Dynamo Lights Brightness PROBLEM

Don Kelly said:

Thanks for the info. You did a good job. I have done a few rough
calculations as below

The electrical efficiency is about 60% in either case (theoretically the
same) and any change in overall efficiency is in the mechanical side.

For the Bottom bracket machine which appears to be the one that the test
results were for:
No capacitor: internal impedance at 190Hz =3.8+j8.07 ohms and total
impedance magnitude is 17.74 ohms so the internal voltage is about 9.5 V
With capacitor the total impedance drops to about 15.8 ohms and the internal
voltage generated is still 9.5V. Electrical efficiency in both cases is
about 76%

For the Union machine:
Zinternal =7.9+j7.4 ohms (190Hz)so total circuit impedance is about 21.3
ohms without the capacitor so, using 0.53A or 6.4V the internal generated
voltage is about 11.3V
With capacitor the total impedance drops to 19.9 +j1 ohms or a magnitude of
19.93 ohms and the internal voltage is about 12V.
The difference may be some drop in demagnetising effect or a slight change
in speed. Electrical efficiency about 60%

I used the higher inductance as this is the one that actually counts (direct
axis vs quadrature axis inductance).

Actually, your bike generators are, electrically, rather conventional
synchronous machines with permanent magnet fields and the internal voltage
is speed dependent, with the current depending on the load and the internal
impedance. It is simply a lousy "voltage source " which is identical to a
lousy "current source".
However in the case of the bottom bracket machine, the inductance is
dominant and rises with speed as well so the total circuit impedance rises,
tending to limit the current somewhat. Note that at twice speed and a 12 ohm
load the current will be about 0.85A and the power about 8.6 watts. Not
really constant current. However there may be other factors involved such as
the demagnetising effect of the current actually reducing the voltage a bit.
That depends on the actual characteristics of the field magnets.
If you put a 24 ohm load on, at the original speed, then you'll get about
7.9V,0.33A and 2.6 watts. Lower current and power. At twice the speed the
internal voltage will be doubled as will the inductance so the output would
be 14.2V (probably a bit less) and current 0.6A for 8.4watts (actually less
current in practice as bulb is hotter). The Union machine will likely have a
wider current range due to its lower inductance.

These figures are based on the data given and other factors come into play.
Some other experiments for you to while away spare time that you could spend
riding!.

:-) Oh, I've wasted plenty of riding time on these things!

Here you go. The article starts on page 4, I think. Some of the
figures don't print properly in this PDF, but were fine in the printed
version.

http://www.ihpva.org/HParchive/PDF/hp49-1999.pdf

- Frank Krygowski

In message <[email hidden]>, dated Wed,
6 Sep 2006, David Kerber <ns_dkerber@ns_ids.net> writes

Quoted message said:

In article <[email hidden]>, [email hidden]
says...

...

Quoted message said:

Quoted message said:

No, the torque required to turn a generator at a given speed into a
resistive load is a function of the *power* (Voltage * Current), not the
current alone. If you short it, your voltage goes to near zero, so the
power does too.

That would be true if the generator had zero-impedance windings. It
doesn't. When you short-circuit the generator, all the power is
dissipated in the resistance of the winding.

That's why I said *near* zero. It still doesn't use much true power, so
the actual mechanical load isn't very high.

Frank Krygowski reported:

Union bottle generator: 7.9 Ohms, 5.45 mH (up to 6.2 mH)

The designed output is 6 V at 0.5 A. We don't know the frequency, so
we'll neglect the inductance (but at 100 Hz it's about 3.8 ohms), and
just note that the voltage across the 7.9 ohms is 3.95 V. Compared with
the 6 V output, that's far from negligible.

When short-circuited, we have the same 9.75 V generator, but now seeing
a load of 7.9 ohms (or 8.7 ohms if we include the inductance). That
gives a current of 1.23 A, and a power of 12 W, **four times** the
normal output power.
--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

In article <[email hidden]>, [email hidden]
says...

....

Quoted message said:

Frank Krygowski reported:

Union bottle generator: 7.9 Ohms, 5.45 mH (up to 6.2 mH)

The designed output is 6 V at 0.5 A. We don't know the frequency, so
we'll neglect the inductance (but at 100 Hz it's about 3.8 ohms), and
just note that the voltage across the 7.9 ohms is 3.95 V. Compared with
the 6 V output, that's far from negligible.

When short-circuited, we have the same 9.75 V generator, but now seeing
a load of 7.9 ohms (or 8.7 ohms if we include the inductance). That
gives a current of 1.23 A, and a power of 12 W, **four times** the
normal output power.

