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Understanding PWM of Motor (current problem)

A

amdx

Jan 1, 1970
0
Hi All,
I've been having a discusion about PWM of DC motors on an electric vehicle
forum.
I keep seeing the idea that battery current is different than motor current
when using a
PWM controller. As quoted below.

http://www.4qdtec.com/pwm-01.html This site says this;
"You should see from the above that, if the drive MOSFET is on for a 50%
duty cycle, motor voltage is 50% of battery voltage and, because battery
current only flows when the MOSFET is on, battery current is only flowing
for 50% of the time so the average battery current is only 50% of the motor
current! "

I understand that 50% duty cycle would cause a motor voltage equivalent to
50% of the B+.
And full current would flow for 50% of the time (ignoring inductance caused
rise and fall time).

Quote;
"so the average battery current is only 50% of the motor current!"

But where are they getting the other 50% of the current?
If you say it is from the collapsing magnetic field of the field and
armature, then
I would say, but the battery had to build that field to begin with so
battery current during the
50% on time was had to be higher than the full cycle motor current.
Ok, I'm going to stop now, because I'm not sure I communicating!

This site has waveforms in Section #12 although my thought is he has the
500hz
and 20khz labels reversed.
http://www.picotech.com/applications/pwm_drivers/#chap12

Your help appreciated, Mike

M

[email protected]

Jan 1, 1970
0
Hi All,
I've been having a discusion about PWM of DC motors on an electric vehicle
forum.
I keep seeing the idea that battery current is different than motor current
when using a
PWM controller.   As quoted below.

http://www.4qdtec.com/pwm-01.html This site says this;
"You should see from the above that, if the drive MOSFET is on for a 50%
duty cycle, motor voltage is 50% of battery voltage and, because battery
current only flows when the MOSFET is on, battery current is only flowing
for 50% of the time so the average battery current is only 50% of the motor
current! "

I understand that 50% duty cycle would cause a motor voltage equivalent to
50% of the B+.
And full current would flow for 50% of the time (ignoring inductance caused
rise and fall time).

Quote;
"so the average battery current is only 50% of the motor current!"

But where are they getting the other 50% of the current?
If you say it is from the collapsing magnetic field of the field and
armature, then
I would say, but the battery had to build that field to begin with so
battery current during the
50% on time was had to be higher than the full cycle motor current.
Ok, I'm going to stop now, because I'm not sure I communicating!

This site has waveforms in Section #12 although my thought is he has the
500hz
and 20khz labels reversed.http://www.picotech.com/applications/pwm_drivers/#chap12

Your help appreciated, Mike

i think that part you are missing is the CATCH DIODE. When the switch
switches off, the current continues to flow through the diode. So
with the right inducantce, and a 50% DF, the current in the load flows
all the time, the current through the battery flows 1/2 time so the
average output current is 2x the input current and the output voltage
is 1/2 the input voltage. Input power = output power. This is
simplifed and ignores losses of course.
Mark

A

amdx

Jan 1, 1970
0
Hi All,
I've been having a discusion about PWM of DC motors on an electric vehicle
forum.
I keep seeing the idea that battery current is different than motor
current
when using a
PWM controller. As quoted below.

http://www.4qdtec.com/pwm-01.html This site says this;
"You should see from the above that, if the drive MOSFET is on for a 50%
duty cycle, motor voltage is 50% of battery voltage and, because battery
current only flows when the MOSFET is on, battery current is only flowing
for 50% of the time so the average battery current is only 50% of the
motor
current! "

I understand that 50% duty cycle would cause a motor voltage equivalent to
50% of the B+.
And full current would flow for 50% of the time (ignoring inductance
caused
rise and fall time).

Quote;
"so the average battery current is only 50% of the motor current!"

But where are they getting the other 50% of the current?
If you say it is from the collapsing magnetic field of the field and
armature, then
I would say, but the battery had to build that field to begin with so
battery current during the
50% on time was had to be higher than the full cycle motor current.
Ok, I'm going to stop now, because I'm not sure I communicating!

