SMPS for tube pedals

Started by snailspacejase, July 10, 2023, 02:42:56 PM

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Rob Strand

#40
Quote from: printer2 on June 03, 2024, 12:27:19 PMYes it could be a lot of reasons why it might have blown, from poor design to bad parts. First heard of the diode blowing from this video.

https://youtu.be/qB8FpGTyVTw?t=755

Think this may be the schematic.
Unfortunately the devil is in the details of with these type of issues. If we look at what we know:
- the diode fails
- the diode gets hot
- the output current is relatively low compared to the diode forward current rating.
- the diode failed at the high voltage setting in the video.
  (Unfortunately the high voltage setting is also when the output current
   is highest in the video.)

I'm thinking the problem is related to diode reverse-recovery.

The design itself uses a transformer, unlike the some of the transformerless designs in the thread.  That usually helps reduce stresses in the parts.

As to why the US3MC fails and the UF4007 does not.   That's not so obvious.

Something else about the video.  It looks like the converter was set to a voltage
where the output no longer increases.   This is like the switchmode equivalent of
regulator dropout.   Some designs can stress the parts under those conditions as
some parts get driven too hard trying to keep the output voltage regulated (which
it can't do).   It's best to choose adjustable pots so that can't happen.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

printer2

Quote from: Rob Strand on June 04, 2024, 08:14:51 PM
Quote from: printer2 on June 03, 2024, 12:27:19 PMYes it could be a lot of reasons why it might have blown, from poor design to bad parts. First heard of the diode blowing from this video.

https://youtu.be/qB8FpGTyVTw?t=755

Think this may be the schematic.
Unfortunately the devil is in the details of with these type of issues. If we look at what we know:
- the diode fails
- the diode gets hot
- the output current is relatively low compared to the diode forward current rating.
- the diode failed at the high voltage setting in the video.
  (Unfortunately the high voltage setting is also when the output current
   is highest in the video.)

I'm thinking the problem is related to diode reverse-recovery.

The design itself uses a transformer, unlike the some of the transformerless designs in the thread.  That usually helps reduce stresses in the parts.

As to why the US3MC fails and the UF4007 does not.   That's not so obvious.

Something else about the video.  It looks like the converter was set to a voltage
where the output no longer increases.   This is like the switchmode equivalent of
regulator dropout.   Some designs can stress the parts under those conditions as
some parts get driven too hard trying to keep the output voltage regulated (which
it can't do).   It's best to choose adjustable pots so that can't happen.


I am a toddler when it comes to SMPS's, I can follow an explanation of how the circuit works but have only read about a couple of topologies and that was a number of years ago. That said, the diode reverse recovery was something I thought might be an issue. The UF4007 has the same spec. as the US3MC, is the UF4007 much faster than it is listed at or is the US3MC slower than it should be? Or is it something else? I don't know. I did realize it was my supply that was shutting down due to overcurrent rather than the module.

Just a thought. The bipolar version of the module has an extra diode and capacitor both running half wave. Do you think the diode will have a better chance of survival if I make the output circuit a full wave rectifier? With two diodes passing current to the capacitor both would be getting half the current. While the total heat dissipated is the same for the same board area the spot temperature at the diode should be lower. Unless there is something I am missing.
Fred

Rob Strand

#42
Quote from: printer2 on June 05, 2024, 02:38:14 PMI am a toddler when it comes to SMPS's, I can follow an explanation of how the circuit works but have only read about a couple of topologies and that was a number of years ago. That said, the diode reverse recovery was something I thought might be an issue.
The fact the diode overheats with load currents much less than the forward rating is a key piece of info to blame reverse recovery.   I probably should know the answer but I don't play with SMPS's every day so it often takes me half a day to reset my mind.

QuoteThe UF4007 has the same spec. as the US3MC, is the UF4007 much faster than it is listed at or is the US3MC slower than it should be? Or is it something else? I don't know.
Not all diodes are alike.   The recovery process is complicated and the reverse recovery figure combines at least two effects.  Those two effects might be different for each diode.   Given the converter is right on the point of diode failure a factor of two difference in the details could save it!

