Motor Controller Flyback

I've been asked to include a motor controller on a board I am designing. T he motor is from Pololu

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The controller is a VNH5019 from ST

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ons/en.CD00234623.pdf

It seems they recommend a capacitor to absorb the flyback current on switch ing to prevent power from flowing back into the PSU. I have no idea how t o size this cap. They recommend 500 uF for each "10 amps of current". It seems to me the capacitance would also depend on the inductance of the moto r since that would determine the energy stored in the magnetic field and so the energy stored in the cap.

Is there something about motors that make the inductance not a factor, like it approximately scales with the current??? Also, wouldn't the capacitor size be related to the square of the current???

The motor is 5.5 amps stalled, so I suppose I could put a 470uF part on the board and be happy. I just don't like winging it when I build things. I' d like to show the reasoning for selecting this part.

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Reply to
Ricketty C
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The motor is from Pololu

tions/en.CD00234623.pdf

ching to prevent power from flowing back into the PSU. I have no idea how to size this cap. They recommend 500 uF for each "10 amps of current". I t seems to me the capacitance would also depend on the inductance of the mo tor since that would determine the energy stored in the magnetic field and so the energy stored in the cap.

ke it approximately scales with the current??? Also, wouldn't the capacito r size be related to the square of the current???

he board and be happy. I just don't like winging it when I build things. I'd like to show the reasoning for selecting this part.

The energy of the winding is related to the square of the current as you wr ite

But, the current is continuous, and during the freewheeling period, the dea dtime, the current is constant (your PWM is much higher than the winding ti me constant)

So the energy fed to the rails during deadtime is proportional to the windi ng current

Also, the deadtime is normally very small, so only limited energy is freewh eeled to the rail (deadtime is normaly less than 1% of the switching cycle)

In average half the deadtimes, the current is freewheeled to GND, so you ar e left with 0.5% of the energy from the freewheeling to the rails. The moto r takes energy to run, so I would not expect to see any increase of rail vo ltage due to this. I have run motor drives at sub 100W with less than 10uF rail capacitance, so I do not think this is a real problem.

Maybe I am mistaken, but it seems to me this is not a real problem

Try to do the math. For your example a 10A current, freewheeled for 1us onl y produces 20mV of ripple on the 500uF cap. Note that the winding inductanc e is high, so the winding and thus the motor torque will be blind to a 20mV voltage increase for 1us

I would say you could run with a least 10 times lower capacitance with no p roblem a all

Cheers

Klaus

Reply to
Klaus Kragelund

. The motor is from Pololu

itching to prevent power from flowing back into the PSU. I have no idea h ow to size this cap. They recommend 500 uF for each "10 amps of current". It seems to me the capacitance would also depend on the inductance of the motor since that would determine the energy stored in the magnetic field an d so the energy stored in the cap.

like it approximately scales with the current??? Also, wouldn't the capaci tor size be related to the square of the current???

the board and be happy. I just don't like winging it when I build things. I'd like to show the reasoning for selecting this part.

Sorry, I'm not following much of what you are saying.

write

Later I realized there is also the energy of the rotor that has to be spun down if it is turning. But if that is the case the current is a lot less t han 10 amps. I believe the motor is closer to 2 amps when in normal use.

eadtime, the current is constant (your PWM is much higher than the winding time constant)

I think you are referring the times between pulses when they motor is undri ven electrically. I am thinking of when the motor is being stopped. Then the H-bridge just opens up and the motor has to come to a halt.

ding current

Yes, if the voltage does not change much and in particular if the RPM does not change much, then the energy is proportional to the product of current and time... so constant power.

wheeled to the rail (deadtime is normaly less than 1% of the switching cycl e)

Yes, but this design is intended to reverse direction periodically. That w ould seem to be worse case. The fly back voltage on the motor is the rever se polarity of the applied voltage and the MOSFETs are then turned on in re verse so the diode drop isn't even there. They likely ramp down the speed in all cases, so maybe this is not an issue... other than in failure modes.

