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

[LScamper]

Salvo - like they say a picture is worth 1000 words!

It would be nice, if you have time, to see the time delay from turn on to when the converter starts to supply load current. Is the thermistor at the lowest resistance at that time?

Also would be interesting to see how fast the load current ramps up.

Also would be interesting to see input current when applying load after the converter is powered up and the thermistor has had a chance to cool down. (A small load so it won’t destroy your converter!)

[Salvo]

What was missing in previous calculations was actual ac line and converter ac input resistance. In my case, it's about 1.7 ohm. My test setup and ac power source is about 10 feet away from the main breaker. That means I don't have a long ac cable run to my test station. In the OP's case, I think we can add 1.5 ohm resistance to a peak current calculation.

That means if the converter is turned on when the ac is at it's peak, max current is:

With 2 ohm thermistor: I = 170V / (1.5 + 2 ohm) = 49 A
With 5 ohm thermistor: I = 170V / (1.5 + 5 ohm) = 26 A

The 2nd cycle is 78% of the first, so 38A for 2 ohm and 20A for 5 ohm.

Since the diode bridge has a 300A surge rating, it is well protected with the 2 ohm thermistor.

Sounds like PowerMax doesn't understand the failure mechanism. I don't believe the issue is a fast on/off/on sequence. You don't see large surge currents when the cap bank is partially charged. In my testing, I had to wait 10 min between turn-ons to ensure the caps are discharged to less than 1 V.

Let's compare a hot turn-on to a cold one.

The thermistor instantaneous power consumption of a cold turn-on is:

P_cold = I^2 * R = 26A^2 * 5 ohm = 3,380 W

The thermistor instantaneous power consumption of a hot turn-on is (note: cap bank is at 75V & thermistor at 0.03ohm):

P_hot = ((170V - 75V)/1.53ohm)^2 * 0.03 ohm = 116 W

The thermistor is not getting stressed that much during a hot turn-on.

That is not the problem.

Sal

[BFL13]

Interesting for sure. (needs an edit for that A to be an R) It would be helpful to scale those results (even a WAG) to my 100 amper since the dispute has been to use above or below 3R and I have a choice of 5Rs and 2Rs in my collection of spare thermistors.

PowerMax is going to look into what might be causing my problem, but they don't have much to work with since there is such a miniscule failure rate on their converters.

Meanwhile, thanks to all the good advice here, and what I have tested, I believe I personally can stay in business by not doing any more hot restarts and by clamping on first.

[Salvo]

Had a little time on my hands this morning. I used another method to measure converter turn-on current. I inserted a precision 1 ohm, 50W resistor in the "hot" side of the ac power going into the converter. CH1 of scope is measuring incoming ac voltage. CH2 which is inverted measures ac voltage on the other side of the 1 ohm resistor. The scope is set-up so that a third trace adds CH1 & CH2. Since CH2 is inverted, the third trace measures voltage across the resistor. CH1 & CH2 are both at 20V/Div. The third trace is scaled 20V/Div, or 20A/Div. The 10:1 scope probe grounds are connected to ac neutral. The ground connection of the scope's ac power plug is cut off. An isolation transformer also works, but don't have one.

I ran the test about 5 times to try and get a turn-on max current. This occurs when turn-on coincides with the ac voltage is at its peak (170V).

The scope plot shows:

1st cycle max current is about 32A
2nd cycle max current is about 25A
3rd cycle max current is about 17A
4th cycle max current is about 15A

We can also get an approximate ac resistance that limiting the current. At around 150V, 32A are going into the discharged cap bank. The actual turn-on did not start at 170V, but more like 150V.

R = 150V / 32A = 4.7 ohm

Subtracting the 1 ohm current sense resistor and the 2 ohm thermistor, converter input resistance and ac line resistance account for 1.7 ohm.

Sal

[Salvo]

What gets protected by the thermistor?

"Surge-Gard inrush current limiting NTC thermistors are ideal for switching power supplies where the low impedance of the charging capacitor exposes the diode bridge rectifier to an excessively high current surge at turn-on," said Larry Howanitz, Vice President, EIS & Sensors, API Technologies. "They offer maximum current protection when the power supply is turned on, allowing the design engineer to select lower peak current rated diode bridge rectifiers for use in the switching power supply."


http://news.thomasnet.com/fullstory/Inrush-Current-Limiting-Thermistor-reduces-circuit-failures-2000...

No mention of magnetics, capacitors, or what ever requiring protection.

Sal

[BFL13]

Thanks for the detailed explanation. So I think we are spinning our wheels now until I hear more from PM. I will update if anything new learned. Thanks for all the help everybody.

[ken_white]

BFL13 wrote:
Getting back to this, where did that 50a come from anyway? (I see the 60 number for part is not a rating just an identity)

If I want to use my 2R thermistor, then if I get my amps to go with the 170 from where that 50 came from, then I would need to use 85a.
Why can't I use 85 instead of 50? If I did, would I get 60 cycles instead of 100 before it failed? Cycles of what anyway?

The 170v comes from peak volts RMS on 120v line voltage, ? , so that means every time you do a cold start, the voltage could be anything from peak to zip which means in rush may be small or great each time.

So whatever thermistor I pick, I could do several starts and it would work every time but that proves nothing, because it may be that I never caught it at peak inrush, but the next time I could and it could blow? So just because it works great after a repair, that proves nothing?

Meanwhile we do see my thermistors blowing but no harm to the diode bridge, or at least not so you would notice. Which brings us back to the spec about handling stress for 100 cycles. A cycle of what? Does that mean you get 100 starts and then anytime after that your diode bridge might fail? Converters are supposed to last years and years so that can't be right.

(I have not forgotten the comments that the production 100amper design for the straight converter version may not work quite right for the same thing used with variable voltage where perhaps higher inrushes might occur. They might need two different designs, no idea. Might hear something about that from them or not. )

LOL and please look at the data sheet link I posted and follow my explanation.

The 60 in the part number is used to identify the maximum or peak recurrent reverse voltage that can be supplied to the diode bridge -> 600 volts peak or 420 volts RMS.

The 50 amps of current I used was an estimate of a safe value of current that can flow for 100 cycles of a 50 Hz sine wave.

This information is found on the graph labeled as Maximum Non-Repetitive Peak Forward Surge Current located at the bottom of sheet 3/4.

As a side note, the current flowing into the capacitor is proportional to the change in voltage across the capacitors terminals in respect to the change in time (dv/dt).

This means maximum instantaneous current will flow into the capacitor when the voltage starts at zero volts (maximum dv/dt)and will end when the capacitor is fully charged at 170 volts (zero dv/dt).

If you use the peak voltage equation to find the current, as in C*170V/4.166ms, you are actually calculating the average value of the current over that period of time, and not the peak current.

The peak current is almost 2 times higher than the average.

If you want the best reliability, use the 5 ohm part and follow the procedures I outlined.

If you want to gamble and possibly cause premature diode bridge failure, or some other series connected part failures, use the 2 ohm part.

I am tired of Salvo's posts and arrogance so I am done posting in this thread.

Take care,

Ken

[Salvo]

Not that difficult. Put your camera in move mode and record the action. You can easily acquire the transient at a 5 ms/div sweep rate.

My scope plot of Parallax converter (way back from page 28).



Vert: 20V/div, Horz: 5 ms/div

The waveform has a linear ramp, charging to about 75V in 5 ms during the first cycle. Since the capacitor charge rate is constant we can deduct that the current is also constant (through most of the cycle). The steeper the slope, the greater the charging current. Subsequent charge cycles have lower current.

The current is:

I = C * dV/dt
I = 2400 uF * 75V/5 ms
I = 36A

The 2 ohm thermistor limits current to 36A.

Sal

DryCamper11 wrote:
I might start watching things with my scope to see if I could figure out what was happening that was causing the failure. However transients are hard to find.

[DryCamper11]

BFL13 wrote:
I cannot remember a failure from a cold start with any of the different size thermistors, only with a hot restart.However, since the converter runs ok after the thermistor has blown, which I have seen with a hot restart and the lid off so I can watch the thermistor go up in smoke; then with the cold starts and the lid on so I can't see the thermistor, I cannot state for sure that the thermistor did not blow after a cold start and just keep running so nothing seemed wrong.

Hmmmmm. In all of this discussion about resistance, we're talking about the zero power resistance, i.e. the cold start resistance. None of that is particularly relevant if it's a hot start problem. If I was trying to just solve the problem, and wanted to save money (as I always do) I'd be inclined to put some 2 or 3 ohm thermistors in series and tack in a time delay bypass relay (or just be really careful with hot restarts, then see what happens. My junk box is big enough I've got all those items in there. 🙂

If I was more interested in the electrical puzzle, then I might start watching things with my scope to see if I could figure out what was happening that was causing the failure. However transients are hard to find.

