powerCircuitSimulation(rsA = 8200, rsB = 220000, rsC = 1000, rsD = 33000, rsE = 56000, rsF = 5600, rsG = 3300, rsH = 58.5, rsI = 1000, rsJ = 120, trK = 130, trL = 100, trM = 130, tlA = 5, tlB = 10, tlC = 10, tlD = 5, tlE = 5, tlF = 5, tlG = 10, tlH = 5, tlI = 5, tlJ = 5, tlK = 5, tlL = 10, tlM = 5, each = 50, seed = NA)each times.
each * n rows and with columns:
| rsA | numeric |
value of rsA |
rsB |
| numeric | value of rsB |
| rsC | numeric |
value of rsC |
rsD |
| numeric | value of rsD |
| rsE | numeric |
value of rsE |
rsF |
| numeric | value of rsF |
| rsG | numeric |
value of rsG |
rsH |
| numeric | value of rsH |
| rsI | numeric |
value of rsI |
rsJ |
| numeric | value of rsJ |
| trK | numeric |
value of trK |
trL |
| numeric | value of trL |
| trM | numeric |
value of trM |
tlA |
| numeric | value of tlA |
| tlB | numeric |
value of tlB |
tlC |
| numeric | value of tlC |
| tlD | numeric |
value of tlD |
tlE |
| numeric | value of tlE |
| tlF | numeric |
value of tlF |
tlG |
| numeric | value of tlG |
| tlH | numeric |
value of tlH |
tlI |
| numeric | value of tlI |
| tlJ | numeric |
value of tlJ |
tlK |
| numeric | value of tlK |
| tlL | numeric |
value of tlL |
tlM |
| numeric | value of tlM |
$$V = \frac{136.67(a+\frac{b}{Z(10)})+d(c+e)\frac{g}{f}-h}{1+d\frac{e}{f}+b[frac{1}{Z(10)+0.006(1+\frac{13.67}{Z(10)})]+0.08202a}}$$ where $$a = \frac{Z(2)}{Z(1)+Z(2)}$$ $$b=\frac{1}{Z(12)+Z(13)}(Z(3)+\frac{Z(1)Z(2)}{Z(1)+Z(2)})+Z(9)$$ $$c=Z(5)+Z(7)/2$$ $$d=Z(11)\frac{Z(1)Z(2)}{Z(1)+Z(2)}$$ $$e=Z(6)+Z(7)/2$$ $$f=(c+e)(1+Z(11))Z(8)+ce$$ $$g=0.6+Z(8)$$ $$h=1.2$$ with $Z(1),\ldots,Z(10)$ resistances in $\Omega$ of the 10 resistances and $Z(11),Z(12),Z(13)$ are the $h_{FE}$ values of three transistors.
pistonSimulation,
simulationGroup
powerCircuitSimulation(seed=123, each=3)
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