I've been asked to provide a dead switch in the power path to be able to sh ut down all current to the motor. I guess they are considering a failure in the motor controller chip. Then the current to the motor is just shut o ff from before the H-bridge. I was thinking the 500 uF cap would protect t he fail safe MOSFET.

are left with 0.5% of the energy from the freewheeling to the rails.

I don't follow this. When the switches are cut off, the intrinsic diodes a ct as freewheeling diodes and conduct the energy back out the way it came i n. This is a brushed motor circuit where the H-bridge is used to drive for ward or reverse. Are you thinking of something else?

of rail voltage due to this. I have run motor drives at sub 100W with less than 10uF rail capacitance, so I do not think this is a real problem.

nly produces 20mV of ripple on the 500uF cap. Note that the winding inducta nce is high, so the winding and thus the motor torque will be blind to a 20 mV voltage increase for 1us

problem a all What about when the motor is stopped suddenly?

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Ricketty C

ng. The motor is from Pololu

switching to prevent power from flowing back into the PSU. I have no idea how to size this cap. They recommend 500 uF for each "10 amps of current" . It seems to me the capacitance would also depend on the inductance of th e motor since that would determine the energy stored in the magnetic field and so the energy stored in the cap.

, like it approximately scales with the current??? Also, wouldn't the capa citor size be related to the square of the current???

on the board and be happy. I just don't like winging it when I build thing s. I'd like to show the reasoning for selecting this part.

u write

n down if it is turning. But if that is the case the current is a lot less than 10 amps. I believe the motor is closer to 2 amps when in normal use.

deadtime, the current is constant (your PWM is much higher than the windin g time constant)

riven electrically. I am thinking of when the motor is being stopped. The n the H-bridge just opens up and the motor has to come to a halt.

inding current

s not change much, then the energy is proportional to the product of curren t and time... so constant power.

eewheeled to the rail (deadtime is normaly less than 1% of the switching cy cle)

would seem to be worse case. The fly back voltage on the motor is the rev erse polarity of the applied voltage and the MOSFETs are then turned on in reverse so the diode drop isn't even there. They likely ramp down the spee d in all cases, so maybe this is not an issue... other than in failure mode s.

When the motor is turned off, the BEMF creates a voltage equal to the suppl y voltage, if you were running at nominal speed just before turning off.

When it slows down, the BEMF just falls of proportional to the RPM

If you break the motor, this can be done with breaking resistors, which is done if you need to slow the rotor down fast. If you reverse the direction, then you will see up to double the voltage on the rail

shut down all current to the motor. I guess they are considering a failur e in the motor controller chip. Then the current to the motor is just shut off from before the H-bridge. I was thinking the 500 uF cap would protect the fail safe MOSFET.

u are left with 0.5% of the energy from the freewheeling to the rails.

act as freewheeling diodes and conduct the energy back out the way it came in. This is a brushed motor circuit where the H-bridge is used to drive f orward or reverse. Are you thinking of something else?

You have a motor that is not connected to anything. it will conduct in the appropiate freewheeling diode to the rail.

e of rail voltage due to this. I have run motor drives at sub 100W with les s than 10uF rail capacitance, so I do not think this is a real problem.

only produces 20mV of ripple on the 500uF cap. Note that the winding induc tance is high, so the winding and thus the motor torque will be blind to a

20mV voltage increase for 1us

no problem a all

See above

Cheers

Klaus

Reply to
Klaus Kragelund

If an efficient driver, like a switching h-bridge, drives a motor, and the demand is to decelerate a real spinning load, energy is extracted from the motor and load, and dumped into the power supply. Some systems may blow up of there's nothing but supply capacitors to absorb that energy.

