What passes as Pulse Width Modulation in DC Motor Control?

[Phil Allison]

"John Fields"
"Phil Allison"

The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.



** Completely wrong.

PWM of the DC supply to a motor has the same effect on available torque
and
free running rpm as varying the DC voltage to that motor.


---
Not true,

** It is totally true.

since lowering the voltage into the motor in order to reduce
its speed will also lower its stall torque and cause excessive power
dissipation and heating to occur if the lowered stall torque is
exceeded.

** Same goes for PWM drive too.

PWM'ing the motor will cure that problem by allowing the higher stall
torque available using the higher voltage

** Total BOLLOCKS.

The available torque is the SAME as as with the equivalent DC voltage
input.

Fields is blathering.



....... Phil

 

[Sjouke Burry]

John Fields wrote:

On Wed, 28 Jan 2009 10:35:14 +1100, "Phil Allison"
philallison@tpg.com.au> wrote:

"John Fields"

The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.


** Completely wrong.

PWM of the DC supply to a motor has the same effect on available torque and
free running rpm as varying the DC voltage to that motor.


---
Not true, since lowering the voltage into the motor in order to reduce
its speed will also lower its stall torque and cause excessive power
dissipation and heating to occur if the lowered stall torque is
exceeded.

PWM'ing the motor will cure that problem by allowing the higher stall
torque available using the higher voltage while using the width of the
pulse to determine how far the armature can travel in one PWM period.

That is, how fast it goes.

JF
You have just demonstrated that you know

nothing of electricity and motors.
Please study some, before you try to
correct people.
Or they will see you as a troll.

 

[Jon Slaughter]

"John Fields" <jfields@austininstruments.com> wrote in message
news:m4n1o4pq2ucj5nkcm5htc6nomtbtpkd7ov@4ax.com...

On Wed, 28 Jan 2009 10:35:14 +1100, "Phil Allison"
philallison@tpg.com.au> wrote:


"John Fields"


The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.



** Completely wrong.

PWM of the DC supply to a motor has the same effect on available torque
and
free running rpm as varying the DC voltage to that motor.


---
Not true, since lowering the voltage into the motor in order to reduce
its speed will also lower its stall torque and cause excessive power
dissipation and heating to occur if the lowered stall torque is
exceeded.

A motor has maximum torque at stall. Lowering the voltage reduces speed and
increases torque.

The reason has to do with the back emf. When the motor is stalled there is
no back-emf to resist current and hence large current can flow and hence a
large torque(since it is proportional to the current). As current flows a
back-emf counters it reducing the overall current.

If say, you lower speed by loading the motor so it can't turn then it will
overhead it. (this is easily demonstrated by jaming a fan and watching it
burn up)

But PWM is different!!! It doesn't load the motor to lower the speed but
reduces the current!! Hence at low speeds there is low average current but
high peak current since the rpms are low.

e.g., suppose the motor draws 1A at stall.

If we PWM at a duty cycle of d then d will control the speed(it will be
approximately proportional assuming no loading effect).

At, say, d of 1/100 which the motor turns slowly it will draw 1A but only
for 1/100 of the cycle. The average current is 10mA. This is definitely not
enough to get the motor to speed up.

What happens is you are "pulsing" the motor with high peak currents but low
average currents. An example is turning a bicycle wheel by your hand. To
keep it going fast you have to "pulse" and keep it up.. you can only get it
to go so fast though. Eventually it's inertia and your hand speed keep it
from going any faster.

If you grabed the wheel for only 1us and turned it with a huge force it
would be the same as some weak kid turning it continuously with a small
force. You might cause it to go fast quickly but only for that small time
frame.. for the rest of the time it is not getting any force(unlike with the
kid).

So even at stall speeds, while we are drawing a large current, because it is
using PWM the average current is low. (the peak current is still important
for practical matters though)

You have to realize that the "impedence" of the motor depends inversely on
the angular velocity. It is independent of the duty cycle. The PWM basically
prevents enough average current to cause it to spin up to speed(again, even
though high peak currents occur).

