LTspice diode commutation

Question

I have modeled an ideal diode rectifier fed from a 230V/50Hz grid with a slim DC link capacitor and a constant power load (1kW) coupled with a small resistive load. For Cdc=14uF i get a phase input current that oscillates between 0A and 8A while it is dampened. However for Cdc=15uF (or 16uF, 19uF, 20uF) the current never gets to zero (which actually greatly improves the harmonic content and also the power factor).

The only reason I can think of for the input line current to become 0A is diode commutation. However, apparently this does not happen for certain Cdc, Lg value combinations. Why could that be?

The schematic of my system:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Pictures of the different input currents are presented below (V(n001) is the upper diode to earth voltage).

The netlist of the model is given below.

Lgra L1 N001 {Lg} Rpar=100k
Lgrb L2 N002 {Lg} Rpar=100k
Lgrc L3 N003 {Lg} Rpar=100k
XX1 N001 N002 N003 Vdc 0 diode_rectifier
R2 Vdc 0 {Rload}
Vga2 L1 COM SINE(0 {Vac} {fg} 0 0 0) AC 1 0
Vgb2 L2 COM SINE(0 {Vac} {fg} 0 0 120) AC 1 120
Vgc2 L3 COM SINE(0 {Vac} {fg} 0 0 -120) AC 1 -120
R3 COM 0 10Meg
B1 Vdc 0 I={1000/V(Vdc)}
C1 Vdc 0 {Cdc} Rser=14m

* block symbol definitions
.subckt diode_rectifier Va Vb Vc V+ V-
D1 Va V+ D
D2 Vb V+ D
D3 Vc V+ D
D4 V- Vc D
D5 V- Va D
D6 V- Vb D
.ends diode_rectifier

.model D D
.lib C:\Users\NVA\Documents\LTspiceXVII\lib\cmp\standard.dio
.param Vac = 230V
.param fg=50Hz
.param Lg =100u
.param Rg =1m
.param Cdc =15u

Without a schematic people will need to guess what you are doing. Without the .asc file people will not be able to try this themselves in ltspice. — PlasmaHH, Nov 28 '17 at 16:45
That is why I uploaded the netlist. How can I upload the .asc file? — Nikos Vandoros, Nov 28 '17 at 16:52
Schematic and netlist should be enough. But the schematic is a lot easier to read than the netlist. It also helps to label the nodes you're going to plot so we don't have to guess which node is `n001`. — The Photon, Nov 28 '17 at 17:14
Please, add a snapshot of the LTspice schematic. As @ThePhoton said, delving into a netlist without the corresponding schematic is way too awkward. You should show a reasonable amount of effort to make your question "palatable" for people here, if you want to be helped. The prospective of losing time on a netlist while you could add a schematic in a breeze is not something that would induce people to answer. — LorenzoDonati4Ukraine-OnStrike, Nov 28 '17 at 17:29
What is B1 there to model? It's drawing about 3 A while the resistor is only drawing ~100 mA. And it's a negative resistance element so it's likely to cause some unexpected behavior if you don't think carefully about what to expect. — The Photon, Nov 28 '17 at 18:11
B1 is a behavioral current source. It simulates a vector controlled AC motor as it produces constant power. — Nikos Vandoros, Nov 28 '17 at 18:20
@NikosVandoros, I don't know much about that, but are you sure that perfect instantaneous constant power is a good model? Or should it just be something that averages out to constant power over a few cycles of the AC? — The Photon, Nov 28 '17 at 18:38
I think it is pretty accurate. But what are you proposing exactly? — Nikos Vandoros, Nov 28 '17 at 18:43
Try adding this to your subcircuit: `.model d d ron=0.1 roff=0.1g vfwd=0.7 epsilon=0.1 revepsilon=0.1`. This adds a higher quadratic region to the knee. Or, you could try this: `.model d d Cjo=10p`, which will make use of the internal diode, or use any other model from the available ones. By default LTspice uses the ideal model. — a concerned citizen, Nov 29 '17 at 07:14
@aconcernedcitizen using the first diode model solves my issues, but the output is too ideal. I mean the rectifier does not even function as one anymore. — Nikos Vandoros, Nov 29 '17 at 09:02
Sure, but what's the timestep in your oscilograms? You croped the picture which deleted the x scale and numbering. — winny, Nov 29 '17 at 09:35
@winny It is 0-20msec with 2ms for each vertical dashed line — Nikos Vandoros, Nov 29 '17 at 10:30
That information needs to go in the picture, not search though the comments to find that. — winny, Nov 29 '17 at 10:50

score 1 · Answer 1 · answered Nov 29 '17 at 18:23

1

There are two problems here: 1) The grounding situation you have is making the simulation unstable
2) Your constant power load is causing numerical instabilities

To fix problem 1) get rid of the 10M resistor and put the ground on the load side

To fix problem 2) look at the equation below, what happens when it gets near zero? \$ 1000/0 = \infty \$ and the solver is not going to like that, so its going to try and change the timestep but it still wont be able to solve the problem. You seemed to skirt around the problem by adding in a 10M resistor to ground on the load, but the circuit is still numerically unstable.

 B1 Vdc 0 I={1000/V(Vdc)}

How to solve this? Use a constant power load that the numerical simulation can handle:

Source: Linear

Here is the way I ran it, I used a constant power load with a B source and also moved the ground to the load:

answered Nov 29 '17 at 18:23

Voltage Spike

75,799
36
80
208

thanks for the reply. The reason I used the ground both on the grid side and the load side is that this givs me accurate measurements of my Vdc. In your simulation you can see that the Vdc oscillates between 126mV and 148mV. This cannot be the case, since Vdc is a rectified voltage with an ideal average value of 1.35*230*sqrt(3)=537.8V That is also the reason that Vdc never gets close to zero. Thus, I(B1) is never close to infinity. – Nikos Vandoros Nov 30 '17 at 11:12
Maybe I wasn't quite clear on that, it never reaches zero, but it has a hard time reaching zero becuase of numerical problems. To go into detail on this would require several pages of discussion. Regardless, you'll be better off using the other version of the current source. If this sovled your problem, please mark the answered box. – Voltage Spike Nov 30 '17 at 16:15

LTspice diode commutation

1 Answers1