This site has waveforms in Section #12 although my thought is he has the
500hz
and 20khz labels
reversed.http://www.picotech.com/applications/pwm_drivers/#chap12

Your help appreciated, Mike

i think that part you are missing is the CATCH DIODE.

No, I realize it's there and is passing the current caused by the
collapsing
magnetic field in the motor.
When the switch switches off, the current continues to flow through the
diode.

When you say "the current" I'm assuming you mean the current from the
collapsing field?
So with the right inducantce, and a 50% DF, the current in the load flows
all the time, the current through the battery flows 1/2 time so the
average output current is 2x the input current and the output voltage
is 1/2 the input voltage. Input power = output power. This is
simplifed and ignores losses of course.
Mark

What is DF?
Your saying the same thing as the EV group. Now I need to understand
where the (non battery supplied) current is coming from.
Mike

W

Winfield Hill

Jan 1, 1970
0
i think that part you are missing is the CATCH DIODE.

Perhaps it'll be helpful to think about power flow.
If a dc-dc converter consumes no power, and it
delivers power at half the voltage of an input,
it must consume half the current from the input.
Consider, it it consumes the same current, on
the average, then where would the unused half
of the power go?

Lately I've been working with the "bus converter"
concept. Use PWM converters with a fixed duty
cycle, like 63 or 75%, and transform DC voltages
around, without power loss, to speak of anyway.

For example, in two stages I go from a 24V 3.3A
source to a supply delivering +4V, -12V at 5A.
Both source and destination are 80W. OK, the
source may have to provide 85 watts, but who's
worrying about the extra 5 watts?

When you're thinking about motors, the really
interesting stuff happens at start-up, when you
need high torque = high current, delivered to
the motor, but not taken from the power supply.
Where does it come from? The supply designer
knows the high-current buck stops at the low-
resistance low-ac-loss inductor and the low esr
bypass capacitors. 50A at a 10% duty cycle is
only 5A average -- if you have a capacitor that
can deliver 50A for the other 90% of the time.

B

Bob Eld

Jan 1, 1970
0
amdx said:
Hi All,
I've been having a discusion about PWM of DC motors on an electric vehicle
forum.
I keep seeing the idea that battery current is different than motor current
when using a
PWM controller. As quoted below.

http://www.4qdtec.com/pwm-01.html This site says this;
"You should see from the above that, if the drive MOSFET is on for a 50%
duty cycle, motor voltage is 50% of battery voltage and, because battery
current only flows when the MOSFET is on, battery current is only flowing
for 50% of the time so the average battery current is only 50% of the motor
current! "

I understand that 50% duty cycle would cause a motor voltage equivalent to
50% of the B+.
And full current would flow for 50% of the time (ignoring inductance caused
rise and fall time).

Quote;
"so the average battery current is only 50% of the motor current!"

But where are they getting the other 50% of the current?
If you say it is from the collapsing magnetic field of the field and
armature, then
I would say, but the battery had to build that field to begin with so
battery current during the
50% on time was had to be higher than the full cycle motor current.
Ok, I'm going to stop now, because I'm not sure I communicating!

This site has waveforms in Section #12 although my thought is he has the
500hz
and 20khz labels reversed.
http://www.picotech.com/applications/pwm_drivers/#chap12

Your help appreciated, Mike

Your on the right track. Fifty percent duty cycle means the voltage is 50 %
and also the average current is 50% as you mentioned. So, the power is 1/4
or 25%. That is to say both the motor speed and the motor torque are reduced
by 1/2.

It's wrong to say that the motor current is 50% of the battery current. The
average currents of the battery and the motor are the same. No current is
gained or lost unless some form of dynamic breaking is in the mix. The 50%
number is relative to the Maximum current for a given load. The situation is
made more complicated by acceleration, torque required, back emf, speed and
breaking and other issues.

In general, think of the PWM acting like a simple analog rheostat dropping
the voltage available to the motor. Both the voltage and the current are
reduced.

A

amdx

Jan 1, 1970
0
Bob Eld said:
Your on the right track. Fifty percent duty cycle means the voltage is 50
%
and also the average current is 50% as you mentioned. So, the power is 1/4
or 25%. That is to say both the motor speed and the motor torque are
reduced
by 1/2.
Ok so far.
It's wrong to say that the motor current is 50% of the battery current.
The
average currents of the battery and the motor are the same.

