Flywheel Diode Overheating

Thread Starter

Matt H

Joined May 2, 2017
13
Hello, I'm a first time poster that has been lurking here for a while. I'm a mechanical engineer that has been finding himself doing more and more electrical work and embedded programming over the last year or two. I have hit a roadblock on one of my projects and am hoping that someone can help me better understand what is happening.

Some of our previous engineers were notorious for cutting corners, making unsafe circuits, and leaving little to no documentation on projects. I am currently trying to revise one of the boards that they worked on in the past. I have managed to fix the pots for the amplifier and AtoD circuits, have removed the excess capacitors from the rectifier circuit (because why wouldn't you store more than 5x what you need?), and have added optical isolation between the AC and DC signals (for some reason, they thought tying the DC common into the AC neutral with caps storing over 2000uF was a good idea), but am stuck on the main driver circuit.

I am trying to drive a fairly large electromagnet (half transformer) that moves a vibratory assembly (the electromagnet has springs opposing it to reset the assembly's position). The vibration is controlled from an IC that receives feedback from an accelerometer. Prior to working on this project, the electromagnet was driven by an IRS21864 chip with four schottky diodes and two IGBTs (~$70 worth of components for a simple on-off action that doesn't require polarity switching). After adding in the optical isolator, I have been trying to drive the circuit using a single N-channel Mosfet and a flywheel diode.

Unfortunately, I have been left with no datasheet on the electromagnet and am waiting to hear back from the manufacturer on if they can provide one since apparently it was a custom order from a small company. I have attached images of both the electromagnet and my current driver circuit. The electromagnet is powered by rectified mains (170V here in the US) through PWM. The older unit in the lab is currently drawing under 100W under load and we have a 3A fuse in the unit and I plan to limit the maximum PWM to keep the driver under 200W. The PWM frequency is set to 60Hz to allow the springs in the unit time to recover and produce vibration.

When I first ran the current circuit, I noticed that the temperatures on the Mosfet (Q1) and Flywheel Diode (D2) were both increasing. After adding the ferrite bead (FB1), the Mosfet temperature appeared to be well under control. While the Mosfet's temperature is fine, the temperature across the flywheel diode continues to rise and I am completely unsure as to why. When the duty cycle of the PWM is increased, the Flywheel Diode overheats faster; I have been turning the circuit off when the outer temperature of the diode and heatsink read 60C on my IR thermometer. To my understanding, when the Mosfet turns off, the current in the electromagnet (P6) is redirected through the Flywheel Diode and back into the electromagnet coil to safely discharge. The diode that I started with had Vf = 1.6V @ 10, and Io = 10A, so I figured I might just need to make it bigger and tried other diodes with Vf = 1.8V @ 10, and Io = 36A and then one with Vf = 1.8V @ 30, and Io = 99A (I have listed digikey links with the info on each and the datasheets below).

1st: https://www.digikey.com/product-detail/en/rohm-semiconductor/SCS210KGC/SCS210KGC-ND/3902856
2nd: https://www.digikey.com/product-detail/en/microsemi-corporation/APT10SCD120B/APT10SCD120B-ND/3770182
3rd: https://www.digikey.com/products/en?keywords=APT30SCD120B

To my surprise, there was no detectable performance difference between these three diodes and they all three would reach the same temperatures at approximately the same time when tested at the same duty cycle (with or without the ferrite bead). I feel like I have to be missing something very obvious, but am unsure what is causing them to get this hot in the first place, let alone why increasing their size had no effect.

After this, I tried adding an RC snubber circuit on the return path between the Flywheel Diode and the HV circuit. As expected, the resistors bled off a large portion of the energy and greatly reduced the turnoff time of the circuit, but built up heat very quickly. I tried this with a spare 1uF cap and five 3W 5k resistors in parallel (this should equal a 1k 15W resistor) and again with five 100k 3W resistors in parallel (equal to a 20k 15W resistor)

I am currently considering ordering a unidirectional TVS diode and placing it between the Flywheel Diode and HV circuit (with the cathode towards the Flywheel Diode) to act as a snubber. Since Vds on the Mosfet is 600V, and I would like to be able to use this circuit globally, I am going to assume that the maximum rectified input voltage would be 355V, so I would want the breakdown voltage of the TVS diode to be .8*600-355 = 125V at the minimum. Should one of the below TVS diodes help, or am I on the wrong track?