QuoteI did realize it was my supply that was shutting down due to overcurrent rather than the module.
When your PSU current limits it will lower the output voltage from the set value.  That could set-up a situation similar to setting the SMPS output voltage to something it cannot meet - a similar condition to what happens in the video except you are reducing the input instead of increasing the output.

QuoteJust a thought. The bipolar version of the module has an extra diode and capacitor both running half wave. Do you think the diode will have a better chance of survival if I make the output circuit a full wave rectifier? With two diodes passing current to the capacitor both would be getting half the current. While the total heat dissipated is the same for the same board area the spot temperature at the diode should be lower. Unless there is something I am missing.
The way a boost converter works is power is transferred on only one of the switch cycles.  It has to work that way for either polarity output.   The way you get a negative polarity output is to flip the connection of the winding (The diode doesn't actually get flipped.  It usually gets moved to the other side of the winding).

I have a feeling the extra diode winding + diode + cap would do little.   With such a small dummy load on the negative polarity output you would just get a little more loss on the negative output diode but the positive side would pretty much do what it did before - perhaps an insignificant fraction less.

A general trend with switching devices is throwing bigger parts at a problem doesn't always fix the problem.   You can imagine someone building an SMPS, frying the diode, then updating the design with a high current diode.   In reality it can actually make the true problem worse.  Higher rating parts are slower.   If the problem is related to switching losses the losses will get worse but perhaps the larger package diode handles the losses and gives the appearance of fixing the problem.   A better solution might be to stick with a smaller rating diode which is faster.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

printer2

Quote from: Rob Strand on June 05, 2024, 08:57:02 PM
QuoteJust a thought. The bipolar version of the module has an extra diode and capacitor both running half wave. Do you think the diode will have a better chance of survival if I make the output circuit a full wave rectifier? With two diodes passing current to the capacitor both would be getting half the current. While the total heat dissipated is the same for the same board area the spot temperature at the diode should be lower. Unless there is something I am missing.
The way a boost converter works is power is transferred on only one of the switch cycles.  It has to work that way for either polarity output.   The way you get a negative polarity output is to flip the connection of the winding (The diode doesn't actually get flipped.  It usually gets moved to the other side of the winding).

I have a feeling the extra diode winding + diode + cap would do little.   With such a small dummy load on the negative polarity output you would just get a little more loss on the negative output diode but the positive side would pretty much do what it did before - perhaps an insignificant fraction less.

A general trend with switching devices is throwing bigger parts at a problem doesn't always fix the problem.   You can imagine someone building an SMPS, frying the diode, then updating the design with a high current diode.   In reality it can actually make the true problem worse.  Higher rating parts are slower.   If the problem is related to switching losses the losses will get worse but perhaps the larger package diode handles losses and gives the appearance of fixing the problem.   A better solution might be to stick with a smaller rating diode which is faster.

I took a look at the schematic after reading your explanation and it seems obvious. I looked through Digikey's selection and not a lot of fast high voltage diodes that will fit. Found the following, physically small but doable.

https://www.digikey.ca/en/products/detail/vishay-general-semiconductor-diodes-division/VS-E7FX0212-M3-I/22145760

Fred

Rob Strand

Quote from: printer2 on June 06, 2024, 04:10:39 PMI took a look at the schematic after reading your explanation and it seems obvious. I looked through Digikey's selection and not a lot of fast high voltage diodes that will fit. Found the following, physically small but doable.

https://www.digikey.ca/en/products/detail/vishay-general-semiconductor-diodes-division/VS-E7FX0212-M3-I/22145760
You did well just the same.  Modern diode technology and it certainly looks like it's made for the job.  You can die from information overload selecting parts these days!
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

printer2

Quote from: Rob Strand on June 06, 2024, 05:09:56 PMYou did well just the same.  Modern diode technology and it certainly looks like it's made for the job.  You can die from information overload selecting parts these days!

Took half a day trying to find one that ticks off all the boxes. And now that I am ordering, what else should I pick up? So much for a quick use of someone else's technology.
Fred

Rob Strand

Quote from: printer2 on June 07, 2024, 10:56:23 AMTook half a day trying to find one that ticks off all the boxes. And now that I am ordering, what else should I pick up? So much for a quick use of someone else's technology.
Don't feel alone, it's a common occurrence.  Sometimes you get find a good part but it's 35wks lead time.  You can see why companies stick to using the same chips/parts.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

markusw

#47
Quote from: Rob Strand on July 14, 2023, 08:53:06 PMInteresting the timing cap value is 390pF.  The trace I had for the through-hole version quoted 100pF to 270pF
which seemed a bit low.  The 390pF (say 330pF to 470pF) seems more likely.

