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Comment by benjamincburns | Hacker News Reader

You'd actually need 9 of them. Capacitor networks don't behave like resistor networks. Placing caps in series increases their voltage tolerance but reduces their overall capacitance. You'd need three parallel sets of three of these caps in series to get the voltage and capacitance you're after.

Edits: It turns out that 3*3=9 these days, not 6.

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You can use a DC/DC to boost the voltage.

Sure. In fact you'd want to use a high-efficiency buck/boost converter to take advantage of the entire discharge curve. However you're only storing about 20% of the energy at 2.7V.

Edit: I'll show my work.

Farads are Joules (energy) per Volt (electric potential) squared. That means that the formula for energy stored (E) in a capacitor is E=CV^2.

A 1200F cap charged to 2.7V stores 8478 Joules of energy. A 1200F cap charged to 6V stores 43,200 Joules of energy. That means that at a charge of 2.7V there's ~80% less energy stored in the capacitor than at 6V. Incidentally this ratio holds no matter value of the capacitor.

The point is that you don't need to round up. 2.7 volts is pretty close to 3, so you need pretty close to 2x2=4. Bump it to five to compensate for the rounding and look it matches the actual math. Half of your initial suggestion.

Caps tend to explode when they fail, and high energy caps explode violently. You always, always, want to round down when spec'ing caps, not up.

But forget all that. Lets do it differently and fix our math toward the author's original comparison, the 1500mAh battery, not his weird, somewhat arbitrary choice, of a 6V-rated cap which drains to 2.5V after ~10 hours. And instead of talking about electricity, let's talk about what we're really after: energy.

A 1.5Ah battery stores about 22.68kJ of energy when fully charged to 4.2V (1.5Ah×3600C×4.2V=22,680J). From above, At 2.7V, a 1200F cap stores 8.478KJ of energy. 22.68kJ/8.478kJ = 2.675. Meaning we need at least 3 of these caps to meet or exceed the energy storage of a 1500mAh battery.

So we're both wrong in the context of the original comparison. And more importantly, so is the author.

Edit: This 3000F supercap rated for 2.7V would get us damn close to the 22.68kJ requirement, coming in at 21.87kJ of total storage: http://www.mouser.com/ProductDetail/Maxwell-Technologies/BCA...

Edit again:

I messed up again! The battery energy capacity is actually less than 22.68kJ, because the energy equation I used assumes that the voltage stays constant during the battery's discharge cycle. It doesn't. For a constant-current discharge of a LiPoly battery, the mean voltage is something like 3.4-3.6V (estimating from the discharge voltage curve, assuming discharge stops at 2.7V). So really we're talking something closer to 19kJ of energy stored within the battery. By the figures in her submission, her cap would weigh 262.6g and would occupy 30.8m^2 (no figures on thickness to calculate volume). Titanium density is 4.5g/cm^3. Assuming titanium dominates the density of her material, it'd be about 58cm^3 in volume - something like 10cm by 5.8cm by 1cm. If the density matched that of copper, it'd be about half that size.

Whoa, whoa, nowhere did I suggest applying extra voltage. I was saying you could use them at 2.7, or 2.6 if you want a safety margin. You don't have to use pure RLC and drop down to 2 volts.

It is amusing that the author screwed up.

>Sure. In fact you'd want to use a high-efficiency buck/boost converter to take advantage of the entire discharge curve. However you're only storing about 20% of the energy at 2.7V.

That doesn't have to be an issue. There are high efficiency joule thief IC's that'll take a voltage input around 0.3v.

At the cost of power which will increase your necessary total capacitance.

Digital circuits require some minimal voltage in order for their internal transistors to switch. This is because semiconductors only conduct in the presence of an electric field with a high enough potential to excite and free up the electrons hanging out near their junctions. That means that if your voltage drops below, say 1.8V (common for most low-power digital circuits these days), you won't be able to use your device any more.

There are some very efficient buck/boost converters out there. Like 95% and above efficient. Typically the cost of the power to use a buck/boost dc-dc converter is far outweighed by the ability to convert more of the stored charge into a potential that's suitable to operate the circuit.

Put differently, picture a barrel with a tap on it's wall somewhere near the bottom, and a different barrel with a conical-shaped bottom which tapers into a drain-like tap. Which is more useful? The latter is what you get from a good buck/boost converter.

I've learned something from your comments on this thread, and thanks for taking the time, but you've lost me with the barrels. Pressure depends on height of the water and is independent of the container shape.

benjamincburns was referring to a barrel with a tap "somewhere near the bottom". The point being that it's very difficult to get stuff out near the flat bottom of the barrel, especially if you can't move the barrel itself. The other option with buck/boost converters is if the barrel's bottom tapers into your tap, so that there's no way for any liquid to avoid gravity's call and leave through the tap.

That makes sense. I was over-thinking it.

Thanks for the compliment! As for the question, see what Elessar said.

Also:

In the typical water/electric circuit analogy the volume of water is synonymous to the amount of electrical energy stored (Joules). The pressure of the water is synonymous to the electric potential (Volts). Pipe diameter can be thought of as resistance (ohms), flow rate through a pipe (meters per second) can be thought of as current, and flow volume (cubic meters per second or litres per second) as power (watts).

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