Storing power in CAPs may not be as bad an idea as you think.

The magic formula is:

1 farard = 1 amp-second

You need 3 amps for 0.2 second or 0.6 amp-seconds or 0.6 Farad

Now, I am not going to claim that 0.6 Farad is a small number, but it isn't 
all that huge either.

Looking at the DigiKey website, I find a 9 Farad, 28 Volt supercap with an 
equivalent series resistance of 60 mOhm (this means it only 'loses' 180 
mVolt internally when discharging at 3A). The part number is 
B48611A5903Q012. Unfortunately it is neither in stock, nor is a price 
listed.

A more realistic choice might be:

EMHSR-0002C5-005R4 (589-1008-ND), which is a 2.5 Farad, 5.4V CAP with an ESR 
of 130 mOhm. Five of these in series (with appropriate balancing resistors) 
is the same as 0.5 Farad at 27 volts and an ESR of 650 mOhm. At a three amp 
draw you will have about 2V or drop inside the caps.

This item is priced at Digikey at

6.06 each,
5.55 in 10's
4.54 in 100's
2.93 in 1000's

Yet another choice:

Store the energy in a lower voltage supercap and convert it to 24V when you 
need it.

Assuming a 2.7 Volt device, you will need:

24V / 2.7V * 0.6 Farards = 5.3 Farads to store the same energy.
Of course the current draw will be 24V/2.7V * 3A = 27A (assuming a perfectly 
efficient converter)

This is a bit on the high side for current and rather low on voltage (tough 
to make a converter from such a low voltage that is in any way efficient)

So, what if we used two caps in series to get 5.4V at half the capacitance? 
A particular device to look at is:
ESHSR-0025C0-002R7, which is 25 F at 2.7 V with an DC ESR ( at 2.7A) or 30 
mOhm ($4.28 in 100's at Digikey).

This gives us a total energy storage of 50F * 2.7V, which is the same as 
5.6F at 24V, or more than 9 times the 0.6 F required. We have to use such a 
high capacity device to get a high enough current rating (as opposed to 
storage capacity).

Putting the two caps in series gives us 12.5 F at 5.4V and 60 mOhm ESR. A 
100% efficient up-converter would need to run at about 13.5A which would 
result in a voltage loss in the caps of 0.060 ohm * 13.5A = 1.00 V, which is 
high, but not impossibly so. A real up-coonverter, designed to keep costs in 
line, might pull nearly 20A.

Going forward with the 20A assumption, we get an internal loss of 0.06 * 
20.0 = 1.2V. This of course means we have an internal _power_ loss of  1.2V 
* 20.0A = 24W going on here. Assuming a 0.2 second discharge each pulse will 
generate 24 W*0.2 seconds = 24 Ws of heat. In addition, we have to charge 
the capacitor back up, which will take another 24Ws. This gives us 48Ws per 
pulse of heat, and this sounds pretty scary. However, if the repetition rate 
is reasonable, then this isn't so bad. I one pulse per minute, you would 
have 48Ws/60 seconds or 0.8 W of dissipation total (0.4 W per device). This 
seems reasonable.

The most amazing thing: these things are TINY, just 25mm high and 16mm in 
diameter!

Bob Ammerman
RAm Systems



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