The SMD board has an SMD inductor.  From what I can work out it's 12mm x 12mm x 8mm high.  Wurth sometimes have different inductor heights for the same footprint and the inductors have different specs.
From what I can work out it's a,

Wurth (WE), PD SMT Shielded Power Inductor, 744770222
Isat    1.49A
RDC    247m ohm typ.,  390m ohm max

Here's the through-hole trace from the web with a few fixes,



There's no heatsink on the MOSFET.


Any idea why the current sensing resistor was omitted?
Could it be added to replace the fuse? Not as a part of the reverse-polarity protection of course, only to protect the circuit in case of a short. 
Would a current sensing resistor change noise level? Probably not, I guess.

Regards,
Markus

markusw

Quote from: markusw on June 10, 2024, 06:22:06 AMAny idea why the current sensing resistor was omitted?
Could it be added to replace the fuse? Not as a part of the reverse-polarity protection of course, only to protect the circuit in case of a short. 
Would a current sensing resistor change noise level? Probably not, I guess.

One question I can answer myself  :icon_redface:
The current sensing resistor won't help in case of a short since there is a path to gnd via the diode.  ;)
The other questions remain.
Markus

Rob Strand

Quote from: markusw on June 10, 2024, 12:58:50 PMOne question I can answer myself  :icon_redface:
The current sensing resistor won't help in case of a short since there is a path to gnd via the diode.  ;)
The other questions remain.
That's pretty much it!

Quote from: markusw on June 10, 2024, 06:22:06 AMAny idea why the current sensing resistor was omitted?
Could it be added to replace the fuse? Not as a part of the reverse-polarity protection of course, only to protect the circuit in case of a short. 
Would a current sensing resistor change noise level? Probably not, I guess.
The current limit on these devices only limits the switch peak current; the app note sets the limit to the built-in switch current limit.    I don't think it does a good job of limiting the output current and won't protect against short circuits.  So it's more parts, lower efficiency due to more voltage drop across the sense resistor, then ultimately not full protection.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

markusw

#50
Quote from: Rob Strand on June 10, 2024, 11:03:37 PMThe current limit on these devices only limits the switch peak current; the app note sets the limit to the built-in switch current limit. 

Thank you very much for your help!
As I understand the current limiting resistor is calculated via the saturation current of the inductor. Therefore, it should prevent saturation of the inductor. Wouldn't this improve the design?
Not sure what you mean with "the app note sets the limit to the built-in switch current limit."  :icon_redface:

Quote from: Rob Strand on June 10, 2024, 11:03:37 PMSo it's more parts, lower efficiency due to more voltage drop across the sense resistor, then ultimately not full protection.
I did a LTSpice simulation of the Kingsley SPMS with 320V output, 130k load (~2.5mA) and a 0.3 ohm current sensing resistor (limiting peak current to ~1A). Dissipation of the current sensing resistor in average is only roughly 15 mW versus 0.8W of the load.



Rob Strand

Quote from: markusw on June 11, 2024, 01:25:22 AMAs I understand the current limiting resistor is calculated via the saturation current of the inductor. Therefore, it should prevent saturation of the inductor. Wouldn't this improve the design?
Not sure what you mean with "the app note sets the limit to the built-in switch current limit."  :icon_redface:
It limits both the inductor current and the switch current.   You can limit to whatever breaks first, or lower currents if you want to impose a different limit.    The MC34063 application note (and some other parts like the TL497) just highlight the maximum current is set by the switch (build into the part).

Under normal conditions the boost converter limits it's own current by design (the peak inductor current is set by the on-time and the inductor value).   The idea is not to rely on the current limit to set the operating peak current.

Quote from: markusw on June 11, 2024, 01:25:22 AMI did a LTSpice simulation of the Kingsley SPMS with 320V output, 130k load (~2.5mA) and a 0.3 ohm current sensing resistor (limiting peak current to ~1A). Dissipation of the current sensing resistor in average is only roughly 15 mW versus 0.8W of the load.
It seems quite low but believable.  I haven't checked by any hand calculations.

