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Slaughter on PWM of DC Motors

P

Phil Allison

Jan 1, 1970
0
** Just in case anyone missed it and needs a good laugh:

From SEB, 1-29- 2009


"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.

--------------------------------------------------------------------------------
 
J

Jon Slaughter

Jan 1, 1970
0
High peak current but low average current through an inductor?

Do you not understand the concept of PWM? There is no current flowing
through the inductor for some percentage of the cycle hence the average must
be equal or lower. Now it may be the case in some instances that the
inductor limits the inrush current significantly so the peak current is
close to the average but in general that isn't the case.. (at least that
isn't the point as we were talking about average behavior). For very low
duty cycle it may be that 1/10% is virtually the same as 0% and the average
current ~= peak current.

In any case you have to forgive Phil "FUCKWIT" Allision for leading you
astray. He doesn't have much else to do in the outback and the heat has
fried more than a few of his braincells.

Just do a google search on him and you can learn how he got the name
FUCKWIT.
 
C

Clifford Heath

Jan 1, 1970
0
Jon said:
Do you not understand the concept of PWM? There is no current flowing
through the inductor for some percentage of the cycle

You really *are* a dill, who apparently doesn't understand inductance.
 
J

Jon Slaughter

Jan 1, 1970
0
Clifford Heath said:
You really *are* a dill, who apparently doesn't understand inductance.

Your a fucking moron. If you actually cared to read what was posted you
might be able to pass middle school.


VCC ---- switch ---- inductor ---GND

Your telling me when the switch is open that current is flowing through the
inductor!!! Go learn some basic electronics! You apparently don't understand
basic electronics nor basic logic.

(morons like you are why humans will become extinct. Luckily I can add you
to my ignore filters.)
 
C

Clifford Heath

Jan 1, 1970
0
Jon said:
Your telling me when the switch is open that current is flowing through the
inductor!!!

If the switch has just been opened, and there was current flowing,
then yes, it's still flowing. Inductors do that, it's why they blow
up (semiconductor) switches.
 
N

Nobody

Jan 1, 1970
0
Your a fucking moron. If you actually cared to read what was posted you
might be able to pass middle school.


VCC ---- switch ---- inductor ---GND

Your telling me when the switch is open that current is flowing through the
inductor!!!
Yep.

Go learn some basic electronics! You apparently don't understand
basic electronics nor basic logic.

You should take your own advice.

In the above circuit, opening the switch will simply turn it into a spark
gap, across which current continues to flow.

[This is true whether the switch is a physical switch or a semiconductor,
the main difference being that a physical switch might recover from being
turned into a spark gap; a semiconductor won't.]

As Clifford says: "Inductors do that".

You cannot instantaneously change the current through an inductor.
It WILL maintain its current, regardless of how much resistance you put in
the way. More resistance just means a higher voltage; as high as is
necessary to preserve the current, even if that means vapourising the
"switch".

This is a common way of generating high voltages. Generating kilovolts
from a few volts is entirely practical.
(morons like you are why humans will become extinct. Luckily I can add you
to my ignore filters.)

Using an ignore filter for anyone who tells you you're wrong is a good way
to keep being wrong.

In this specific case, it could even result in *you* becoming "extinct".

If you doubt what people are telling you, you could try the following
experiment (note: I hereby disclaim any and all liability for subsequent
death or injury).

Construct your circuit with a car battery as the power supply, a nice big
inductor with really thick wire, and a "switch" consisting of two pieces
of metal (without any insulation). Hold one piece of metal in each hand,
touch the two together for a few seconds, then pull them apart.

Tell us how you get on. Or, rather, arrange for your next of kin to tell
us.
 
J

Jon Slaughter

Jan 1, 1970
0
Nobody said:
Your a fucking moron. If you actually cared to read what was posted you
might be able to pass middle school.


VCC ---- switch ---- inductor ---GND

Your telling me when the switch is open that current is flowing through
the
inductor!!!
Yep.

Go learn some basic electronics! You apparently don't understand
basic electronics nor basic logic.

You should take your own advice.

In the above circuit, opening the switch will simply turn it into a spark
gap, across which current continues to flow.

[This is true whether the switch is a physical switch or a semiconductor,
the main difference being that a physical switch might recover from being
turned into a spark gap; a semiconductor won't.]

As Clifford says: "Inductors do that".

