What passes as Pulse Width Modulation in DC Motor Control?

http://electronicdesign.com/Articles/Print.cfm?ArticleID=6199

MOSFETS! This is PWM controller with speed regulation and the better circuit
for slow rotor speeds??

I'm probably wrong! :c)
 
"Rich" <notty@emailo.com> wrote in message
news:6u9midFdv7tpU1@mid.individual.net...
http://electronicdesign.com/Articles/Print.cfm?ArticleID=6199

MOSFETS! This is PWM controller with speed regulation and the better
circuit
for slow rotor speeds??

I'm probably wrong! :c)
You need to start with the basics first:

http://en.wikipedia.org/wiki/H-bridge

If you don't understand what most of that stuff is in the first place then
it can get confusing. It helps to understand the basic's before worrying
about practical details.
 
Thanks. I think what you wrote is beginng to help me understand that
particular circuit. I need to ponder what you said until I fully grasp
things.

As a novice, I think I just happened to pick on a tricky circuit to
explain.
I was looking for a PWM circuit that had speed regulation and I thought
that
perhaps that circuit was such a circuit. It now seem I have the answer I
have been seeking - this is not a PWM controller that will drive well at
slow speeds, unlike PWM that tends to do that quite well.
No, it might be fine at slow speeds? I didn't see anything that showed other
wise.

Slow speeds just me's low voltage or current to the motor.

Using PWM that means choosing a low duty cycle(so very low current actually
gets through on average).

For the bjt circuit it means low average current.






Just one more thing: What do you think the reference input is meant to be
where it is marked "Input voltage from computer"?
Well, looks like it is the voltage that sets the motor current. Do you know
anything about op amps? If not you should learn them before trying to learn
something that uses them.

If you understand mosfets it should be pretty easy to see what's going on.
(specially since most of the ideas are labeled)
 
On Tue, 27 Jan 2009 13:38:51 -0000, "Rich" <notty@emailo.com> wrote:

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

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?

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?

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

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.


I don't do motors but there may be insight for someone in my writing.

I have been working with PWM the past two months as a hobbyist using a
PIC after this group SEB recommended the PIC approach. It is a simple
device yet a marvelous compliment reflecting the genius of modern
engineers. Many of my future projects will utilize the PIC now that I
have passed through the learning curve.

PWM creates a controlled average voltage to a device by "varying the
duty cycle" using any fixed frequency. What varies is how long the
voltage is on. If you chose one hertz for the PWM frequency (which
could be any freq depending on the app) and then you pulsed the PWM ON
for 1/100th of a second then the load would get 1% of the available
voltage/energy. If the pulse was ON for 99/100th of a second then OFF
for 1/100th of a second then the load has 99% of the available
voltage/energy.

This approach has advantages in that load gets good torque if needed
with most of the energy to the device load and not lost like across a
variable resistance.

This biggest advantage for "me" is PWM can be generated ideally by a
Computer/PIC output. It transforms a dumb device/circuit into one
that can monitor and correct for environmental changes. I use one
diode and one capacitor on the output of the PWM to get a smooth +4v
to +8v controlled voltage sweep. This drives a circuit that requires
only one milliamp.

I have never worked in electronics so go easy gentlemen if some of my
terminology is weak.


* * * *

Christopher

Temecula CA.USA
http://www.oldtemecula.com
 
"Jon Slaughter" <Jon_Slaughter@Hotmail.com> wrote in message
news:6CNfl.923$Lr6.920@flpi143.ffdc.sbc.com...
"Rich" <notty@emailo.com> wrote in message
news:6u9midFdv7tpU1@mid.individual.net...
http://electronicdesign.com/Articles/Print.cfm?ArticleID=6199

MOSFETS! This is PWM controller with speed regulation and the better
circuit
for slow rotor speeds??

I'm probably wrong! :c)


You need to start with the basics first:

http://en.wikipedia.org/wiki/H-bridge

If you don't understand what most of that stuff is in the first place then
it can get confusing. It helps to understand the basic's before worrying
about practical details.
H -bridge is a *switch* arrangement.

Does not that imply PWM?

In the bjt circuit, is not the transistor acting as a kind of variable
resistance? Therefore not PWM?
 
"Rich" <notty@emailo.com> wrote in message
news:6u9oghFecftrU1@mid.individual.net...
"Jon Slaughter" <Jon_Slaughter@Hotmail.com> wrote in message
news:6CNfl.923$Lr6.920@flpi143.ffdc.sbc.com...