That is what I think, the waveform may be different but the average or
effective
value is the same. But the EVers are saying if you use two shunts, one to
measure
battery current and one to measure motor current the numbers will be quite
different.
No current is gained or lost .

That's what I think.

Now how do you explain what Mark had to say?
"i think that part you are missing is the CATCH DIODE. When the switch
switches off, the current continues to flow through the diode. So
with the right inducantce, and a 50% DF, the current in the load flows
all the time, the current through the battery flows 1/2 time so the
average output current is 2x the input current and the output voltage
is 1/2 the input voltage. Input power = output power. This is
simplifed and ignores losses of course."

I can't make this work, but I see others give the same logic.

Still searching, Mike

A

amdx

Jan 1, 1970
0
i think that part you are missing is the CATCH DIODE.

Perhaps it'll be helpful to think about power flow.
If a dc-dc converter consumes no power, and it
delivers power at half the voltage of an input,
it must consume half the current from the input.
Consider, it it consumes the same current, on
the average, then where would the unused half
of the power go?

Lately I've been working with the "bus converter"
concept. Use PWM converters with a fixed duty
cycle, like 63 or 75%, and transform DC voltages
around, without power loss, to speak of anyway.

For example, in two stages I go from a 24V 3.3A
source to a supply delivering +4V, -12V at 5A.
Both source and destination are 80W. OK, the
source may have to provide 85 watts, but who's
worrying about the extra 5 watts?

When you're thinking about motors, the really
interesting stuff happens at start-up, when you
need high torque = high current, delivered to
the motor, but not taken from the power supply.
Where does it come from? The supply designer
knows the high-current buck stops at the low-
resistance low-ac-loss inductor and the low esr
bypass capacitors. 50A at a 10% duty cycle is
only 5A average -- if you have a capacitor that
can deliver 50A for the other 90% of the time.

Win, go to bed, get a good nights rest, get up and have your coffee.
Sorry, I'm picking on you a little cause I know you can explain it better
than you have so far.
After your coffee, explain a PWM motor speed controller driving a motor
with
inductance. I'm using a 300 amp 48 volt motor control, so I don't know
if it has capacitors large enough to deliver that 300 amps for the other 90%
of the time.
Mike
PS. If you see a guy on an electric gokart holding an oscilloscope dragging
an extension cord,
That's me

P

Phil Allison

Jan 1, 1970
0
"Bob Eld"
It's wrong to say that the motor current is 50% of the battery current.
The
average currents of the battery and the motor are the same.

** No way.

In a correctly functioning PWM, DC motor controller - the motor has an
almost steady current flow ( due to the free-wheeling diode) while the
battery experiences a PWM current draw.

During " on " periods, motor and battery currents are the same.

So with a 50% duty cycle, the average battery current is one half that of
the motor.

...... Phil

A

amdx

Jan 1, 1970
0
No, the peak currents are the same. With a PWM assuming
- the presence of a freewheel diode or equivalent
- 50% PWM
- switching frequency is high

then the average motor current will be twice the average battery
current.

The PWM turns on and current is drawn from the controller/battery
through the motor. Call this I1

The PWM turns off the current continues to flow through the motor
winding and freewheel diode.

Is this current from the collapse of the magnetic field in the motor?
Over several PWM cycles this will ramp up the current until it reaches a
steady state. Where the average current drawn during the on time
matches that in the freewheel diode during the off time. So the current
in the motor winding is I1*ton/tduty + I1*toff/tduty. The current from
the battery is I1*ton/tduty (1/2 that of the motor).

Where is the extra current coming from?
The effect is commonly referred to as current multiplication.

Another way of looking at is the energy from the battery is Ibat*Vbat,
the energy in the motor side is Imot*Vmot. Since Vmot is Vbat/2 then
Imot must be about Ibat*2 or something is getting awfully hot.

You must run the PWM fast enough to avoid significant decay in the motor
inductance but that's easier than it used to be. The bigger subtle
point is that the wire inductance from the power source (I keep calling
it a battery) must be compensated for in the controller in order to
avoid power spikes in the power devices.

