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\f0\b\fs24 \cf0 Display:\
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\b0 60 divisions (1 for each second)\
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Each division has 5 segments (300 segments per revolution)\
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Suppose the clock spins at 3000 RPM\
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50 revolutions per second\
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20,000 microseconds per revolution\
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15,000 segments per second\
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66.666666666667 microseconds per segment\
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0.018 degrees per microsecond\
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1066.66666666666667 CPU clock cycles per segment\
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The display is split up in to seconds, each with 5 subdivisions
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(segments). That's a total of 300 segments per revolution.
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If the propeller spins at 2000RPM, that's 33.3 revolutions per second,
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or 30ms (30,000μs) per revolution. That means we'll be drawing 10,000
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segments per second, which is 100μs per segment. With a clock speed
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of 16MHz, this is 1600 cycles per segment, which is plenty.
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The diode (D14) across the fan's power connections is there because if
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the power acrss the fan breaks (due to the unreliable nature of the
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brushes), the motor in the fan has coils, which act like an inductor
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and will produce a back EMF (a huge negative voltage across the power
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connections) as the magnetic field collapses. This won't be good for
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the arduino and could cause sparks on the brushes. The diode simply
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shorts the negative voltage.
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The capacitor (C1) and resistor (R14) are there to smooth the power
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supply from the unreliable brushes. The capacitor would discharge
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fairly slowly (due to the resistance of the circuit), but will charge
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very quickly. Potentially, it will charge so quickly that it'll pull
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too much current from the power supply (i.e., short the power supply
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and trip it). So the resistor limits this. Unfortunately, the
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resistor will also have a potentiometer effect (with the resistence of
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the main circuit). 10Ω was chosen as a value due to these rough
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workings: Lets say the arduino circuit takes 500mA. If we aim to lose
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1V across the resistor, that's 1V / 0.5A = 2Ω (from V=IR). The 100μF
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was a guess (from Dad), but "PCB" Mat suggested something larger, like
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2200μF. So we went with 1000μF, which appears to power the board
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after power-off for a couple of revolutions. The factors here are
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that a capacitor that is only able to hold a small charge won't be
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able to maintain a current for a reasonable amount of time when the
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power breaks. If it's too large, it will take ages to charge and
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effectively short the power (save for the resistor) while it does.
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