On Thu, 28 Aug 2003 14:43:06 -0700, Eric R Snow
wrote:

The motor I tested that comes closest to what I want is 2.25 long and
1.5 dia.. At 10,000 rpm it was putting out 9 volts @ 1.5 amps. It was
$10.00 at H.F.. It came complete with a drill chuck, batteries,
charger and a drill motor case wrapped around it. For 8 bucks they
have one with a keyed chuck. If two were connected in series then spun
to 15,000 rpm max then with a cheap regulator the batteries could be
charged and the lights lit. Now, if a similar sized motor came along
with a higher voltage output so only one would be needed then that
would be great. I'm curious to see what Ax-Man has.

They had several different sizes, including a couple of 24-volt
motors. Prices ranged between $5 and $10 for the smaller ones --
sizes like you describe and a bit larger. In general, the larger
the diameter you can use the more volts per rev you'll probably get
because there's more room for magnets and peripheral velocity is
greater for given RPM. I didn't see any 90-volt PM motors this time.

If you could find a power seat motor at a junkyard, that might work
very well. I have one here that's 2.56" dia and would be a little
under 4" long if one cut the wormgear drive off or made a new endcap.
It turns about 113 RPM thru 44:1 worm reduction, which makes me think
it would generate pretty close to 12 volts at about 5000 RPM. The
armature resistance is only 0.175 ohms so IR drop would be very low.
I think there's be no problem getting several amps out of it. It
draws about 15 amps before it starts slowing down noticably.

AxMan had some windshieldwiper motors with gearbox attached, but they
wanted way too much for them -- $19.95, I think, and you don't need
gear reduction.

Jim Pentagrid is correct that the power will be much less if the
speed and field are the same. But you can either increase the speed
or field to make the voltage the same.

Dan


(Dan Caster) wrote in message m...
It will be good for the same amount of current and therefore the same
amount of wattage if the voltage is the same. What are you trying to
do and what kind of motor are you thinking of using? Induction motors
will work well especially if you use the grid to excite them.

Dan

In article ,
says...

Though I've never tried it, I'd expect one would make a
very good 3-phase generator. They certainly do a good job
of pumping up the bus voltage on the servo amp when
delivering negative torque to an overhauling load.

I agree, but it may be necessary to substitute diodes for the internal
elex.

I assume you're talking about substituting diodes for the
output stage of the amp here. Obviously if you're using the
motor as a generator, there's no need to actively commutate
the motor windings, but you do need to rectify the motor's
output if you want DC.

If the brushless motor has internal commutation electronics they may
need to be circumvented with diodes. Maybe some use external amps?


Sounds like we're imagining different classes of motor
here. When I hear "brushless motor" I think servo, where
the motor typically includes only the field windings, Hall
sensors to provide rotor position feedback for commutation,
and an encoder for position and velocity feedback to the
servo control loop. The power and control circuitry is
external to the motor.

I forget there are other motors for less demanding apps
that integrate the drive electronics in the motor package
with only the power leads, and perhaps speed control lines,
brought out. I suspect this is the sort of motor you're
talking about when you refer the "internal elex"?

Rectifying the output of the servo type would be trivial,
though the issue of voltage regulation you mentioned
elsewhere remains. As you say, not so simple for the
integrated motor.

Ned Simmons

On Fri, 29 Aug 2003 00:51:51 +0100, wrote:
If we now run this machine as a generator AT THE SAME SPEED it
will generate the SAME 80v EMF. Note that this 80v is an open circuit
voltage which is the result of the rate at which armature conductors
cross the flux linkages, it is independent of the value of the
armature resistance. However it has to pass through the armature
resistance before reaching the output terminals At the maximum
permissible 1A load current, 20v will be lost leaving a net loaded
output voltage of only 60v = 60w i.e 60% of the rated input power of
the same machine used as a motor.

Right, but not 60% of its input power as a generator. Efficiency is still
80%, you've just arbitrarily limited the amount of input power you're
allowing to be applied to the generator input shaft. Efficiency is still
always the ratio of output power to input power, and now the input
power is at the shaft instead of at the winding terminals.

Gary

In article , Gary Coffman says...

On Fri, 29 Aug 2003 00:51:51 +0100, wrote:
If we now run this machine as a generator AT THE SAME SPEED it
will generate the SAME 80v EMF. Note that this 80v is an open circuit
voltage which is the result of the rate at which armature conductors
cross the flux linkages, it is independent of the value of the
armature resistance. However it has to pass through the armature
resistance before reaching the output terminals At the maximum
permissible 1A load current, 20v will be lost leaving a net loaded
output voltage of only 60v = 60w i.e 60% of the rated input power of
the same machine used as a motor.

Right, but not 60% of its input power as a generator. Efficiency is still
80%, you've just arbitrarily limited the amount of input power you're
allowing to be applied to the generator input shaft. Efficiency is still
always the ratio of output power to input power, and now the input
power is at the shaft instead of at the winding terminals.