About the only thing you can say is that at low speeds you have high peak
currents and vice versa. PWM is simply controlling the average current.

 

[Dan Coby]

amdx wrote:

"Dan Coby" <adcoby@earthlink.net> wrote in message
news:N5adnf0AQc5yiB3UnZ2dnUVZ_hadnZ2d@earthlink.com...
amdx wrote:
"Jamie" <jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote in
message news:heNfl.27335$2o3.14283@newsfe10.iad...
Rich wrote:

Can anyone please tell me what passes as PWM in Motor Control?

Now that it's all sorted out.
Assuming a PWM 50% duty cycle,
If I put an ampmeter in the positive lead from the battery
and a second ampmeter in series with the motor,
Will the two ampmeters read the same?
No. That may seem to be a little strange but it is true.

Lets take an example. We have an ideal motor (i.e. no resistance losses,
etc.) which is being driven by an ideal PWM (i.e. a PWM with no losses).
Also let us assume that the PWM is being powered by 20 volt DC supply and
the PWM is switching the motor drive between 20 volts and 0 volts. And
we are driving the motor with a 10% duty cycle and that the average motor
current is 3 amps. We will also assume that the PWM frequency is high
enough that there no significant changes in the motor speed and current
during a cycle. (This is usually a reasonable assumption. PWMs are
usually run at a high enough frequency to make this true. See below.)

The average voltage at the motor is 2 volts (20v times 10%). The power
in the motor is 3 amps times 2 volts = 6 watts. Since there are no losses
with our ideal motor and ideal PWM, then the power from the 20 volts power
supply must also be 6 watts. The power from the supply is only provided
while the PWM is turned on (i.e. during the 10% on part of the cycle).
During the on part of the cycle we have to be providing 60 watts of power
(6 watts / 10% = 60 watts). This means that we are getting 3 amps (60
watts
divided by 20 volts during the 'on' cycle. (So in this sense the
currents are equal since we are putting 3 amps into the motor and we are
drawing 3 amps from the power supply during the 'on' cycle.) HOWEVER the
ammeter connected to the power supply will indicate the average current
from the supply. Since we are supply 3 amps for 10% of the cycle and
0 amps for 90% of the cycle, we will an average current of 0.3 amps.

So the average motor current will be 3 amps and the average power supply
current will be 0.3 amps.

This leaves a few questions:

1) What is supplying the 3 amps of motor current during the 90% part of
the duty cycle when the PWM circuit is 'off'?

The PWM is switching the drive between 20 volts and 0 volts. During the
'off' part of the cycle, the 3 amps is being pulled from the 0 volt side
of the PWM (i.e. the motor is pulling current from the circuit ground).


2) Is the motor's current and speed really constant the entire PWM cycle?

In theory, no. There is always a small amount of speed and current ripple
but the amount of ripple can be made very small by increasing the PWM
frequency. Since we are assuming an ideal motor (i.e. no resistance
losses)
then our motor looks like it has a back EMF and a series inductance.
the back EMF of the motor will be equal to the 2 volt average voltage at
the motor. (A real motor with a non-zero winding resistance would have
a lower back EMF since there would be some voltage drop across the winding
resistance.)

During the 'on' part of the PWM cycle, there is an 18 volt difference (20
volt
power supply versus a 2 volt back EMF) between the motor's back EMF and
the
power supply. This voltage difference will try to increase the current
flowing
into the motor. The inductance of the motor will limit the increase in
current.
But there is a small (usually very small) increase in current during the
'on'
part of the cycle. This increased current will try to accelerate the
motor.
However the inertia of the motor and its connected load will oppose the
acceleration but there will be a small change in speed.

During the 'off' part of the cycle, there is -2 volt difference between
the
motor's back EMF and the drive voltage from the PWM circuit. This -2
volts
will try to decrease the motor current. The decrease in current will
decrease
the motor's torque and motor will start to slow down. Once again, the
motor's
inductance will oppose the change in the motor's current. The inertia of
the
motor and its connected load will oppose the change in speed. The rate of
decrease in current speed will be 1/9 (18 volts version 2 volts) of the
rate
of increase in speed and current during the on part of the cycle but it
also
lasts 9 times as long. Thus the total increases and decreases will
balance.

