Capacitor loop with no resistance

Question

I recently started using circuit simulation software that is warning me that I have a "capacitor loop with no resistance" and then refuses to run any simulations until I correct the problem. I can trigger this warning simply by creating a circuit with a battery and a capacitor tied together. I don't understand the prohibition of connecting a circuit this way as it does produce measurable effects in the capacitor.

Can anyone help me to understand what this warning may mean and why I should avoid circuits that are connected this way?

Answer 1

Guesstimate the resistances you would have in the real circuit, and insert resistors in the sim circuit to represent them.

A little more explanation of why the simulator reports an error without them may be useful : the instantaneous current when the battery and capacitor are connected is independent of the capacitance; it is purely the voltage divided by the resistances in series. Which, with ideal components, is the well known "divide by zero " error.

The capacitance will have some value of "ESR" = equivalent series resistance, which you may find from datasheets. (Some datasheets model ESR as loss, or "tan delta" instead) This page allows you to select capacitors by ESR; values from 0.003 ohms to 1.6 kilohm with 0.015 to a few ohms being common.

The battery will also have an ESR which will increase as the battery drains; again, see the datasheet. A fully charged NiCd or NiMh may have 0.1 ohm or more ESR, up to tens of ohms for a small button cell.

For the wiring or PCB traces, you can find the resistance in ohms per meter for approximately the right wire, (e.g. 0.1 ohm/m for 26SWG) and estimate the length of wire required (say 10cm, or 0.01 ohm).

Any one of these will break the loop and produce a circuit that can be simulated; modelling all of them will produce a more accurate simulation (may not change the results very much)

There may also be better simulation models for components such as capacitors, that allow you to set the ESR (and other parameters) and other features of real components.

Answer 2

There are two 'ideal' circuits simulators cannot simulate: shorted capacitors and open ended inductors.

Say, have a capacitor charged to 1V. Then it short-circuited, it could discharge with infinite current. Even if computer can operate with infinities, this could produce totally impractical result. So simulators simulators refuse to work with such circuits.

To avoid warning do not short-circuit capacitors, do not connect them directly to power source, do not connect them in parallel (replace them with one of summary capacitance). In case to simulate real circuit having ones, add resistances in series, as others answered.

Open ended inductor is similar. It could produce infinite voltage on its ends to discharge, which is also totally impractical. Avoid this also.

Answer 3

Learning to respect that all parts have ESR is the 1st to understanding how real parts work. The TL78L05 needs a cap with very low ESR to be stable but not zero. Then you get low ripple and near zero idle current in an LDO. The key is understanding the cap is a critical part of the design. Zero ESR in a cap means infinite current to a Voltage source (hence error) and zero ripple feedback to an LDO (unstable).

If you want to analog electroncs, get familar with the ESR in all common parts and understand why it is the most critical quality factor for most power designs.

Consider the effects and cost of 5,50,500, or 5000 milliOhm in simulation of transient loads.