Completely discharge a capacitor on an active load - Page 1
Forum section: electronics / beginners

author: raff5184
author_rank: Regular Contributor
timestamp: January 30, 2020, 05:11:24 pm
content: Hi,is it possible,only using passive componentsor components that absorb a minimum amount of energy, to completely discharge a capacitor on an active load that needs a minimum voltage of say 2V to operate?For example, if the capacitor has a voltage rating of 5V, can I use only the energy stored when the voltage across the capacitor is between 2V and 5V (leaving the energy from 0V to 2V unused) or can I do better?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 30, 2020, 05:25:50 pm
content: Sure, use nothing. Let the "active load" discharge the capacitor as it will. Presumably the load will continue draining it below 2V, so I don't really know how anything should go "unused"...If you have more specific scales of time and current in mind here, in terms of what counts as "absorb", "minimum energy", "completely", "discharge", "minimum voltage" and so on, a more useful answer may be possible.Tim

author: raff5184
author_rank: Regular Contributor
timestamp: January 30, 2020, 05:47:08 pm
content: Thanks.I think that I failed to mention that the active load (for example a PCB) has to keep working even when the voltage across the capacitor goes below the minimum voltage (2V) of the active load. What I see is that the capacitor gets drained but the active load does not work because the voltage is not enough. Is there something to "push up" that remaining charge in the capacitor so that the load can actually use it?In terms of "absorption" I was thinking of something passive (not an amplifier or similar) because I don't have a supply (e.g. a battery) that absorbs currents in the order of few microAmps. My only supply is the capacitor

author: Mr Evil
author_rank: Regular Contributor
timestamp: January 30, 2020, 07:42:20 pm
content: You can't do that with passive components, but it's possible for a buck-boost converter to do it, powered from the capacitor itself. For instance theLTC3119could output a constant 2V from the initial 5V input all the way down to 0.25V.Is it really worth it though? Once the capacitor has drained from 5V to 2V, it has only 16% of its energy left.

author: Berni
author_rank: Super Contributor
timestamp: January 30, 2020, 07:48:37 pm
content: Quote from: raff5184 on January 30, 2020, 05:11:24 pmYep put a large value resistor across said capacitor. Its a passive component that absorbs a minimum amount of energy (high resistance value) and operates down to 0VBut yeah all electronic loads have a minimum voltage. In practice this issue is solved by introducing a additonal power supply in series to effectively bring the ground terminal of the electronic load down to a negative voltage below ground. So it doesn't matter if the load can only drop a minimum of 2V if the load terminal is held at -3V for example. But yes this is an active component. It must be active because the load dropping extra voltage causes extra power to be absorbed by it. This power has to come from somewhere, hence a power supply to provide it.

author: raff5184
author_rank: Regular Contributor
timestamp: January 30, 2020, 08:50:48 pm
content: Quote from: Mr Evil on January 30, 2020, 07:42:20 pmYes this is what I need, powering from the capacitor is fine too, but if it worth or not well it depends on how much it absorbs compared to my circuit (I still don't know).Quote from: Berni on January 30, 2020, 07:48:37 pmI thought about doing something like this. But I don't know how to choose the additional power supply, if it can be another capacitor, how much smaller than the other it can be. Also I guess at that point I need a cell voltage balance circuit, right?

author: Berni
author_rank: Super Contributor
timestamp: January 31, 2020, 06:06:38 am
content: Quote from: raff5184 on January 30, 2020, 08:50:48 pmIt can be any power supply that can produce at least -2V and supply as much current as the electronic load will consume (Its in series so all the current has to pass trough the supply). It could also be a capacitor provided it has enough capacitance to do the job.I think its better you explain the whole story and what this load is for. Often the solution can be made a lot simpler by taking a different approach and forum members can only suggest one when they have the big picture of what you are actually trying to do.