[Rack] Electric door strike working again. Use your Pin or key to get in.
h.zeller at acm.org
Mon Jul 15 03:56:54 UTC 2013
On 14 July 2013 20:41, Jake <jake at spaz.org> wrote:
>> The motor Scotty found takes about 300mA at 6V when free-running (so
>> for no-load 'return', maybe 500mA with load). It runs about 1 second
>> for the amount of time we need. That is about 1.8 Joules.
>> Say we can charge up a capacitor to 18 V (peak voltage of a bit more
>> than 12 V AC) and drain it down to 4 V (when this particular motor
>> just works), then we'd need about 20'000 microfarad ( 2 * 1.8 joules /
>> ((18 - 4) volt)^2 ). The calculation would be similar for other
>> motors found in the pile.
> there are at least ten 1000 microfarad capacitors rated at 50 volts in the
> pile, and i can certainly find more. But i think the math is more
> complicated. But since it's based on voltage squared, one 50 volt 1000 uF
> capacitor is the same energy as a 20,000 uF capacitor rated at 11.1 volts.
It is not that rating, that is important, but the voltage difference
you end up with. The energy is calculated from the difference of the
charged voltage minus the end voltage.
If you charge up the capacitor to 50 volt, then you need less, but you
might fry a 6volt motor (and you might not get that from a 12V AC
supply; that is why I calculated with the 18V max (about the peak
voltage in an 13V AC)
The rating on the capacitor is only telling you when it starts to
explode if you give it more voltage :)
> anyway that motor sounds hella weak. The motor i already mounted on the
> door is rated 24 volts and probably can take a good amp or two at least,
> which is a lot more power.
Yep, the motor is rated 24V volt, 330mA, so it has 4x more power, and
even more if you drive it at 36. But then, we need to drive the buzzer
with a different voltage.
With the right gearing/pulley size, this should easily do it (if it is
1 inch pull@ 5 newton, then we only need 0.13 Joule mechanical power).
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