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Reply to
jlarkin

ning. The motor is from Pololu

n switching to prevent power from flowing back into the PSU. I have no id ea how to size this cap. They recommend 500 uF for each "10 amps of curren t". It seems to me the capacitance would also depend on the inductance of the motor since that would determine the energy stored in the magnetic fiel d and so the energy stored in the cap.

or, like it approximately scales with the current??? Also, wouldn't the ca pacitor size be related to the square of the current???

t on the board and be happy. I just don't like winging it when I build thi ngs. I'd like to show the reasoning for selecting this part.

you write

pun down if it is turning. But if that is the case the current is a lot le ss than 10 amps. I believe the motor is closer to 2 amps when in normal us e.

he deadtime, the current is constant (your PWM is much higher than the wind ing time constant)

ndriven electrically. I am thinking of when the motor is being stopped. T hen the H-bridge just opens up and the motor has to come to a halt.

winding current

oes not change much, then the energy is proportional to the product of curr ent and time... so constant power.

freewheeled to the rail (deadtime is normaly less than 1% of the switching cycle)

at would seem to be worse case. The fly back voltage on the motor is the r everse polarity of the applied voltage and the MOSFETs are then turned on i n reverse so the diode drop isn't even there. They likely ramp down the sp eed in all cases, so maybe this is not an issue... other than in failure mo des.

ply voltage, if you were running at nominal speed just before turning off.

s done if you need to slow the rotor down fast. If you reverse the directio n, then you will see up to double the voltage on the rail

Ok, I assume you know what you are talking about, but I thought an inductor did not limit the voltage, rather it would create a voltage proportional t o the dI/dt which can be very high if the switch is opened. So the cap all ows the motor to spin down and the inductance to drop current more slowly w ithout a large voltage jump.

I take it the BEMF does not work the same way as the inductance?

o shut down all current to the motor. I guess they are considering a fail ure in the motor controller chip. Then the current to the motor is just sh ut off from before the H-bridge. I was thinking the 500 uF cap would prote ct the fail safe MOSFET.

you are left with 0.5% of the energy from the freewheeling to the rails.

es act as freewheeling diodes and conduct the energy back out the way it ca me in. This is a brushed motor circuit where the H-bridge is used to drive forward or reverse. Are you thinking of something else?

e appropiate freewheeling diode to the rail.

Yes, and that voltage will be the same polarity as the applied power was be fore the switch connecting the PSU is opened.

ase of rail voltage due to this. I have run motor drives at sub 100W with l ess than 10uF rail capacitance, so I do not think this is a real problem.

us only produces 20mV of ripple on the 500uF cap. Note that the winding ind uctance is high, so the winding and thus the motor torque will be blind to a 20mV voltage increase for 1us

h no problem a all

I guess I'll have to take your word for it as I don't even have a mental im age of what the motor is doing once the circuit is opened.

But I don't think I'll be building this after all. Looks like I wasn't the only one wondering why we were building yet another ventilator and we are merging with another group who is further along and have their hardware rea dy to start testing. They picked a very similar controller that is not 3.3 volt compatible. The CPU is the AVR from the Arduino Uno, so it's a 5 vol t device anyway. I would have tried to steer them toward an ARM based modu le like Teensy or similar and just used the 3.3 volt compatible VNH5019 con troller that has a bit more going for it.

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Reply to
Ricketty C

That's not quite the whole picture though. A disconnect a steadily running DC motor from its supply and it will generate about the supply voltage, not more. It will spin a long time like that because there's no current flowing. The voltage doesn't suddenly become enough to arc over and force current to flow.

However, while current is flowing, there is energy stored in the magnetism of the motor's inductance. When you disconnect it, that energy

*will* go somewhere; current in an inductor doesn't stop flowing instantaneously. So you'll get a voltage spike that could arc over the switch and pulse a high voltage into the power supply.

The energy that causes damage is not the rotational energy of the motor, as sort-of implied by JL's post.

CH

Reply to
Clifford Heath

Sort of implied? A handful of pm motor with a reasonable load can be spun up in a few seconds to store hundreds of joules of rotational energy. The energy stored in inductances is trivial in comparison.

If you want to decelerate that spinning mass in less time than windage, that energy has to go somewhere. With a synchronous switcher or an h-bridge, it has to go back into the power supply. Conservation of energy always works.

