[SOLVED] Capacitor not fully discharging in Qucs

[s.jayaram88]

In qucs I rigged up a simple circuit which charges a capacitor and then discharges it. The capacitor charges fully but when discharging it does only to half i.e., if the capacitor was charged to 1 V, it discharges only to 0.5 V. Why is this?? Pls help me with an answer. this is the circuit.
<Qucs Schematic 0.0.16>
<Properties>
<View=0,0,901,800,1,0,0>
<Grid=10,10,1>
<DataSet=switch_rc.dat>
<DataDisplay=switch_rc.dpl>
<OpenDisplay=1>
<Script=switch_rc.m>
<RunScript=0>
<showFrame=0>
<FrameText0=Title>
<FrameText1=Drawn By:>
<FrameText2=Date:>
<FrameText3=Revision:>
</Properties>
<Symbol>
</Symbol>
<Components>
<Vdc V1 1 200 240 18 -26 0 1 "1 V" 1>
<Switch S1 1 330 170 -26 11 0 0 "off" 0 "1 ms" 0 "0" 0 "1e12" 0 "26.85" 0>
<GND * 1 410 350 0 0 0 0>
<R R2 1 410 280 15 -26 0 1 "1 kOhm" 1 "26.85" 0 "0.0" 0 "0.0" 0 "26.85" 0 "US" 0>
<R R1 1 250 170 -26 15 0 0 "1 kOhm" 1 "26.85" 0 "0.0" 0 "0.0" 0 "26.85" 0 "US" 0>
<Switch S2 1 410 210 11 -26 0 1 "off" 0 "5 ms" 0 "0" 0 "1e12" 0 "26.85" 0>
<C C1 1 520 250 17 -26 0 1 "1 uF" 1 "0" 1 "neutral" 0>
<.TR TR1 1 740 130 0 65 0 0 "lin" 1 "0" 1 "20 ms" 1 "1001" 0 "Trapezoidal" 0 "2" 0 "1 ns" 0 "1e-16" 0 "150" 0 "0.001" 0 "1 pA" 0 "1 uV" 0 "26.85" 0 "1e-3" 0 "1e-6" 0 "1" 0 "CroutLU" 0 "no" 0 "yes" 0 "0" 0>
</Components>
<Wires>
<200 170 200 210 "" 0 0 0 "">
<200 170 220 170 "" 0 0 0 "">
<280 170 300 170 "" 0 0 0 "">
<410 240 410 250 "" 0 0 0 "">
<520 280 520 310 "" 0 0 0 "">
<410 310 520 310 "" 0 0 0 "">
<200 270 200 310 "" 0 0 0 "">
<200 310 410 310 "" 0 0 0 "">
<410 310 410 350 "" 0 0 0 "">
<410 170 410 180 "" 0 0 0 "">
<360 170 410 170 "" 0 0 0 "">
<520 170 520 220 "" 0 0 0 "">
<410 170 520 170 "" 0 0 0 "">
<520 170 520 170 "oupput" 550 140 0 "">
</Wires>
<Diagrams>
</Diagrams>
<Paintings>
</Paintings>

And I have attached the snapshots too

 

[albbg]

This is because you set both switches (S1 and S2) to be initially off (open), then S1 is set to be on (closed) after 1 ms and S2 after 5 ms. So after 1ms the capacitor will start to charge to 1V and after 5ms will start to discharge to 1V*R1/(R1+R2) = 0.5V because when both S1 and S2 are closed, the two resistor are in parallel one each other.
You have to set S1 initially on (see parameter "init" in the dialog box) ans S2 initially off. Set the transition time ("time" in the dialog box) of S1 large enough to completely charge the capacitor (let's say 5 ms). Set the transiotion time of S2 at the value you want the capacitor starts to discharge.

 

[s.jayaram88]

Thanx a lot!! I got it solved :grin: