Converter In-Rush Thermistors etc, UPDATE -4

[BFL13]

UPDATE on 17 Nov, Update 2 on 22 Nov, Update 3 on 16 Jan 14,
Update 4 on 19 Oct 14.
---------------

In a recent thread there was mention of how you can wreck your gizmo (in my case a converter) by turning it off and restarting it again before the capacitors have finished unloading----if I was understanding that discussion.

So I am worried I am going to wreck my nice PowerMax 100 amp converter doing that.

Background is that last year I had one of these and had the top off so I could adjust its voltage with the internal pot they have. Somehow at one point after some weeks of using it, the in-rush current thermistor burst into flames and that was that.

PowerMax sent me a new "beefier" thermistor which they said would also be used in the newer 100 ampers. I got that in and had that going for a while and then I think when starting and stopping and restarting right away (maybe that was it, not sure) that thermistor over-heated and fell to bits.

I am wondering if it was operator error in starting/ stopping sequence that did for both thermistors.

I do see on the voltmeter that after you shut off the converter (when disconnected from battery), its voltage tapers down for a good time rather than drops straight to zero, and from the discussion mentioned above, I understand that is the capacitors "unloading." So now I am wondering if I blew the new in-rush thermistor last year by restarting the converter under full load of 100amps too soon?

The in-rush thermistor seems to be on the 120v input side while the capacitors that unload are on the DC side some distance from the thermistor on the circuit board so I am not sure how they relate.

BTW PowerMax was kind enough to help me out replacing my ruined converter and I now have a fancy new prototype ( I guess) version with an external voltage pot this time. I promised not to take the top off and poke around inside it 🙂

However, the way I do my recharge, it can happen that say the Honda pops its breaker due to other loads added besides the converter, and I restart the Honda before the converter capacitors unload. Even with a normal start, the converter sometimes makes a snap sound like my big inverter will when disconnecting and connecting, which is capacitors according to folks here who have commented on that. (except how can the converter ones snap on start-up if they are already unloaded?)

I have been using this new 100amper for a while now and nothing has gone wrong. But I am scared to death of screwing up my nice converter by some sort of mystery operator error if that is what it was. I can't avoid "doing it again" when I don't know what I did!

Thanks

[Salvo]

Interesting, the PD patent shows a surge current thermistor of 1 ohm.


Sal

[Salvo]

The reason for looking at the smaller thermistor is because it has no, or very little ceramic material surrounding it. The thermistor that we're dealing with has a time constant of 134 s. It's slower due to the attached mass. Still, the small bead thermistor has a sizable time constant.

I could go on with this discussion, but it's really a moot point that we're arguing. The turn-on surge current is not the problem! Regardless if we're using a 2 ohm or 5 ohm device, the energy dissipated in the thermistor is only 35J. The part fails at 1200J. It isn't failing during turn-on. Your argument that the 5 ohm part is better during turn-on is not valid. That is not the problem.

Sal

[BFL13]

DryCamper11 wrote:

BFL13 wrote:
Interesting about the way the "mass" part of the thermistor works.

So what is the implication between these two, which are both 5R but 20a and 25a where both those amps are well above steady state 13a. The obvious difference is in the greater mass of the MS35, but would that be better? (I have managed to blow both types with hot restarts)

http://www.ametherm.com/datasheetspdf/MS355R025.pdf

http://www.ametherm.com/datasheetspdf/SL325R020.pdf

The MS35 5R025 is wider and thicker. Yes, it has the same resistance, but it can handle 600 Joules instead of 200. Note that the capacitance numbers are also exactly 3 times greater. They get the same resistance, but higher energy/capacitance numbers by using different semiconductor material so the net junction volume is greater, but the net resistance is the same. More junction volume with the same energy dissipated means less heating per junction region and less damage, so the ratings can be higher.

I don't know if I answered the question you asked, but it was my best shot. 🙂

I am looking at the 2R025s again after reading the latest on all this.
The above says the 355R025 and the 325R020 have the same material but the data sheet says they are both "material (for Beta and curve) H"

the 322R025 is material G though. No idea what is G or H.

http://www.ametherm.com/datasheetspdf/SL322R025.pdf

anyway, the 2R025 has higher capacitance than the 5R020 and more Joules so that is supposed to make it better for inrush, but it does have lower R but is made of G instead of H (is that meaningful?).

