2 ms·

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

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.

Dylan16807 13y ago
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.
- benjamincburns 13y ago
  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/BCAP3000-P270-K04/?qs=sGAEpiMZZMuDCPMZUZ%252bYl4vDGR228uRDhgVZbUDv3h0%3d 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.
  - Dylan16807 13y ago
    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.
s_baby 13y ago
>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.

Dylan16807 13y ago

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.

benjamincburns 13y ago
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/BCAP3000-P270-K04/?qs=sGAEpiMZZMuDCPMZUZ%252bYl4vDGR228uRDhgVZbUDv3h0%3d 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.
- Dylan16807 13y ago
  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.

s_baby 13y ago

>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.