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

[ken_white]

LScamper wrote:
...B field is controlled by amp-turns --- This is true for DC. For transient and AC conditions the inductance, if designed correctly, keeps the current down and the core from saturating. It is the volt-seconds, that is the number of turns, the saturation flux of the magnetic material, and the size of the core, that determine if the core will saturate. V-S = NBA. Current is not involved.

If the core saturates, which is controlled by the current, then harmonics are produced in the secondary which distorts the output waveform.

Oh, which also changes the inductance and volt-second relationship...

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...

[Salvo]

You are dealing with a thermal transient during turn-on. The pulse of the transient is much smaller than the thermal time constant of the thermistor. The thermistor has no time to cool down during the transient. That means your argument is not correct. Resistance has no effect on how hot the device gets during the transient.

Resistance does have an effect on steady state use. The greater the resistance the hotter the device gets. The 5 ohm device will always be hotter than the 2 ohm device. And we know that heat kills.

Just to clear up another misconception, the surge current calculation used in the tutorial is not correct. They use

I_surge = V_max / R_thermistor

The ac has an appreciable source impedance as does the converter input filter. The ac will not be at 170V when pulling 30A or whatever. The Honda gen that the OP is using also has bandwidth limitations. I doubt it can correct for the increased load (or in this case surge current) within 4 ms. The ac will have a significant voltage drop. That means the surge current is a lot less that these calculations show.

Sal

DryCamper11 wrote:

It doesn't matter how much resistance you are charging the capacitor through - the total energy lost due to that resistance (energy dissipated in that resistance) will be the same.

However, the resistance does have a great effect on the power dissipated, i.e., on the rate at which the 1/2 CV**2 energy is dissipated. The energy dissipated in the resistor appears as heat. The heat is transferred to the surroundings. As the resistance of the thermistor goes up, the current goes down and the capacitor charges more slowly. The same amount of energy is dissipated in the resistor, but it gets dissipated more slowly, and the resistor/thermistor has longer to transfer its heat to the surroundings, so it doesn't get as hot. That's why the resistance calculations produce a "minimum" resistance. If you use more resistance, you just slow things down - decreasing the maximum inrush current and allowing more time for heat to be dissipated. If you go below the minimum, you risk damaging the thermistor as the energy is dissipated too quickly for the device to rid itself of that heat without damage.

[LScamper]

ken white wrote:

"Look at the B-H curves for the various magnetic materials and you will see the B field is controlled by the amp-turns which changes the permeability of the core and induced voltage.

We are talking transient conditions and not steady state...

Why do you think current has no effect?"

B field is controlled by amp-turns --- This is true for DC. For transient and AC conditions the inductance, if designed correctly, keeps the current down and the core from saturating. It is the volt-seconds, that is the number of turns, the saturation flux of the magnetic material, and the size of the core, that determine if the core will saturate. V-S = NBA. Current is not involved.

[BFL13]

ken white wrote:

BFL13 wrote:
The production 100a model uses MS35-5R025

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

Sounds like you have a solution. I am done posting since you have an answer.

🙂

Not quite. That's the big one in the photo I posted that is in my prototype, so I figure it is back to my operating procedures that must be at fault as was discussed in this thread. They are not having troubles with the production models.

[ken_white]

BFL13 wrote:
The production 100a model uses MS35-5R025

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

Sounds like you have a solution. I am done posting since you have an answer.

🙂

[BFL13]

The production 100a model uses MS35-5R025

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

[ken_white]

LScamper wrote:
ken white wrote:

"Not concerned about saturating the core of the transformer, or its kVA rating, when allowing 375 amps to flow?

If saturated, the surge current will flow longer than 1/4 cycle and will persist for more cycles too... "

? How does high current saturate the core, volt - seconds stay the same?

Look at the B-H curves for the various magnetic materials and you will see the B field is controlled by the amp-turns which changes the permeability of the core and induced voltage.

We are talking transient conditions and not steady state...

Why do you think current has no effect?

[LScamper]

ken white wrote:

"Not concerned about saturating the core of the transformer, or its kVA rating, when allowing 375 amps to flow?

If saturated, the surge current will flow longer than 1/4 cycle and will persist for more cycles too... "

? How does high current saturate the core, volt - seconds stay the same?

[ken_white]

Salvo wrote:
I see you found the diode.

It has a peak surge rating of 350A.

Your thermistor needs to be larger than 170V/350A = 0.49 ohm.

I haven't heard any rational reasoning for going greater than 2 ohm.

Note, the surge current is for an extremely short time; 4 ms max. The charging pulses that follow the initial pulse have significantly less current.

Just to review:

1. The desired resistance of the thermistor has nothing to do with the size of the capacitor bank.

2. If this is your standard peak-charge circuit, the diode bridge is the weakest link and may need inrush current surge protection.

3. The energy the thermistor consumes during turn-on is equal to 1/2CV^2. It doesn't matter if the thermistor has 1 ohm or 5 ohm, they will consume equal energy.

4. The size of the capacitor bank determines the minimum energy spec for the thermistor.

Sal

So you are not concerned about saturating the core of the transformer, or its kVA rating, when allowing 350 amps to flow?

If saturated, the surge current will flow longer than 1/4 cycle and will persist for more cycles too...

[DryCamper11]

I've read the thread up to here, and thought it might help to make some brief comments about the NTC resistance and its effect on the charging of a capacitor bank. It's been posted here that a charged capacitor has energy stored equal to 1/2 CV**2. It's also been posted that the thermistor needs to dissipate energy of 1/2 CV**2. Both are correct. The total energy required to charge a capacitor through a resistor (the resistor is the thermistor in our case) is CV**2 (not 1/2 CV**2). Half of that CV**2 energy is dissipated in the resistor/thermistor, and half gets through and is stored in the capacitor.

