Flywheel diode...

[Externet]

Hi.
The 'flywheeling' diode installed in parallel to an electric DC motor conducts the spike generated when the magnetic field collapses at the instant a mosfet speed controller turns off, and can be many, many times per second.

The current generated by the motor at the turn-off instant has a certain duration; How to calculate the duration in function of current, voltage, inductance... ?

The current generated by the motor at the turn-off instant recirculates inside the motor thanks to the presence of the freewheeling diode; does such current pulse produces a braking effect or actually helps to propel ?

How to calculate the power content from such 'back emf' turn-off pulse ?

Miguel

 

[dknguyen]

The duration of the pulse is can be calculated by V = L*di/dt to find the di/dt which is the rate the current decreases at. V is the peak voltage of the spike induced by the inductance (probably what you clamped it down to with your flyback diodes). You need to use all the inductance, including motors and wires and everything else the current flows through along the current path. Whether the current actually reaches zero before your next switching time is another matter

You now have current, voltage, time, and frequency. Take a good *guess* what the waveform will look like to find the power. It's not an exact science. It might be a exponential decay, it might be a linaer one. Take a conservative guess.

Great question. I was thinking about it 2 weeks ago. And after staring at the circuit and running through it, this is what I have concluded:

When you load the motor terminals down to zero voltage, it brakes the motor since the motor is acting like a generator trying to drive a very heavy load. If you did this to give the flyback current a path to flow to supress the voltage spike (generated from the inductance forcing the current to continue to flow and only change gradually and increasing the voltage to do so), once the energy from the inductance of the windings was dumped, the motor woudl start to slow down due to this breaking action (the BEMF of the motor would now start acting as a generator since all the energy from the winding inductance is gone). But...stare at your circuit carefully...the flyback diode isn't driving your motor terminals to 0V...it's just clamping the motor terminal voltage- probably to the supply voltage + the diode voltage drop which is about equal to the motor's back maximum EMF (ie. maximum speed, maximum operating voltage of the motor). So once the energy from the inductance fo the windings is dumped and the voltage spike has been supressed and the motor temrinal voltage has fallen back down to be equal to the and the supply voltage (which is about equal to the BEMF), the diodes turn off and don't driive the motor terminal voltage any lower so the motor coasts. It does not slow down because the terminal voltage is the BEMF voltage which means the motor is acting like a generator driving no-load, so no braking action.

So it all has to do with CLAMPING the motor terminal voltage to it's back EMF voltage (which is about the same as the supply voltage). If you clamped it to something lower than the BEMF, then you would start to get braking action. If you clamped it to higher than the BEMF, you would get no braking action but you would get a higher voltage spike.

Flyback diodes are wired to only give the current a path to flow IF the voltage spike exceeds the motor back EMF. In this way, they supress the voltage spike to be withing operating limits of the motor and electronics, but not to slow the motor down.

 

[dknguyen]

Of course, if you run the motor at lower than 100% PWM to get slower speeds the BEMF will be lower than the supply voltage (and the motor's maximum BEMF for that circuit). So the voltage spike will be higher than the back EMF, but since it is still clamped to the supply voltage (which is presumeably withing operating limits of the motor), then we don't care because no damage will occur. This is the same case as if you clamped the motor voltage to be above it's BEMF- no braking action but higher voltage spike.

 

[Externet]

OK, understood a good amount.

If the freewheeling diode is removed from the circuit, the mosfet suffers from the spike and may get damaged; but if the mosfet was constructed in such way/specifications/sturdiness that would not get harmed by the spike, there would be no braking effect; is that right ?

If the back emf pulse was captured, absorbed, stored in a capacitor instead of recirculated trough the 'generating' motor; there would still be a braking effect because there is a load placed to such spike; is that correct ?

If that accumulation of back emf energy from repeated spikes stored in such capacitor was re-routed back to recharge/reinforce the battery supply, the still existing braking effect would at least provide a useful 'recharge' of the battery instead of wasted as heat in the generator+diode; is that right ?

Miguel

 

[dknguyen]

MOSFETs undergo avalanche breakdown when their voltage is exceeded by too much. They start conducting. They can survive this if the energy is not too high and some MOSFETs are rated for this. During avalanche breakdown, the MOSFET clamps the voltage across it to the breakdown voltage.

If you relied on this and the MOSFET could survive avalanche, then the breakdown voltage would have to be higher than the rail supply (or else the MOSFET wouldn't work in the circuit). During avalanche the voltage would be clamped to this breakdown voltage which would be higher than the supply voltage (and thus the maximum back EMF voltage). So the spike is limited to a voltage higher than the back EMF voltage, and you get no braking action.

