davis testcircuit from bug #240, show achieved precision

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rlar 2018-02-24 19:33:05 +01:00
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* switch as negative resistance oscillator
* compare the transient simulation of this
* switch based relaxation oscillator
* with the analytic solution
* show off the precision which can be achieved
* even though switching the time-constant by 7 orders of magnitude
I1 1 0 -100u
C1 1 0 1n
SW1 1 0 1 0 SWITCH1
.MODEL SWITCH1 SW VT=2.5 VH=2.0 RON=1 ROFF=10MEG
.option method=trap
.control
* two e-t/T shapes
* one is tau1 = 10Meg*1nF = 10ms
* rising from 0 to 100u*10Meg = 1000V
* second is tau2 = 1Ohm*1nF = 1ns
* falling down to 100u*1 = 100uV
* first upper switch point t1 is
* 4.5v = 1000v * (1 - e^-t/tau1)
* first crossing of 0.5 point is t0
* first crossing of 4.5 point is t1
*
* tr and tf, 0.5 to 4.5 and back to 0.5
* 4.5 = 0.5 + (1000 - 0.5) * (1 - e(-tr/tau1))
* tr = tau1 * log((1000 - 0.5)/(4.5 - 0.5))
* 0.5 = 100u + (4.5 - 100u) * e(-tf/tau2)
* tf = tau2 * log((4.5 - 100u)/(0.5 - 100u))
let tau1 = 10ms
let tau2 = 1ns
let t0 = -tau1 * log(1 - 0.5/1000)
let t1 = -tau1 * log(1 - 4.5/1000)
let tr = tau1 * log((1000 - 0.5)/(1000 - 4.5))
let tf = tau2 * log((4.5 - 100u)/(0.5 - 100u))
let Tperiod = tr + tf
tran 10us 300us uic
let len = length(time)
let gold = vector(len)
let steps = vector(len)
let kk = 0
repeat $&len
let tt = time[kk] - t0
let tt = tt lt 0 ? tt : tt - Tperiod * floor(tt/Tperiod)
let tt = tt + t0;
let gold[kk] = (tt lt t1) ? 1000*(1-exp(-tt/tau1)) : (100u + (4.5-100u)*exp(-(tt-t1)/tau2))
let steps[kk] = time[kk] - time[vecmax(kk-1, 0)]
let kk = kk + 1
end
plot v(1) gold
plot (v(1) - gold) ylimit -1mV 1mV
* the first discharge in more detail:
plot v(1) gold xlimit 45.095e-6 45.110e-6
* compare the golden first discharge time
* with the timesteps choosen in the following print
* the last step is 1ns before the ideal switch,
* the following is 100ps after the ideal switch
* one can readily see the timestep truncation in action
* when aproaching the discharge time point
print (time - t1) steps
.endc
.end