[Salvo]

DryCamper11 wrote:

Salvo wrote:
Tell me again why do you want 5 ohm?

I gave my answer above.

1) It reduces inrush current max.

Not an issue. Turn on energy is 35J. The part is stressing out at 300J. Turn-on resistance is not the problem. Can't you see that?

2) It reduces di/dt on inductive components (which are unknown).

Your arguments are sounding like the boy who keeps crying wolf. We're talking about a 4 ms transient! Tell me what's supposed to happen to the emi filter that you're afraid of?

3) It reduces dv/dt on capacitive components (which are unknown).

Again cry wolf. We're talking electrolytic caps. I've never heard that dV/dt is an issue. Nippon Chemi-Con, a major manufacturer of electrolytics doesn't mention dV/dt problems. Ripple current is the biggest issue with these caps.
Al electrolytics

The pdf does mention inrush current. They say it can be 100 times greater than typical operation, but due to the short transient time period, they're harmless.


4) We don't know the minimum resistance since we don't know much about the circuit, but given that there is no "maximum resistance" calculation for thermistors, it's better to increase resistance than decrease it when faced with a failing thermistor.

:h I fail to see that logic. If the part is blowing up, then it's better to give the next part even more resistance???

I would rather conduct an analysis why it's failing. We know it's not getting overstressed at turn-on. 35J won't overstress the device. Where does that leave us? The only thing that's left is continuous operation. The converter is pulling 13Arms. Power dissipation of our thermistor is P = I^2 * R. The greater the thermistor resistance the greater the power dissipation. A 2 ohm thermistor will have a lower resistance and subsequent lower power dissipation than the 5 ohm part.

I would turn this around and ask what benefit you think there is by reducing the resistance? I listed the possible benefits that I'm aware of from such a reduction, and why I think those benefits aren't as great as the benefits from increased resistance.

I haven't seen any benefits list. You got a cry wolf list. Iota doesn't have a thermistor. Are their emi inductors failing? Are their cap banks failing? Come on, lets get real!

I've produced to the diode bridge spec and shown the actual max surge current requirement. Please do that to your list.

Sal

[ktmrfs]

OK, Like many other folks, I've had more than enough of Salvo aka "karnak the magnificent" over the years and his responses. I'm done with this post.

BFL: If you have any questions, PM me.

[BFL13]

I cannot remember a failure from a cold start with any of the different size thermistors, only with a hot restart.

However, since the converter runs ok after the thermistor has blown, which I have seen with a hot restart and the lid off so I can watch the thermistor go up in smoke; then with the cold starts and the lid on so I can't see the thermistor, I cannot state for sure that the thermistor did not blow after a cold start and just keep running so nothing seemed wrong.

The converter with the big fat thermistor did not start in Nov after not being used since Sep. It worked ok in Sep, but taking the lid off in Nov, there was the thermistor as shown in the photo. I have no idea when that happened or how many times the converter worked ok while the thermistor fell apart to the point the converter would not start at all.

I have seen smoke after hot restarts that came from the thermistor. Sometimes after the smoke event the converter has died with the thermistor burning right up. Other times there is just a little smoke and a brief red spot and the converter keeps running at full amps. I have never seen smoke after a cold restart.

The general advice which I am taking, is to just not do any hot restarts, and also to clamp on first, since clamping on second is similar to a hot restart.

My suspicion now is that if I do that, it doesn't matter much which size thermistor I put in there, so I might as well use these ones I have on hand and not buy any more of a different size. Unfortunately I have 5Rs and 2Rs so if I do need to cover for 3R, then I would have to put two of the 2Rs in series to make a 4R if I want to use them or else it is wasted money ($3.00 each !!! 🙂 )

EDIT--there is also the problem of the hit and miss just what part of the a/c cycle you happen to start up in, so it may not be possible to draw any conclusions from observations of smoke/no smoke based on whether it was a cold or hot start.

[DryCamper11]

Salvo wrote:
Tell me again why do you want 5 ohm?

I gave my answer above.

1) It reduces inrush current max.
2) It reduces di/dt on inductive components (which are unknown).
3) It reduces dv/dt on capacitive components (which are unknown).
4) We don't know the minimum resistance since we don't know much about the circuit, but given that there is no "maximum resistance" calculation for thermistors, it's better to increase resistance than decrease it when faced with a failing thermistor.

