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inquiry about the results of the serial capacitors in cadence

IanX

hello. everyone. I am doing the simulation in Cadence(IC6.1.6). Pls see the schematic below. What I want to know is the voltage of each capacitor. In each two circuit,  a 10V vdc , 2u cap and 8u cap. 1Ω resistor are used.  Tran simulation is used by setting 100u. I set initial volatge of all capacitor equal to 0.

After runing the simulation, the c1 always equals to 5V no matter what the value of the two capacitor are in the upper circuit. If I leave the initial voltage of the capacitor blank, the C1 equal to 0V

In the lower circuit, the C2t and C1b equals to 3.5V when 2 resistors are included. 

And by calculation the C1 should be 2V, and the the same with c2t. I don't know why the results are so strange. I have disucessed with many people. but I still could not fix it. Could someone help me? 

 

Parents

  • For the top circuit, the issue is that the DC condition cannot be met. You have no resistance in the circuit, and when the capacitors are removed for the DC, it cannot possibly have both of them at zero volts yet with a DC voltage of 10V across the pair. This is why the simulator reports:

        C1: Initial condition computed between nodes in1 and c1 is in error by 5 V.
        C2: Initial condition computed for node c1 is in error by 5 V.            

    The voltage is coming about from the "rforce" value that is used to force a voltage (or current) on the circuit for initial conditions - and so you have a potential divider with two rforce resistors at 1 ohm each. 

    By the way, the top circuit has the second capacitor as 8n rather than 8u, but that doesn't matter here.

    In the bottom circuit, the issue is that the 1ohm series impedance with the source is too similar a value to the rforce value - and hence that is affecting the solution. If I change the circuit to this and use the rforce option to set it to 1mOhm (don't set it smaller than this, as it could adversely affect convergence):

    // forum example
    V1 (in1 0) vsource dc=10
    R1 (in1 in2) resistor r=1
    C1 (in2 c1) capacitor c=2u ic=0
    C2 (c1 0) capacitor c=8u ic=0

    myOpts options rforce=1m
    dc dc
    tran tran stop=1m

    Then I get the plot below. Note that the 1mOhm impedance is still only a 1000 times different from the series 1ohm impedance (note I only had a single resistor, rather than two). With this I get 2.006V at node c1.

    Regards,

    Andrew.

Reply

  • For the top circuit, the issue is that the DC condition cannot be met. You have no resistance in the circuit, and when the capacitors are removed for the DC, it cannot possibly have both of them at zero volts yet with a DC voltage of 10V across the pair. This is why the simulator reports:

        C1: Initial condition computed between nodes in1 and c1 is in error by 5 V.
        C2: Initial condition computed for node c1 is in error by 5 V.            

    The voltage is coming about from the "rforce" value that is used to force a voltage (or current) on the circuit for initial conditions - and so you have a potential divider with two rforce resistors at 1 ohm each. 

    By the way, the top circuit has the second capacitor as 8n rather than 8u, but that doesn't matter here.

    In the bottom circuit, the issue is that the 1ohm series impedance with the source is too similar a value to the rforce value - and hence that is affecting the solution. If I change the circuit to this and use the rforce option to set it to 1mOhm (don't set it smaller than this, as it could adversely affect convergence):

    // forum example
    V1 (in1 0) vsource dc=10
    R1 (in1 in2) resistor r=1
    C1 (in2 c1) capacitor c=2u ic=0
    C2 (c1 0) capacitor c=8u ic=0

    myOpts options rforce=1m
    dc dc
    tran tran stop=1m

    Then I get the plot below. Note that the 1mOhm impedance is still only a 1000 times different from the series 1ohm impedance (note I only had a single resistor, rather than two). With this I get 2.006V at node c1.

    Regards,

    Andrew.

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