Dynamo Lights Brightness PROBLEM

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.

Quoted message said:

You DID put the capacitor in SERIES, didn't you?

How about: µ⏀␀⑀⒀Ⓚ─╀␂=
Jobst Brandt

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

Jobst Brandt

In message <[email hidden]>, dated Fri, 1 Sep
2006, [email hidden] writes

Quoted message said:

How about: How about: ?

That comes out as '=C2=B5' in some e-mail clients. I don't think it will
catch on. (;-)

B5 is indeed 'μ', but not all e-mail clients will render it. I don't
understand C2 (194), which is 'Â'. 'u' works all the time.
--
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

dvt said:

Helpful hint: shouting at the world (using all caps) is not likely to
generate lots of helpful replies. You were probably frustrated at the
time, a sentiment we all understand, but it sounds a bit juvenile.

WHY DO YOU THINK THIS IS TRUE?

--
TOM SHERMAN - BEHIND THE CHEDDAR CURTAIN

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

Quoted message said:

David Kerber said:

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

Quoted message said:

In message <[email hidden]>, 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.

My power engineering courses were a long time ago, but I don't believe
you can increase the power transfer into a resistive load (the filament)
by adding a capacitance to a circuit; the cap just acts as an
*additional* load (even though it's reactive load).

Well, I tried it a few years ago. IANA Electrical Engineer but I had
one helping me.

We tested three different generators and found similar results with
all. Here's some data for a Soubitez bottom-bracket generator. I'll
just give the values for 14 mph, 12 ohm load (i.e. standard generator
headlight bulb), with and without some capacitance. The capacitance
was chosen to give max power boost at about 12 mph, IIRC, and was 100
microFarads. (What's the ASCII abbreviation for "micro"? I'll go with
"mu".)

RMS Voltage: With 100 muF: 7.3 V
without: 6.4 V

Current: With 100 muF: 0.6 A
without: 0.53 A

Power: With 100 muF: 4.4 W
without: 3.4 W

Efficiency: With 100 muF: 42%
without: 39%

The efficiency figures are the least reliable, BTW. Our method of
measuring did not take into account the losses resulting from the
interface between the rubber tire and the generator roller surface, and
those can be considerable.

I'll also note, the September 1995 issue of "Electronics World +
Wireless World" (a British magazine) has a letter to the editor, plus
response, on p. 770 that deals with this issue. For their own reasons,
they used a resistance of 24 Ohms instead of 12 Ohms, with and without
a 50 microFarad capacitor. Above 10 mph, they got more power with the
capacitor than without. A graph on that page shows their results.

So you can get more out of a bike generator by adding capacitance. But
it's not much more, it varies quite a bit with speed, and we judged it
not worth the trouble.

- Frank Krygowski

It appears that the current and power measurements were for the bulb. So
what has the capacitor done? It raised the voltage at the load and hence
there was more power to the load. Fair enough. The efficiency figures are,
as indicated, questionable. In the case where the capacitor is included,
there will be a higher total current from the generator and more electrical
losses. In addition to the extra power to the load, there will be extra
losses and the power source has to provide both.

In this case, I have to agree with your conclusion-why bother?

Out of curiosity, what was the frequency at which you took the
measurements?.

Generally in power system applications, maximum power transfer is not a
meaningful thing except as a situation to avoid due to inefficiency and
stability problems. Having a source impedance which is much lower than the
load impedance is more beneficial.

--

Don Kelly [email hidden]
remove the X to answer

Quoted message said:

Don Kelly said:

It appears that the current and power measurements were for the bulb.

Well, we actually used 12 Ohms worth of power resistors for the load.
But yes, voltage, current and power were all as delivered to the load.

Quoted message said:

So
what has the capacitor done? It raised the voltage at the load and hence
there was more power to the load. Fair enough. The efficiency figures are,
as indicated, questionable. In the case where the capacitor is included,
there will be a higher total current from the generator and more electrical
losses. In addition to the extra power to the load, there will be extra
losses and the power source has to provide both.

