Challenge is...
1 ~ 8bit processor (eg any 8bit cpu attiny, atmegaxxx)
1 fully charged 1000uf. (Could be energy pumped from small solar cells)
Then what can be "done" with that energy storage, eg could it be run for minutes? Even hours?!, could it sleep for possibly a day or longer using sleep techniques / hacks... to allow even more energy efficient devices.
1000uF, rated at 630V, would last a long time:
q=CV
Ic = dq/dt = C dV/dt
dt = C dV / Ic
T = ? dt = (C/Ic) ? dV
Roughly [unless my math is off, somebody check it :-)], for a 50mA load on a 1000uF cap charged to 630V,
T = (1000*10^-6 F / 0.05A) * 630V = 12.6 sec.
OTOH, if load is reduced to 50uA, and charging from solar cells at 12V,
T = (1000*10^-6 F / 0.00005A) * 12V = 240 sec.
For an ATTiny in powerdown mode (not very useful), you could get about 15 minutes from a 1mF capacitor charged to +5v before the processor stops from undervoltage.
With it doing anything more useful than being a 1MOhm resistor, you are looking at far far less time.
Now if you have a 1F supercap charged to +5v, now you are talking.
oric_dan:
q=CV
Ic = dq/dt = C dV/dt
dt = C dV / Ic
T = ? dt = (C/Ic) ? dVRoughly [unless my math is off, somebody check it :-)], for a 50mA load on a 1000uF cap charged to 630V,
T = (1000*10^-6 F / 0.05A) * 630V = 12.6 sec.
You are assuming that a linear regulator is used to reduce the 630V to 5V for the ATtiny. If instead you use a switching regulator:
E = 1/2 C * V^2 = 0.5 * 1000*10^-6 * 630^2 = 198.45 joules
Suppose the switching regulator is on average 75% efficient over the input voltage range 0 to 630V. Then there are about 150 joules available to power the ATtiny.
5V @ 50mA = 250mW
150 joules used up @ 250mW = 600 seconds.
oric_dan:
q=CV
Ic = dq/dt = C dV/dt
dt = C dV / Ic
T = ? dt = (C/Ic) ? dVRoughly [unless my math is off, somebody check it :-)], for a 50mA load on a 1000uF cap charged to 630V,
T = (1000*10^-6 F / 0.05A) * 630V = 12.6 sec.
You are assuming that a linear regulator is used to reduce the 630V to 5V for the ATtiny. If instead you use a switching regulator:
E = 1/2 C * V^2 = 0.5 * 1000*10^-6 * 630^2 = 198.45 joules
Suppose the switching regulator is on average 75% efficient over the input voltage range 0 to 630V, when loaded with 50mA @ 5V. Then there are about 150 joules available to power the ATtiny.
5V @ 50mA = 250mW
150 joules used up @ 250mW = 600 seconds.
So is there a variant of the 7805 which accepts a 630V input? 
Right, I wasn't thinking about the multiplicative-effect on current [actually inverse-effect] of having an efficient stepdown ckt. If 100% efficient, then the actual "constant" discharge current from the cap would be only 50mA * (5V/630V) = 0.397 mA, or the discharge time would be 12.6 * (630V/5V) = 1588 sec. [hmm, missing a factor of 2 yet, duh].
A more realistic example for Arduino, consider "fully charged" to mean ~10V at the input to the Arduino voltage regulator, while the board is consuming about 50mA, and that it will stop working when the input cap reaches about 5V (regulator output ~2V at that point), you an use the equation from How do I calculate how fast a capacitor will discharge? - Electrical Engineering Stack Exchange
t = C*(Vo-V)/I
= .001 * (10-5) / .05
= 0.1 s
westfw:
A more realistic example for Arduino, consider "fully charged" to mean ~10V at the input to the Arduino voltage regulator, while the board is consuming about 50mA, and that it will stop working when the input cap reaches about 5V (regulator output ~2V at that point), you an use the equation from How do I calculate how fast a capacitor will discharge? - Electrical Engineering Stack Exchange
t = C*(Vo-V)/I
= .001 * (10-5) / .05
= 0.1 s
That's why I mentioned earlier, with the cap charged to 12v from solar cells, you'd get a few minutes runtime if you drew only 50uA load current. All in all, the whole thing is fairly marginal.
And these are the considerations for developing an electric car using supercapacitors - you need capacitors of many Farads to operate at thousands of volts (using of course, ceramic microstructure dielectric) and inherently needing to use switchmode converters to adapt this to the working voltage of the motors and for recharging.
If you allow for a five times ratio of operating voltage, say charging to 2 kV and discharging to 400V, you can use 96% of the energy stored (except of course, for losses).
considerations for developing an electric car
It all boils down to an easily measured data-sheet value for "Energy Density", which can be expressed in WattHours per kilogram or WattHours per liter. Capacitors have (at least theoretically) certain advantages WRT charge/discharge rates and number of cycles, but those are pretty irrelevant if the energy density isn't there. Here's a quote from the wikipedia article on supercapacitors:
As of 2013 the commercial available effective energy densities of supercapacitors range from around 0.5 to 15 Wh/kg. For comparison, an aluminum electrolytic capacitor stores typically 0,01 to 0.3 Wh/kg while a conventional lead-acid battery stores typically 30 to 40 Wh/kg and modern lithium-ion batteries about 100 to 265 Wh/kg. That means, supercapacitors can store 10 to 100 times more energy than electrolytic capacitors but only one tenth of batteries.
Which is extremely promising.
If commercially available energy densities of (carbon) supercapacitors are only one tenth that of modern lithium-ion batteries which have little promise of further development (even when replaced by much cheaper sodium-ion technology), we should have modest hope of supercapacitor dominance in the next ten years at least (Moore's Law).