You cannot instantaneously change the current through an inductor.
It WILL maintain its current, regardless of how much resistance you put in
the way. More resistance just means a higher voltage; as high as is
necessary to preserve the current, even if that means vapourising the
"switch".

NO ONE FUCKING SAID ANYTHING ABOUT INSTANTAONUOUS CURRENT! HOW LONG DOES
THAT SPART LAST? 10 MINS?!?! read the **** what I said instead of jumping on
Paul "FUCKWIT" Allison's bandwagon with the rest of the fuckwits.

You fucking morons supprise me every day...

you have no idea what the thread was about or have read even what allison
posted but because of your ego you gotta act like a moron.

I would explain it but what's the use.. you don't have enough brain cells to
understand. But if I don't explain it you and others will just say I don't
know what I'm talking about. (um, the discussion was about PWM which has to
do with peak current verses average current and if you actually cared to
read the posts to get the context you might figure that out... but I doubt
it. That so called spark isn't going to last long enough to make much of a
difference. Also inductors tend to have snubbers to prevent that so the
switch's are not ruined... hence the so called spart your talking about has
nothing to do with it.)

Anyways, I know you were just trying to look smart but your just a dumbass.

Another moron to add to my ignore list!!
 
J

John Devereux

Jan 1, 1970
0
Jon Slaughter said:
Nobody said:
Do you not understand the concept of PWM? There is no current flowing
through the inductor for some percentage of the cycle

You really *are* a dill, who apparently doesn't understand inductance.

Your a fucking moron. If you actually cared to read what was posted you
might be able to pass middle school.


VCC ---- switch ---- inductor ---GND

Your telling me when the switch is open that current is flowing through
the
inductor!!!
Yep.

Go learn some basic electronics! You apparently don't understand
basic electronics nor basic logic.

You should take your own advice.

In the above circuit, opening the switch will simply turn it into a spark
gap, across which current continues to flow.

[This is true whether the switch is a physical switch or a semiconductor,
the main difference being that a physical switch might recover from being
turned into a spark gap; a semiconductor won't.]

As Clifford says: "Inductors do that".

You cannot instantaneously change the current through an inductor.
It WILL maintain its current, regardless of how much resistance you put in
the way. More resistance just means a higher voltage; as high as is
necessary to preserve the current, even if that means vapourising the
"switch".

NO ONE FUCKING SAID ANYTHING ABOUT INSTANTAONUOUS CURRENT! HOW LONG DOES
THAT SPART LAST? 10 MINS?!?! read the **** what I said instead of jumping on
Paul "FUCKWIT" Allison's bandwagon with the rest of the fuckwits.

You fucking morons supprise me every day...

you have no idea what the thread was about or have read even what allison
posted but because of your ego you gotta act like a moron.

I would explain it but what's the use.. you don't have enough brain cells to
understand. But if I don't explain it you and others will just say I don't
know what I'm talking about. (um, the discussion was about PWM which has to
do with peak current verses average current and if you actually cared to
read the posts to get the context you might figure that out... but I doubt
it. That so called spark isn't going to last long enough to make much of a
difference. Also inductors tend to have snubbers to prevent that so the
switch's are not ruined... hence the so called spart your talking about has
nothing to do with it.)

Sorry Jon.

PWM is not "about peak current vs average current", not in a context
where you can call the load an inductor.

There is a reason why everyone is jumping on you!

The whole point about feeding an inductive load with a PWM voltage is
that the PWM is "smoothed out", with the current ramping up according
to the *averaged value* of the PWM voltage (and the value of the
inductance). And the catch diode or "snubber" is essential to this
process, it it not just a detail.

It is only when the inductance is *negligible* - when the load is
resistive - that the load sees the current pulses.
 
J

Jon Slaughter

Jan 1, 1970
0
John Devereux said:
Jon Slaughter said:
Nobody said:
On Fri, 06 Feb 2009 20:16:22 -0600, Jon Slaughter wrote:

Do you not understand the concept of PWM? There is no current flowing
through the inductor for some percentage of the cycle

You really *are* a dill, who apparently doesn't understand inductance.

Your a fucking moron. If you actually cared to read what was posted you
might be able to pass middle school.


VCC ---- switch ---- inductor ---GND

Your telling me when the switch is open that current is flowing through
the
inductor!!!