So there is some small changes in motor current and speed during the PWM's
period. However if the PWM frequency is 100 kHz and the motor and its
connected
load weigh 10 pounds, the changes in speed will be very very small.


A couple of final notes:

Please note that PWM circuits are not magical, even though we are only
drawing an average of 0.3 amps from the power supply, we are drawing
6 watts. This is the same 6 watts that is being provided by the motor
to its load. We are not getting something for nothing.

In much of the previous analysis, it is very important that we are looking
at a motor being driven by a PWM. If, instead of motor, we are driving a
simple resistor, then the current through the resistor and the current
from the power supply would be exactly the same. The resistor does not
store any energy. However the motor stores energy in both its inductance
and the inertia of the motor and its connected load.


Dan

Hi Dan,
I new the answer, I just wanted to see the responses to the inquiry.
It took me a while to get my head around it. When someone finally
pointed out Power in = Power out. Duh! So if we have 20v battery pack
flowing .3 amps that's 6 watts. If we have a 50% duty cycle
the average voltage is 10 volts so the current must be .6amps.
You described it well, especially where that current comes from.

There are a few more complications.

Several things happen with the power into a motor. A real motor has several
types of losses. There is power lost into heating the resistance of the motor
windings. There is some power lost in circulating currents in the magnetic
core and other magnetic losses There are friction losses in the bearings in
the motor and its attached load. And finally there is the actual power which
is delivered to the load, i.e. the real usable power from the motor.

With an ideal motor we can ignore all of the losses so we consider the power
into the motor being equal to the power out. (That is not a very good assumption
but it makes the analysis much easier.)

In the earlier discussion, I mentioned the 'back EMF' of the motor without
describing any of its details. The term EMF is short for 'electromotive force'.
It is basically a voltage that is created as the motor turns. It is usually
called 'back EMF' since it opposes the current flow into the motor. The faster
the motor turns, the higher the back EMF. For ideal motors, the back EMF is
proportional to speed. With an ideal motor, the power out of the motor is
equal to the back EMF times the motor current. (With non ideal motors, you
have to subtract off the losses in the magnetics, friction, etc.)

In the earlier discussion, you should note that with our ideal motor, the back
EMF was equal to the average voltage from the PWM circuit. If you increase
the duty cycle of the PWM to 50% then the average voltage will increase to
10 volts (as you said). The motor will respond to this increase in voltage
by increasing its speed until the back EMF is also 10 volts (or a little less
with a non ideal motor). This means that the motor speed increases by a factor
of 5.

For many loads that are connected to motors, the increase in speed will also
increase the power required to drive the load at the higher speed. This
higher power required for the load will mean that the power delivered to the
motor must also increase. How the power requirements of the load vary with
speed depends upon the nature of the load. If the power required to drive the
load at a 5 times faster speed also increases by a factor of 5 then will need
to deliver 30 watts (6 watts times 5) to the load. This means that the motor
current stay the same (3 amps). The current from the power supply will increase
to 1.5 amps (30 watts divided by 20 volts).

For many types of loads, the power requirements will increase with speed at a
rate higher higher than simple proportionality. For instance if the motor is
driving an electric car, the power requirement increase rapidly with speed
since a higher speed implies both a larger distance per unit time and higher
drag forces (and power is force times distance per unit time).

The final result is that increasing the duty cycle will increase the motor
speed but probably not reduce the power requirements.


Dan

 

[Phil Allison]

"Jon Slaughter"

" Thesis on DC Motors"
---------------------------

A motor has maximum torque at stall. Lowering the voltage reduces speed and
increases torque.

The reason has to do with the back emf. When the motor is stalled there is
no back-emf to resist current and hence large current can flow and hence a
large torque(since it is proportional to the current). As current flows a
back-emf counters it reducing the overall current.

