Using Energy Lost In A Flyback Diode

Thread Starter

ElectronicMotor

Joined May 1, 2016
53
Hey Guys
I have been developing an electric buggy, using a brush-less three phase DC motor (BLDC).
The advantage of a BLDC is of course efficiency and control.
I have built the buggy, the MOSFET drivers (x3) and the PIC controller, and it really goes .
I got the circuit from a magazine for a high power DC motor speed controller X 3.
It is an alternator with an external fan on the belt pully, and goes so fast that I don't have to dry my hair.
After removing some spike suppression zener diodes and capacitors that kept catching fire, I noticed the main flyback protection diodes getting so hot that I could only run this ripper for about one minute.
And these diodes are huge, in a similar package to a bolt on TO-220, but twice as big.
So, I ask, 'All that flyback current that gets suppressed by these diodes and wasted as heat, surely can be harnessed in the next switching cycle'.
Any ideas ?
 

Thread Starter

ElectronicMotor

Joined May 1, 2016
53
A basic circuit for a 3 phase motor controller, with a flyback protection diode on each winding. These are driven sequentially, to make the armature rotate. The protection diodes waste energy, but protect the MOSFETs from high voltage spikes produced when the inductive field coil is turned on or off suddenly. When switched off (MOSFET off) the inductance wants to continue delivering current through an almost infinite resistance, producing a high voltage spike, that can 'tunnel' through the MOSFET substrate. This current forward biases the diode, and allows it to short itself out in a very quick, high current exponential decay. This is a waste of good energy, I think ?
3 phase brushless motor.jpg
 

#12

Joined Nov 30, 2010
18,224
Look at it this way, the diode is just directing the current back to the power supply. The best way to reduce the heat is to use a synchronous rectifier scheme with a mosfet instead of a diode. But first, calculate how much energy it will cost to drive the mosfet gates.

Well...DickCappels just beat me to it with an IGBT setup.
 

Thread Starter

ElectronicMotor

Joined May 1, 2016
53
I'me not sure it is ? The diodes get HOT (cook an egg hot), so there must be a lot of energy being spent when the current crosses the 0V6 forward bias 'barrier'. Some of it may return to the power supply, but when I put a wet finger on the 'heat sink' back plane of the diode, and it sounds like my jug is boiling, then there is a LOT of energy being wasted.
Snubber energy recovery sounds like a good R & D.
 

crutschow

Joined Mar 14, 2008
34,285
The flyback diode simply allows the current to keep flowing through the motor inductance when the PWM signal is off.
All the energy, except that dissipated in the forward drop of the diode, goes back to the motor to help keep it going.
So the only energy loss you can minimize is that dissipated in the diode drop.

Going to a diode with a lower drop, such as a Schottky will help.

For even lower loss you could use a MOSFET as a synchronous diode as #12 suggested, since a MOSFET can have a very low voltage drop when conducting, but that requires a signal to the MOSFET gate so it turns ON only when the control MOSFET is OFF.
This generally requires a small delay between the two signals.
 
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Thread Starter

ElectronicMotor

Joined May 1, 2016
53
I get your point. The current continues to drive the winding after the FET is switched off, so is not necessarily wasted, except through the junction barrier voltage of the diode. This would explain why the feedback sensor seems to want to fire each coil slightly out of phase. The coil is still energized after the MOSFET is switched off. I will compensate for this if this is the case. A Schottky will definitely reduce wasted heat energy, and possibly change the phase delay slightly.
Anyway, more R & D, but now I have a path. Thanks.
 

MrAl

Joined Jun 17, 2014
11,396
Hi,

The heating is caused by both the current and the voltage drop in the diode, as well as the time it is conducting. The time can be reduced substantially by simply adding more than one diode in series for each snubber. For example two diodes reduces the conduction time by about 50 percent, three diodes by about 66 percent, four diodes by 75 percent, etc. The power dissipation is proportional to the duty cycle, so if you reduce the on time by 50 percent the power heating effect goes down by 50 percent and even that change might be enough to keep them from overheating.

