Ideal capacitor power in rectifier simulation - Page 1
Forum section: electronics / projects

author: Digibin
author_rank: Regular Contributor
timestamp: April 09, 2016, 07:21:28 pm
content: I'm simulating (LTSpice) a mains full wave rectifier bridge in order to experiment with LC smoothing networks rather than just smoothing caps (which I haven't got to yet). I've added some postprocessing .meas statements to calculate the power factor of the rectifier and the power loss in the load. The network is just the four rectifier diodes, smoothing cap and load resistor.Looking at the results my AC source is supplying 1338VA of apparent power and 580W of real power, giving a power factor of 0.4. But the (real) power dissipation in the load resistor is only 414W, so where's the rest of the 580W real power from the source? So I looked at the real power in the capacitor as well, which comes out as 168W accounting for the difference. My simulation is using ideal components, so my question is how can an ideal capacitor dissipate power? The ideal capacitor should be storing energy during the sharp input current spikes and releasing energy to the load for the rest of the half cycle, with the net energy flow being zero. Is there an issue with my simulation or is there some quirk of LTSpice I'm not aware of?

author: orolo
author_rank: Frequent Contributor
timestamp: April 09, 2016, 08:08:49 pm
content: Try.meas C1_power AVG V(V_load)*I(C1) from (end/2) to endwith a timestep of 1e-6s. Note that the capacitor stores energy at the beginning, which is not released into the load. Also, a timestep not fine enough would give approximation errors. Have you checked how much energy is lost in the diodes?

author: Digibin
author_rank: Regular Contributor
timestamp: April 09, 2016, 09:10:57 pm
content: Quote from: orolo on April 09, 2016, 08:08:49 pmGreat, yep, finer timestep fixed it. The real power delivered by the source now almost matches the power in R1. Still a 2W discrepancy, perhaps this is dissipated in the diodes but they are ideal so shouldn't be consuming any power. Power in C1 is down to 1mW.Thanks for the help!

author: orolo
author_rank: Frequent Contributor
timestamp: April 09, 2016, 11:49:20 pm
content: Glad I could help. You'll probably find that the rest of the energy is dissipated in the diodes: even if they're ideal, the voltage drop while conducting will cause some loss.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 10, 2016, 04:17:39 am
content: SPICE "ideal" diodes follow the Shockley function without series resistance or reverse recovery. That's as reasonably "ideal" as one should hope to expect from any diode you'd want to simulate.An ""ideal"" diode (meaning, zero forward drop, zero reverse current) would never simulate because it's an absolute discontinuity: unrealistic both in regards to real diodes, and to numerical solutions that depend upon having continuous derivatives to produce a solution.(You can make arbitrarily "ideal" diodes by using a switch (both in simulation and real life), but the difference is, it takes time to decide whether voltage is positive, or current is negative; which leads to finite forward and reverse recovery times. So there's still mathematical continuity on a fine enough timescale -- which SPICE needs to dip down to, to make sure it's getting it right -- which makes this okay. Basically, what happens at DC is meaningless; what happens at AC is where things get interesting.)Tim

author: Digibin
author_rank: Regular Contributor
timestamp: April 10, 2016, 09:00:08 pm
content: On a separate but related point, I'm having trouble with Fairchild encrypted spice models. I'm trying to import the model for the FCP190N60, which is available on the product pagehere. It's provided in encrypted form (attached). Do I just use the model in the same way as an un-encrypted model? Because LTSpice can't find the .model -m1: Can't find definition of model "fcp190n60e_3p".