If say, you lower speed by loading the motor so it can't turn then it will
overhead it. (this is easily demonstrated by jaming a fan and watching it
burn up)

But PWM is different!!! It doesn't load the motor to lower the speed but
reduces the current!! Hence at low speeds there is low average current but
high peak current since the rpms are low.

e.g., suppose the motor draws 1A at stall.

If we PWM at a duty cycle of d then d will control the speed(it will be
approximately proportional assuming no loading effect).

At, say, d of 1/100 which the motor turns slowly it will draw 1A but only
for 1/100 of the cycle. The average current is 10mA. This is definitely not
enough to get the motor to speed up.

What happens is you are "pulsing" the motor with high peak currents but low
average currents. An example is turning a bicycle wheel by your hand. To
keep it going fast you have to "pulse" and keep it up.. you can only get it
to go so fast though. Eventually it's inertia and your hand speed keep it
from going any faster.

If you grabed the wheel for only 1us and turned it with a huge force it
would be the same as some weak kid turning it continuously with a small
force. You might cause it to go fast quickly but only for that small time
frame.. for the rest of the time it is not getting any force(unlike with the
kid).

So even at stall speeds, while we are drawing a large current, because it is
using PWM the average current is low. (the peak current is still important
for practical matters though)

You have to realize that the "impedence" of the motor depends inversely on
the angular velocity. It is independent of the duty cycle. The PWM basically
prevents enough average current to cause it to spin up to speed(again, even
though high peak currents occur).

About the only thing you can say is that at low speeds you have high peak
currents and vice versa. PWM is simply controlling the average current.


--------------------------------------------------------------------------------

 

[John Fields]

On Thu, 29 Jan 2009 11:20:54 +1100, "Phil Allison"
<philallison@tpg.com.au> wrote:

"John Fields"
"Phil Allison"

The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.



** Completely wrong.

PWM of the DC supply to a motor has the same effect on available torque
and
free running rpm as varying the DC voltage to that motor.


---
Not true,


** It is totally true.


since lowering the voltage into the motor in order to reduce
its speed will also lower its stall torque and cause excessive power
dissipation and heating to occur if the lowered stall torque is
exceeded.

** Same goes for PWM drive too.


PWM'ing the motor will cure that problem by allowing the higher stall
torque available using the higher voltage


** Total BOLLOCKS.

The available torque is the SAME as as with the equivalent DC voltage
input.

Fields is blathering.

---
I suggest you try an experiment.

Get a DC motor which you can't keep from rotating by pressing your thumb
and forefinger against its shaft with its rated voltage applied, and
then decrease the voltage into it until you can stop it.

Then, while keeping the input voltage constant, decrease the time the
input voltage is applied to the motor on a cycle-to-cycle basis while
you're trying to stop it with your thumb-to-forefinger brake until the
average input current is the same as in the DC case.

I'd be willing to bet that as slow as you can PWM the motor you won't be
able to stop it.

JF

 

[Phil Allison]

"John Fields"
"Phil Allison"

The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.


** Completely wrong.

PWM of the DC supply to a motor has the same effect on available torque
and free running rpm as varying the DC voltage to that motor.



** It is totally true.

Fields is blathering.

I suggest you try an experiment.

** I suggest YOU stop posting DAMN LIES !!

What you are now DISHONESTLY alluding to is ** NOT** " PWM drive".

Discontinuous pulses of current is ** NOT ** " PWM drive" .

As you very well know.

Asshole !!




........ Phil

 

[Joseph Sroka - 10.2.8]

In article <6ucd9aFeplb8U1@mid.individual.net>, "Phil Allison"
<philallison@tpg.com.au> wrote:

<SNIP>

So, PhilAss, write a better explanation, if you can.

AND, make it understandable to people who need/want BASICS.

I bet you can't.

--- Joe

--
Delete the second "o" to email me.

 

[Phil Allison]

"Joseph Sroka - 10.2.8"
"Phil Allison"

So, PhilAss, write a better explanation, if you can.

** You miss the point entirely - fuckhead.

I can and HAVE written good explantions of PWM drive.

One can easily find them on the net too.