Ok, now that makes sense. I was hugely under-estimating the coil
impedence of the generator.

--
Remove the ns_ from if replying by e-mail (but keep posts in the
newsgroups if possible).

Thanks- I haven't correlated the graphs with my estimates which do not take
all into account and appear to be off. The general shape of current vs
speed curves is flatter than I estimated which implies a higher inductance
than measured.
a)Measurement of inductance with an inductance meter may be a bit iffy
because of the perversity of ferromagnetic devices. Short circuit and open
circuit tests similar to those used for larger machines may give a
different result because armature reaction is not negligable and affects
the apparent inductance.
To sort this out requires more sophisticated tests which aren't worth while.

Again- good job- you made your point.

--

Don Kelly [email hidden]
remove the X to answer
----------------------------
<[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Don Kelly said:

Thanks for the info. You did a good job. I have done a few rough
calculations as below

The electrical efficiency is about 60% in either case (theoretically the
same) and any change in overall efficiency is in the mechanical side.

For the Bottom bracket machine which appears to be the one that the test
results were for:
No capacitor: internal impedance at 190Hz =3.8+j8.07 ohms and total
impedance magnitude is 17.74 ohms so the internal voltage is about 9.5 V
With capacitor the total impedance drops to about 15.8 ohms and the
internal
voltage generated is still 9.5V. Electrical efficiency in both cases is
about 76%

For the Union machine:
Zinternal =7.9+j7.4 ohms (190Hz)so total circuit impedance is about 21.3
ohms without the capacitor so, using 0.53A or 6.4V the internal generated
voltage is about 11.3V
With capacitor the total impedance drops to 19.9 +j1 ohms or a magnitude
of
19.93 ohms and the internal voltage is about 12V.
The difference may be some drop in demagnetising effect or a slight
change
in speed. Electrical efficiency about 60%

I used the higher inductance as this is the one that actually counts
(direct
axis vs quadrature axis inductance).

Actually, your bike generators are, electrically, rather conventional
synchronous machines with permanent magnet fields and the internal
voltage
is speed dependent, with the current depending on the load and the
internal
impedance. It is simply a lousy "voltage source " which is identical to
a
lousy "current source".
However in the case of the bottom bracket machine, the inductance is
dominant and rises with speed as well so the total circuit impedance
rises,
tending to limit the current somewhat. Note that at twice speed and a 12
ohm
load the current will be about 0.85A and the power about 8.6 watts. Not
really constant current. However there may be other factors involved such
as
the demagnetising effect of the current actually reducing the voltage a
bit.
That depends on the actual characteristics of the field magnets.
If you put a 24 ohm load on, at the original speed, then you'll get about
7.9V,0.33A and 2.6 watts. Lower current and power. At twice the speed the
internal voltage will be doubled as will the inductance so the output
would
be 14.2V (probably a bit less) and current 0.6A for 8.4watts (actually
less
current in practice as bulb is hotter). The Union machine will likely
have a
wider current range due to its lower inductance.

These figures are based on the data given and other factors come into
play.
Some other experiments for you to while away spare time that you could
spend
riding!.

:-) Oh, I've wasted plenty of riding time on these things!

Here you go. The article starts on page 4, I think. Some of the
figures don't print properly in this PDF, but were fine in the printed
version.

http://www.ihpva.org/HParchive/PDF/hp49-1999.pdf

- Frank Krygowski

In message <sHLLg.512208$Mn5.109648@pd7tw3no>, dated Thu, 7 Sep 2006,
Don Kelly <[email hidden]> writes

Quoted message said:

Measurement of inductance with an inductance meter may be a bit iffy
because of the perversity of ferromagnetic devices.

I think that's a valuable contribution. Magnetic devices have been known
for years to exhibit reluctance, so it's a small step of intention to
exhibit perversity. This is obviously measured in murphys, but a formal
definition is required.
--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

Quoted message said:

John Woodgate said:

Quoted message said:

What's the ASCII abbreviation for "micro"? I'll go with "mu".)

Quoted message said:

'u'. but 'mu' is understood.

How about: How about: µ

not ASCII.

--

Bye.
Jasen

John Woodgate said:

Yes, it's difficult but it can be done. But I doubt that ferrite
permanent magnets are used in cycle dynamos. They would probably be too
big.

I've looked, they are. the magnet was about the size of 1/2 a "D" cell.
it had 4 poles. (this was in a friction drive dynamo)

Bye.
Jasen

John Woodgate said:

In message <[email hidden]>, dated Thu,
17 Aug 2006, David Kerber <ns_dkerber@ns_ids.net> writes

Quoted message said:

Yes, you are. A filament is a nearly pure resistance load, so the pf
is already equal to 1.

