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Step 6Performance Calculations
At this stage lets look at how we'd expect the torch to perform. Performance wasn't part of my goals, so this is for interests sake only. Of course if the performance of the torch were critical, these calc's would have been done as a first step to determine the feasability of the project, and the component values.
My discalimer: I'm a scructural, not electrical engineer, so I may have made multiple mistakes here. Please let me know if I did!
The energy stored in a charged capacitor is given by
E = 0.5 x C x V2
where C is the capacitance and V is the voltage
Here
E = 0.5 x 1 x 52
= 12.5 joule
1 joule is one watt of power for one second.
We've got 12.5 joules to play with, that would get us 1 watt for 12.5 seconds
Ignoring the internal resistance of the capacitor (not insignificant, in reality will affect the result), the LED/resistor is consuming power
P = V x I
= 5.0 x 0.03
= 0.15 watt
Therefore 12.5 joule will power 0.15 watt for 12.5/0.15 = 83.3333333 seconds.
Hmmm... not exactly inspiring stuff, and will get worse when you included the capacitor internal resistance... but OK to muck around with. Of course the LED will continue to give off light as the voltage fades, gaining a bit of time, but it won't be as bright. On the plus side, recharges are (virtually) free and quick.
AHA! But how quick?
The charging profile of a capacitor is generally calculated in terms of a "time constant". The capacitor will increase in charge by 63.8% every time constant. The attached images shows the cycles and came from here
eg. Assuming starting at zero, after one time constant, charge will be 63.8% of max, after two will be 63.2% + 63.2% of remaining 36.8% = 86.5% and so on, 3 time constants = 95%, 4 = 98%, 5 = 99+%.
the time constant is given by
s = R x C
Here the internal resistance of the capacitor is 30 ohms.
s = 30 x 1 = 30 seconds
So it will take 2.5 minutes or so to be pretty much fully charged... and for an 83.3 second use time thats a 35% duty cycle...
My discalimer: I'm a scructural, not electrical engineer, so I may have made multiple mistakes here. Please let me know if I did!
The energy stored in a charged capacitor is given by
E = 0.5 x C x V2
where C is the capacitance and V is the voltage
Here
E = 0.5 x 1 x 52
= 12.5 joule
1 joule is one watt of power for one second.
We've got 12.5 joules to play with, that would get us 1 watt for 12.5 seconds
Ignoring the internal resistance of the capacitor (not insignificant, in reality will affect the result), the LED/resistor is consuming power
P = V x I
= 5.0 x 0.03
= 0.15 watt
Therefore 12.5 joule will power 0.15 watt for 12.5/0.15 = 83.3333333 seconds.
Hmmm... not exactly inspiring stuff, and will get worse when you included the capacitor internal resistance... but OK to muck around with. Of course the LED will continue to give off light as the voltage fades, gaining a bit of time, but it won't be as bright. On the plus side, recharges are (virtually) free and quick.
AHA! But how quick?
The charging profile of a capacitor is generally calculated in terms of a "time constant". The capacitor will increase in charge by 63.8% every time constant. The attached images shows the cycles and came from here
eg. Assuming starting at zero, after one time constant, charge will be 63.8% of max, after two will be 63.2% + 63.2% of remaining 36.8% = 86.5% and so on, 3 time constants = 95%, 4 = 98%, 5 = 99+%.
the time constant is given by
s = R x C
Here the internal resistance of the capacitor is 30 ohms.
s = 30 x 1 = 30 seconds
So it will take 2.5 minutes or so to be pretty much fully charged... and for an 83.3 second use time thats a 35% duty cycle...
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Jul 12, 2008. 4:06 PMUnit042
says:
That's some heavy duty thinking there. How about adding a 0.47uF (47uf?) cap in parallel with the main storage cap to (significantly) decrease apparent internal resistance? (see solarbotics.net for more info on working with super caps)
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