Technically, capturing the spike would not produce braking action. Braking action is only produced when you try and capture energy from the back EMF of the motor but to get to that energy you have to capture the energy in the spike from the inductive windings first. If you captured all the energy from the windings and kept on capturing energy, you would then be pulling energy from the back EMF and then the motor would brake. So as long as you only captured energy when the terminal voltage was higher than the supply voltage, you would not cause braking.

The group of devices that recover energy dumped by the winding inductance during switching is called are called LOSSLESS SNUBBER. They supress the voltage spike by siphoning away and storing the energy dumped by the winding inductance rather than burning off. THe other kinds (like the flyback, or RC, and RCD snubbers are dissipative snubbers) simply burn off this energy. The increased complexity often does not warrant the use of a lossless snubber.

If you're trying to stop the motor by drawing power the back EMF and storing it, then this is regenerative braking which is different than trying to recover the energy dumped from the winding inductance during switching. Again, the increased complexity often does not warrant the use of regen braking.

 

[Externet]

Very clear, thank you.

Not trying to stop the motor by drawing the power from the back emf; just trying to find out how much is wasted, as in a 40 KHz 36 V 50A PWM speed controlled DC motor.

If something about 1% corresponds to wasted bemf, fine, but if the figure is over ~5%; some circuitry deserves to be implemented.

Not trying to implement regenerative braking either. For a series-wound; non permanent magnet motor may get too elaborated. For a permanent magnet DC motor should pose not much complexity, is that right ?

I will read about the lossless snubbers to learn and evaluate implementation or not.

So... If a mosfet driver is 'bulletproof - failproof' to bemf, not using the freewheeling diode should improve efficiency, right ?

Miguel

 

[dknguyen]

Very clear, thank you.

Not trying to implement regenerative braking either. For a series-wound; non permanent magnet motor may get too elaborated. For a permanent magnet DC motor should pose not much complexity, is that right ?

It's still much more complicated than just throwing in a diode and you might not gain enough energy back to make it worth it which usually seems to be the deciding factor other than cost.

So... If a mosfet driver is 'bulletproof - failproof' to bemf, not using the freewheeling diode should improve efficiency, right ?

Miguel

No, the spike is going to burn the energy off somehow- it can either be through the diode or the MOSFET. The MOSFET in avalanche is much more likely to fail than the diode in forward conduction. And using the MOSFET also makes it heat up more (in addition to it's normaly heating) whereas if you use a diode the diode takes the heat.

All real MOSFETs actually have a parasitic diode that can do the job of the flyback diode. They tend to be able to handle higher current (on par with the MOSFET itself), but slower. So you might still need a flyback diode, RC snubber or some other snubber that is fast to take care of the initial spike (if the voltage drop of the external snubbers is higher than the parasitic diode, then the parasitic diode will eventually kick in. If not, the external snubber does all the work since the parasitic diode never turns on). Without these external snubbers the avalanche mode of the MOSFET will try and handle the spike until the parasitic diode turns on.

 

[mneary]

1) The flywheel diode does not do any braking. It is backwards from the polarity needed to catch back emf. It only captures the spike due to inductance.
2) The MOSFET's parasitic diode is in the wrong place to act as a flywheel diode. It does not protect the MOSFET from avalanche.
3) Some MOSFETs are constructed with zener diodes which do protect the transistor.

 

[dknguyen]

2) The MOSFET's parasitic diode is in the wrong place to act as a flywheel diode. It does not protect the MOSFET from avalanche.

If there is a high-side and low-side transistor pair in a half-bridge configuration it will (like in an H-bridge or BLDC motor driver). The parasitic diode of one MOSFET will protect the other MOSFET from avalanche.

But yeah, it won't if the transistor is on it's own like in an SMPS. Everything I said earlier applies to half-bridges where each MOSFET is part of a low-side and high-side pair like what you usually find in motor drives.

But I noticed you said "the flyback diode in parallel with the motor" so some of what I have said does not apply including how the flyback diodes clamp the motor terminal voltage to the power supply voltage. To do this with a component in parallel with the motor you need to use a TVS diode rated for the clamping voltage you desire, since you can't use rail clamp diodes in this case. Using the proper TVS diode in parallel with the MOSFET will also protect the MOSFET from avalanche because the TVS diode will avalanche before the MOSFET does (and is more likely to survive the avalanche).

 

[mneary]

Everything I said earlier applies to half-bridges

That wasn't clear to me.

 

[dknguyen]

That wasn't clear to me.

When I see motors I tend to think bi-directional which usually involves a half-bridge.