The alternator has internal resistance and inductance. To match its
impedance for maximum power transfer, you need a resistor AND a
capacitor. The capacitor resonates with the inductance, but the
resistances apply heavy damping.

As I indicated a while back, doing this can seriously damage your
permanent magnet.
--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

How can this seriously damage the permanent magnet? The only possible
significant damage I can think of is demagnetization, but I don't see
how the circuit operation would create this. Demagnetizers, at least
for soft materials, work by cycling the cores magnetically in
decreasing BH loops, until the remanent magnetization is tolerably
small. I don't see how this operation would happen in a bike light
system.

Ken

Quoted message said:

Don Kelly said:

Thanks- that is enough info to determine the inductance but not the
resistance of the generator.

We measured the resistances and inductances directly. We had the
instruments.

Union bottle generator: 7.9 Ohms, 5.45 mH (up to 6.2 mH)

Soubitez bottom bracket generator: 3.8 Ohms, 6.76 mH (down to 4.9 mH)

Briefly, a standard 3 Watt generator bulb (assuming only headlight, no
taillight) is 12 Ohms. A 2.4 Watt bulb used with a 0.6 Watt taillight
in parallel has a combined R of 12 Ohms.

Bike generators are, roughly speaking, constant current devices. Open
circuit, their output current is zero, and their output voltage is
roughly proportional to their rpm (up to a certain limit).

- Frank Krygowski

If the internal resistance of the dynamo source is about 6 ohms and the
load resistance applied (the bulb) is 12 ohms or more, the source is
more akin to a Thevenin source (a voltage source) than a Norton source
(a current source). NOT a constant current device, but a constant
voltage device. For a true (zero resistance) Thevenin source with a
small series resistance in addition to the load (your measured values
of 3.8 and 7.9 ohms) the voltage will be only about 40 percent higher
with the load removed, than with the load. But the power will be zero
because the current into an open circuit is zero.

For a current source the open-circuit voltage would become much higher
(theoretically infinite) with the load removed.

For the generators you cite, the situation is more like a matched load,
where about teh same power is dissipated in the source internal
resistance and in the load resistance. In this case the dissipations
are within about 50%, which practically is pretty well matched.

Sorry for all the dweeby stuff, but I have fun with this as well as
with bikes!

Ken

In message <[email hidden]>,
dated Sun, 10 Sep 2006, [email hidden] writes

Quoted message said:

How can this seriously damage the permanent magnet? The only possible
significant damage I can think of is demagnetization, but I don't see
how the circuit operation would create this. Demagnetizers, at least
for soft materials, work by cycling the cores magnetically in
decreasing BH loops, until the remanent magnetization is tolerably
small. I don't see how this operation would happen in a bike light system.

The output current of the dynamo tries to do exactly that. The 'cycling'
in decreasing hysteresis loops occurs as you slow down your own
'cycling', of course. But the root cause of the risk to the magnet is
the INCREASED load current. We assume that the magnet is designed to
resist the demagnetizing effect of the 0.5 A (or whatever is the design
value), but if the load current is increased above that value, the
magnet wasn't designed with that value of H in mind.
--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
There are benefits from being irrational - just ask the square root of 2.
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

jasen said:

Quoted message said:

John Woodgate said:

> What's the ASCII abbreviation for "micro"? I'll go with "mu".)

Quoted message said:

'u'. but 'mu' is understood.

How about: How about: µ

not ASCII.

--

Bye.
Jasen

What's TRULY (yes, I'm shouting) clear in ASCII (not shouting, it's an
acronym) is "micro."

In message <[email hidden]>, dated
Sun, 10 Sep 2006, [email hidden] writes

Quoted message said:

If the internal resistance of the dynamo source is about 6 ohms and the
load resistance applied (the bulb) is 12 ohms or more, the source is
more akin to a Thevenin source (a voltage source) than a Norton source
(a current source). NOT a constant current device, but a constant
voltage device.

The generators obviously aren't either a constant-voltage OR a
constant-current source. 'More akin' doesn't help understanding. And you
haven't taken the inductances into account.
--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
There are benefits from being irrational - just ask the square root of 2.
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

In article <[email hidden]>

Quoted message said:

jasen said:

Quoted message said:

John Woodgate writes:

>> What's the ASCII abbreviation for "micro"? I'll go with "mu".)

> 'u'. but 'mu' is understood.

How about: How about: µ

not ASCII.

