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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

  • hello, Tom. Thank you. Though I don't know where to type the command line you given, I found the older version manual from the internet. It's said that The Spectre simulator sets initial conditions on a node by attaching a voltage source through a resistor. The default value of this resistor is 1, but you can control the value through the options parameter rforce. Just like you said, I consider that the capacitor equal to a resistor (set by the rforce) when the initial condition is calculated. Is it right?

    And Andrew, I think the reason why the voltage across the resistors equal to 6V and 4V in the second circuit is that the initial voltage in c1t = 7.5V and c1b=5V c2t =2.5V. And from this initial voltage, the steady voltage equal to 3.5V. I could not explain why the c1 always =5V in the upper circuit. I can understand the initial voltage is 5V. I think the voltage should change according the capacitance.

     So I did another simulation. 

     

    1. I still do the 'tran analyses'. I use two capacitor in parallel , and one terminal with ground. Initial voltage is 5V, 2V cross each capacitor . And plot the node between between. I can see the curve from 5V to almost 0V if the tran stop time is long enough. 
    2. Another simulation: The same with previous one besides adding a resistor between two capacitor. I measure two terminal of the resistor. One curve is from 4V and one curve is from 3V. And the final value is almost 0V if the tran stop time is long enough. It is quite strange . I can not understand how it work. the Spectre simulator is quite interesting.

     

Reply

  • hello, Tom. Thank you. Though I don't know where to type the command line you given, I found the older version manual from the internet. It's said that The Spectre simulator sets initial conditions on a node by attaching a voltage source through a resistor. The default value of this resistor is 1, but you can control the value through the options parameter rforce. Just like you said, I consider that the capacitor equal to a resistor (set by the rforce) when the initial condition is calculated. Is it right?

    And Andrew, I think the reason why the voltage across the resistors equal to 6V and 4V in the second circuit is that the initial voltage in c1t = 7.5V and c1b=5V c2t =2.5V. And from this initial voltage, the steady voltage equal to 3.5V. I could not explain why the c1 always =5V in the upper circuit. I can understand the initial voltage is 5V. I think the voltage should change according the capacitance.

     So I did another simulation. 

     

    1. I still do the 'tran analyses'. I use two capacitor in parallel , and one terminal with ground. Initial voltage is 5V, 2V cross each capacitor . And plot the node between between. I can see the curve from 5V to almost 0V if the tran stop time is long enough. 
    2. Another simulation: The same with previous one besides adding a resistor between two capacitor. I measure two terminal of the resistor. One curve is from 4V and one curve is from 3V. And the final value is almost 0V if the tran stop time is long enough. It is quite strange . I can not understand how it work. the Spectre simulator is quite interesting.

     

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