But nothing quite as hilarious as this one from the Slaughter man.


-------------------------------------------------------------

"Jon Slaughter"

" Thesis on DC Motors"
---------------------------

A motor has maximum torque at stall. Lowering the voltage reduces speed and
increases torque.

The reason has to do with the back emf. When the motor is stalled there is
no back-emf to resist current and hence large current can flow and hence a
large torque(since it is proportional to the current). As current flows a
back-emf counters it reducing the overall current.

If say, you lower speed by loading the motor so it can't turn then it will
overhead it. (this is easily demonstrated by jaming a fan and watching it
burn up)

But PWM is different!!! It doesn't load the motor to lower the speed but
reduces the current!! Hence at low speeds there is low average current but
high peak current since the rpms are low.

e.g., suppose the motor draws 1A at stall.

If we PWM at a duty cycle of d then d will control the speed(it will be
approximately proportional assuming no loading effect).

At, say, d of 1/100 which the motor turns slowly it will draw 1A but only
for 1/100 of the cycle. The average current is 10mA. This is definitely not
enough to get the motor to speed up.

What happens is you are "pulsing" the motor with high peak currents but low
average currents. An example is turning a bicycle wheel by your hand. To
keep it going fast you have to "pulse" and keep it up.. you can only get it
to go so fast though. Eventually it's inertia and your hand speed keep it
from going any faster.

If you grabed the wheel for only 1us and turned it with a huge force it
would be the same as some weak kid turning it continuously with a small
force. You might cause it to go fast quickly but only for that small time
frame.. for the rest of the time it is not getting any force(unlike with the
kid).

So even at stall speeds, while we are drawing a large current, because it is
using PWM the average current is low. (the peak current is still important
for practical matters though)

You have to realize that the "impedence" of the motor depends inversely on
the angular velocity. It is independent of the duty cycle. The PWM basically
prevents enough average current to cause it to spin up to speed(again, even
though high peak currents occur).

About the only thing you can say is that at low speeds you have high peak
currents and vice versa. PWM is simply controlling the average current.

--------------------------------------------------------------------------------









........ Phil

AND, make it understandable to people who need/want BASICS.

I bet you can't.

--- Joe

--
Delete the second "o" to email me.

 

[Jasen Betts]

On 2009-01-27, Rich <notty@emailo.com> wrote:

Can anyone please tell me what passes as PWM in Motor Control?

turning the power on and off at a high rate such that the motor does
not slow aprecciably during the off periods or speed up appreciably
during the on periods.

Is it limited to supplying the motor with a square wave, where a set and
constant level of voltage is switched on or off to the motor?

no. eg thyristors are used for speed control of power tools.

Or can other waveforms consitutute PWM control of a dc motor? Such as where
voltage does not drop to zero, but has periodic peaks having square,
triangular, sawtooth or some other waveform?

PWM in general turns the power on and off.

Can PWM control consist of dc pulses that rise and fall in voltage in a
sinusoidal fashion?

I would not call that PWM.

At the moment I am believing that PWM is just on/off of a contant
level of dc voltage. But wonder if that belief is correct or not. Thanks.

drop the bit about "contant level of dc" and I'd be inclined to agree.

 

[Jasen Betts]

On 2009-01-27, John Fields <jfields@austininstruments.com> wrote:

On Tue, 27 Jan 2009 14:08:30 -0600, "Jon Slaughter"
Jon_Slaughter@Hotmail.com> wrote:


Since the power delivered is proportional to the speed(ideally) and hence
proprotional to the current, we would expect the speed to be reduced!!!

this is quite amazing!!!!! We can reduce the speed without costing any
power!!! If we used a resistor we would waste power in the resistor!!!!!

The method described is exactly what is done except a voltage controlled
switch, e.g. a mosfet, is used instead of our hand.

Note that if we do it too slow the motor will be jerky but if we do it
really fast it will seem continuous. This is because the motor has inertia
and for the switch was off it will still move as fast assuming it is not off
for too long.

---
Nice post.

The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.

