iter. # 17 took 0:00+ 0:00+ 0:00 with norm = 29.0 at n( 30, 14).
Iter. # 18 took 0:00+ 0:00+ 0:00 with norm = 29.9 at n( 31, 14).
Iter. # 19 took 0:00+ 0:00+ 0:00 with norm = 31.9 at n( 31, 15).
Iter. # 20 took 0:00+ 0:00+ 0:00 with norm = 26.0 at n( 30, 16).
+------> 1 x nodes/character maximum voltage = 9.73 kT/q = 0.252 V
! at 31, 28 going > the local dV = 11.26 kT/q = 0.292 V
! elec. conc. = 0.16E+08 nio = 0.25E+18 /cm**3
! 2 y nodes/character minimum voltage = -34.42 kT/q = -0.892 V
! elec. conc. = 0.67E-11 nio = 0.10 /cm**3
V
gggggggggggggggggggggggggggggggg
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
sNNNN-- -Nd
sNN-- -Nd
sN- nd
sN -d
s- -d
s- d
s d
s d
s d
s 88888888888 d
s 8999999999999999998 d
s 8ppppppppppppppppppp9 d
s 9pppppppppppppppppppp8 d
8ppppppppppppppppppppp9
9pppppppppppppppppppppp8
9ppppppppppppppppppppppp9
889ppppppppppppppppppppppppp9
ppppppppppppppppppppppppppppp988
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
Adding a node to the left of ( 32, 28) because:
Er=0.10469E+06 V/cm vr=-.71552E+08 cm/s v/E= -683.4 cm^2/V-s hr=0.40E-01 um
Dr= 18.1 cm^2/sec 2D/v=0.50676E-02 um h*E= 16.325 kT dV= 15.939 kT
A node was added at the right side, so now xnodes= 33 and ynodes= 78.
Iter. # 21 took 0:00+ 0:00+ 0:00 with norm = 32.8 at n( 31, 16).
Iter. # 22 took 0:00+ 0:00+ 0:00 with norm = 4.09 at n( 27, 22).
Iter. # 23 took 0:00+ 0:00+ 0:00 with norm = 2.28 at n( 27, 22).
Iter. # 24 took 0:00+ 0:00+ 0:00 with norm = 2.39 at n( 28, 26).
Iter. # 25 took 0:00+ 0:00+ 0:00 with norm = 2.97 at n( 28, 26).
Iter. # 26 took 0:00+ 0:00+ 0:00 with norm = 5.93 at n( 30, 30).
Iter. # 27 took 0:00+ 0:00+ 0:00 with norm = 16.4 at n( 32, 30).
Iter. # 28 took 0:00+ 0:00+ 0:00 with norm = 6.56 at n( 31, 33).
Iter. # 29 took 0:00+ 0:00+ 0:00 with norm = 4.67 at n( 32, 34).
Iter. # 30 took 0:00+ 0:00+ 0:00 with norm = 4.68 at n( 32, 33).
Iter. # 31 took 0:00+ 0:00+ 0:00 with norm = 5.79 at n( 32, 32).
Iter. # 32 took 0:00+ 0:00+ 0:00 with norm = 13.6 at n( 32, 31).
Iter. # 33 took 0:00+ 0:00+ 0:00 with norm = 7.48 at n( 32, 31).
Iter. # 34 took 0:00+ 0:00+ 0:00 with norm = 0.614E-01 at n( 32, 32).
Iter. # 35 took 0:00+ 0:00+ 0:00 with norm = 0.437E-05 at V( 33, 7).
Iter. # 36 took 0:00+ 0:00+ 0:00 with norm = 0.182E-05 at V( 33, 7).
Iter. # 37 took 0:00+ 0:00+ 0:00 with norm = 0.753E-06 at V( 33, 7).
The Newton solution converged in 37 iterations which took 0:02.
==== Shark v2.0 REB 1988 charge map ====
The following is the charge map for equilibrium:
g = resistive gate
i = insulator, or oxide
n = n-type
p = p-type
blank = depletion region
+------> 1 x nodes/character maximum voltage = 10.41 kT/q = 0.270 V
! at 33, 31 going v the local dV = 10.85 kT/q = 0.281 V
! elec. conc. = 0.16E+08 nio = 0.25E+18 /cm**3
! 2 y nodes/character minimum voltage = -34.42 kT/q = -0.892 V
! elec. conc. = 0.67E-11 nio = 0.10 /cm**3
V
ggggggggggggggggggggggggggggggggg
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
sNNNN-- -NNNd
sNN-- -NNd
sN- --Nd
sN -Nd
s- -Nd
s- Nd
s -d
s -d
s -d
s 88888888888 -d
s 8999999999999999998 d
s 8ppppppppppppppppppp8 d
s 9ppppppppppppppppppp9 d
8ppppppppppppppppppppp8
9ppppppppppppppppppppp9
9ppppppppppppppppppppppp8
889pppppppppppppppppppppppp98
pppppppppppppppppppppppppppp988
pppppppppppppppppppppppppppppp999
ppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppp
==== Shark v2.0 REB 1988 starting transient ====
Outputting the simulation results for t = 0.00 (psec).
Iter. # 1 took 0:00+ 0:00+ 0:00 with norm = 0.311E-06 at V( 33, 7).
The Newton solution converged in 1 iterations which took 0:00.
==== Shark v2.0 REB 1988 charge map ====
The following is the charge map for equilibrium:
g = resistive gate
i = insulator, or oxide
n = n-type
p = p-type
blank = depletion region
+------> 1 x nodes/character maximum voltage = 10.41 kT/q = 0.270 V
! at 33, 31 going v the local dV = 10.85 kT/q = 0.281 V
! elec. conc. = 0.16E+08 nio = 0.25E+18 /cm**3
! 2 y nodes/character minimum voltage = -34.42 kT/q = -0.892 V
! elec. conc. = 0.67E-11 nio = 0.10 /cm**3
V
ggggggggggggggggggggggggggggggggg
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
sNNNN-- -NNNd
sNN-- -NNd
sN- --Nd
sN -Nd
s- -Nd
s- Nd
s -d
s -d
s -d
s 88888888888 -d
s 8999999999999999998 d
s 8ppppppppppppppppppp8 d
s 9ppppppppppppppppppp9 d
8ppppppppppppppppppppp8
9ppppppppppppppppppppp9
9ppppppppppppppppppppppp8
889pppppppppppppppppppppppp98
pppppppppppppppppppppppppppp988
pppppppppppppppppppppppppppppp999
ppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppp
Outputting the simulation results for t = ****** (psec).
Total CPU time used: 0 days, 00:00:02
The Makefile is set up for a typical MacOS box, but there are versions of
the makefile for a lot of different platforms. I just kept making new ones
as I moved to new systems.
$ makewill build everything, and then you're ready to run the code.
Homebrew - all of the following tools can be installed with Homebrew. If it's not already installed on you laptop, it's easy enough to go to the website, run the command to install it, and then continue to the next step. If you are on linux, expect to be installing some scripts and RPMs, but everything is available there as well - just not as convenient.
A gfortran or f77 compiler. I have installed with:
$ brew cask install gfortranand you should get just gfortran from the cask, and be ready to go.
If you have any previous versions of gfortran on your box, as I did, you
might want to ininstall them first with:
$ brew uninstall --force gfortranand then do the insall, above. It all works pretty seamlessly.