1st: https://www.digikey.com/product-detail/en/littelfuse-inc/5KP120A/1294-5KP120A-CHP/4864269
2nd: https://www.digikey.com/product-detail/en/micro-commercial-co/15KP120A-TP/15KP120A-TPCT-ND/2776339

Thanks for taking the time to go through this. If anyone is able to shed some light on why this is happening and where I should go from here, I would be very appreciative. I am running out of time on this project and am to the point where I am just getting frustrated with it and not making progress.

TL;DR: Flywheel diode (D2) is overheating under load. I have tried an RC snubber without success and am considering a TVS snubber. The load is an electromagnet on rectified mains.

Schematic.jpg

Transformer.jpg
 

MaxHeadRoom

Joined Jul 18, 2013
28,576
It appears you are driving an AC device with pulsating DC?
If so there is your problem.
What is the resistance of the device?
OR it could just be a simple electromagnet.
Max.
 

dl324

Joined Mar 30, 2015
16,788
Welcome to AAC!

Thank you for using paragraphs, attaching a schematic, and giving the TL;DR summary.

Can you post more of the drive circuitry for the MOSFET? From what you posted, you're going to have issues with slow turn off.

How much current is the MOSFET switching?
 

MaxHeadRoom

Joined Jul 18, 2013
28,576
I am trying to drive a fairly large electromagnet (half transformer) that moves a vibratory assembly (the electromagnet has springs opposing it to reset the assembly's position).
The load is an electromagnet on rectified mains.
I have seen a few applications that require a vibrator and a AC coil/armature was used across the AC supply, this results in a 50hz/60hz vibration.
Max.
 

crutschow

Joined Mar 14, 2008
34,201
The use of a larger diode won't make a big difference in its dissipation since that's determined by the forward drop times the current through the diode.
Larger didoes will have a slightly lower drop but will still heat up nearly the same as a smaller diode unless the current is approaching the diode's rating.
The dissipation is due to its carrying the inductor current when the MOSFET is off until the current decays to zero.
This current is obviously high enough to cause significant heating in the diode.
To calculate that we need to know the current and the duration through the diode.

One solution is simply to mount the diode on a heatsink.

Another is to use a higher power snubber resistor.

And another is to use a MOSFET as a synchronous rectifier, driven by another opto isolator with an opposite phase signal (with sufficient non-overlap) from the logic.
The MOSFET has a low forward drop and thus will dissipate much less than a diode.
Of course that will dissipate all the energy in the inductor.
 
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Thread Starter

Matt H

Joined May 2, 2017
13
It appears you are driving an AC device with pulsating DC?
If so there is your problem.
What is the resistance of the device?
OR it could just be a simple electromagnet.
Max.
This is actually pretty standard in the industry. Since electromagnets and solenoids are basically just coils, the main difference between AC and pulsed DC is just that when using AC, the direction of the current flow reverses. You do have to be careful not to increase the duty cycle past the required rating though (VRMS = 170V while the coil is rated for 120V). We actually have two units that use the same coil despite one running on 120VAC with a timer block and the other running on 24VDC with a duty cycle.

The resistance of the electromagnet is only 1.2 Ohms, though this will increase while the unit is running due to it being an inductive load. Unfortunately, I have no idea what the actual inductance is and am hoping to hear back from the manufacturer and I don't have a (working) function generator to measure it.


I have seen a few applications that require a vibrator and a AC coil/armature was used across the AC supply, this results in a 50hz/60hz vibration.
Max.
Since the load is driven by a PWM on rectified mains, the frequency is set by the PWM, though in this case, 60Hz works great.

Welcome to AAC!

Thank you for using paragraphs, attaching a schematic, and giving the TL;DR summary.

Can you post more of the drive circuitry for the MOSFET? From what you posted, you're going to have issues with slow turn off.

How much current is the MOSFET switching?
Thanks.