So back to the question whether that R number matters for the in-rush stage or just for the steady stage. We have no idea AFAIK what the inrush amount is for calculating minimum R. We do know the spec for the diode bridge 25a 60 where Ken used 170/50 to get that 3.4R

so is the 2R025 out of play unless put in series as 4R, or could the single 2R025 be ok after all?

[ken_white]

DryCamper11 wrote:

Salvo wrote:
Yes, we're only talking about the actual thermistor, not the adjacent material. It has a large time constant. Take a look here.

I'm not sure why you're pointing to a bead thermistor. They are used for measuring temperature, not for controlling current inrush.

I once designed and built a device that measured how fast an elevator was rising/falling. It used two tiny bead thermistors in a tube connected to a thermos bottle. As the bottle went up/down in the elevator, air would flow in/out of the tube to fill or exit the bottle. The bead thermistors were positioned in line so that one or the other would be cooled more than the one behind as air pressure change due to elevation change caused air to flow into the bottle (elevator down) or air to leave the bottle (elevator up). A differential amp with a bridge circuit allowed measurements of only a few feet per second of elevator speed.

Encased thermistors have 2.5 s max time constant. ....
I've provided a link showing thermistor time constant is max 2.5s.....

You're still looking at time constant for the entire thermistor. That's not what's relevant. It's the time constant for heat to leave the semiconductor active region and flow into the bulk semiconductor material on either side thereof. We're talking micron size regions and heat flow distances. That has little or nothing to do with the quoted time constant of the bulk device. That time constant is factored into the Joule rating and the capacitance rating of the device. It isn't specified directly.

The capacitance rating and the voltage set the Joules that the device is able to absorb. That energy appears as heat in the active region of the semiconductor junction. It will move from those hot spots in milliseconds. If the heat isn't allowed the millisecond(s) time it takes to diffuse away from that region, it is damaged and the device fails.

The device time constant that is specified in the ratings is different. That number is used to determine how long it takes before the device has heated to the low resistance value (or recovered to the high resistance value specified at zero power after power is removed.) That time constant has nothing to do with the surge ratings.

Yep, he just doesn't get it...

[DryCamper11]

Salvo wrote:
Yes, we're only talking about the actual thermistor, not the adjacent material. It has a large time constant. Take a look here.

I'm not sure why you're pointing to a bead thermistor. They are used for measuring temperature, not for controlling current inrush.

I once designed and built a device that measured how fast an elevator was rising/falling. It used two tiny bead thermistors in a tube connected to a thermos bottle. As the bottle went up/down in the elevator, air would flow in/out of the tube to fill or exit the bottle. The bead thermistors were positioned in line so that one or the other would be cooled more than the one behind as air pressure change due to elevation change caused air to flow into the bottle (elevator down) or air to leave the bottle (elevator up). A differential amp with a bridge circuit allowed measurements of only a few feet per second of elevator speed. I could hold it near my feet, raise it to my head and read off the rate of elevation change. Those devices are amazing.

Encased thermistors have 2.5 s max time constant. ....
I've provided a link showing thermistor time constant is max 2.5s.....

You're still looking at time constant for the entire thermistor. That's not what's relevant. It's the time constant for heat to leave the semiconductor active region and flow into the bulk semiconductor material on either side thereof. We're talking micron size regions and heat flow distances. That has little or nothing to do with the quoted time constant of the bulk device. The time constant I'm referring to isn't specified in any ratings. It is factored into the Joule rating and the capacitance rating of the device.

The capacitance rating and the voltage set the Joules that the device is able to absorb. That energy appears as heat in the active region of the semiconductor junction. It will move from those hot spots in milliseconds. If the heat isn't allowed the millisecond(s) time it takes to diffuse away from that region, it is damaged and the device fails.

The device time constant that is specified in the ratings is different. That number is used to determine how long it takes before the device has heated to the low resistance value (or recovered to the high resistance value specified at zero power after power is removed.) That time constant has nothing to do with the surge ratings.

[DryCamper11]

BFL13 wrote:
I got lost with the thermistor/filter capacitor relationship. Is the MS355R025 superior for that compared with the SL325R020, where I have 2400uF of capacitors?

They should both work. The MS355R025 can handle 41,250uF. The SL325R020 can handle 13,893uF (@ 120VAC). You only have 2400uF. As to which is best, the MS355R025 will be stressed less.

It was mentioned there is a minimum thermistor spec wrt the capacitors, but if both are above minimum does it matter which is used?

Not really. One will cost less and the other will be stressed less.

[Salvo]

DryCamper11 wrote:

The thermistor has no time to cool down during the transient.