Yes and for his conditions the input shaft speed is indeed
specified as being the "SAME SPEED" as the motor speed on
the nameplate - he made that even clear enough that it sunk
in for me. So for his given boundary conditions the explaination
was quite enlightnening for me.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

On 29 Aug 2003 10:30:24 -0700, jim rozen wrote:
In article , Gary Coffman says...
On Fri, 29 Aug 2003 00:51:51 +0100, wrote:
If we now run this machine as a generator AT THE SAME SPEED it
will generate the SAME 80v EMF. Note that this 80v is an open circuit
voltage which is the result of the rate at which armature conductors
cross the flux linkages, it is independent of the value of the
armature resistance. However it has to pass through the armature
resistance before reaching the output terminals At the maximum
permissible 1A load current, 20v will be lost leaving a net loaded
output voltage of only 60v = 60w i.e 60% of the rated input power of
the same machine used as a motor.

Right, but not 60% of its input power as a generator. Efficiency is still
80%, you've just arbitrarily limited the amount of input power you're
allowing to be applied to the generator input shaft. Efficiency is still
always the ratio of output power to input power, and now the input
power is at the shaft instead of at the winding terminals.

Yes and for his conditions the input shaft speed is indeed
specified as being the "SAME SPEED" as the motor speed on
the nameplate - he made that even clear enough that it sunk
in for me. So for his given boundary conditions the explaination
was quite enlightnening for me.

Yeah, I'd picked up on his same speed and same current (torque)
parameters. What I objected to was the statement that *efficiency*
as a generator was only 64%. That was using inputs and outputs
incorrectly, ie input as a motor and output as a generator. You can't
do that when calculating efficiency. It is always power out divided
by power in *at the same instant*. To use figures from two different
modes of operation doesn't give efficiency.

Note also that it isn't mandatory to follow his limits of same RPM
or same current either. Those were just his arbitrary choices.
Increasing either speed (voltage) or torque (current) will allow
greater input power, and greater output power, up to the thermal
limits of the motor/generator.

Gary

Gary sez:

" Note also that it isn't mandatory to follow his limits of same RPM
or same current either. Those were just his arbitrary choices.
Increasing either speed (voltage) or torque (current) will allow
greater input power, and greater output power, up to the thermal
limits of the motor/generator."

I thought Jim Pentagrid covered that very well myself. His choices were
made in accordance with preserving the parameters addressed in the original
question. As Jim Rozen stated, boundary conditions were specified at the
outset.

Bob Swinney

On Sat, 30 Aug 2003 04:30:57 +0100, wrote:

On Fri, 29 Aug 2003 22:49:30 GMT, (Gary Coffman)
wrote:

On 29 Aug 2003 10:30:24 -0700, jim rozen wrote:
In article , Gary Coffman says...
On Fri, 29 Aug 2003 00:51:51 +0100, wrote:
If we now run this machine as a generator AT THE SAME SPEED it
will generate the SAME 80v EMF. Note that this 80v is an open circuit
voltage which is the result of the rate at which armature conductors
cross the flux linkages, it is independent of the value of the
armature resistance. However it has to pass through the armature
resistance before reaching the output terminals At the maximum
permissible 1A load current, 20v will be lost leaving a net loaded
output voltage of only 60v = 60w i.e 60% of the rated input power of
the same machine used as a motor.

Right, but not 60% of its input power as a generator. Efficiency is still
80%, you've just arbitrarily limited the amount of input power you're
allowing to be applied to the generator input shaft. Efficiency is still
always the ratio of output power to input power, and now the input
power is at the shaft instead of at the winding terminals.

Yes and for his conditions the input shaft speed is indeed
specified as being the "SAME SPEED" as the motor speed on
the nameplate - he made that even clear enough that it sunk
in for me. So for his given boundary conditions the explaination
was quite enlightnening for me.

Yeah, I'd picked up on his same speed and same current (torque)
parameters. What I objected to was the statement that *efficiency*
as a generator was only 64%. That was using inputs and outputs
incorrectly, ie input as a motor and output as a generator. You can't
do that when calculating efficiency. It is always power out divided
by power in *at the same instant*. To use figures from two different
modes of operation doesn't give efficiency.

Note also that it isn't mandatory to follow his limits of same RPM
or same current either. Those were just his arbitrary choices.
Increasing either speed (voltage) or torque (current) will allow
greater input power, and greater output power, up to the thermal
limits of the motor/generator.

Gary


PLEASE read the original post carefully! The post did NOT state
that the efficiency as a generator was only 64%.

The postscript to the post commented that increased power could be
obtained by increased speed.

Increased current is not permissible because the load current is
chosent to be equal to the rated full load current when operating as a
motor. Currents higher than this will normally result in overheating.

Reading carefully is definitely helpful! Confusion arises from the
"same speed" assumption. I agree to the same-current constraint
but the speedlimit should be no-load speed which the motor presumably
can handle without damage. If the motor operated as a generator at
no-load speed then generated EMF is rated terminal voltage, terminal
voltage is that minus IR drop or 80%. We now have same current
hence same losses and same torque, same power in and power out
hence same efficiency, terminal voltage is 80% of nameplate rating
when run as a generator.

Both Gary and Jim allude to thermal limit hence max current. I
concur. They both also noted that spinning the motor faster as a
genny makes more volts (more power in, more power out) while loss is
proportional only to current. In this case it would actually be
*more* efficient, the situation being analogous to operating a motor
at 20% above rated voltage with same torque and current. In Jim's
example, if it were spun 20% faster than no-load speed then it would
produce rated terminal voltage at rated current. I doubt if 20%
overspeed would present a problem other than reducing brush and
bearing life a bit.