That is all of the main drive circutiry at present. The signal from the IC comes through the optocoupler at R15.

I did observe the slow turnoff time as I increased the size of the flyback diode. I ended up just increasing the load on the vibratory assembly and being okay though. The IC receives feedback data from an accelerometer and tries to maintain a constant "strength" by upping the duty cycle to maintain output. Would this slow turnoff time be something that would contribute to the diode heat?
 

MaxHeadRoom

Joined Jul 18, 2013
28,576
Since electromagnets and solenoids are basically just coils, the main difference between AC and pulsed DC is just that when using AC, the direction of the current flow reverses.
The resistance of the electromagnet is only 1.2 Ohms, though this will increase while the unit is running due to it being an inductive load. Unfortunately, I have no idea what the actual inductance is and am hoping to hear back from the manufacturer and I don't have a (working) function generator to measure it.
The final inductive reactance will be much different between (pulsating) DC and AC.
Max.
 

Alec_t

Joined Sep 17, 2013
14,263
That electromagnet looks quite chunky. I would guess its inductance as >100mH. Simulation shows that, with 60Hz PWM at 25% duty cycle, the peak-to-peak current ripple through 100mH would be about 4A on top of a 26A standing current. That ripple is the source of the required vibrations, but is outweighed by the standing current. Surely there ought to be a more efficient way of generating vibrations? The big standing current accounts for much of the heat dissipated in the freewheel diode.
 

MaxHeadRoom

Joined Jul 18, 2013
28,576
A job to see if it has laminated cores, but I would be tempted to try it on a Variac for straight AC usage for a 60Hz vibrator.
Max.
 

Thread Starter

Matt H

Joined May 2, 2017
13
Crutschow: That makes a lot of sense about the diode ratings. I thought that a better rated diode would generate less heat, but it is still trying to dissipate the same amount. I have been using a fairly large heatsink on all of these, so the heatsink alone won't be enough especially since I can't add extra ventilation. The best option may be to use an extra Mosfet like you suggested. We already have a driver chip from another board that handles the on-off crossconduction times. Dissapating the energy in the inductor is fine with me since the accelerometer provides PID feedback and will compensate for a steady output and the coil itself is already rated for 120V continuous without problems.


I went ahead and hooked my multimeter up to the transformer on one of the lower duty settings and was measuring 7.5A. My meter is only rated for 10A with brief periods over 20A. At first, this didn't make much sense to me since the unit is pulling much less current than that, then I read Alec's post and realized that the bulk of that is standing current.

Since the current is so high, I'm guessing that means that means that adding a TVS diode would do nothing to help the problem?

Due to the amount of inventory that we are sitting on, changing the vibration method itself is out of the question. I definitely feel like this coil isn't efficient by any means though.

I'm going to try to get back down to my desk in a bit and will post what I have of the schematic for the driver on the old design and ask about parts that I think are redundant on it.
 

Thread Starter

Matt H

Joined May 2, 2017
13
Here is the schematic for the original driver that was used for the driver. I have also linked the datasheet below and the example schematic from it.

Is there any point to using diodes D2, D3, D4, or D5? I don't see any point in having them doubled up in parallel (especially since I don't remember there being heat issues). I'm also not sure what D4 or D5 would offer at all. I suppose that D2 or D3 would provide a flywheel path if there is a timing issue though.

The only other difference between the two that I spot is the capacitor in the top right corner of the datasheet schematic. Since the +HV connection is from rectified power already, I'm assuming that this is just a .1uF or 1uF ceramic cap?


OriginalDriver.jpg

DatasheetSchematic.jpg
 

crutschow

Joined Mar 14, 2008
34,201
I went ahead and hooked my multimeter up to the transformer on one of the lower duty settings and was measuring 7.5A. My meter is only rated for 10A with brief periods over 20A. At first, this didn't make much sense to me since the unit is pulling much less current than that, then I read Alec's post and realized that the bulk of that is standing current.

Since the current is so high, I'm guessing that means that means that adding a TVS diode would do nothing to help the problem?
You could add a power resistor in series with the diode to help dissipate the inductive energy and eliminate the standing current.
You'd have to experientially determine the correct resistor value.
To high a resistance and the kickup voltage could zap the MOSFET.
 

