14
14
The diode (D14) across the fan's power connections is there because if
15
the power across the fan breaks (due to the unreliable nature of the
15
the power acrss the fan breaks (due to the unreliable nature of the
16
16
brushes), the motor in the fan has coils, which act like an inductor
17
17
and will produce a back EMF (a huge negative voltage across the power
18
18
connections) as the magnetic field collapses. This won't be good for
19
the Arduino and could cause sparks on the brushes. The diode simply
19
the arduino and could cause sparks on the brushes. The diode simply
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20
shorts the negative voltage.
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22
The capacitor (C1) and resistor (R14) are there to smooth the power
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23
supply from the unreliable brushes. The capacitor would discharge
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24
fairly slowly (due to the resistance of the circuit), but will charge
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25
very quickly. Potentially, it will charge so quickly that it'll pull
26
too much current from the power supply (i.e., short the power supply
27
and trip it). So the resistor limits this. Unfortunately, the
28
resistor will also have a potentiometer effect (with the resistance of
29
the main circuit). 10Ω was chosen as a value due to these rough
30
workings: Lets say the Arduino circuit takes 500mA. If we aim to lose
31
1V across the resistor, that's 1V / 0.5A = 2Ω (from V=IR). The 100μF
32
was a guess (from Dad), but "PCB" Mat suggested something larger, like
33
2200μF. So we went with 1000μF, which appears to power the board
34
after power-off for a couple of revolutions. The factors here are
35
that a capacitor that is only able to hold a small charge won't be
36
able to maintain a current for a reasonable amount of time when the
37
power breaks. If it's too large, it will take ages to charge and
38
effectively short the power (save for the resistor) while it does.
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26
too much current from the power supply. So the resistor limits this.
27
Unfortunately, the resistor will also have a potentiometer effect
28
(with the resistence of the main circuit). 10Ω was chosen as a value
29
due to these rough workings: Lets say the arduino circuit takes 100mA.
30
if we aim to lose 1V across the resistor, that's 1V / 0.1A = 10Ω (from
31
V=IR). The 100μF was a guess! The problem with the capacitor is that
32
if it's can only hold a small charge, it won't be able to maintain a
33
reasonable charge when the power breaks. If it's too large, it will
34
take ages to charge (and effectively short the power, save for the
35
resistor, while it charges). 100μF seemed like a good value,
b'\\ No newline at end of file'
added
removed
notes
