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I was running the following LTSpice simulation to see if a capacitor would divide a DC signal,

enter image description here

this results in the following

enter image description here

As you can see, it works fine and obeys the cap divider formula of C1/C1+C2. Now, for this to work, I had to change the simulation settings so that "all external DC sources start at 0V".

So essentially, I had made an AC signal out of it hence why this worked. But let's say I changed the 1.8V signal to a 3V signal, surely the capacitive divider will still work but it sees it as an AC change.

So my question is, why do people say that a capacitive divider doesn't work for DC? Can't you simply use a switch or something so that you still use a DC voltage source, it's just the capacitors see it as an AC change?

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You are using an idealized capacitor model. This is the simplest model for a real-world capacitor that is accurate for the problem you are addressing with your simulation.

Keep in mind, when doing simulations, that it is up to you to determine that your model is accurate enough. The real world is for all intents and purposes infinitely complex, simulation models are of necessity finite, and the tool doesn't know what you're trying to find out.

schematic

simulate this circuit – Schematic created using CircuitLab

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If you had an osciloscope with a high input impedance you could probably see this behaviour in a real circuit. You'll not be able to draw any current from the divider, as the capacitors will reach equilibrium after releasing their charges.

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Real capacitors have leakage current. The voltage divides across the capacitors depending on their relative leakage, not on their relative capacitance. For a true dc situation the capacitance is irrelevant.

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You applied a Step Input which has a broad continuous spectrum, thus the C divider is very effective, initially. But over time, unlike a battery in which many orders of magnitude higher capacitance the RC decay time constant ratio will likely not be the same as the C ratio so be sustained.

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