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What passes as Pulse Width Modulation in DC Motor Control?

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.

Reply to
Jon Slaughter
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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

Reply to
Dan Coby

"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.

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

Reply to
Phil Allison

"John Fields" "Phil Allison"

** 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

Reply to
Phil Allison

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.
Reply to
Joseph Sroka - 10.2.8

"Joseph Sroka - 10.2.8" "Phil Allison"

** You miss the point entirely - f*****ad.

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

Reply to
Phil Allison

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.

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

PWM in general turns the power on and off.

I would not call that PWM.

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

Reply to
Jasen Betts

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.

Reply to
Jasen Betts

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?

Reply to
Rich

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

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Scope, Spectrum, Spectrogram, Sound Level Meter FREE Signal Generator Science with your sound card!

Reply to
Bob Masta

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.

Reply to
Jon Slaughter

--- 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

Reply to
John Fields

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

Thanks, Rich

Reply to
Rich Grise

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

Good Luck! Rich

Reply to
Rich Grise 

--
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.
Reply to
John Fields

"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

Reply to
Phil Allison

** John knows what he posted is wrong and is now just playing games with words.

He has never admitted error in his whole life and it not about to break the habit.

Being the psychotic asshole that he is.

...... Phil

Reply to
Phil Allison

Perhaps I wasn't clear.. The controller I spoke of (didn't give a circuit link) is a commerical unit ("mine" only because I bought it!) that uses a rectangular PWM drive signal applied to SCRs that chop the incoming mains supply. That PWM drive is controlled by a feedback circuit that compares the actual motor current (and maybe back EMF or something) to a setpoint. So the duty cycle of the PWM is modulated as needed to maintain the desired setpoint.

Now, it's true that this circuit is applying the PWM to the sinusoidal mains, but it should be clear that due to the feedback the exact nature of the mains waveform is not very important... it could just as well be random noise or an opera aria. Assuming the waveform has a reasonable RMS voltage and there are minimal SCR switching losses, the waveform should have little (if any) effect on efficiency.

Best regards,

Bob Masta DAQARTA v4.51 Data AcQuisition And Real-Time Analysis

formatting link
Scope, Spectrum, Spectrogram, Sound Level Meter FREE Signal Generator Science with your sound card!

Reply to
Bob Masta 

--
You must have me confused with another John, since I make mistakes every
once in a while and gladly own up to them when they\'re pointed out to
me.   I even own up to them myself if I catch them before someone else
does after I\'ve posted.
Reply to
John Fields

"John Fields"

** John knows what he posted is wrong and is now just playing games with words.

He has never admitted error in his whole life and it not about to break the habit.

Being the psychotic, lying asshole that he is.

...... Phil

Reply to
Phil Allison

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