bzr branch
http://bzr.ed.am/elec/propeller-clock
39
by edam
updated notes and scematic (adding capacitor+resistor and diode) |
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DISPLAY |
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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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SCEMATIC NOTES |
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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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41
by edam
modified schematic and updated notes |
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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. |