I would turn this around and ask what benefit you think there is by reducing the resistance? I listed the possible benefits that I'm aware of from such a reduction, and why I think those benefits aren't as great as the benefits from increased resistance.

[DryCamper11]

BFL13 wrote:
I don't care what R it is, I just want one that won't blow! 🙂

So remind me again - when does it blow?
1) During steady state operation? 2) During cold start? 3)Hot start?

I thought it was 2) - cold start. I thought I read that care was being taken to avoid hot starts, and it was never failing except at startup - which led me to think we were dealing with a 2) cold start issue.

Let's briefly discuss some of the options:

In general, changing the resistance (the "zero power" resistance, which is the cold start resistance) is how you limit the charging current. It protects diode bridges, transformer windings, etc. from too much inrush current. It limits inrush current

Changing the max steady state current limit is how you protect the thermistor against damage during steady state operation.

Changing the Joules rating (which is the same as the capacitance rating if you know the charge voltage since joules=1/2 CV**2) is how you prevent the device from being damaged during cold start.

(Above when I say "changing" I mean selecting a thermistor having a different rating.)

Hot start protection is handled a couple of ways. One is to choose one with a low thermal time constant. It will cool more quickly after shut down. None will cool instantly, so the most common way to deal with that issue is to add a relay. Once the thermistor has done its work at limited inrush current, it gets bypassed, no current flows through it, and it cools so it's instantly ready for another cold start. If it fails at hot start, this is the best option. You just bypass the thermistor with a time delay relay.

Now, why have some been recommending 5 ohms, while others do not? From the above, one would think that the proper response would be to select a thermistor having a higher Joules rating. The problem there is that the selected thermistor seems to have a more than sufficient Joules rating. All of those selected do, and all seem to have way more than required.

Nonetheless, the thermistor is still failing, and as I understood it, it was failing at startup. If that is the case, then what options are left? One can certainly select a higher resistance - after all the resistance calculation produces a minimum, and higher resistance is allowed. In general, the higher resistance reduces stress on the device at startup, all things being equal. Of course, "all things" aren't equal. You have to select another thermistor. It will have different specs. But as long as the Joules rating is the same or higher, cold startup is likely to produce less stress on the device. It also produces less stress on all parts of the circuit. Current flow is reduced. Rate of change of voltage is reduced, which may help out if the problem is inductively caused.

OTOH, as long as the device installed has specs sufficient for the circuit conditions it faces, it should work. A 5 ohm may have a longer time constant, which may result in more problems with hot start. It may have higher steady state resistance, which causes more power to be dissipated in it. So the selection between 5 ohms and 2 ohms is a tradeoff. That's always the case with electronics.

The problem here is we don't know why the device is failing, given that all those installed seem to have sufficiently high specs for what we know about the circuit. I still think that the simplest test is to go with higher resistance (and higher Joule rating). It should produce better performance with respect to the other circuit components, which remain unknown to us, by reducing inrush current.

Basically, a thermistor is a type of "soft start" circuit and the higher the resistance when cold, the "softer" the startup. I don't see any advantage to lower resistance except possibly with regard to thermal time constant/hot start issues.

There may be some steady state improvement by using a lower resistance, but in general, if the steady state current spec of the thermistor is met, then it's met and problems are unlikely. The steady state of the circuit is much easier to understand, even when we don't know the details of the other circuit components. I'm willing to accept that the other (unknown) circuit components could affect the dynamics of startup in a way that causes the thermistor to blow. I find it much harder to believe they would cause steady state thermistor failure when we know the steady state current and the steady state current rating of the thermistor.

If hot start is the problem, then greater care with respect to operation will help - wait longer after turn off. If it turns out that using more care to avoid hot starts works when a higher resistance/ higher Joule rating thermistor is installed, then adding a time delay relay might be the next step. It will ensure the device is cold when a restart is needed, without requiring any waiting period. There is a reason why such relays are often added to the circuit design to bypass the thermistor - even though they add expense.

[BFL13]

Salvo, no call for being so nasty.

In any case, ISTR he already did estimate 30a to go with the 170 and came up with 5.7R, while Ken used 50a to come up with 3.4R. My previous post seeks clarification on where the 30 or 50 or any other number comes from. 85 would let you use 2R. Seems to have to do with the diode bridge specs.

On that Salvo used the inrush limit spec of 300 to get 1/2 an ohm for R, so there is a conflict over which part of the diode bridge's specs to use to go with 170v peak to get Min R.