This raises the clamp voltage a little but usually the voltage does not have to be clamped at the exact same voltage as the winding. Even twice as high is usually acceptable as long as the MOSFET's can handle that. This gives rise to other methods such as actual snubber circuits, not just diodes. That would mean a diode, capacitor, and bleeder resistor where the bleeder resistor might connect to the power supply +Vcc, but regardless how it is connected it has to be sized carefully because if the bleeder resistor value is too high i can allow a higher than desired voltage transient and thus burn out the MOSFET.

Also, it is a little unusual to see just three transistors driving the windings. Normally a 3 phase bridge would be used and that would take care of energy recovery as well as driving the windings. Some snubbers might still be used, but then they would be low energy snubbers.
That could very well be part of the problem. For example if one of the phases is being pulled low, the other two can not have a return to +Vcc if they are out of phase, which they should be. A simpler example is with a transformer with a center tap where the center tap goes to +Vcc. If you pull one winding down, you can NOT clamp the other winding to +Vcc because it would SHORT OUT the winding and that means lots and lots of current. The clamp voltage would have to be twice the supply voltage, or use a dynamic clamp that can clamp at whatever voltage it gets to (like around twice the supply voltage). The dynamic clamp would be low energy.

We could look into this further, but a simple simulation would show how this works pretty quickly.
 
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ian field

Joined Oct 27, 2012
6,536
Hey Guys
I have been developing an electric buggy, using a brush-less three phase DC motor (BLDC).
The advantage of a BLDC is of course efficiency and control.
I have built the buggy, the MOSFET drivers (x3) and the PIC controller, and it really goes .
I got the circuit from a magazine for a high power DC motor speed controller X 3.
It is an alternator with an external fan on the belt pully, and goes so fast that I don't have to dry my hair.
After removing some spike suppression zener diodes and capacitors that kept catching fire, I noticed the main flyback protection diodes getting so hot that I could only run this ripper for about one minute.
And these diodes are huge, in a similar package to a bolt on TO-220, but twice as big.
So, I ask, 'All that flyback current that gets suppressed by these diodes and wasted as heat, surely can be harnessed in the next switching cycle'.
Any ideas ?
Might be as simple as tethering all back emf diodes back to the battery.
 

tcmtech

Joined Nov 4, 2013
2,867
To be honest that circuit is a very poor one for driving a BLDC motor. What you should have is a full three phase H bridge circuit.

One thing I see everyone overlooking is that while one phase is conducting the other two can be at various stages of being either in or out of phase with it and thus have their winding sets working as second and third sets of windings on a transformer with a half wave short through their respective diodes.

The best way to set your drive up would be to either have the three windings configured in a Delta or Wye connection with a proper three-phase H-bridge driving them with proper bypass diodes in parallel with each switching device.

Odds are you will see a substantial current reduction in your drive while still producing the same RPM's and torque.
 

Thread Starter

ElectronicMotor

Joined May 1, 2016
53
I was in fact surprised at how warm these diodes got. This is the first time I have built a BLDC driver, and this is an alternator from an old Datsun. The field winding's typically pull about 30 odd amps from a 12V lead acid battery (I think, my meter doesn't go above 20A) . It is wound in a star configuration, and I have pulled out the central connection so that there are four terminals, three winding, and one neutral, which is tied to V+. The others are driven to V-. I have done some research and have found that there is no advantage in using a full bridge driver. If it helps, there are 12 armature poles (6 north and 6 south), and 18 field poles (9 north and 9 south). So there are 18 steps. I will look into this further, but I cannot see a reason to shift from a half-bridge. The circuit shown is conceptual, there are four 169 amp MOSFETs on heatsinks on each winding. Thanks guys.
 
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kubeek

Joined Sep 20, 2005
5,794
The full bridge driver basically has a mosfet instead of the recuperation diode, which makes the losses a lot smaller.
 

MrAl

Joined Jun 17, 2014
11,396
I was in fact surprised at how warm these diodes got. This is the first time I have built a BLDC driver, and this is an alternator from an old Datsun. The field winding's typically pull about 30 odd amps from a 12V lead acid battery (I think, my meter doesn't go above 20A) . It is wound in a star configuration, and I have pulled out the central connection so that there are four terminals, three winding, and one neutral, which is tied to V+. The others are driven to V-. I have done some research and have found that there is no advantage in using a full bridge driver. If it helps, there are 12 armature poles (6 north and 6 south), and 18 field poles (9 north and 9 south). So there are 18 steps. I will look into this further, but I cannot see a reason to shift from a half-bridge. The circuit shown is conceptual, there are four 169 amp MOSFETs on heatsinks on each winding. Thanks guys.
Hello again,

Well the main point is that you might not be able to tie one phase to +Vcc with a diode while another phase is tied to ground and the common point is tied to +Vcc. The turning point would be the way the windings are arranged on the core metal. This is because as one winding is grounded, the other winding could be trying to put out double the +Vcc voltage. Because the diodes get SO hot, this is what it sounds like is happening.