I guess from a design perspective you have to ask what are you protecting against.   It becomes a bit of a value judgement.

Things like design issues should be fixed before relying on protection, eg. the diode fault in this thread.    In normal operation for a tube there's not a lot to go wrong and short the supply rails.    In other applications the parts in the load circuit can fail and short and then wreak havoc on the supply.   For cases like that I have been very grateful designers have put in supply protection because it limits the damage and helps debug the fault.   In other cases the supply or load has failed and shorted out a whole lot of parts.   In one case I had to fix the fault and rewind the inductor then make sure I got it all right before power up.  That's where you wish some form of safety valve was put in place to minimize the damage.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

markusw

#52
Quote from: Rob Strand on June 13, 2024, 07:08:17 PMIt limits both the inductor current and the switch current.  You can limit to whatever breaks first, or lower currents if you want to impose a different limit.    The MC34063 application note (and some other parts like the TL497) just highlight the maximum current is set by the switch (build into the part).

Under normal conditions the boost converter limits it's own current by design (the peak inductor current is set by the on-time and the inductor value).  The idea is not to rely on the current limit to set the operating peak current.

Again, thank you very much for your help!

Without the current sensing resistor and a 330p timing cap the LTSpice sims predict slightly above 0.8 A peak current (roughly the same as with the 0.3 ohm sensing resistor), with a 470p cap peaks are around 1.2A without the current sensing resistor (all with 320V output voltage and 130k load). 
Without the current sensing resistor there is a ~3A peak at start up, that is prevented by current sensing resistor.
Don't know whether this only a LTSpice artifact.




Rob Strand

#53
Quote from: markusw on June 15, 2024, 05:15:53 AMWithout the current sensing resistor and a 330p timing cap the LTSpice sims predict slightly above 0.8 A peak current (roughly the same as with the 0.3 ohm sensing resistor), with a 470p cap peaks are around 1.2A without the current sensing resistor (all with 320V output voltage and 130k load). 
Without the current sensing resistor there is a ~3A peak at start up, that is prevented by current sensing resistor.
Don't know whether this only a LTSpice artifact
You could be on to something there.   The fact the current limit helps reduce the start-up current is hint it is real.  Power up surges are a source of components failing.   Silicon power devices will handle somewhat higher than rated currents for short periods.   It's when those currents are too high it can cause failures in the field, not always today but some time in the future.   A lot of modern devices implement soft starting.  (I've got some simulations but they are on another computer.)

Something you can try on spice is change the power source delayed pulse.   Then you can see what happens at power up.  You can also experiment with the risetime of the pulse to see how it affects the start-up current.   That gives an idea what is real and what the cause is.

In a real circuit you might have a wall-wart with 20 ohm output impedance feeding a 1000uF cap when the main is switched on there is a natural soft start.   However, if you have 2A a switch mode power supply wall-wart the output impedance is lower and the start-up isn't as soft.   Yet another scenario is the wall-wart is already powered-up then you plug it into the DC converter.   The different power scenarios can affect what currents flow at power up.

For the boost converter you can get a start-up current pulse from the inductor through the diode to the output cap.   That current pulse cannot be fixed with with the current limit (other than the current limit resistor adding some series resistance and limiting crazy high currents.  Normally the inductor DC resistance will be higher.)
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

markusw

Quote from: Rob Strand on June 15, 2024, 08:19:17 PMYou could be on to something there.   The fact the current limit helps reduce the start-up current is hint it is real.  Power up surges are a source of components failing.   Silicon power devices will handle somewhat higher than rated currents for short periods.   It's when those currents are too high it can cause failures in the field, not always today but some time in the future.   A lot of modern devices implement soft starting.  (I've got some simulations but they are on another computer.)

Something you can try on spice is change the power source delayed pulse.   Then you can see what happens at power up.  You can also experiment with the risetime of the pulse to see how it affects the start-up current.   That gives you can give you an ideal what is real an what the cause is.

Thanks!
You mean delaying the start of the supply voltage and/or slowly ramping it up?

R.G.