It's not the entire thermistor that cools down. It is the tiny active region(s) that must transfer heat from that region into the adjacent material. That transfer has a time constant that is much much less than the bulk time constant of the entire device. You can damage the thermistor without producing significant heating of the entire device if enough current is passed through it in a short time.

Yes, we're only talking about the actual thermistor, not the adjacent material. It has a large time constant. Take a look here.

Encased thermistors have 2.5 s max time constant. The transient is only 4 ms. There is no cooling during the transient. It doesn't matter if you got a 1 ohm or 10 ohm thermistor, both will get to the same temperature following a turn-on transient.

That means your argument is not correct. Resistance has no effect on how hot the device gets during the transient.

We disagree. The resistance of the thermistor controls how much heat (energy=1/2 CV**2) is dissipated in the semiconductor active region(s) of the thermistor per unit time. Given the small size of the semiconductor region, the total energy dissipated as heat has sufficient time to diffuse out of that region before the region is damaged even though the entire device does not have time to cool significantly.

Again, I've provided a link showing thermistor time constant is max 2.5s. The greater the thermistor thermal capacity, the greater the time constant. That's something I was eluding to quite a few pages ago. The big capacity thermistors do not respond fast enough in lowering their resistance. 2.5s is an eternity when the converter is up and running drawing 13A. This is the time when we see the tremendous heat stress. The greater the thermistor resistance the greater the stress.

Resistance does have an effect on steady state use. The greater the resistance the hotter the device gets.

No, the greater the resistance, the longer the time it takes to dissipate the 1/2 CV**2 energy (during the initial charge cycle of the capacitor). Power is energy divided by time, and the energy remains constant, while increasing resistance increases the time over which that energy must be dissipated.

You're not reading my comment correctly. I'm not talking about turn-on, but rather steady state operation. The converter is drawing 13A continuous.

In steady state, we are no longer interested in the initial (zero power) resistance - we are interested in the steady state resistance, which is close to zero.

The thermistor resistance gets close to zero after one time constant. Until then, there's some major cooking going on.

Sal

[BFL13]

I got lost with the thermistor/filter capacitor relationship. Is the MS355R025 superior for that compared with the SL325R020, where I have 2400uF of capacitors?

It was mentioned there is a minimum thermistor spec wrt the capacitors, but if both are above minimum does it matter which is used?

[DryCamper11]

BFL13 wrote:
Interesting about the way the "mass" part of the thermistor works.

So what is the implication between these two, which are both 5R but 20a and 25a where both those amps are well above steady state 13a. The obvious difference is in the greater mass of the MS35, but would that be better? (I have managed to blow both types with hot restarts)

http://www.ametherm.com/datasheetspdf/MS355R025.pdf

http://www.ametherm.com/datasheetspdf/SL325R020.pdf

The MS35 5R025 is wider and thicker. Yes, it has the same resistance, but it can handle 600 Joules instead of 200. Note that the capacitance numbers are also exactly 3 times greater. They get the same resistance, but higher energy/capacitance numbers by using different semiconductor material so the net junction volume is greater, but the net resistance is the same. More junction volume with the same energy dissipated means less heating per junction region and less damage, so the ratings can be higher.

I don't know if I answered the question you asked, but it was my best shot. 🙂

[BFL13]

Interesting about the way the "mass" part of the thermistor works.

So what is the implication between these two, which are both 5R but 20a and 25a where both those amps are well above steady state 13a. The obvious difference is in the greater mass of the MS35, but would that be better? (I have managed to blow both types with hot restarts)

http://www.ametherm.com/datasheetspdf/MS355R025.pdf

http://www.ametherm.com/datasheetspdf/SL325R020.pdf

[Salvo]

Why the transformer discussion? The transformer is way out in left field. It has nothing to do with the problem at hand. The transformer does not see any turn-on surge. It is isolated from the ac circuit.

If the manufacturer is still using the same thermistor then there are two possibilities (operator error is not one of them):

1. they haven't found the problem yet.
2. you got a defective converter.

Perhaps the soft-start is not functional or the pwm controller chip is turning on when the cap bank voltage is still too low.

Sal

[DryCamper11]

Salvo wrote:
You are dealing with a thermal transient during turn-on.

Agreed.

The pulse of the transient is much smaller than the thermal time constant of the thermistor.