I think mine runs at 16Khz, Power source, mine is a battery or 4-12volt
car batteries.
Mike

B

Bob Eld

Jan 1, 1970
0
Phil Allison said:
"Bob Eld"

** No way.

In a correctly functioning PWM, DC motor controller - the motor has an
almost steady current flow ( due to the free-wheeling diode) while the
battery experiences a PWM current draw.

During " on " periods, motor and battery currents are the same.

So with a 50% duty cycle, the average battery current is one half that of
the motor.

..... Phil

You are right, I misunderstood the question. The PWM is simply a "DC
transformer." The Power is constant (excluding losses). So if the voltage is
cut in half, the current has to double, Duh! If the input voltage is twice
the output voltage, the input current has to be half of the output current.
Was that the question?

A

amdx

Jan 1, 1970
0
The extreme case is a superconductive inductor, like the main field
electromagnet in an MRI machine. This is a big inductor with literally
zero resistance. They connect a power supply to it for an hour maybe,
build up the current, then they short the inductor leads and remove
the power supply, literally load it back on the truck. The current
keeps circulating in the coil for years.

The motor has inductance. The power supply is applied to it part of
the time, and some current I flows from the supply into to the
inductor when they're connected. Then some fets are switched around to
disconnect the power supply and short the motor/inductor. Current I
continues to flow in the shorted inductor (the motor windings) because
that's what inductors do; power supply current is zero in this state.

Ignoring resistive losses, the ratio of average motor current to power
supply current can be very large. The motor of course has resistance,
and if it's spinning and driving a load it's doing work, so those
"losses" keep the current multiplication ratio down. Motor current
will, in practice, droop a bit while the motor is shorted, due to
various losses.

John
Thanks John,
I guess the problem I have is that energy stored in the Magnetic field had
to come from the battery. Right?
Please look at http://www.picotech.com/applications/pwm_drivers/#chap12
part 2 with the motor driven at 20khz.
My thinking is the ratio between the on time current and the off time
current should be higher. The graph shows the
on time current is only slightly larger than the off time current. It seems
to me, the on time current would need to be
higher to run the motor and store the energy in the magnetic field for the
off time current.
Please straighten out my thinking on this.

Also, assuming 50% duty cycle;
If I put a shunt in the battery circuit and a shunt in the motor circuit
and use analog meters for their averaging effect,
is it your opinion that the battery meter will read 1/2 of the current the
Thanks, Mike

M

[email protected]

Jan 1, 1970
0
No, I realize it's there and is passing the current caused by the
collapsing
magnetic field in the motor.

When you say "the current"  I'm assuming  you mean the current from the
collapsing field?

Yes. The energy is stored in the mag field and the current tries to
continnue to flow...think of it as if the electronis in an inductor
have intertia...

> So with the right inducantce, and a 50% DF, the current in the load flows
What is DF?

DF = Duty Factor = Duty Cycle...same thing

Mark

A

amdx

Jan 1, 1970
0
Yes. The energy is stored in the mag field and the current tries to
continnue to flow...think of it as if the electronis in an inductor
have intertia...

What is DF?

DF = Duty Factor = Duty Cycle...same thing

Mark

Thanks, Mark
Mike

A

amdx

Jan 1, 1970
0
John Larkin said:
Sure. But it's *stored* in the magnetic field, so once it's there, the
battery isn't needed to keep it there.

If the frequency is low, like in the 500 Hz case, the current ramp up
and ramp down are big slopes. The rampup is the power supply charging
the inductor, and the rampdown is caused by losses in the
inductor/motor, caused by resistance in the windings and especially
any real work being done.

At higher frequencies, the slopes *appear* to be flatter because we're
operating in shorter time slices. When the battery is disconnected,
the circulating current decays only a little, so when we switch back
on to top off the current, we only need a little up-slope. The higher
the frequency, the less ripple in the inductor current. Less ripple
means that the power supply "on" current is very close to the inductor
circulating "off" current.

If the frequency is reasonably high, so that motor inductive ripple is
low, then yes, close to 1/2.

