Re: That Lowe's flashlight, again

On Jul 18, 9:59 am, Peter Cole <peter_c...@xxxxxxxxxxx> wrote:
SMS wrote:
Peter Cole wrote:
SMS wrote:

A bridge rectifier is only 81.2% efficient (maximum with zero-voltage
drop diodes)

You keep repeating this, but it's not true. You misunderstand (and
misquote) a reference on rectification.

I think where you're confused is that you're looking at the AC input
power and the full wave rectified power, but not the DC power. Indeed,
with perfect diodes, you'd have 100% efficiency, but you won't have DC
yet! When you convert to DC, using capacitors to filter out the ripple,
you are at 81.2% (assuming perfect diodes and capacitors).

You are mixing up voltage and power. "Rectification efficiency" (the
source of your 81.2%) is a measure of the output (DC) to input (AC)
voltage. It has nothing to do with power losses/efficiencies.

After you rectify AC, you do have DC ("time-varying" DC). "AC" means
"alternating current", which means the current is reversing direction.
This can not happen (by definition) with a diode in the circuit.

The principle of superposition states that a time-varying DC signal can
be treated equivalently to a constant component plus an AC component,
but that principle only applies to "lumped, linear, time-invariant"
circuits (resistors, capacitors, inductors). Diodes are non-linear.

If you're able to run the LED directly from the the rectified AC power
from the dynamo, without over-voltaging it, then you could get by
(put two LEDs in series, as you want to keep them under 4.5 volts).

Again, LEDs (and diodes in general) are (approximately) constant voltage
devices. The ideal diode has a discontinuity in its resistance (slope of
V/I), changing from infinite to zero when it passes through the forward
voltage threshold. It is this discontinuity that makes them non-linear
devices. You can't "over-voltage" a diode, LED or otherwise, you can
only "over-current" it.

The only reason to stack 2 or more LEDs in a generator driven
application is to improve the efficiency slightly by matching the
generator load more closely to its designed load (incandescent) and
making the real rectification power (loss) a smaller fraction of the
load power.

An LED operating at 3.5V and 1.0A will be dissipating 3.5W. If the
waveforms are time-varying DC (like FW rectified sinusoid), all those
numbers are RMS (root mean squared). If the rectifier is a Schottky
bridge, the voltage drop will be approximately 0.6V, the power 0.6W. The
non-linearity of the diodes makes computing the actual power tricky, the
shapes of the voltage and current waveforms are complex, but since both
the rectifiers and the load are diodes, an efficiency calculation based
on the ratio of the forward drops would be very close.

In reality, the almost all LED lights run off pure DC. Many flashlights
use DC-DC converters between the LED and the power source so they can
run off of a variety of different voltages.

The classical way to analyze a source/load circuit is to model both as
circuits made from ideal components (equivalent circuits). The
equivalent circuit for a generator is an ideal (sinusoidal, variable
frequency, variable amplitude) voltage source with a series resistance
(winding resistance) and inductance. The LED (and rectifier) equivalent
circuit must be separately modeled before and after the transition
(infinite resistance/zero resistance). The net is, that for most of the
time (cycle), the generator is driving an equivalent 0 ohm load (short
circuit), the current being limited by the series resistance and
inductance. The inductive component being significant in that it limits
the *rate* which the current can increase during the cycle, which limits
its peak value. Since the voltage (internal ideal) and frequency both
rise together with shaft speed, the net is somewhat self-regulating.

The reason incandescent bulbs may burn out at high generator speeds is
that they have a bad dynamic resistance characteristic. Their effective
resistance goes up with power. If you think of the series inductance and
resistance as "ballast", the ratio of the ballast to the load changes as
the bulb heats up, putting (relatively) more power to the load,
increasing the efficiency (load to ballast power ratio). This is similar
to solid state "thermal runaway".

Batteries have relatively low internal resistance, so the equivalent
circuit is an ideal (constant) voltage source in series with a small
resistance. When driving a short circuit load (diode after transition)
the currents can become large. For this reason, battery-driven LED
circuits must be current limited. Additionally, it is often desirable
(to optimize battery life) to operate at multiple, selectable power
levels. For LED loads, this means variable current regulation. Boost
converters may also be needed if the battery voltage is below the LED
operating voltage.

Suitably paired LEDs and generators need no regulation, either for
current limiting or voltage boosting. Current limiters could be used to
lessen the generator mechanical load selectably, but I don't really see
the advantage.

Sorry for the two-sentence comment on a long post. But I nominate
Peter Cole's post, above, for the "most cogent post on generators &
LEDs" to date.

- Frank Krygowski
.

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