Given that a stalled motor is electrically similar to an inductor with
a series resistance, and that torque is proportional to the current
through the motor; it's easy to show that PWM and proportionately reduced
voltage both givwe the same current, so there's no way that that claim is
correct.

 

[Rich]

"Rich" <notty@emailo.com> wrote in message
news:6uarovFe018lU1@mid.individual.net...

http://electronicdesign.com/Articles/Print.cfm?ArticleID=6315

What does anyone think the reference input is meant to be here it is
marked "Input voltage from computer"?

I don't understand why the input is meant to come from a computer.

Op amp "31" I take to be a differential amplifier.

Before I go any further is this correct?

A DIFFERENCE amplifier is a special case of DIFFERENTIAL amplifier. It
outputs the actual difference between the inputs only. ie. it has a gain of
1. A DIFFERENTIAL amplifier would output an AMPLIFIED difference.

A

"The current through the motor armature is caused to pass through a
resistance (rm/10) that is, for example, approximately 0.1 as large as the
ohmic resistance of the motor. The voltage across this resistance is then
amplified by a factor of approximately 10, and the resulting voltage is
added to a second voltage in a *differential* amplifier. This second
voltage is the voltage as measured across the two brushes of the motor."

(That says (21) is a DIFFERENTIAL amplifier.)

B
"The output of this amplifier is compared to the reference voltage (provided
externally to the circuit, which determines the speed of rotation of the
motor) in another *differential amplifier*. The output difference is used to
control the output of a power output stage that drives the motor. In this
way, the reference voltage is compared to the back-EMF and the motor is
caused to run at a constant speed set by the reference voltage. To soften
the switch from driving to not driving, a sawtooth waveform is superimposed
on the reference voltage."

(That says (31) is a DIFFERENTIAL amplifier.)

C
"In the schematic, the voltage across the motor is measured (amp 12),
*multiplied by minus one* and fed to one input of a *difference* amplifier
(amp 21) At the same time, the voltage across resistor rm/10 is measured and
multiplied by approximately minus ten (amp 11). This output is fed to the
other input of amp 21.

(That I think makes (12) a DIFFERENCE amplifier.)
(That makes (21) a DIFFERENCE amplifier)

D
"The output of amp 21 is then equal to the back-EMF of the motor (reconnect
the motor to the output stage and adjust the gain and stiffness controls to
suit your application). This output is fed into one input of a
*differential* amplifier (amp 31) and compared to a reference voltage
(provided externally). The output of this amplifier is the error signal and
is used to drive the output stage (amp 32, BC337, and BC327) to keep the
motor running at the speed at which the back-EMF equals the reference
voltage."

(That says (31) is a DIFFERENTIAL amplifier.)

ALLAM BELLS!! In "A" (21) is a DIFFERENTIAL amplifier. In C" (21) is a
DIFFERENCE amplifier.

I think (12) is a DIFFERENCE amp because it has a gain of -1.

I think (31) is definitely a DIFFERENTIAL amplifier.

I think (21) should be a DIFFERENCE amplifier

There is an error in circuit in article 6315. (31) should be marked
DIFFERENTIAL amplifier.What do you folks think?

Is (12) a DIFFERENCE amplifier?

 

[Bob Masta]

On Wed, 28 Jan 2009 10:24:06 -0600, "Jon
Slaughter" <Jon_Slaughter@Hotmail.com> wrote:

The easiest way to do PWM with arbitrary function without being to ambiguous
is to use periodic functions that are 0 after a portion of the period.

Actually, there is a much easier way, which is
what my previously-mentioned industrial DC motor
controller does: Monitor the output current (or
power, RPM, etc), compare to the setpoint, and
let feedback take care of the details. This
allows you to use any arbitrary waveform, even one
that is non-periodic.

Best regards,


Bob Masta

DAQARTA v4.51
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
FREE Signal Generator
Science with your sound card!

 

[Jon Slaughter]

"Bob Masta" <N0Spam@daqarta.com> wrote in message
news:4981a930.807047@news.sysmatrix.net...