That's true. In my notes, I see that we had difficulty with some sort
of instability when measuring generator torque with the capacitor in
the circuit. That is, the torque measurement fluctuated - I can't say
why. The torque values (and, hence, efficiency values) probably
contain more than the normal amount of "eyeballing."

Quoted message said:

Out of curiosity, what was the frequency at which you took the
measurements?.

Frequency of the generator's AC output is proportional to road speed,
and can be surprisingly high. For this generator, 12 MPH gave about
190 Hz, IIRC. We tested it (and other generators) from about 8 MPH to
about 25 MPH.

- Frank Krygowski

In message <[email hidden]>, dated
Sat, 2 Sep 2006, [email hidden] writes

Quoted message said:

In my notes, I see that we had difficulty with some sort of instability
when measuring generator torque with the capacitor in the circuit. That
is, the torque measurement fluctuated - I can't say why.

The current is increased, and this means that the rotor needs more
torque to pass the pole-pieces. If you short-circuit the generator and
try to turn it by hand, you will find it very difficult.

The increased current also increases the de-magnetizing field applied to
the permanent magnet, and will cause its strength to deteriorate over
time.
--
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

John Woodgate said:

In message <[email hidden]>, dated
Sat, 2 Sep 2006, [email hidden] writes

Quoted message said:

In my notes, I see that we had difficulty with some sort of instability
when measuring generator torque with the capacitor in the circuit. That
is, the torque measurement fluctuated - I can't say why.

The current is increased, and this means that the rotor needs more
torque to pass the pole-pieces. If you short-circuit the generator and
try to turn it by hand, you will find it very difficult.

??

Not so. I can feel some increase in resistance with output shorted,
but it's minor.

- Frank Krygowski

In message <[email hidden]>, dated
Sun, 3 Sep 2006, [email hidden] writes

Quoted message said:

Not so. I can feel some increase in resistance with output shorted,
but it's minor.

I expect that the winding impedance is reducing the effect.
--
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

"John Woodgate" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

The increased current also increases the de-magnetizing field applied to
the permanent magnet, and will cause its strength to deteriorate over
time.

You can demagnetize ferrites?

In message <uxFKg.5616$Hr1.3070@clgrps12>, dated Sun, 3 Sep 2006, Homer
J Simpson <[email hidden]> writes

Quoted message said:

"John Woodgate" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

The increased current also increases the de-magnetizing field applied to
the permanent magnet, and will cause its strength to deteriorate over
time.

You can demagnetize ferrites?

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.
--
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

"John Woodgate" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message 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.

Would have been my assumption price wise.

Thanks- that is enough info to determine the inductance but not the
resistance of the generator. I should also have asked for the open circuit
voltage and the DC resistance of the generator to get a better handle on the
model. Also I have been assuming the capacitor is in parallel with the load
resistance. Is it?

Out of curiosity - why 12 ohms?

As for fluctuations in torque due to the capacitor- there shouldn't be any
other than double frequency components which would have an average of 0.
Fluctuations of this nature will also be present without the capacitor. As
for demagnetisation, that is unlikely as the generator probably can handle
heavier loads and also a more leading pf will reduce demagnetisation.

--

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

Quoted message said:

Don Kelly said:

It appears that the current and power measurements were for the bulb.

Well, we actually used 12 Ohms worth of power resistors for the load.
But yes, voltage, current and power were all as delivered to the load.

Quoted message said:

So
what has the capacitor done? It raised the voltage at the load and hence
there was more power to the load. Fair enough. The efficiency figures
are,
as indicated, questionable. In the case where the capacitor is included,
there will be a higher total current from the generator and more
electrical
losses. In addition to the extra power to the load, there will be extra
losses and the power source has to provide both.

That's true. In my notes, I see that we had difficulty with some sort
of instability when measuring generator torque with the capacitor in
the circuit. That is, the torque measurement fluctuated - I can't say
why. The torque values (and, hence, efficiency values) probably
contain more than the normal amount of "eyeballing."

Quoted message said:

Out of curiosity, what was the frequency at which you took the
measurements?.

