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For some reason, every time that I include a capacitor in this circuit LTSpice transient analysis is very jumpy.

Is there a better way to use the output of a bridge rectifier for this purpose?

I tried putting a resistor in parallel with the capacitor instead, but I was getting a singular matrix error.

Additionally, once the simulation gets to a few microseconds, the transient analysis starts to take forever and effectively stops the simulation.

Edit: The voltage with the blue trace is the voltage across the capacitor.

Any thoughts?enter image description here

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    \$\begingroup\$ It's probably not helping matters that you seem to just be using the default diode model, but that's likely not the cause of this issue. Still, I'd suggest changing them to an actual diode model, perhaps 1N4007s or SS14s. \$\endgroup\$
    – Hearth
    Nov 4, 2023 at 20:12

3 Answers 3

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You are not measuring the voltage across the capacitor as the ground is on the wrong node. The output node may be floating (depends on the model associated with "D") which will cause problems and probably why the simulation is having problems converging. Use a real diode model. Due to the 1 kHz frequency, a Schottky diode should be used.

Moving the ground node to the bottom of the capacitor will solve your issues as shown below. A 100 uF capacitor was used to speed up the simulation.

enter image description here

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  • \$\begingroup\$ I added the ground to the node underneath the capacitor and that fixed the problem. This is why I love these forms- Thank you for the answers, these were both very informative and I appreciate the time. \$\endgroup\$ Nov 4, 2023 at 20:28
  • \$\begingroup\$ Hi! I'm a total newbie and most of this electronics stuff flies over my head, but this one is simple enogh for me to understand. Why does it matter where you attach a ground, if the rest of the circuit seems to be completely detached from it? In fact, why do it at all? Or is it somehow implied that the power source is also connected to it, like maybe it's a connection to the power grid or a grounded diesel generator? \$\endgroup\$
    – Vilx-
    Nov 5, 2023 at 7:29
  • \$\begingroup\$ Also, why does the "out" line not have a return line? How can that work? \$\endgroup\$
    – Vilx-
    Nov 5, 2023 at 7:30
  • \$\begingroup\$ @Vilx- the ground icon in LTspice has only one purpose: It sets the potential of the attached node as 0 V. \$\endgroup\$
    – tobalt
    Nov 6, 2023 at 15:31
  • \$\begingroup\$ So the behavior in real life would be different? \$\endgroup\$
    – Vilx-
    Nov 6, 2023 at 18:11
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You should measure the voltage across the capacitor rather than the capacitor relative to one side of the AC input. Also, it's bad practice to use automatically named nodes- always name them yourself. When you edit the schematic, the node numbers already present may change, which can cause you to plot the wrong voltage.

I would also suggest always using a real component model, rather than the default. The default may be very odd compared to a real discrete component- for example it could be for a tiny structure on an IC chip.

enter image description here

Above is a 1 second simulation of the circuit with the values you used and 1N4148 diodes. The time constant of 100 ohms and 11,000uF is 1.1 second so it will only reach a fraction of the final voltage in a second.

It's possible to leave the input voltage grounded and plot the difference between the two sides of the capacitor but I would suggest always grounding the most logical part of the circuit. Sometimes there are small anomalies from taking a difference.

In the case of the above circuit, if I wanted to plot the AC input, I could add a trace that plots V(Vx)-V(Vy) because I've named those nodes.

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Assume capacitor is charged to Vp. During the period input sine wave is not at peak voltage neither diode conducts. This means both input lines are virtually open circuit looking from the capacitor pins. An open circuit reference used as gnd level is not applicable. To be able to read something meaningful with your circuit you can add a high value resistor parallel to one of the diodes connected to your reference point, but this has no practical use. In one of my projects I had to detect sync signal, setting reference point to negative line of capacitor and parallel resistor on one of the diodes, just the same as yours but connection is reversed.

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