On Wed, 28 Jan 2009 10:24:06 -0600, "Jon
Slaughter" <Jon_Slaughter@Hotmail.com> wrote:

The easiest way to do PWM with arbitrary function without being to
ambiguous
is to use periodic functions that are 0 after a portion of the period.

Actually, there is a much easier way, which is
what my previously-mentioned industrial DC motor
controller does: Monitor the output current (or
power, RPM, etc), compare to the setpoint, and
let feedback take care of the details. This
allows you to use any arbitrary waveform, even one
that is non-periodic.

But that isn't PWM and it isn't as efficient as rectangular PWM.

As my post proved that any function is equivalent to PWM. i.e., whatever
waveform is being generated by your feedback can be done using rectangular
PWM and since most likely(I didn't look at your circuit) your dissipating
power continuously in the component that is controlling the current your
wasting that power. It would only be efficient if the component has
extremely low power dissipation but chances are mosfets would be even more
efficient.

 

[John Fields]

On Thu, 29 Jan 2009 13:34:24 +1100, "Phil Allison"
<philallison@tpg.com.au> wrote:

"John Fields"
"Phil Allison"


The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.


** Completely wrong.

PWM of the DC supply to a motor has the same effect on available torque
and free running rpm as varying the DC voltage to that motor.



** It is totally true.

Fields is blathering.



I suggest you try an experiment.


** I suggest YOU stop posting DAMN LIES !!

What you are now DISHONESTLY alluding to is ** NOT** " PWM drive".

Discontinuous pulses of current is ** NOT ** " PWM drive" .

As you very well know.

Asshole !!

---
Well, let's see...


If I have a circuit that looks like this:



.. +V
.. |
.. [MOTOR]
.. |
..+-----+ D
..| OSC |-----G
..+-----+ S
.. |
.. GND

and the signal going into the gate looks like this:

_ _
__| |______________________________| |___________


then discontinuous pulses of current will flow through the motor, yes?

Moreover, if I change the width of the pulse, but keep the period of the
signal the same, like this:

________________ _____________
__| |_______________|

then discontinuous pulses of current will flow through the motor, yes?

Now, for the same load on the motor, the torque presented to the load
will be the same in both cases since the current will be the same, but
the speed will increase in the second case because the current is
allowed to flow for a longer time.

Is that not Pulse Width Modulation of the supply voltage?

JF

 

[Rich Grise]

On Thu, 29 Jan 2009 01:28:00 +0100, Sjouke Burry wrote:

John Fields wrote:
On Wed, 28 Jan 2009 10:35:14 +1100, "Phil Allison"
"John Fields"

The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.

** Completely wrong.

PWM of the DC supply to a motor has the same effect on available torque and
free running rpm as varying the DC voltage to that motor.
---
Not true, since lowering the voltage into the motor in order to reduce
its speed will also lower its stall torque and cause excessive power
dissipation and heating to occur if the lowered stall torque is
exceeded.

PWM'ing the motor will cure that problem by allowing the higher stall
torque available using the higher voltage while using the width of the
pulse to determine how far the armature can travel in one PWM period.

That is, how fast it goes.

You have just demonstrated that you know
nothing of electricity and motors.
Please study some, before you try to
correct people.
Or they will see you as a troll.

No, actually, John is right.
Are you Phalluson junior?

Thanks,
Rich

 

[Rich Grise]

On Wed, 28 Jan 2009 19:37:41 -0600, John Fields wrote:

On Thu, 29 Jan 2009 11:20:54 +1100, "Phil Allison"
"John Fields"
"Phil Allison"

The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.
** Completely wrong.

PWM of the DC supply to a motor has the same effect on available torque
and
free running rpm as varying the DC voltage to that motor.

Not true,

** It is totally true.

since lowering the voltage into the motor in order to reduce
its speed will also lower its stall torque and cause excessive power
dissipation and heating to occur if the lowered stall torque is
exceeded.

** Same goes for PWM drive too.

PWM'ing the motor will cure that problem by allowing the higher stall
torque available using the higher voltage

** Total BOLLOCKS.

The available torque is the SAME as as with the equivalent DC voltage
input.