What's TRULY (yes, I'm shouting) clear in ASCII (not shouting, it's an
acronym) is "micro."

Why is 'abbreviation' such a long word?

----------------------------
<[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

John Woodgate said:

In message <[email hidden]>, dated
Sun, 10 Sep 2006, [email hidden] writes

Quoted message said:

If the internal resistance of the dynamo source is about 6 ohms and the
load resistance applied (the bulb) is 12 ohms or more, the source is
more akin to a Thevenin source (a voltage source) than a Norton source
(a current source). NOT a constant current device, but a constant
voltage device.

The generators obviously aren't either a constant-voltage OR a
constant-current source. 'More akin' doesn't help understanding. And you
haven't taken the inductances into account.

Strictly speaking, bike generators are neither constant current nor
constant voltage. Strictly speaking, that's also true of batteries -
because their internal resistance affects their apparent voltage output
to varying degrees, depending on current draw. And if we wish to speak
even more strictly, we could include their lowering voltage over time
(needing recharge, or perhaps replacement).

IOW, nothing is perfect. Welcome to the universe!

But there is definite usefulness in thinking of batteries as constant
voltage. Almost everyone learns the bulk of their electronics theory
by assuming that, and for the bulk of practical problems, it's a
perfectly adequate assumption.

Similarly, stating that a bike generator is approximately constant
current does have validity, within limits; and it's certainly more
valid than saying it's constant voltage! Here's one reason why:

As I've described, with certain generators, it's possible to drive two
identical headlamp bulbs connected in _series_. If the generator
design is such that the drive doesn't slip, and if the bike speed is
high enough (above 14 mph for mine) the generator will light both lamps
to normal brightness. The power output of the generator will be
approximately double it's normal value. Current output will be
depressed only slightly.

See http://www.blayleys.com/articles/lights/page2.htm or
http://www.peterwhitecycles.com/wiringinstructions.asp for some
details.

If generators were constant voltage, that setup wouldn't be possible.
Current would be cut in half, and each bulb would glow only dimly. If
generators were constant voltage, you'd instead connect two headlights
in parallel. That absolutely does not work - try it!

Another minor point: Generator headlamps and taillamps are normally
connected in parallel. It's fairly well known that, if you blow your
headlamp bulb, your incandescent taillamp will last only a few seconds.
Why? All the 0.5 amp is pumped through that low-power (i.e. high
resistance) bulb. Voltage output rises as the generator
instantaneously tries to maintain its constant current. The bulb
blows.

If the generator were constant voltage, the bulb would feel no
difference.

Yes, I realize "constant current" is a simplification, but it's good
enough to be useful, and points out important differences between bike
generators and batteries.

- Frank Krygowski

Speed---> voltage induced
Current--->torque
At a given speed, the internal voltage will be nearly constant. Internal
impedance exists- you can treat the device as a voltage source with a series
internal impedance (as it actually is) or as a current source shunted by
this impedance. There are some demagnetising amp turns under load so that
the flux density decreases as load increases resulting in a decrease of the
internal voltage. This can be expressed in terms of an equivalent inductance
in addition to the leakage inductance that an inductance meter can measure.
Extra internal impedance or the equivalent tends to limit the current
variation with load or speed leading to the nearly constant current
characteristic due to the dominant effect of the internal impedance.

You simply have a high impedance speed dependent (variable- not constant)
voltage source. The difference between this and a battery is that the
battery internal voltage is not speed dependent and its impedance is low so
it looks more like a constant voltage source .
If the headlamp burns out, the tail lamp will see more current as the
terminal voltage will rise. That is true. It is not a case of the generator
trying to pump 0.5A through the bulb although it may look that way.

Consider a 10V source with a 10 ohm internal resistance (to simplify by
ignoring inductance) supplying a 10 ohm load. The load voltage and current
will be 5V, 0.5A
Now increase the load to 20 ohms. The load voltage will be 7.5V and current
0.75A.
In the case of the generator, the armature reaction will actually reduce the
internal voltage so the output will be lower- possibly 6V and 0.6A. No more
than that, excluding speed effects.
--

Don Kelly [email hidden]
remove the X to answer

Quoted message said:

In article <DZ2Ng.526182$Mn5.110669@pd7tw3no>,

Don Kelly said:

At a given speed, the internal voltage will be nearly constant.
Internal impedance exists- you can treat the device as a voltage
source with a series internal impedance (as it actually is) or
as a current source shunted by this impedance. There are some
demagnetising amp turns under load so that the flux density
decreases as load increases resulting in a decrease of the
internal voltage. This can be expressed in terms of an equivalent
inductance in addition to the leakage inductance that an
inductance meter can measure. Extra internal impedance or the
equivalent tends to limit the current variation with load or
speed leading to the nearly constant current characteristic due
to the dominant effect of the internal impedance.