John
Ok, I think I have it, when I said to myself, the voltage across the
motor should reverse
polarity during the off time, why don't I see it in the graph. Then I
realized I do see it, but it's
clamped to the diode drop.
I see why responses kept going back to power in = power out. With a 50%
duty cycle the voltage
is effectively 1/2 so the current has to be double. Intuitively, It seems
like the current should be higher
during the on time (in comparison to the off time) to run the motor and
store the energy for off time.
And looking at the the graph referenced before, the current is higher, but
not as much as I would think,
especially for this 30% duty cycle.
Thanks, Mike

F

Frithiof Jensen

Jan 1, 1970
0
amdx said:
Hi All,
I've been having a discusion about PWM of DC motors on an electric vehicle
forum.
I keep seeing the idea that battery current is different than motor
current when using a
PWM controller.

A PWM controller behaves like a variable transformer: The input power has to
match the output power (ignoring losses), so the voltage and current on the
output can be different from those present on the input. The product of
voltage and current will be the same.

I think you lot are getting mixed up over the pulsed current in the versus
the rms-current flowing in the motor

S

Spehro Pefhany

Jan 1, 1970
0
A PWM controller behaves like a variable transformer: The input power has to
match the output power (ignoring losses), so the voltage and current on the
output can be different from those present on the input. The product of
voltage and current will be the same.

I think you lot are getting mixed up over the pulsed current in the versus
the rms-current flowing in the motor

What's a versus, and why is there pulsed current flowing in it?

Best regards,
Spehro Pefhany

J

JosephKK

Jan 1, 1970
0
Ok so far.

That is what I think, the waveform may be different but the average or
effective
value is the same. But the EVers are saying if you use two shunts, one to
measure
battery current and one to measure motor current the numbers will be quite
different.

That's what I think.

Now how do you explain what Mark had to say?

I can't make this work, but I see others give the same logic.

Still searching, Mike

The one good concept presented here is that battery power ~= motor
power.

Keeping that in mind; motor current is proportional to torque, and
motor CEMF is proportional to motor speed. Net motor voltage is
effective applied voltage (battery voltage * DF) - CEMF.

Vehicle friction losses are minor except at very low speed, typically
drag dominates. Drag has v^2, v^3, and v^4 components, with the
higher powers becoming dominant at higher speeds. Remember that power
~= torque * speed.

Since is real vehicles operating at lower effective voltage also means
operating at a lower speed. And none of it is very linear, thanks to
drag.

Operating at steady speeds, the power required goes up by a curve
similar to an exponential. (In the general case, in specially designed
vehicles there are speeds with atypical drag minimums.)

So at rated max speed (max power) we have effective full voltage and
full current and all the power is being eaten up fighting drag.

A 1/4 decrease in power should translate to about 78% of top speed,
86% voltage and 88% current and at full battery voltage 75% current.

A 1/2 decrease in power should translate to about 60% of top speed,
70% voltage and 73% current and at full battery voltage 50% current.

More decreases result in similar arithmetic.

HTH

J

JosephKK

Jan 1, 1970
0
Ok, I think I have it, when I said to myself, the voltage across the
motor should reverse
polarity during the off time, why don't I see it in the graph. Then I
realized I do see it, but it's
clamped to the diode drop.
I see why responses kept going back to power in = power out. With a 50%
duty cycle the voltage
is effectively 1/2 so the current has to be double. Intuitively, It seems
like the current should be higher
during the on time (in comparison to the off time) to run the motor and
store the energy for off time.
And looking at the the graph referenced before, the current is higher, but
not as much as I would think,
especially for this 30% duty cycle.
Thanks, Mike

I seem to have gotten some of my sums wrong in my previous post, i did
not pay proper attention to speed proportions and CEMF.

J

JosephKK

Jan 1, 1970
0
What's a versus, and why is there pulsed current flowing in it?

Best regards,
Spehro Pefhany

versus is a comparison (often numerical)

J

JosephKK

Jan 1, 1970
0
ALERT! Improper use of language!

"Versus" is a preposition.

...Jim Thompson

Dawgs, i love my pop-off chain. So handy when some d^2V/dt^2 insists
on jerking it.

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