Thread Starter

Matt H

Joined May 2, 2017
13
I would have thought you could have left Q2 on continuous and just switched Q1?
Max.
I honestly don't see why this wouldn't work. I guess it doesn't hurt to try both ways and see.


You could add a power resistor in series with the diode to help dissipate the inductive energy and eliminate the standing current.
You'd have to experientially determine the correct resistor value.
To high a resistance and the kickup voltage could zap the MOSFET.
If I'm thinking through this right, that would have to be one hell of a resistor. If we could get away with a 1k resistor at 170V (and the the voltage off the inductor could spike to ~x10 that), it would be dissipating ~30W.
 

Alec_t

Joined Sep 17, 2013
14,263
I would have thought you could have left Q2 on continuous and just switched Q1?
If you do that you get a big standing current, which is undesirable. The first circuit in post #12 works fine to eliminate the standing current, but note that bootstrap cap C16 does nothing useful there (and if it were connected to Q1 source, LTspice says spurious high frequency oscillations occur).
 

Thread Starter

Matt H

Joined May 2, 2017
13
If you do that you get a big standing current, which is undesirable. The first circuit in post #12 works fine to eliminate the standing current, but note that bootstrap cap C16 does nothing useful there (and if it were connected to Q1 source, LTspice says spurious high frequency oscillations occur).
Makes sense. What is your take on the diodes D2, D3, D4, and D5? I don't see any point in D4 or D5 and am mixed about the necessity of D2 or D3.
 

Sensacell

Joined Jun 19, 2012
3,420
Seems like the driver should provide a symmetrical drive waveform- with zero net DC component.

Drive it with an H_Bridge, with symmetrical duty cycle, this would prevent the flux from "ratcheting up"

Vary the frequency to match the mechanical resonance frequency - this will reduce the power required.
 

Tonyr1084

Joined Sep 24, 2015
7,829
Two comments from the peanut gallery here:

1) You're hitting a coil with a square wave. The rise time and fall time are on the order of "Instantaneously". This is different from a sine wave, where the voltage (starting at zero and at peak acceleration) rises toward peak voltage, at which time the acceleration stops and begins to reverse direction. From there it accelerates across the zero threshold and begins to decelerate as it approaches the bottom peak of the sine wave. AC is like swinging a baby in a baby-swing. Accelerating in one direction then slowing down to a stop then accelerating in the opposite direction. Repeat and rinse (ok forget the rinse cycle), just repeat. Imagine what would happen if you swung the baby digitally: Slam the baby forward then slam the baby back. The baby is not going to be happy about that. Neither will your circuit be happy.

2) You have a magnetic field. It collapses. The flyback diode takes that nearly full rush of electrons (electrical current). Depending on the voltage and the current you're going to see X number of watts. NO MATTER WHAT DEVICE YOU PUT ON IT it's going to see the same watts. Putting larger and larger devices on it isn't going to make it run cooler. You still have the same amount of energy to dissipate. If you use a device rated at 10 amps and approach 10 amps the device will survive but get as hot as it will get. If you put a device that can handle 20 amps - it's still going to dissipate the same wattage (heat). If you put a 200 amp device on it - you're still going to see the same wattage being dissipated. The 10 amp versus the 200 amp device - well, the 200 will last longer. But shouldn't run any cooler.

This has been a comment from the peanut gallery. Right or wrong - this is how I'm seeing the problem. As to what the solution is - I think I'd best leave it up to the experts.
 

Tonyr1084

Joined Sep 24, 2015
7,829
You have an average breakdown voltage from the coil and a max amperage. Whatever those numbers are you calculate wattage. The only way to lower the amount of heat (watts) is to lower the voltage or lower the current. Maybe (from the peanut gallery) a resistor in series with the flywheel diode will make it run cooler. Keep the resistance low so as to not adversely affect the rate of collapse in the magnetic field.

[edit]
However, the resister plus the diode - wattage dissipated - will still be the same. Just that each device will be dissipating some share of the heat. If this is inside a poorly ventilated cabinet the temperature will still rise at the same rate.
 
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