Medium microfarad low voltage pulse capacitor - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 22, 2018, 07:25:31 am
content: Low uF pulse rated capacitor's are widely available of film type or film /foil type of Polypropylene dielectric.But I'm having difficulty in sourcing capacitance in 100uf to 500uF and voltage range around 16 to 35V, pulse rated capacitors.The one's available are photo flash type that are upto 330V or 450V, bulky in size, expensive and also have limited discharge cycles.What capacitor would be suitable for these, will low ESR electrolytic capacitors do the job?Thanks in advance!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 22, 2018, 07:32:11 am
content: Polymer is very much analogous to low-voltage film, in terms of ripple current and energy density. Dissipation factor or ESR are comparable to a crappy MKT made with absolutely minimal metallization, say.If you need more pulse current, stack up a lot of ceramic. It's going to be expensive, which is your hint that you should maybe try to look for a different way to solve this problem (say at a higher impedance, where film is fine).Incidentally, "pulse rated" means something very different for film versus photoflash. The former have low ESR for low losses, the latter have high ESR to limit short circuit current.Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 22, 2018, 09:33:47 am
content: Quote from: T3sl4co1l on September 22, 2018, 07:32:11 amI wonder if ceramic are available, at the capacitance that I'm looking at and connecting in parallel would yield 10's of capacitors which might not be feasable.Looks like I've walked into a dead end.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 22, 2018, 01:36:34 pm
content: The ceramic you will find in that range of values are multiple parts soldered onto a lead frame. Same as doing it yourself but somehow ten times more expensive...Large value ceramics tend to be unusually expensive and poorly stocked, too, so you might consider using even more, like a hundred pieces. Which would be a fair amount of board area, so it might be double sided, or spread onto two or more PCBs connected with headers, or something.Also, mind the C(V) curve, make sure you have enough of them to get the energy you require, or whatever.If you need energy in a small space, go for higher voltages, period. Few components are sized proportional to CV^2 (i.e., energy), tending more towards CV (charge) for whatever reason. The difference between, say, 50V and 500V is so dramatic, it is well worth adding two power converter stages (if they're small and efficient, which is quite possible with the availability of power GaN nowadays) just to take advantage of that.Or, like I said, rearrange the load so it can use the higher voltage directly. Say this were an LED strobe, you wouldn't run a single 100W "light engine" (30V 3A or whatever), you'd use the same number of dice in one long series string at 300V (probably having to use a lot of smaller LEDs since you can't find an all-series module?).Tim

author: drussell
author_rank: Super Contributor
timestamp: September 22, 2018, 01:45:07 pm
content: What kind of currents are we talking about here? What is your application?With some more information about what you're intending to do we can probably make some better recommendations.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 25, 2018, 07:37:07 am
content: Quote from: T3sl4co1l on September 22, 2018, 01:36:34 pmJust trying to wrap my head around that CV^2 vs CV statement. I generally tend to calculate the energy stored by a capacitor as 0.5 * C * V^2. So, is that not the practical way to go about things vs the CV equation you posted.So if a practical 1uF cap is charged to 100V how much would the energy stored in it. Do i use 0.5 * C * V^2 or use C * V.And incase I use C*V, do I use the Q * V to find the energy stored in the cap.How do I practically verify this that 0.5 * C * V^2 is not what the energy the cap stores?

author: drussell
author_rank: Super Contributor
timestamp: September 25, 2018, 08:29:29 am
content: Quote from: ZeroResistance on September 25, 2018, 07:37:07 amThechargeon the plates is \$CV\$Theenergystored in the cap is \${1 \over 2} {CV^2}\$ due to the resistance, regardless of how low or high the resistance is.I believe thatT3sl4co1lwas talking about theenergy densityof available capacitors and was trying to point out that it seems to him that forenergy density, the physical size of the cap seems to be more related to thechargethan theenergystored. (Energy per unit volume, density, rather than energy itself.)\$U = {energy \over volume} = \frac{1}{2}\epsilon _{0}E^{2}\$