Again, agreed, assuming we are talking about the bulk thermistor time constant - but we aren't. I thought about discussing this, but didn't think it was worth getting into that detail. An NTC thermistor is typically made of a sintered semiconductor material pressed into a disk. The active semiconductor regions in the thermistor are actually quite small.

The thermistor has no time to cool down during the transient.

It's not the entire thermistor that cools down. It is the tiny active region(s) that must transfer heat from that region into the adjacent material. That transfer has a time constant that is much much less than the bulk time constant of the entire device. You can damage the thermistor without producing significant heating of the entire device if enough current is passed through it in a short time.

That means your argument is not correct. Resistance has no effect on how hot the device gets during the transient.

We disagree. The resistance of the thermistor controls how much heat (energy=1/2 CV**2) is dissipated in the semiconductor active region(s) of the thermistor per unit time. Given the small size of the semiconductor region, the total energy dissipated as heat has sufficient time to diffuse out of that region before the region is damaged even though the entire device does not have time to cool significantly.

Resistance does have an effect on steady state use. The greater the resistance the hotter the device gets.

No, the greater the resistance, the longer the time it takes to dissipate the 1/2 CV**2 energy (during the initial charge cycle of the capacitor). Power is energy divided by time, and the energy remains constant, while increasing resistance increases the time over which that energy must be dissipated.

In steady state, we are no longer interested in the initial (zero power) resistance - we are interested in the steady state resistance, which is close to zero. We aren't talking about steady state design here and in the initial charge cycle, it is the zero power resistance (5 ohm vs. 2 ohm) that is critical. That's why the calculations all include C = the capacitance and produce a minimum resistance - the resistance needed to ensure that the energy dissipated in the device is spread over a sufficient period that heat at the semiconductor junctions of the thermistor can escape those junctions without excessively heating the semiconductor material to the point it is damaged. More resistance is OK. Less is not. That's why the number calculated is a "minimum."

The 5 ohm device will always be hotter than the 2 ohm device.

The 5 ohm device will dissipate the same energy in the initial charge cycle as a 2 ohm device, but it will be spread out over time. The temperature of the active region(s) of the device will be lower as the heat in that region can move out of the active region before it is damaged. We're not talking about bulk heating of the entire device, but localized heating.

In steady state (which is not what we are concerned about), the temperature of the device is a function of the steady state resistance. I agree that if the steady state resistance of the 5 ohm thermistor is greater, then its temperature will be higher. However, provided the device is rated for the steady state current that it actually sees, it will happily survive that higher temperature. The problem encountered here is not with steady state failure. It is with surge/turn on failure, and that is where the zero power resistance (2 ohms vs. 5 ohms) is critical. To put it another way: the maximum temperature of the regions that are sensitive to heat damage (heat produced by current flow through those regions) will be less for the 5 ohm device at surge/turn than for a 2 ohm device because current flow is less.

[ken_white]

BFL's question has been answered so arguing is moot...

[ken_white]

LScamper wrote:
ken white wrote:

"The volt-second relationship is only valid for continuous current flow during steady state.

If you exceed the kVA rating, with a large impulse current, then my guess is the core will saturate - it is being operated outside of design specifications... "

First, I left out that core saturation is due to not enough magnetizing inductance that keeps the unloaded current down. It has nothing to do with load current.

If you exceed the kVA rating with a large impulse current you may melt the windings but you will not saturate the core if you have enough V-S!

I spent 35+ years in pulse-power. I have designed hundreds of high voltage pulse transformers. V-S is the king, VS=NBA. Current is not involved with core saturation.

Couple the under damped capacitor current, with the transformer in-rush current, and the increased second order voltage effect, and then get back to me about not having a change in flux or saturation.

EDIT: My guess is there will be a DC bias occurring between the half cycles during the transient state since the magnitude of both the primary and secondary voltages will be unbalanced due to the very large, changing, impulse current.

However, you are correct that magnetizing current has a large effect on core saturation, but DC bias and increased voltage does too.

[LScamper]

ken white wrote:

"The volt-second relationship is only valid for continuous current flow during steady state.

If you exceed the kVA rating, with a large impulse current, then my guess is the core will saturate - it is being operated outside of design specifications... "

First, I left out that core saturation is due to not enough magnetizing inductance that keeps the unloaded current down. It has nothing to do with load current.

If you exceed the kVA rating with a large impulse current you may melt the windings but you will not saturate the core if you have enough V-S!

I spent 35+ years in pulse-power. I have designed hundreds of high voltage pulse transformers. V-S is the king, VS=NBA. Current is not involved with core saturation.