Ian Smith <
[email protected]> writes:
> On Mon, 06 Nov 2006, Internet Physics Police <> wrote:
>>
>> Ian Smith <[email protected]> writes:
>
>> > dynamos have regulators to dump the excess power somewhere safe, but
>> > cheap dynamos won't, and may be prone to blowing bulbs.
>>
>> Regulators can either connect the generator to ground or switch it
>> off. Switching should be preferred here, since connecting the
>> generator to ground (dumping the power) produces power and heat from
>> mechanical work done on the generator.
>
> The problem is when you have a dynamo generating, say, 6W, and bulbs
> in the circuit that can only carry 3W without blowing. You can't
> switch off, because you need that 3W, so you have to arrange
> something.
The switching in this sort of regulator happens very quickly and the
signal is usually fed through a low-pass filter (capacitor and
inductor) so the load sees a very smooth voltage. Light bulbs are
slow to heat up compared to the spped of most electronics, so average
power regulation is all that is needed. Indeed, the light bulb is
already being fed AC current.
Switching power supplies are very common nowadays. You will find them
in computers, iPods (which use an even more sophisticated switching
conversion to inrease voltage), and even some headlamps (torches). I
would be surprised if the better bicycle lighting systems used
anything else.
Linear regulators which "dump" current would get quite hot at speed.
>> > If you have two bulbs in parallel, and one blows, the dynamo will
>> > try and put all the current through the one remaining bulb, even if
>> > that means a sudden dramatic voltage increase.
>>
>> Actually, if you have two bulbs in parallel and one blows, the current
>> drops significantly because the resistance of the new (single-bulb)
>> load is more than twice that of the original (parallel-bulb) load.
>
> The current through the remaining bulb increases (for a moment, at
> least, then it tends to drop to zero).
While the remaining bulb sees an increase in current, the generator
actually produces less current, contradicting your assertion that
current would tend to remain constant. Here's why:
Assume the initial voltage is V0 and each bulb has resistance R0.
After the first bulb burns out, the voltage across the new bulb
increases and so does its resistance. Even though the remaining bulb
may never reach a steady state condition, let's pick voltage and
resistance values V1 and R1 at any point after the first bulb fails.
R1 will be greater than R0, because the filament is hotter.
A simple generator cannot produce more power as the load resistance in
increased. The load resistance has more than doubled, going from R0/2
to R1, so we can evaluate the most favorable case, where the
generator's power output is constant.
Before the first failure, current can be calculated using Ohm's law:
C0 = V0/(R0/2) = 2V0/R0
Afterward we have
C1 = V1/R1
Power can be expressed as resistance times current squared. Equating
power before and after yields:
C1^2 * R1 = C0^2 * R0/2
Thus: C1 * sqrt(R1) = C0 * sqrt(R0/2)
so: C1 = (sqrt(R0/R1)/sqrt(2)) * C0
Since R0/R1 is less than zero, we can put an upper bound on C1:
C1 < 0.707 * C0
In other words, the current generated drops significantly after the
first bulb burns out.
What I am trying to explain is that the generator does not "try and
put all the current through the one remaining bulb." On the contrary,
immediately after the burnout, the generator begins to increase its
voltage as current output decreases. As the energy stored inductively
in the generator's coils quickly fades, the current drops to no more
than 0.707 of its original amount, and continues to drop as the light
bulb's temperature increases. The bulb, on the other hand, sees an
increasing voltage, allows more current through itself than before,
and quickly burns out.
Lastly, don't think that the inductance of the generator, which tends
to maintain current output over very short periods of time, will
effect the light bulb. The inductive energy is too small to heat the
light bulb significantly.