Frequency of the generator's AC output is proportional to road speed,
and can be surprisingly high. For this generator, 12 MPH gave about
190 Hz, IIRC. We tested it (and other generators) from about 8 MPH to
about 25 MPH.

- Frank Krygowski

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

Quoted message said:

In message <[email hidden]>, dated
Sat, 2 Sep 2006, [email hidden] writes

Quoted message said:

In my notes, I see that we had difficulty with some sort of instability
when measuring generator torque with the capacitor in the circuit. That
is, the torque measurement fluctuated - I can't say why.

The current is increased, and this means that the rotor needs more
torque to pass the pole-pieces. If you short-circuit the generator and
try to turn it by hand, you will find it very difficult.

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.

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

Johnny Sunset aka Tom Sherman said:

dvt said:

Helpful hint: shouting at the world (using all caps) is not likely to
generate lots of helpful replies. You were probably frustrated at the
time, a sentiment we all understand, but it sounds a bit juvenile.

WHY DO YOU THINK THIS IS TRUE?

WHAT? I CAN'T HEAR YOU!

--
DAVE
AHEAD OF (?) THE CHEDDAR CURTAIN

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

Quoted message said:

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

Quoted message said:

In message <[email hidden]>, dated
Sat, 2 Sep 2006, [email hidden] writes

Quoted message said:

In my notes, I see that we had difficulty with some sort of instability
when measuring generator torque with the capacitor in the circuit. That
is, the torque measurement fluctuated - I can't say why.

The current is increased, and this means that the rotor needs more
torque to pass the pole-pieces. If you short-circuit the generator and
try to turn it by hand, you will find it very difficult.

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.

Check out Thévenin's Theorem and Norton's Theorem.
--
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

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)

Inductance measurements were done using an impedance bridge. They
varied with angular position of the generator shaft... whether the
generator was in one of it's "notches" or held in a different position.
Ultimately, we decided it didn't matter much which value we used; we
were measuring just to calculate a roughly appropriate capacitor size,
and if it was off 30%, it didn't matter much for our purposes.

Quoted message said:

I should also have asked for the open circuit
voltage and the DC resistance of the generator to get a better handle on the
model.

Well, I think I've got those open circuit voltages somewhere...

Quoted message said:

Also I have been assuming the capacitor is in parallel with the load
resistance. Is it?

Nope. In series.

Quoted message said:

Out of curiosity - why 12 ohms?

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).

When given a resistive load, they will do their darndest to put out
their rated current. Most bike generators are designed to produce 0.5
Amp. But their rating is invariably stated as 6 Volt, 3 Watt. That
rating depends on having a 12 Ohm resistance in the load.

It's interesting that you can get more power out of a generator by
giving it more resistance. For example, seeing a 24 Ohms load, the
generator will try its darndest to put out 0.5 Amp. To do that, it
will generate 12 volts, and produce 6 watts. Same generator, twice the
power. (This only works if the speed is high enough.)

Problem is, most generators won't succeed at that job, because their
drive wheels need about twice the torque as usual. They'll slip. One
reason I like the Soubitez bottom bracket generator is that it can pull
this off without slipping. So, of course, can the hub generators like
the SON (or Schmidt). Bottle generators usually can't do it.

Quoted message said:

As for fluctuations in torque due to the capacitor- there shouldn't be any
other than double frequency components which would have an average of 0.
Fluctuations of this nature will also be present without the capacitor. As
for demagnetisation, that is unlikely as the generator probably can handle
heavier loads and also a more leading pf will reduce demagnetisation.

The fluctuations we saw weren't at anything like double the frequency.
They had a period measured in (by memory) a second or so - i.e.
frequency of 1 Hz or less.

To make this clear: We measured reaction torque by having the
generator mounted in a sort of gimbal arrangement, with a long torque
arm pressing on a digital scale. (We used a balloon between the arm
and scale to absorb vibrations.) Anyway, the measuring system worked
well in "normal" mode, but when we used it with the capacitor in the
circuit, scale readings (i.e. torque readings) varied quite a bit.

