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I use a capacitor and a resistor to be a battery,the simulation show me the current does flow into it(charge it).The reason that i believe the current does flow into the capacitor is because that the current is positive for flowing into the component,and negative is for flowing out off the component in this software(virtuoso cadence).However,the capacitor voltage did't rise up ,why?it does't make any sense.

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Battery model

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Resistor and capacitor current and capacitor voltage

enter image description here After changing the capacitor from 1 to 1p,the wave become like a square wave.

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  • \$\begingroup\$ Per the answer below you need to increase the pulse width a LOT and decrease the value of the capacitor by a million or so. Your numbers were working against you. For any expected result you must use rational values. The program will not tell you what is out of range in terms of charge time. IRL you would also have capacitor leakage to account for. \$\endgroup\$ – user105652 May 5 '18 at 5:24
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Since the voltage of a capacitor can be calculated by

$$u(t)=\frac{1}{C}\int i dt$$

Let's look at your simulation settings. Your current is milliamp scale, which is to the \$10^{-3}\$ order. And for your pulse, I guess it is several nanoseconds. Therefore, for each pulse, your voltage will increase \$10^{-9}\$V. That's why it is not increasing in your simulation setting.

Given the voltage formula, there are basically 3 ways to solve your problem:

  1. Increase the time that you applied your current. For example, applied a square wave.
  2. Increase the current flown into the capacitor.
  3. Decrease the capacitance. 1 farad is fairly a large number for a capacitor, try 1 nF or something like that.
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  • \$\begingroup\$ I have modify something as you said,but the voltage still doesn't rise up.1.i increase the time from 80ms to 1s,but the voltage is still a constant (2V). By the wave,why should i increase the time by applying a square wave? 2.I use PZT to charge it,so i can't increase the chargeing current,because it is meaningless to me. 3.I decrease the capacitance from 1 to 1p,but the wave become a square wave. \$\endgroup\$ – XM551 May 5 '18 at 8:05
  • \$\begingroup\$ It helps to know OP is trying to model a battery (simple way) to see if it charges, or not. If you're curious, you can look in OP's previous questions, though, it's true, OP could have said what purpose it is for. \$\endgroup\$ – a concerned citizen May 5 '18 at 8:06
  • \$\begingroup\$ @aconcernedcitizen OP?why do you say OP?does it have relation with operational amplifier? \$\endgroup\$ – XM551 May 5 '18 at 8:24
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    \$\begingroup\$ @XM551 No, OP = "Original Poster", or "Original Post", or something along those lines. It's a neutral point of view to address someone when you want to avoid the ambiguous "he" or "she" (when you simply don't know), or the long "the one who answered the question". \$\endgroup\$ – a concerned citizen May 5 '18 at 8:28
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    \$\begingroup\$ @XM551: The OP (original poster) is you. \$\endgroup\$ – Transistor May 5 '18 at 8:58

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