In the case of that bottom bracket generator the OP
physically measured 3.8 ohms and 6.72mH.

But he also did a separate experiment, where the output
was seen to increase from 6.4V to 7.3V when a 100uF
capacitor was placed in series.

If it assumed that the internal voltage, the resistance,
and the inductance remained reasonably constant across the
experiment, then the inductance works out to about 7mH.

7mH is not far off the physical measurement of 6.72mH.

Umm... all sums assume a sinewave output, something that
is not neccessarily true for small PM alternators.

[snip]

Quoted message said:

Consider a 10V source with a 10 ohm internal resistance (to
simplify by ignoring inductance) supplying a 10 ohm load. The
load voltage and current will be 5V, 0.5A Now increase the load
to 20 ohms. The load voltage will be 7.5V and current 0.75A.

Small typo there? At a 20-ohm load the load voltage will
only rise to 10*20/30 and the current will drop to 10/30.

--
Tony Williams.

----------------------------
"Tony Williams" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

In article <DZ2Ng.526182$Mn5.110669@pd7tw3no>,

Don Kelly said:

At a given speed, the internal voltage will be nearly constant.
Internal impedance exists- you can treat the device as a voltage
source with a series internal impedance (as it actually is) or
as a current source shunted by this impedance. There are some
demagnetising amp turns under load so that the flux density
decreases as load increases resulting in a decrease of the
internal voltage. This can be expressed in terms of an equivalent
inductance in addition to the leakage inductance that an
inductance meter can measure. Extra internal impedance or the
equivalent tends to limit the current variation with load or
speed leading to the nearly constant current characteristic due
to the dominant effect of the internal impedance.

In the case of that bottom bracket generator the OP
physically measured 3.8 ohms and 6.72mH.

But he also did a separate experiment, where the output
was seen to increase from 6.4V to 7.3V when a 100uF
capacitor was placed in series.

If it assumed that the internal voltage, the resistance,
and the inductance remained reasonably constant across the
experiment, then the inductance works out to about 7mH.

7mH is not far off the physical measurement of 6.72mH.

Umm... all sums assume a sinewave output, something that
is not neccessarily true for small PM alternators.

[snip]

Quoted message said:

Consider a 10V source with a 10 ohm internal resistance (to
simplify by ignoring inductance) supplying a 10 ohm load. The
load voltage and current will be 5V, 0.5A Now increase the load
to 20 ohms. The load voltage will be 7.5V and current 0.75A.

Small typo there? At a 20-ohm load the load voltage will
only rise to 10*20/30 and the current will drop to 10/30.

--
Tony Williams.

The physically measured inductance using an inductance meter only measures
the leakage reactance at a negligably low current (varying with position as
it is salient pole) which is generally much lower than the total direct axis
reactance of the machine. I did calculations based on the given reactance
and these estimations indicated a wider variation of current with speed or
load resistance than his given curves. To me, this indicates that there is
more going on than what can be represented by a leakage reactance and Dc
resistance behind a speed dependent voltage-- i.e. demagnetization effect
(depending on the characteristics of the actual permanent magnet material
used). is not negigable In models, this is treated by an increase in
reactance above leakage reactance (in a large machine this will far exceed
leakage reactance but in this device-who knows). The apparent internal
voltage based on the data does change when the capacitor is added. Less so
for the bottom bracket machine than the other (better magnet?). An
inductance meter is inherently inadequate for measuring the "inductance" of
a synchronous machine. If the magnet's BH curve is nearly flat topped, one
could assume no demagnetisation effect- but is it?

Data for his published curves may have enough information to actually
construct a model which gives a more realistic "inductance" than what the
inductance meter gives. However, I have lost his reference and even with
the curves- it would be still a crude estimation.

It is true that the behaviour of the generator more closely resembles a
"constant current source" than a "constant voltage source" because it is a
"variable voltage "source. The inductive impedance also varies with speed so
that where this is dominant, the current change with a given load and
varying speed will be small. This is what is happening. A true constant
current source would produce the same current (at the same speed) for any
reasonable load variation. This is not not happening.

As for sine wave output- you are quite right.

As for the example that I gave- you are also quite right- and I am
embarrassingly wrong! Even with the wine I had consumed at the time I should
have seen that increasing the resistance reduces the current.

Anyhow it is fun even if it is not particularly useful.
--

Don Kelly [email hidden]
remove the X to answer