Yep.

Go learn some basic electronics! You apparently don't understand
basic electronics nor basic logic.

You should take your own advice.

In the above circuit, opening the switch will simply turn it into a
spark
gap, across which current continues to flow.

[This is true whether the switch is a physical switch or a
semiconductor,
the main difference being that a physical switch might recover from
being
turned into a spark gap; a semiconductor won't.]

As Clifford says: "Inductors do that".

You cannot instantaneously change the current through an inductor.
It WILL maintain its current, regardless of how much resistance you put
in
the way. More resistance just means a higher voltage; as high as is
necessary to preserve the current, even if that means vapourising the
"switch".

NO ONE FUCKING SAID ANYTHING ABOUT INSTANTAONUOUS CURRENT! HOW LONG DOES
THAT SPART LAST? 10 MINS?!?! read the **** what I said instead of jumping
on
Paul "FUCKWIT" Allison's bandwagon with the rest of the fuckwits.

You fucking morons supprise me every day...

you have no idea what the thread was about or have read even what allison
posted but because of your ego you gotta act like a moron.

I would explain it but what's the use.. you don't have enough brain cells
to
understand. But if I don't explain it you and others will just say I
don't
know what I'm talking about. (um, the discussion was about PWM which has
to
do with peak current verses average current and if you actually cared to
read the posts to get the context you might figure that out... but I
doubt
it. That so called spark isn't going to last long enough to make much of
a
difference. Also inductors tend to have snubbers to prevent that so the
switch's are not ruined... hence the so called spart your talking about
has
nothing to do with it.)

Sorry Jon.

PWM is not "about peak current vs average current", not in a context
where you can call the load an inductor.

There is a reason why everyone is jumping on you!

The whole point about feeding an inductive load with a PWM voltage is
that the PWM is "smoothed out", with the current ramping up according
to the *averaged value* of the PWM voltage (and the value of the
inductance). And the catch diode or "snubber" is essential to this
process, it it not just a detail.

It is only when the inductance is *negligible* - when the load is
resistive - that the load sees the current pulses.


That simply is not true. If the switch did not fully turn off then it is not
switching and would be working in the linear mode(assuming a mosfet). That
means high power dissipation.

It might be true that for an inductive load the switch doesn't turn off as
sharply but that is irrelevant for the point of the discussion which no one
in this group cared to follow(which is not just the post that Phil "FUCKWIT"
Allison posted but a who thread in another group).

If the switching frequency is fast then the effects of the transitions due
come into play more(obviously because they take up more time in the cycle
than for long periods) but that is a non-ideal case when the discussion was
about how PWM worked.

But as usual people want to take stuff out of context so they can act
intelligent.


"
You really *are* a dill, who apparently doesn't understand inductance.

Your a fucking moron. If you actually cared to read what was posted you
might be able to pass middle school.


VCC ---- switch ---- inductor ---GND


Your telling me when the switch is open that current is flowing through the
inductor!!! Go learn some basic electronics! You apparently don't understand
basic electronics nor basic logic.
"

NOTE, did I say anything about when the switch is opening? No!!!!!!! But
these guys think it that knowing an an inductor tries to keep the current
through it from changing(known as faradays law) is some expert knowledge.

And when you have snubbers it is not the case that you have inductive
kickback(at least it reduces it) so the current through the switch durring
the transition is reduces so the transitions are sharper like that of a
resistive load.

But in any case no one cares about the context nor the facts. I guess Phil
"FUCKWIT" Allison has several accounts he likes to post to.

If you don't believe what I have said, simply measure the instantaneous
current through the switch. It isn't always on like you assume(at least
significantly). It is 0 when the switch is off and non-zero when the switch
is on. Durring the transitions the current might be high or not depending on
the snubbing.

The instantanous current may not have square transitions but so what? That
isn't the point... nothing in the real world of electronics is perfect. It
doesn't have square transistions with a resistive load either.

Ideally given a snubbing circuit(which most power mosfets have built in just
for that reason) the switch will have relatively sharp transistions assuming
the switching frequency isn't extreme.
 
J

John Devereux

Jan 1, 1970
0
Jon Slaughter said:
[...]
Sorry Jon.

PWM is not "about peak current vs average current", not in a context
where you can call the load an inductor.

There is a reason why everyone is jumping on you!

