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In electronics I have encountered the topic of filters in power supply. One of the concepts mentioned is about the bleeder resistor, where a resistor is connected in parallel to a filter so that the capacitor is able to discharge any leftover charge when the equipment is turned off.

What really questions me is that although I have researched a lot, there is no source that says that there is ALWAYS a leftover charge retained in a capacitor when an equipment is turned off or that a bleeder resistor is always necessary when using a filter. Yes, I know that they are important for safety precautions. I just want to know, is this always the case? Will there always be a leftover charge? Will someone always get a shock or be in a dangerous situation when handling something involving a filter without a bleeder resistor? I want to know if it is something that is absolutely necessary and not it will increase the chances of being safe because I already know that it does.

Sorry if the question sounds stupid, but I am not an expert and just want some sort of precision with this idea explained by someone who knows the concepts really well.

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  • \$\begingroup\$ You don't need a bleed resistor if the circuit itself will drain down the caps fast enough on its own. It's also a good thing to have in a low voltage digital circuit... Not a safety issue, rather if the power is cycled by the user and the caps dont fully drain fast enough, you can get an incomplete reset. In our battery operated products, we generally put a 1k to 10k resistor on the power switch that will short the caps to ground when user slides it to 'off'. \$\endgroup\$
    – Kyle B
    Sep 8, 2020 at 4:10

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There is. Even a capacitor that has been immediately discharged will, over time release more charge from the dieelectric causing it to rise in voltage unless it is bled out. The effect is called dielectric absorption and for a capacitor operating at high voltage capacitors this can be significant.

I'm not sure if all types of capacitors do this, but the big electrolytic types do.

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  • \$\begingroup\$ Interesting phenomenon I was not aware of. Are you sure the voltage rise is due to absorption/soaking and not "dielectric decompression"? (plates mechanically decompress after supply is removed).. I'm not 100% sure myself, but here's more keith-snook.info/capacitor-soakage.html \$\endgroup\$
    – P2000
    Sep 8, 2020 at 4:48
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    \$\begingroup\$ @P2000 I'm not a physicist so I don't know. Dieelecric absorption is just what it is called. I don't actually know what it is. If it was a mechanical thing, wouldn't the charged plates trying to be pushing each other apart and coming back together after wards? I feel it's not a mechanical thing though but I have no idea. Wiki seems to imply it is a dielectric thing and is related to how caps are polarized so is only applicable to polarized caps. Sounds very similar to residual magnetism to be honest. \$\endgroup\$
    – DKNguyen
    Sep 8, 2020 at 5:31
  • \$\begingroup\$ @P2000 Oh wait, the plates would be pulled together when charged due to opposite polarity. For some reason I was thinking identical charge on the plates. \$\endgroup\$
    – DKNguyen
    Sep 8, 2020 at 20:06
  • \$\begingroup\$ Yes, and since C=Q/V and C=epsA/d, a charged cap has voltage V~Qd. As the dielectric relaxes, d increases and so V rises. I'm not sure whether that just partially makes up for the soaking drop, or whether it can peak above it. \$\endgroup\$
    – P2000
    Sep 8, 2020 at 20:14

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