Highest energy cap that's available in hobby volumes .. maybe a few hundred pieces

What's the highest energy cap available and it's source?

The most cost effective?

where energy is the product C * V^2

Reply to
fpga_toys
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** Groper alert !

** How about 1215 joules for 113 UK pounds ?

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....... Phil

Reply to
Phil Allison

1/2 C * V^2 in fact.

Storing energy at high voltage is best since a caps's size is determined by the CV product wheras energy is prop to V^2.

Graham

Reply to
Eeyore

Maxwell has BCAP1200-E270 which is 1200 farads at 2.7 VDC, for about $45.40 in quantity of 1 to 99. That's 4374 joules. A bank of 27 in series will provide 33 farads at 72 VDC, which is enough energy to absorb (or supply)

13 kW for 5 seconds (for EV acceleration and braking) at a cost of $1226. Still not a replacement for batteries, but perfect for high current surges. These ultracaps (BoostCaps) have an ESR of only 0.8 uOhms and peak current of 3750 amps. They are about 3" long and 2.5" dia, and weigh about 3/4 lb. I received this quote recently.

Check

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for more info. They have 650 to 2600 farads in this series.

Paul

Reply to
Paul E. Schoen

i agree with you

Reply to
wittyakash

I think the volume of the required dielectric is proportional to the maximum energy stored, assuming a constant volts per mil dielectric breakdown rating.

Best regards, Spehro Pefhany

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Reply to
Spehro Pefhany

the

Surely the thickness of the dielectric would be prop to voltage ? Hence size > CV whilst energy > 1/2 C*V^2.

This certainly what I've found.

Graham

Reply to
Eeyore

the

Yes, let's say. But capacitance is prop to 1/thickness.

Okay (parallel plates, ignore edge effects)

capacitance = e * A/t where a is the area and t is the distance between plates and e is the permittivity.

t = k * V, where V is the maximum operating voltage and 1/k is the dielectric breakdown voltage in volts/unit thickness

(Max) Energy = (1/2) * V^2 * e * A/t = (1/2) * V * e* A/k

we also have dielectric Volume = t * A = k * V * A

so (Max) Energy/Volume = e/(2*k^2), a constant dependent only on dielectric characteristics

I also think this makes sense intuitively.

Best regards, Spehro Pefhany

--
"it\'s the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
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Reply to
Spehro Pefhany

the

The size proportional to CV is sort of true in electrolytics, but obviously not in film caps. Only a small part of a 'lytic is the actual dielectric, and most of it is carrier foil and electrolyte/spacer.

John

Reply to
John Larkin

As a minor nit ... the 1/2 is necessary only to scale the product into joules units, the base equation is the fundamental equation for capacitance and voltage representing energy.

Reply to
fpga_toys

Reply to
fpga_toys

But you did post to Usenit.

John

Reply to
John Larkin

I sent the Maxwell sales link a request for price and availabilty in hundreds a little over a month ago, and never got a response. Did a distributor respond with that quote?

I've since been looking at surplus high voltage caps in the fractional farad range that are also kilojoule, but are about the same cost and volume. The only advantage is that they can be connected in parallel, and not have to worry about the balancing voltage divider.

Reply to
fpga_toys

Methinks the "minor nit" is your brain capacity ;-)

The 1/2 comes from Integral(C*V)dV

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
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Reply to
Jim Thompson

A big array of series/parallel electrolytics, like photoflash caps, doesn't need equalizing resistors. Electrolytic leakage is nonlinear enough to balance the voltages nicely. 'lytics don't fail suddenly like film caps, they just start leaking more.

Check the surplus houses for electrolytic, film, and oil-filled caps. Cost per joule will be a fraction of new stuff. Around a few kilojoules, things start to get interesting.

John

Reply to
John Larkin

by the

Sure, I agree. Here are a few from the same Nichicon series:

Voltage Cap Diam Height Energy Volume Energy/Volume

400 22 16 25 1.76 5.02624 0.350162348 35 1000 12.5 20 0.6125 2.45421875 0.249570255 25 1000 10 20 0.3125 1.5707 0.19895588 16 2200 12.5 20 0.2816 2.45421875 0.114741198

Of course, on power supplies, going to a universal (worldwide) input off-line switcher means you need to squander a lot of the energy storage capacity of your primary energy storage capacitor (at minimum input voltage and frequency), so power supply caps may end up not being that different in size from that required for a linear supply.

Best regards, Spehro Pefhany

--
"it\'s the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com
Reply to
Spehro Pefhany

Surely you must have some other constraints, such as minimum or maximum voltage, capacitance, size, weight, PCB'ness, discharge rate? All significant.

Reply to
Ancient_Hacker

I received the quote from Frank Dodaro mailto: snipped-for-privacy@maxwell.com

Generally I have only seen surplus capacitors about 5000 uF at 400 VDC,

which are 400 Joules each. They are about 3" dia and 6" long. I paid about $5 each for them. So the new Maxwell parts are about the same price per

Joule, and are much smaller by a factor 10 or so. The voltage divider is a problem. It would probably be a good idea for the manufacturer to have an optional built-in bleeder resistor for this purpose.

Paul Sent via Google because my news server sucks

Reply to
Paul

Oh, joy, a voice of reason. All the old ARRL Handbooks and many other books specify that you should put equalizing resistors across electrolytics and diodes too. With a little reflection you can convince yourself pretty thoroughly that this is a poor idea. Electrolytics and diodes are kinda like zeners- they'll self-klimit the voltage across themselves to a value they can handle. If you add equalizing resistors you're just forcing the undervoltage-formed capacitors to dissipate more power.

Reply to
Ancient_Hacker

Nope ...l didn't specify units, the units less the 1/2 constant are JimFart units, where two JimFars are a Jewel :)

err ... joule ...

Reply to
fpga_toys

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