RC discharge simulation

rc_discharge.ipynb

I got the below expressions for the intervals.

Note that log of capacitor voltage VC divided by starting voltage V0 is always negative (we can assume even with error bounds that this is true (it is because VC drops so much per 5us)).

I assumed [ta, tb] = [t, t] because the clock is very high precision in this context, at 20ppm, resulting in only 100 picoseconds of error per 5 us interval that I'm using. According to my previous formulas, its effect in the total error is absolutely invisible (I did take into account that this error is cumulative).

Fwiw the formula for error caused by time (using 5us steps) is E_t(t) = (E_t_meas * t/(5e-6)) * (-1)/(C*log(VC(t)/V0)) with E_t_meas 100e-12 being the measurement error (20ppm of 5us)

[Ra, Rb] = -t/([Ca, Cb] * log([VCa, VCb]/[V0a, V0b]))     
                                                            
[VCa, VCb]/[V0a, V0b]                                       
= [VCa, VCb] * [1/V0b, 1/V0a]                               
= [VCa/V0b, VCb/V0a]                                
                                                            
[Ra, Rb]                                                           
= -t/([Ca, Cb] * log([VCa/V0b, VCb/V0a]))           
= -t/([Ca, Cb] * [log(VCa/V0b), log(VCb/V0a)])      
= -t/[Cb * log(VCa/V0b), Ca * log(VCb/V0a)]   # Ca/Cb inverted here because logs are known to be negative
= -t * 1/([Cb * log(VCa/V0b), Ca * log(VCb/V0a)])
= -t * [1/(Ca * log(VCb/V0a)), 1/(Cb * log(VCa/V0b))]
= [-t/(Cb * log(VCa/V0b)), -t/(Ca * log(VCb/V0a))]