Pspice help RLC parrallel circuit.

Click For Summary

Discussion Overview

The discussion revolves around troubleshooting a Pspice simulation for an over-damped RLC parallel circuit. Participants explore issues related to initial conditions for voltage and current in the circuit, comparing results from Pspice with those obtained from MATLAB and theoretical calculations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents their calculations for voltage and current in the RLC circuit, noting discrepancies between Pspice and MATLAB results.
  • Another participant suggests that Pspice may not recognize the initial current through the capacitor as a valid condition, leading to a different initial voltage calculation.
  • A participant expresses frustration over the inability to set the initial conditions correctly and notes that attempts to set the capacitor's initial voltage to 90V did not yield the expected results.
  • One participant reports success after modifying their code multiple times, but mentions that the same code does not consistently produce the same results, indicating potential instability or sensitivity in the simulation setup.

Areas of Agreement / Disagreement

Participants express differing views on how Pspice handles initial conditions, particularly regarding the capacitor's current. There is no consensus on a definitive solution to the initial voltage discrepancy, as participants continue to explore various approaches.

Contextual Notes

Limitations include potential misunderstandings of how Pspice interprets initial conditions for capacitors and inductors, as well as the variability in simulation results based on code modifications.

xXOfNiRXx
Messages
13
Reaction score
0
Hello everybody. I have an over-damped RLC parallel circuit. Using techniques taught, I solved for v(t) = 70e^(-10,000t)+20e^(-40,000t) V for t> zero. This indicates that my initial voltage is 90v. Using MATLAB (Part of the assignment) I verified this. I also found that Current as a function of time in my inductor is I(t)= -28e^(-10,000t)-2e^(-40,000t) mA for t>0. Which indicates that at t=0 current is -30mA; through the inductor. Which again, agrees with both my Matlab graph and the answer in the book.

C1 has a capacitance of 10nF and an initial current of -15mA (solved for current, agrees with book)
R1 has a resistance value of 2000ohms
L1's value is 250mH with an initial current of -30mA. (Given in problem)

However, upon running Pspice with this code:

C1 1 0 10e-9F IC = -15e-3A
R1 1 0 2.0e3ohm
L1 1 0 250e-3H IC = -30e-3A
.tran 5e-6 1e-3 UIC
.plot tran v(1)
.plot tran I(L1)
.end

My voltage graph doesn't agree with Matlab or my voltage function. According to Pspice, my initial Voltage at node 1 is only 60 not 90. I'm confused because my current plot for L1 is right on with an initial of -30mA, as it should.

Does anybody see anything wrong with my code? I've tried sing .IC =V(1)=90, but that didn't do anything either. Anyhelp would be greatly appreciated...it's driving me crazy. Thanks in advance.
 
Engineering news on Phys.org
I think the problem is that pspice does not recognize the current through the capacitor as a valid initial condition. You can specify a capacitor node voltage as an initial condition, but not a current. So it ignores the 15 mA flowing through the capacitor and assumes that the 30mA flowing through the inductor is flowing back through the resistor, leading to the 60V initial node voltage. I couldn't find a way to get it to do what you want.
 
Thanks for your reply, appreciate the help. It's an interesting problem. What you said makes sense though. Setting IC = 90V for the capacitor does nothing. Wow, I'm at a loss; I have literally tried everything that I can think of.
 
So, I finally got it to work. However, I changed the code around many-a-time until it finally worked. Curiously enough, the same code doesn't give me the same answer every time! Weird. Here is he code that code:

C1 1 0 10e-9F IC = 90V
R1 1 0 2.0e3ohm
L1 1 0 250e-3H IC= -30e-3A
.tran 5e-6 5.5e-4
.plot tran V(1)
.plot tran I(L1)
.end
 

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
4K
Replies
44
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
1K