The whole point about feeding an inductive load with a PWM voltage is
that the PWM is "smoothed out", with the current ramping up according
to the *averaged value* of the PWM voltage (and the value of the
inductance). And the catch diode or "snubber" is essential to this
process, it it not just a detail.

It is only when the inductance is *negligible* - when the load is
resistive - that the load sees the current pulses.


That simply is not true.

It is true *exactly to the extent* that the load can be considered an
inductor. You can ignore the inductance to an extent, if the PWM
frequency is low enough and the load resistance high enough. But then
it's a resistor!
If the switch did not fully turn off then it is not
switching and would be working in the linear mode(assuming a mosfet). That
means high power dissipation.

It might be true that for an inductive load the switch doesn't turn off as
sharply but that is irrelevant for the point of the discussion which no one
in this group cared to follow(which is not just the post that Phil "FUCKWIT"
Allison posted but a who thread in another group).

If the switching frequency is fast then the effects of the transitions due
come into play more(obviously because they take up more time in the cycle
than for long periods) but that is a non-ideal case when the discussion was
about how PWM worked.

But as usual people want to take stuff out of context so they can act
intelligent.


"
Your a fucking moron. If you actually cared to read what was posted you
might be able to pass middle school.


VCC ---- switch ---- inductor ---GND


Your telling me when the switch is open that current is flowing through the
inductor!!! Go learn some basic electronics! You apparently don't understand
basic electronics nor basic logic.
"

(for the record, I did not post the lines above)
NOTE, did I say anything about when the switch is opening? No!!!!!!! But
these guys think it that knowing an an inductor tries to keep the current
through it from changing(known as faradays law) is some expert knowledge.

And when you have snubbers it is not the case that you have inductive
kickback(at least it reduces it) so the current through the switch durring
the transition is reduces so the transitions are sharper like that of a
resistive load.

But in any case no one cares about the context nor the facts. I guess Phil
"FUCKWIT" Allison has several accounts he likes to post to.

If you don't believe what I have said, simply measure the instantaneous
current through the switch. It isn't always on like you assume(at least
significantly). It is 0 when the switch is off and non-zero when the switch
is on. Durring the transitions the current might be high or not depending on
the snubbing.

It's actually the load *voltage* that "might be high or not", during
the transition. The current does not increase during switching either
way.
The instantanous current may not have square transitions but so what? That
isn't the point... nothing in the real world of electronics is perfect. It
doesn't have square transistions with a resistive load either.

Ideally given a snubbing circuit(which most power mosfets have built in just
for that reason) the switch will have relatively sharp transistions assuming
the switching frequency isn't extreme.

You mean the body diode? That is an an unavoidable consequence of the
mosfet construction AFAIK, I don't think they "put it in" as a
snubber. (And if you're talking about a simple mosfet switch with an
inductive load, I think you will find that the body diode is the wrong
way around to suppress anything!)
 
S

Spehro Pefhany

Jan 1, 1970
0
Your a fucking moron. If you actually cared to read what was posted you
might be able to pass middle school.


VCC ---- switch ---- inductor ---GND

Your telling me when the switch is open that current is flowing through the
inductor!!! Go learn some basic electronics! You apparently don't understand
basic electronics nor basic logic.

For motor PWM you also would want a diode or an active switch to
ground in order to allow the current to continue to flow when the pass
switch is "off". Also, for a DC motor, the inductor behaves as if it
has a voltage source in series, the voltage of which is pretty much
proportional to rotor RPM. Also there is some resistance in the
winding.

If your PWM frequency is too low relative to the motor inductance, the
current troughs during the "off" times will be exessively low and the
peaks will be excessively high, and you'll get major I^2R heating and
inefficiency because of the high RMS currents, despite the average
current looking reasonable.