 

[Ubergeek63]

The diode across the motor actually does help maintain forward motion since it maintains the current flow.

Braking does not occur until the energy in the inductance is depleted so the BEMF can supply the braking energy. Braking current is in the opposite direction of the drive current.

The amount of braking power available is limited by resistance and speed. The best you can do without a full bridge drive is to put a short directly across the motor.

 

[Externet]

Hello.

The circuit of the lossless snubber that I have found is not complex at all, as in page 29 figure 29 here:

https://www.electro-tech-online.com/custompdfs/2008/06/SMPSRM-DPDF.pdf

Which also corresponds to the intended mosfet+motor wiring circuit subject of this chat.

...... Braking current is in the opposite direction of the drive current.....

No.
The current in the motor flows in the same direction while on the energized instant and while generating at the de-energized instant

Miguel

 

[Ubergeek63]

Hello.

The circuit of the lossless snubber that I have found is not complex at all, as in page 29 figure 29 here:

https://www.electro-tech-online.com/custompdfs/2008/06/SMPSRM-DPDF-1.pdf

Which also corresponds to the intended mosfet+motor wiring circuit subject of this chat.



No.
The current in the motor flows in the same direction while on the energized instant and while generating at the de-energized instant

Miguel

Actually you are wrong.

I design multiple horsepower motor drives and displays for a living. The lossless snubber is old news to me.

Do the math yourself. It takes LI/V seconds for the energy in the inductance to dissipate.

For a large motor into the diode that is 2mH*40A/1V=80mS. However for a 110V motor at half speed there will be about 50V on the rotor so it will take 1.6mS. At a 20KHz PWM frequency you have a period of 0.05mS - continuous current in the same direction.

To reverse the torque on the rotor you need to reverse the flow of current through the rotor. You can do this by either shorting the motor and waiting for the 1.6mS and letting the motor act as a generator to supply the braking, or you can force the issue with a bridge drive by applying reverse voltage to the motor.

 

[Externet]

Good ! Now I may learn the details !

First, to make sure we are in the same page, am talking only about DC series-wound non-permanent magnet motor; and the braking effect mentioned here corresponds only to the back-emf current that briefly recirculates at the generating instant when the paralleled flywheel diode conducts. (not at all about deliberately applying reverse power to slow down)

The circuit being nothing else than a motor in series to a mosfet, with some type of lossless snubber to recover/send back to battery/reuse that back emf in the next energizing cycle instead of being wasted as heat.

When you mention ..."Actually you are wrong"... ;
do you refer to my statement ..."The current in the motor flows in the same direction while on the energized instant and while generating at the de-energized instant"... or is it about something else ?

Taking -say a 2mH and 36V and 50A at 40Khz motor driving,
Time = L I / V will be 0.002 x 50 ÷ 36 = 2.7mS for the bemf to decay. Is that correct? You used 1 as voltage; is that for the diode Vf instead ?

Thanks.

 

[Ubergeek63]

The one volt would be the diode voltage. The principle is the same accept that now the field is proportional to motor current instead of constant and you will have an even higher inductance.

As the motor speed goes down it's voltage goes down. As the voltage goes down the current and hence the field flux goes up.

The calculation looks correct. And under those conditions at 40KHz you will see a maximum ripple current in the motor of I=TV/L or 0.45A

 

[Ubergeek63]

Yes it was about the instant thing and the 1V would be the diode drop.

That type of motor is a bit unpredictable since the field flux varies with the current but the principles are similar. Problem is I am not sure how the field voltage behaves. The rotor voltage would be proportional to both the speed and the field current, it looks like there is a square law in there somewhere.

 

[Externet]

Thanks !

The type of DC motor is almost exact to a heavy truck starter motor, running as say, 36V, 50 A, 40KHz.

I do not understand how to use the 0.45 A ripple current to find the power wasted by the bemf. ¿ Is it 0.45 ripple Amperes x 36V = 16 wasted watts, being that 0.88 % of waste. Not much that deserves recovery with a lossless snubber

Instead of using a lossless snubber, will a simple capacitor in parallel with the motor greatly smooth out/average/integrate the pulsed voltage from the switching supply and thus the off instant will greatly diminish or disappear, and there would be nearly no lossy bemf 'generation' between cycles ?

Miguel

 

[Ubergeek63]

BEMF does not waste energy, nor does the induction that keeps the current flowing. To improve efficiency reduce the drop across the FET and diode.

If you use a

IPB100N06S3L-03

for both the FET and the diode you can limit the dissipation to about 7W each (synchronous rectification). With a permanent magnet motor you could to regenerative braking, but I do not know how that motor responds when it freewheels.