"Rich" <notty@emailo.com> wrote in message
news:6u9midFdv7tpU1@mid.individual.net...
http://electronicdesign.com/Articles/Print.cfm?ArticleID=6199

MOSFETS! This is PWM controller with speed regulation and the better
circuit
for slow rotor speeds??

I'm probably wrong! :c)


You need to start with the basics first:

http://en.wikipedia.org/wiki/H-bridge

If you don't understand what most of that stuff is in the first place
then it can get confusing. It helps to understand the basic's before
worrying about practical details.

H -bridge is a *switch* arrangement.

Does not that imply PWM?
To answer my own question:

No, not necessarily. In this case H bridge is used simply to change motor
direction, not as a switching element in PWM.

Circuit "not* PWM.

Okay, the only circuits I can find use tachometer control. I've not seen any
PWM circuits that speed regulate using the monitoring of voltage across
armature, or current through armature.
 
"Rich" <notty@emailo.com> wrote in message
news:6u9pvbFe2g8fU1@mid.individual.net...

Okay, the only circuits I can find use tachometer control. I've not seen
any PWM circuits that speed regulate using the monitoring of voltage
across armature, or current through armature.
I stand corrected: I think this is one:

http://electronicdesign.com/Articles/Print.cfm?ArticleID=5763

Only one I seem to have found. I'll see if I can make it work with my 12v
2.5A DC brushed motor.
 
"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?
Mike
 
Rich wrote:

"Rich" <notty@emailo.com> wrote in message
news:6u9pvbFe2g8fU1@mid.individual.net...

Okay, the only circuits I can find use tachometer control. I've not
seen any PWM circuits that speed regulate using the monitoring of
voltage across armature, or current through armature.


I stand corrected: I think this is one:

http://electronicdesign.com/Articles/Print.cfm?ArticleID=5763

Only one I seem to have found. I'll see if I can make it work with my
12v 2.5A DC brushed motor.
Wow, you finally found a PWM motor, circuit.

btw, that's not reversing.. (regenerative).

http://webpages.charter.net/jamie_5"
 
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
 
"Jamie" <jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote in message
news:a6Pfl.8884$8O.2380@newsfe06.iad...
Rich wrote:


"Rich" <notty@emailo.com> wrote in message
news:6u9pvbFe2g8fU1@mid.individual.net...

Okay, the only circuits I can find use tachometer control. I've not seen
any PWM circuits that speed regulate using the monitoring of voltage
across armature, or current through armature.


I stand corrected: I think this is one:

http://electronicdesign.com/Articles/Print.cfm?ArticleID=5763

Only one I seem to have found. I'll see if I can make it work with my
12v 2.5A DC brushed motor.

Wow, you finally found a PWM motor, circuit.

btw, that's not reversing.. (regenerative).
My original spec was for a bi-directional, PWM, constant speed, and *not* to
rely on a tachometer.

My motors seem to be designed for one direction only, so I'll ditch the
bi-directional requirement.

It also may be that perhaps PWM might not be so critical for slow speed. I
think I'll build two circuits:

http://electronicdesign.com/Articles/Print.cfm?ArticleID=5763

PWM, constant speed, no tachometer.

http://electronicdesign.com/Articles/Print.cfm?ArticleID=6199

Not PWM, consant speed, no tachometer. But it might be perfectly okay.

http://electronicdesign.com/Articles/Print.cfm?ArticleID=6168

I might even do this, just to mess with tachometer. But there will be plenty
of other tachometer-based circuits.

But, in general I seem to have found it hard to discover a circuit that's
PWM with constant speed control, that's *not* tachometer based.
 
Should have written:

PWM, constant speed, no tachometer:

http://electronicdesign.com/Articles/Print.cfm?ArticleID=5763

Not PWM, constant speed, no tachometer. But it might be perfectly okay:

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

Not PWM, constant speed, bi-directional:

http://electronicdesign.com/Articles/Print.cfm?ArticleID=6199

I might even do this, just to mess with tachometer. But there will be plenty
of other tachometer-based circuits:

http://electronicdesign.com/Articles/Print.cfm?ArticleID=6168

Although it's not easy to read the circuits. Designed that way I think.
 
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?
 
On Tue, 27 Jan 2009 13:38:51 -0000, "Rich"
<notty@emailo.com> wrote:

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

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?

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?

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

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.
All of the posts about the definition of PWM being
rectangular pulses of contant amplitude are true,
but nevertheless I suspect that is not what is
going on in most real-world industrial DC motors.

For example, I have a 0.5 HP Balador industrial DC
motor... weighs about 25 pounds. The speed
controller for it (Graham Transmissions Vari Speed
S1000 DC Control) is a dinky little board with a
100 mA transformer and a handful of TO-220 devices
on a short length of aluminum angle-bracket heat
sink.

