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I'm simulating 4 different circuits, they are all power supplies, with loads, filter capacitors, rectifiers made by four 1N4007s, and transformers fed from 120VAC: enter image description here enter image description here enter image description here enter image description here They all have the same problem, which is that the voltage signal at the input of the rectifier has a strange clipping. It took me a while to figure out where it was coming from, I tried several things until I removed the series resistances from the inductor models themselves and the clipping went away, I then added the resistances as resistors in series (R1 and R2) but the clipping came back. The green plot is the voltage measured across L2 alone (series resistance left blank), the pink clipped plot is measured across R2+L2, the inductor plus it's wire resistance: enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here

It makes sense that there is voltage loss in the wire resistance, but shouldn't it look like a sine wave but slightly scaled down instead of just a hard clean clipping at the peaks?

When I disconnect the circuits from the secondaries and connect R2+L2 directly to just R3 to simulate an open transformer measurement, the pink plot is no longer clipped and matches the green plot perfectly, like this:

enter image description here enter image description here

Graph for the 4th circuit, with current plots for D1 and C1, no clipping ¿?

enter image description here

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  • \$\begingroup\$ It would be much more readable by other people if the schematic flowed from left to right. If you plot the current in the diodes and the charging current for the input capacitor you will see why the clipping is there. \$\endgroup\$ – Kevin White Dec 4 '19 at 19:47
  • \$\begingroup\$ Yes, sorry about the orientation, that's how I drew them in my schematic (image), so it's easier for me to compare and follow what I did. I plotted the currents you said but they don't have any visible clipping, I added the pictures to my question \$\endgroup\$ – Raz Dec 4 '19 at 19:53
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    \$\begingroup\$ I wouldn't expect the current to show clipping but it will only flow at the time the source voltage is being clipped because the voltage has to be greater than the voltage on the filter capacitor. That current pulse causes a drop across the resistance of the source (the inductor in this case). \$\endgroup\$ – Kevin White Dec 4 '19 at 19:55
  • \$\begingroup\$ I see, so the capacitor draws current whenever the diode conducts, and that causes a voltage drop across the wire resistance, but because the current draw happens as a short pulse, the drop can only take place for the duration of the pulse, which is during the peaks. \$\endgroup\$ – Raz Dec 4 '19 at 20:06
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    \$\begingroup\$ @Raz What they said. Search "diode conduction angle" and "spreading resistors". || If you draw circuits left to right people will not talk to you at parties or invite you home for dinner :-). ie it bends people's brains badly. || In situations such as you have using a simple load to start will avoid complxities which MAY be caused by load interactions. When you have a C filter cap and R load the cap charges on the transformer peaks then discharges somewhat during the rest of the cycle. In the real world you get bursts of current and can get severe RF interference from the diodes (!). ... \$\endgroup\$ – Russell McMahon Dec 4 '19 at 20:14

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