Let's to back to the simpler circuit with a center tapped transformer. The circuit here is often used as a voltage converter to step up the voltage from some lower value like 12v up to maybe 170v (pulsed).
The typical circuit is where the center tap is connected to +Vcc and the two windings are driven with two transistors which effectively ground each winding when the respective transistor is turned on, and they are driven alternately possibly each for 50 percent of the time or slightly less than that.
So what happens when one transistor turns on. That winding terminal gets connected to ground, and with the center tap connected to +12v that means the other winding terminal goes up to +24v. Now what happens if we try to clamp that second winding to +12v with a forward biased diode. We see a HUGE current flow, which is much more than any clamping current would really be. It's actually shorting out that second winding, with the only resistance being the resistance of the winding wire which is usually small. Looking at where the energy comes from, it comes from the first winding which is connected properly, but because of the high current the transistor must support a much higher current than it needs to and thus the transistor gets hot too.
When that phase turns off, we see a transient appear, which must be snubbed, but that's not nearly as much energy as when the diode conducts. When the other phase then turns on, the same thing happens again only with the other phase and diode, so we see high current again.

This also could happen with a stepper motor that is driven the wrong way.
If you try to drive a unipolar stepper motor using diodes that try to clamp to +Vcc, same problem. The idea then is to JUST use diodes that clamp negative voltages to ground, which then clamps the positive voltages to the right value (2x Vcc) because when a winding tries to go higher than 2x Vcc it causes a negative voltage in the other winding, which then clamps to ground. So the magnetic coupling partially aides the clamping action by referring the energy to the alternate winding.
In the transformer scenario, this would mean only using diodes across each transistor, which then clamp negative voltages to ground, which in turn clamps 2x positive voltages to 2x +Vcc. It must be allowed to go to 2x Vcc or else we partially short out the winding.

In the alternate scenario where we drive with a 3 phase bridge as i previously suggested, all the windings are clamped with a three phase diode bridge rectifier. The common of the three windings is then left open (floating). The drawback here is we need a higher peak voltage (+Vcc) in order to drive the motor to the maximum speed.
We still need low energy snubbers however in order to clamp the fast rising spikes that the transformer action between windings wont have time to transfer.

So the key differences between the three phase bridge driver and the single ended three phase driver is the bridge needs the common to remain floating, and requires a higher voltage to drive the motor to full capacity, while the single ended driver needs clamping diodes across each transistor (in reverse) instead of to +Vcc and low energy snubbers, but can drive the motor to full capacity without the higher power supply voltage. So if you want to stick with the single ended drivers, then place fast diodes across each transistor in reverse polarity and place low energy snubbers on each collector/drain. You might try relying on the MOSFET internal diodes but might look up the specs first.
 
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tcmtech

Joined Nov 4, 2013
2,867
If it helps, there are 12 armature poles (6 north and 6 south), and 18 field poles (9 north and 9 south).

Sounds like one of those really cheap 'Half wound' or even '1/3 wound" units where each phase has a coil set for every other or even every third pole rather than every pole of the rotor.

If it was wound the proper way being there are 12 rotor poles each phase should also have 12 stator poles ( 6 N & 6 S) not 6 (3 N & 3 S or 6 N & 0 S).

Ther should be 36 slots in the stator core which relate to 3 sets of 12 that overlap each other by 1 slot offset from one phase winding to the next.
 

tcmtech

Joined Nov 4, 2013
2,867
Sorry, I only read the first sentence but perhaps the other two off coils are picking up nice big inductive linkage. Man !

Wow! Imagine what you could learn if you read more than just the first sentence when trying to learn how things work.:rolleyes:

BTW if it's an attention span thing there are some really good meds for that now but since this is in the second sentence odds are you will never know I mentioned looking into them. :oops:
 
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