Dynamo Lights Brightness PROBLEM



Don Kelly wrote:
> 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 protected]>, dated Wed,
6 Sep 2006, David Kerber <ns_dkerber@ns_ids.net> writes
>In article <[email protected]>, [email protected]
>says...
>
>...
>
>> >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 protected]>, [email protected]
says...

....

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

>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
 
On 2006-09-01, [email protected] <[email protected]> wrote:
> 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.



--

Bye.
Jasen
 
On 2006-09-03, John Woodgate <[email protected]> wrote:

> 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 wrote:
> In message <[email protected]>, dated Thu,
> 17 Aug 2006, David Kerber <ns_dkerber@ns_ids.net> writes
> >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
 
[email protected] wrote:
> Don Kelly wrote:
> > 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 protected]>,
dated Sun, 10 Sep 2006, [email protected] writes

>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 wrote:
> On 2006-09-01, [email protected] <[email protected]> wrote:
> > 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.
>
>
>
> --
>
> Bye.
> Jasen

What's TRULY (yes, I'm shouting) clear in ASCII (not shouting, it's an
acronym) is "micro."
 
In message <[email protected]>, dated
Sun, 10 Sep 2006, [email protected] writes
>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 protected]>
<[email protected]> wrote:
> jasen wrote:
> > On 2006-09-01, [email protected] <[email protected]> wrote:
> > > 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?
 
John Woodgate wrote:
> In message <[email protected]>, dated
> Sun, 10 Sep 2006, [email protected] writes
> >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
 
----------------------------
<[email protected]> wrote in message
news:[email protected]...
>
> John Woodgate wrote:
>> In message <[email protected]>, dated
>> Sun, 10 Sep 2006, [email protected] writes
>> >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 protected]
remove the X to answer


>
 
In article <DZ2Ng.526182$Mn5.110669@pd7tw3no>,
Don Kelly <[email protected]> wrote:

> 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]
> 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 protected]> wrote in message
news:[email protected]...
> In article <DZ2Ng.526182$Mn5.110669@pd7tw3no>,
> Don Kelly <[email protected]> wrote:
>
>> 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]
>> 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 protected]
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