Fields is blathering.

I suggest you try an experiment.

Get a DC motor which you can't keep from rotating by pressing your thumb
and forefinger against its shaft with its rated voltage applied, and
then decrease the voltage into it until you can stop it.

Then, while keeping the input voltage constant, decrease the time the
input voltage is applied to the motor on a cycle-to-cycle basis while
you're trying to stop it with your thumb-to-forefinger brake until the
average input current is the same as in the DC case.

I'd be willing to bet that as slow as you can PWM the motor you won't be
able to stop it.

Why do you even try to talk sense to Phalluson?

I _know_ from personal experience that this is true - I slapped together
a simple PWM controller once, and just drove the motor with a regulated
current driver, PWM'd of course. It held full torque down to almost zero
RPM.

In fact, I almost built a Prony brake just to test it, but it turned out
that that was unnecessary, because it worked perfectly in the prototype.

Cheers!
Rich

 

[Rich Grise]

On Thu, 29 Jan 2009 08:22:17 +0000, Jasen Betts wrote:

On 2009-01-27, John Fields <jfields@austininstruments.com> wrote:

The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.

Given that a stalled motor is electrically similar to an inductor with
a series resistance, and that torque is proportional to the current
through the motor; it's easy to show that PWM and proportionately reduced
voltage both givwe the same current, so there's no way that that claim is
correct.

OK, Mr. know-it-all, Put your money where your mouth is. Show us.

Good Luck!
Rich

 

[John Fields]

On Thu, 29 Jan 2009 18:56:16 GMT, Rich Grise <rich@example.net> wrote:

On Wed, 28 Jan 2009 19:37:41 -0600, John Fields wrote:
On Thu, 29 Jan 2009 11:20:54 +1100, "Phil Allison"
"John Fields"
"Phil Allison"

The other thing that's nice about PWM'ing a motor is that there's no
loss in torque as speed is lowered as it is when voltage is lowered.
** Completely wrong.

PWM of the DC supply to a motor has the same effect on available torque
and
free running rpm as varying the DC voltage to that motor.

Not true,

** It is totally true.

since lowering the voltage into the motor in order to reduce
its speed will also lower its stall torque and cause excessive power
dissipation and heating to occur if the lowered stall torque is
exceeded.

** Same goes for PWM drive too.

PWM'ing the motor will cure that problem by allowing the higher stall
torque available using the higher voltage

** Total BOLLOCKS.

The available torque is the SAME as as with the equivalent DC voltage
input.

Fields is blathering.

I suggest you try an experiment.

Get a DC motor which you can't keep from rotating by pressing your thumb
and forefinger against its shaft with its rated voltage applied, and
then decrease the voltage into it until you can stop it.

Then, while keeping the input voltage constant, decrease the time the
input voltage is applied to the motor on a cycle-to-cycle basis while
you're trying to stop it with your thumb-to-forefinger brake until the
average input current is the same as in the DC case.

I'd be willing to bet that as slow as you can PWM the motor you won't be
able to stop it.

Why do you even try to talk sense to Phalluson?

I _know_ from personal experience that this is true - I slapped together
a simple PWM controller once, and just drove the motor with a regulated
current driver, PWM'd of course. It held full torque down to almost zero
RPM.

---
Yup.

I once did a variable-speed motor drive for an orbital welder, and at
low RPM the only way to get enough torque to rotate the weldhead, and to
rotate it uniformly and consistently, from weld to weld was to use PWM.

Matter of fact, even at higher RPM, PWM was the only way to get uniform
and consistent welds.
---

In fact, I almost built a Prony brake just to test it, but it turned out
that that was unnecessary, because it worked perfectly in the prototype.

Cheers!
Rich
JF

 

[Phil Allison]

"John Fields"


** I suggest YOU stop posting DAMN LIES !!

What you are now DISHONESTLY alluding to is ** NOT** " PWM drive".

Discontinuous pulses of current is ** NOT ** " PWM drive" .

As you very well know.

YOU LYING ASSHOLE !!



........ Phil