Again, as with the inductance, we were just checking to see if adding
capacitance was possibly worthwhile. Even without precise results, we
learned enough to say capacitance was not worthwhile.

- Frank Krygowski

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!.
In any case, the addition of a capacitor isn't worth the effort and may
actually reduce performance at higher or lower speeds.
--

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

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)

Inductance measurements were done using an impedance bridge. They
varied with angular position of the generator shaft... whether the
generator was in one of it's "notches" or held in a different position.
Ultimately, we decided it didn't matter much which value we used; we
were measuring just to calculate a roughly appropriate capacitor size,
and if it was off 30%, it didn't matter much for our purposes.

Quoted message said:

I should also have asked for the open circuit
voltage and the DC resistance of the generator to get a better handle on
the
model.

Well, I think I've got those open circuit voltages somewhere...

Quoted message said:

Also I have been assuming the capacitor is in parallel with the load
resistance. Is it?

Nope. In series.

Quoted message said:

Out of curiosity - why 12 ohms?

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).

When given a resistive load, they will do their darndest to put out
their rated current. Most bike generators are designed to produce 0.5
Amp. But their rating is invariably stated as 6 Volt, 3 Watt. That
rating depends on having a 12 Ohm resistance in the load.

It's interesting that you can get more power out of a generator by
giving it more resistance. For example, seeing a 24 Ohms load, the
generator will try its darndest to put out 0.5 Amp. To do that, it
will generate 12 volts, and produce 6 watts. Same generator, twice the
power. (This only works if the speed is high enough.)

Problem is, most generators won't succeed at that job, because their
drive wheels need about twice the torque as usual. They'll slip. One
reason I like the Soubitez bottom bracket generator is that it can pull
this off without slipping. So, of course, can the hub generators like
the SON (or Schmidt). Bottle generators usually can't do it.

Quoted message said:

As for fluctuations in torque due to the capacitor- there shouldn't be
any
other than double frequency components which would have an average of 0.
Fluctuations of this nature will also be present without the capacitor.
As
for demagnetisation, that is unlikely as the generator probably can
handle
heavier loads and also a more leading pf will reduce demagnetisation.

The fluctuations we saw weren't at anything like double the frequency.
They had a period measured in (by memory) a second or so - i.e.
frequency of 1 Hz or less.

To make this clear: We measured reaction torque by having the
generator mounted in a sort of gimbal arrangement, with a long torque
arm pressing on a digital scale. (We used a balloon between the arm
and scale to absorb vibrations.) Anyway, the measuring system worked
well in "normal" mode, but when we used it with the capacitor in the
circuit, scale readings (i.e. torque readings) varied quite a bit.

Again, as with the inductance, we were just checking to see if adding
capacitance was possibly worthwhile. Even without precise results, we
learned enough to say capacitance was not worthwhile.

- Frank Krygowski

----------------------------
"David Kerber" <ns_dkerber@ns_ids.net> wrote in message
news:[email hidden]...

Quoted message said:

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

Quoted message said:

In message <[email hidden]>, dated
Sat, 2 Sep 2006, [email hidden] writes

Quoted message said:

In my notes, I see that we had difficulty with some sort of instability
when measuring generator torque with the capacitor in the circuit. That
is, the torque measurement fluctuated - I can't say why.

The current is increased, and this means that the rotor needs more
torque to pass the pole-pieces. If you short-circuit the generator and
try to turn it by hand, you will find it very difficult.

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.

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

If you check basic theory you will find that the torque is related to the
current alone.

On short circuit the current is limited by internal impedance only so it
will be higher than it would be with a load connected (at the same generator
speed). Hence the torque will also be higher.
If the internal impedance is mainly inductive and particularly if it is
relatively low, there can be high current and torque but little power
involved.
In the case of your generator, the short circuit current will be about
double the current (1.1A vs 0.53A) at a 12ohm load and so the torque will
be doubled while the total power will only be 30% higher (1.1A through 3.8
ohms =4.5watts vs 3.5).
--

Don Kelly [email hidden]
remove the X to answer

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.

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