For a given set of conditions - forward current, reverse voltage, and turnoff time - each reverse recovery generates the same amount of heat. The average heat produced in a part depends on how many times you do a reverse recovery per second. That is, raw switching frequency matters. The more often you do a reverse recovery, the hotter your switching things get.

I learned this while in the process of causing a neat, round hole to be melted through the top of a steel cased Motorola TO3 switching transistor. :-( 

The higher the switching frequency, the lower the recovery losses per switching has to be to keep the heat down to the same level.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

Rob Strand

Quote from: markusw on June 16, 2024, 11:30:46 AMYou mean delaying the start of the supply voltage and/or slowly ramping it up?
Yes for the examples I gave.   However, in power electronics you will also come across deliberate schemes where the PWM is limited on startup which lowers start-up current, reduces glitches and of course starts-up slower.

Quote from: R.G. on June 16, 2024, 03:24:53 PMFor a given set of conditions - forward current, reverse voltage, and turnoff time - each reverse recovery generates the same amount of heat. The average heat produced in a part depends on how many times you do a reverse recovery per second. That is, raw switching frequency matters. The more often you do a reverse recovery, the hotter your switching things get.
For the frying diode issue: As output voltage increases the boost converter frequency rises so the losses go up as per your comments.  In the video that mechanism is compounded by the increased output voltage and current.  All three mechanisms are increasing and adding to the stress on the diode.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

R.G.

Quote from: Rob Strand on June 17, 2024, 06:29:34 PMFor the frying diode issue: As output voltage increases the boost converter frequency rises so the losses go up as per your comments.  In the video that mechanism is compounded by the increased output voltage and current.  All three mechanisms are increasing and adding to the stress on the diode.
Yes, that is true.
I've done a small foray into the recovery losses of the two kinds of diodes in play. Honestly, my thumbnailing of the recovery losses from the data sheets don't turn up losses that ought to cause hot diodes. My best guess right now is that there is something odd going on to run up the recovery losses on the diodes.
Not sure what that is right now, but I can't yet reconcile the kinds of recovery losses needed for the heat described with the circuit conditions, even given the higher voltage and current. Really, the currents and voltages are not that high. I'm relating this to the voltages and currents on snubber circuits on the primaries of flyback power converters that produce tens of watts. Something isn't correlating yet.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

R.G.

In my ongoing quest for numbers, I found a couple of references to refresh me:
https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://www.st.com/resource/en/application_note/an5028-calculation-of-turnoff-power-losses-generated-by-a-ultrafast-diode-stmicroelectronics.pdf&ved=2ahUKEwjL3dLz6eOGAxV7lokEHcj5ADEQFnoECBYQAQ&usg=AOvVaw1MEkPifG4MtC-bTcceJyCw
and
https://www.vishay.com/docs/98280/howtocalculatepowerlossesingen5diodes.pdf
as well as a couple of others.
When I get a little free time I'll try to thumbnail the losses to see if the described voltage, current and frequency cause an honest heat rise for the diodes, or if there is obviously something funny.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

Rob Strand

#59
This one is quite good.
https://www.infineon.com/dgdl/an-989.pdf?fileId=5546d462533600a40153559fa625124d

It covers everything and has some numbers.   The diode data is more complete than most diodes.  In the literature you will find reverse recovery is discussed in terms of switch loss and not so much the diode loss but that document covers both.

Quote from: R.G. on June 17, 2024, 09:13:39 PMWhen I get a little free time I'll try to thumbnail the losses to see if the described voltage, current and frequency cause an honest heat rise for the diodes, or if there is obviously something funny.
Agreed.   I plugged in some numbers back when this issue came up and I wasn't convinced;  need to find time to go over it in detail.  The symptoms are quite convincing through.   The problem with reverse recovery is it depends on the circuit.  The values in the datasheet are more values from test cases, not hard numbers to plug into formulas for all circuits.

Something which did seem odd.  Suppose we fry the diode due to overheating at 150degC.  The thermal resistance of the diode was something like 45 degC/W.  So we need 3.3W of diode loss.    I'm not seeing that type of number in the calculations.

Another point was the long tracks on the PCB could have enough inductance to raise the reverse voltage (which is in the doc I posted).  [Actually at the time of the thread I didn't realize there was a transformer until the schematic was posted.  The transformer leakage inductance will contribute.]
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.