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I am very confused about negative voltage in capacitor, it can even light a led as you can see in the circuit.

My questions are;

1-Why capacitor gives ~-2 V in its negative terminal when we tie its positive terminal to ground?

2-How can that LED work by connecting its positive to battery's negative pole and its negative to our negative voltage?

I have been searching this for long time, what i found about first question is this site's "Some Capacitor Theory", why capacitor acts like this?

May you please explain it as simple as you can, because i am interested with electronics as hobby, i really don't understand clearly when it is explained in formulas,mathematics etc

Circuit Image: Circuit Image

Thank you very much.

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    \$\begingroup\$ why does a 1.5V battery measure -1.5V at its negative terminal when you tie its positive terminal to ground? ..... please use the schematic editor to add the schematic to your post \$\endgroup\$ – jsotola Jul 27 at 18:01
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    \$\begingroup\$ Welcome to EE.SE. Please upload the schematic into your question so that we don't have to follow links to understand your question. Use the image upload button on the editor toolbar. \$\endgroup\$ – Transistor Jul 27 at 18:01
  • \$\begingroup\$ added picture, thanks \$\endgroup\$ – afmp Jul 27 at 18:10
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    \$\begingroup\$ It may be a little advanced but have a look at my answer to a related question. \$\endgroup\$ – Transistor Jul 27 at 18:11
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The effect you are seeing is used for ‘flying capacitor’ voltage converters. Basically, the capacitor is storing charge in one polarity, then re-connnected in a different polarity to discharge.

This same principle is used in a now-obsolete LED flasher chip called the LM3909. At the high level, the LM3909 resembles your circuit, except it does something clever: connects the LED across the (-) flying cap to (+) battery, doubling the voltage across the LED. Its ability to run on very low battery for long stretches of time is the stuff of legend.

BONUS: an analysis of the LM3909 here: https://cdn.hackaday.io/files/291791248394336/Discrete%20Version%20Of%20The%20LM3909%20Oscillator%20IC.pdf

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  • \$\begingroup\$ thank you very much! i think that's what i search for! :) \$\endgroup\$ – afmp Jul 27 at 19:09
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That's a very strange circuit there. But...

The first thing to realize is that voltage is always relative, that's why it's also called "potential difference". A voltmeter has two probes, to measure the difference in voltage between two points.

The other thing is that the choice of which part of the circuit you call "0V" is entirely arbitrary. If some part is connected to Earth, that's usually called 0V. Otherwise, the battery's negative terminal is usually 0V. But it doesn't have to be.

So if you charge up a capacitor to some voltage, and then connect the positive terminal of the capacitor to the point you call 0V, then the negative terminal must have a negative voltage. There's nothing deep and meaningful about that; it's all down to which part of the circuit you called 0V.

The LED doesn't care which part of the circuit you called 0V. All it knows is that one of its terminals is more positive than the other. Provided that the voltage across it is the right way round, it will conduct.

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  • \$\begingroup\$ Thank you very much for your helpful explanation, really appreciated, i just wonder why capacitor gives as negative voltage as you said, because i have never seen a rule or something while studying capacitors is it "persistent" to changes or something? \$\endgroup\$ – afmp Jul 27 at 18:03
  • \$\begingroup\$ @afmp One of the main purposes of a capacitor is to store charge. You charge it up, and it stores that charge until you use it. A bit like a rechargeable battery, but a capacitor can charge much faster, and holds much less charge than a battery. \$\endgroup\$ – Simon B Jul 27 at 18:09
  • \$\begingroup\$ Yes, the voltage across a capacitor is 'persistent', until a current flows into / out of the capacitor which changes the voltage, aka changes the charge on the capacitor. \$\endgroup\$ – Neil_UK Jul 27 at 18:11
  • \$\begingroup\$ Persistent, yes. If you charge a capacitor so that there is some voltage over it, this voltage is over the capacitor unless it is charged more or discharged to some other voltage. \$\endgroup\$ – Justme Jul 27 at 18:11
  • \$\begingroup\$ @Neil_UK,@Justme is it okay to conclude this question as: when we apply 0V to charged capacitor, it immediately converts the positive charge to negative charge and it releases it from its negative side as negative voltage? \$\endgroup\$ – afmp Jul 27 at 18:35

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