A moment's consideration indicates that, no, the
controller is *NOT* creating constant-level DC
capable of supplying the current needs of a
half-horse motor. Looking at the block diagram,
it's clear that this is a fancy lamp dimmer.
There is a tiny regulated supply, a zero-crossing
detector and a pulse generator driving a pair of
SCRs, together with current sense, feedback, error
integrator, and comparator.

The motor is getting the (rectified) 60 Hz AC
line, but appropriately chopped to supply the
proper overall power to the motor. In other
words, there is not really anything DC about it.

This may not satisfy a dictionary definition of
PWM, but that term does convey the basic idea of
how a tiny circuit can control a huge current hog
without meltdown. The basic idea is the same: You
are chopping the supply such that the duty cycle
(rather than amplitude) controls the average power
output.

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!
 
"Bob Masta" <N0Spam@daqarta.com> wrote in message
news:49805c3f.2363539@news.sysmatrix.net...
On Tue, 27 Jan 2009 13:38:51 -0000, "Rich"
notty@emailo.com> wrote:

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

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?

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?

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

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.

All of the posts about the definition of PWM being
rectangular pulses of contant amplitude are true,
but nevertheless I suspect that is not what is
going on in most real-world industrial DC motors.
I don't think there is any real definition that says PWM must us a
rectangular pulse. This is just what is the easiest to use and hence most
common. (hence it is almost the definition)

You can obviously do "PWM" with any function. Simply the time it is at low
verses the time it is not gives the duty cycle and it will always be some
fraction of the duty-cycle when compared to the function switching exactly
between the low and hi state. But then any function can be considered PWM.
(the point here is that the output is at some fraction of the input and that
fraction depends on the ratio of the area to the total area)

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.

i.e., f(t)*H(d*P - t). No matter how crazy f(t) is, the time average is a
fraction of d. This is because I clearly defined the off state but not the
on state. With rectangular pulses both states are clearly defined so it is
easy to define the duty cycle. (i.e., it is well defined too)


We can define the duty cycle as int(f(t),t=0..P)/(P*max(f(t))). It is ratio
of the area of f(t) to the total area.

This holds for rectangular pulses and so is as generalization other
functions. (you can see it is almost simply the time-average/max(f(t)))

exp(t) on [0,1] gives 1 - 1/e so we would say the duty cycle ~= 63%. This
means the function would give the equivalent average output as a rectangular
pulse with a duty cycle of 63%.



Another way to think about it that we set g(t) = f(t)/max(f(t)) then we
compute the time-average and we call it's value a duty cycle. A rectangular
pulse using that same vaule will also have the same duty cycle and same
average output.

But does exp(t) seem like a "pulse". If so then the definition of duty cycle
can be used but IMO it seems a bit ambiguous for such functions that don't
clearly transition between the high and low state and are not off for a
clearly defined portion of their period.

I guess the real problem is the definition of "pulse". What is it?
 
Another way to think about it that we set g(t) = f(t)/max(f(t)) then we
compute the time-average and we call it's value a duty cycle. A
rectangular pulse using that same vaule will also have the same duty cycle
and same average output.

But does exp(t) seem like a "pulse". If so then the definition of duty
cycle can be used but IMO it seems a bit ambiguous for such functions that
don't clearly transition between the high and low state and are not off
for a clearly defined portion of their period.
BTW, this says that for any arbitrary function we can just use rectuangular
pulses using the appropriate duty cycle. So if you wanted to use your super
duper f(t) I could just caclulate it using above(which is just calculating
the time average after normalizing) and use rectangular pulses. I'd bet my
circuit would out-perform yours in just about every way.

So if you find someone not using rectangular pulses you know somethings
fishy.
 
"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?
Well, dog-gone they are different (no pun intended). I though they will
surely be the same.

http://hyperphysics.phy-astr.gsu.edu/hbase/Electronic/opampvar.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/Electronic/opampvar6.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/Electronic/opampvar7.html#c1
 
"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.
Thanks, Mike
 
"amdx" <amdx@knology.net> wrote in message
news:e1b1b$4980d68e$d8baf3ed$11900@KNOLOGY.NET...
"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.
Thanks, Mike

Um that simply isn't true. It's not a transformer!

You cut the voltage down you cut the power!!

When you have a simple 2nd grade electronics circuit like

V --- Switch---Light----GND


And you flip the switch really fast does the light stay the brightness? (or
approximately) Does the load change?(exclude temperature)

You have the same load but only power hookup up to it part of the time...
how can you then have the same power as if it was always hooked up?
 
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
 

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