Take a look at the size of the caps needed to store a few hundred joules.

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Reply to
jlarkin

Absolutely correct. I thought you were referring to blowing up the driver by turning off the motor. That can happen, but not from rotational energy.

Reply to
Clifford Heath

Also, to get the energy to go back into the supply (given that the free-wheel voltage will be less than what was being supplied) requires alternating between a brief short-circuit while the current builds in the winding inductance, followed by switching that current back into the supply. Any system that can accurately manage the trick damn well ought to be smart enough not to blow up the supply.

Reply to
Clifford Heath

No, just tweak the duty cycle of the h-bridge. A proper h-bridge (or its unidirectional version, a half-bridge synchronous buck) works like a pair of gears, bidirectionally, boost/buck. It can take 50 volts of back EMF from the motor and charge/explode the caps of a 100 volt supply.

Again, conservation of energy explains what must happen. Given that, we just have to work out the details.

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Reply to
jlarkin

mandag den 8. juni 2020 kl. 13.10.43 UTC+2 skrev snipped-for-privacy@highlandsniptechnology.com:

if you short the motor the energy is dumped in the motor resistance and FETs Rdson

if you turnoff the FETs the motor voltage is rpm * K so less than supply and the motor coasts to a stop

if you keep switching you basically have a buck converter in reverse, iow a boost converter dumping the energy into the supply

Reply to
Lasse Langwadt Christensen

Sure, but that's pretty dramatic.

Yes, eventually.

Right. That's what a straightforward speed control loop will do if you use it to decelerate. Mechanical things can store a lot of energy.

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Reply to
John Larkin

That helps a lot. I understand the effect of the inductance which will cre ate a voltage proportional to the di/dt. The effect of the rotor rotation is more like a transformer with an open winding, no current flow so no forc e and no slowing. The voltage produced will be the same as the voltage app lied to sustain the rotation.

One of the motor controllers I looked at showed a "braking" mode where the two top halves of the bridge were closed creating a shorted loop to continu e the current and I assume letting the resistance of the FETs slow the moto r.

Thanks

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Reply to
Ricketty C

d
b

o

y

,

reate a voltage proportional to the di/dt. The effect of the rotor rotatio n is more like a transformer with an open winding, no current flow so no fo rce and no slowing. The voltage produced will be the same as the voltage a pplied to sustain the rotation.

e two top halves of the bridge were closed creating a shorted loop to conti nue the current and I assume letting the resistance of the FETs slow the mo tor.

resistance of the fets and resistance of the motor

Reply to
Lasse Langwadt Christensen

It doesn't need to be. Depends on the winding resistance

A motor is used to drive something. So often this is a quite short time to standstill. For example, I am working on pumps. It takes not more than about a second from 6000 RPM.

Cheers

Klaus

Reply to
Klaus Kragelund

Not just windage. There is also load and bearing friction to slow the motor. The link to the motor shows it to have a gearbox attached. That provides some friction as well. However, the load rotational inertia is also involved. Characteristics of his load have not been considered... unless I missed that post.

Reply to
John S

In the electrical machinery world, windage refers to all the losses of a motor except the external load.

I had to take two semisters of electrical machines. I learned a lot, actually.

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John Larkin         Highland Technology, Inc 
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Reply to
John Larkin

I learned it as windage and friction of the motor without a load attached. However, we are really on the same page. I accept your definition.

Yes. A valuable course of study. I have learned much about them too. But that does not mean I want to challenge you.

Reply to
John S

I'm fine with discussion, and not everyone has to agree. Everybody can learn stuff.

I did learn a lot in the electrical machinery courses. That's the first time I really understood a transformer or a motor. They didn't mention small stuff, like PM motors and alternators. Half my class was destined to work for utilities, NOPSI or LP&L. The electronics guys didn't even socialize with the power guys.

We hated the labs. Too much smoke, for one thing.

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John Larkin

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