Best regards,
Spehro Pefhany
 
J

J.A. Legris

Jan 1, 1970
0
On Fri, 06 Feb 2009 20:16:22 -0600, Jon Slaughter wrote:
Do you not understand the concept of PWM? There is no current flowing
through the inductor for some percentage of the cycle
You really *are* a dill, who apparently doesn't understand inductance.
Your a fucking moron. If you actually cared to read what was posted you
might be able to pass middle school.
VCC ---- switch ---- inductor ---GND
Your telling me when the switch is open that current is flowing through
the
inductor!!!
Yep.
Go learn some basic electronics! You apparently don't understand
basic electronics nor basic logic.
You should take your own advice.
In the above circuit, opening the switch will simply turn it into a
spark
gap, across which current continues to flow.
[This is true whether the switch is a physical switch or a
semiconductor,
the main difference being that a physical switch might recover from
being
turned into a spark gap; a semiconductor won't.]
As Clifford says: "Inductors do that".
You cannot instantaneously change the current through an inductor.
It WILL maintain its current, regardless of how much resistance you put
in
the way. More resistance just means a higher voltage; as high as is
necessary to preserve the current, even if that means vapourising the
"switch".
NO ONE FUCKING SAID ANYTHING ABOUT INSTANTAONUOUS CURRENT! HOW LONG DOES
THAT SPART LAST? 10 MINS?!?! read the **** what I said instead of jumping
on
Paul "FUCKWIT" Allison's bandwagon with the rest of the fuckwits.
You fucking morons supprise me every day...
you have no idea what the thread was about or have read even what allison
posted but because of your ego you gotta act like a moron.
I would explain it but what's the use.. you don't have enough brain cells
to
understand. But if I don't explain it you and others will just say I
don't
know what I'm talking about. (um, the discussion was about PWM which has
to
do with peak current verses average current and if you actually cared to
read the posts to get the context you might figure that out... but I
doubt
it. That so called spark isn't going to last long enough to make much of
a
difference. Also inductors tend to have snubbers to prevent that so the
switch's are not ruined... hence the so called spart your talking about
has
nothing to do with it.)
Sorry Jon.
PWM is not "about peak current vs average current", not in a context
where you can call the load an inductor.
There is a reason why everyone is jumping on you!
The whole point about feeding an inductive load with a PWM voltage is
that the PWM is "smoothed out", with the current ramping up according
to the *averaged value* of the PWM voltage (and the value of the
inductance). And the catch diode or "snubber" is essential to this
process, it it not just a detail.
It is only when the inductance is *negligible* - when the load is
resistive - that the load sees the current pulses.

That simply is not true. If the switch did not fully turn off then it is not
switching and would be working in the linear mode(assuming a mosfet). That
means high power dissipation.

It might be true that for an inductive load the switch doesn't turn off as
sharply but that is irrelevant for the point of the discussion which no one
in this group cared to follow(which is not just the post that Phil "FUCKWIT"
Allison posted but a who thread in another group).

If the switching frequency is fast then the effects of the transitions due
come into play more(obviously because they take up more time in the cycle
than for long periods) but that is a non-ideal case when the discussion was
about how PWM worked.

But as usual people want to take stuff out of context so they can act
intelligent.

"
You really *are* a dill, who apparently doesn't understand inductance.

Your a fucking moron. If you actually cared to read what was posted you
might be able to pass middle school.

VCC ---- switch ---- inductor ---GND

Your telling me when the switch is open that current is flowing through the
inductor!!! Go learn some basic electronics! You apparently don't understand
basic electronics nor basic logic.
"

NOTE, did I say anything about when the switch is opening? No!!!!!!! But
these guys think it that knowing an an inductor tries to keep the current
through it from changing(known as faradays law) is some expert knowledge.

And when you have snubbers it is not the case that you have inductive
kickback(at least it reduces it) so the current through the switch durring
the transition is reduces so the transitions are sharper like that of a
resistive load.

But in any case no one cares about the context nor the facts. I guess Phil
"FUCKWIT" Allison has several accounts he likes to post to.

If you don't believe what I have said, simply measure the instantaneous
current through the switch. It isn't always on like you assume(at least
significantly). It is 0 when the switch is off and non-zero when the switch
is on. Durring the transitions the current might be high or not dependingon
the snubbing.

The instantanous current may not have square transitions but so what? That
isn't the point... nothing in the real world of electronics is perfect. It
doesn't have square transistions with a resistive load either.

Ideally given a snubbing circuit(which most power mosfets have built in just
for that reason) the switch will have relatively sharp transistions assuming
the switching frequency isn't extreme.

You're still missing the big picture. I cannot find much on the web
about PWM DC motor control theory, but continuous vs. discontinuous
operation in a buck voltage converter illustrates the relevant
principle. Look carefully at the current waveforms shown:

http://en.wikipedia.org/wiki/Buck_converter
 
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