It doesn't matter how much resistance you are charging the capacitor through - the total energy lost due to that resistance (energy dissipated in that resistance) will be the same.

However, the resistance does have a great effect on the power dissipated, i.e., on the rate at which the 1/2 CV**2 energy is dissipated. The energy dissipated in the resistor appears as heat. The heat is transferred to the surroundings. As the resistance of the thermistor goes up, the current goes down and the capacitor charges more slowly. The same amount of energy is dissipated in the resistor, but it gets dissipated more slowly, and the resistor/thermistor has longer to transfer its heat to the surroundings, so it doesn't get as hot. That's why the resistance calculations produce a "minimum" resistance. If you use more resistance, you just slow things down - decreasing the maximum inrush current and allowing more time for heat to be dissipated. If you go below the minimum, you risk damaging the thermistor as the energy is dissipated too quickly for the device to rid itself of that heat without damage.

[BFL13]

Thanks! Another mystery solved. Just need to hear what the production models are using in them these days, and then figure out what I can get away with using in mine. We are getting there. 🙂

[DryCamper11]

BFL13 wrote:

DryCamper11 wrote:
.............. You then plug in the assumed inrush current into a formula for minimum R.

Choosing an R above the minimum is OK. Below the minimum, you risk failure at turn on.

DryCamper11 was going from vague memory there (18 Nov), but now I can tell what he was talking about, wrt those two lesson plans, etc.

This relates to my current puzzle over the meaning of "minimum."

He and Ken seem to contradict the lesson plans where with 5.7 calculated they went with a 5 and with 1.2 calculated, they went with a 1. Ken says with 3.4 to go higher and use the 5.

I looked at the lesson plan here:
http://dkc1.digikey.com/us/en/tod/Ametherm/Choosing_NTC_Thermistor_NoAudio/Choosing_NTC_Thermistor_NoAudio.html
They calculated 5.7 ohms minimum and went with a 5 ohm device. They don't say why they went with less than the calculated "minimum," but read the next page. They calculate 7.94 ohms and select 10 ohms.

AFAICT, the reason they went with 5 ohms in the first example is just that it was close enough to the "minimum" given the tolerances and limits on their assumed numbers used in the calculation. However, when they calculate 7.94 as a minimum, they go up to 10 ohms.

Getting this "minimum" definition clarified would tell me whether, with a 3.4 calculation I should use a 4 or 5 or whether I could use a 2. ( If going with Ken's low size amps where that 3.4 came from)

Can anybody help with that definition of "minimum" business? Thanks.

If you calculate a minimum of 3.4, then you should not use 2 ohms. You should use at least 3.4 ohms.

[ktmrfs]

BFL13 wrote:

DryCamper11 wrote:
.............. You then plug in the assumed inrush current into a formula for minimum R.

Choosing an R above the minimum is OK. Below the minimum, you risk failure at turn on.

DryCamper11 was going from vague memory there (18 Nov), but now I can tell what he was talking about, wrt those two lesson plans, etc.

This relates to my current puzzle over the meaning of "minimum."

He and Ken seem to contradict the lesson plans where with 5.7 calculated they went with a 5 and with 1.2 calculated, they went with a 1. Ken says with 3.4 to go higher and use the 5.

I found in an earlier post that in series, the two 2Rs would be 4R, so that is answered (which is above 3.4 if that comes into play)

Getting this "minimum" definition clarified would tell me whether, with a 3.4 calculation I should use a 4 or 5 or whether I could use a 2. ( If going with Ken's low size amps where that 3.4 came from)

Can anybody help with that definition of "minimum" business? Thanks.

As to two 2R's = One 4R. Maybe, maybe not. Yes, they will have similar cold resistance values. But you also need to insure each 2R can handle the required inrush energy (joules) and and max power disipation. E.g. will get hot enough, but not to hot or to cold at max current.(Kinda like the soup in Goldilocks and the 3 bears). It needs to get hot enough to be low resistance but always be below max power limit.

I didn't go back to the lesson plan, but I suspect when they say "minimum" they are referring to the minimum resistance that will insure the inrush current is below the maximum limit. Therefore, you find a minimum resistance, then find a part that is at least that minumum resistance when cold at lowest operating temperature and lowest value per spec'd resistance tolerance. It can be higher, but not lower. Then determine if that part can handle the joules and power. Keep iterating till you find a part that meets all the requirements.

Remember, the initial resistance of these devices can have a significant variation, look at the tolerance for the initial resistance and determine it's lowest resistance per the tolerance, not the "nominal" resistance. And if your going to be boondocking at high ambient temps plug that into the equation as well. Low temps don't hurt you, resistance goes up. High initial ambient may, initial resistance goes down.

So, as a "hypothetical" example, suppose the calculated "nominal" R value is 5.2 ohms. After doing some checking you find that the values available that meet inrush spec's, power spec's. etc are 4.9ohms, 5.3 ohms and 6 ohms, each with a 10% resistance tolerance

OK, 4.9 is out, to low, below thee 5.2 minimum. 5.3 can range from 4.8 to 5.9 ohms, still to low. 6 ohms can range from 5.4 to 6.6 ohms. Looks like 6 is the one to go with. Although I'd verify that the 6 ohms at high ambient doesn't drop below 5.2.

[BFL13]

DryCamper11 wrote:
.............. You then plug in the assumed inrush current into a formula for minimum R.

Choosing an R above the minimum is OK. Below the minimum, you risk failure at turn on.

DryCamper11 was going from vague memory there (18 Nov), but now I can tell what he was talking about, wrt those two lesson plans, etc.

This relates to my current puzzle over the meaning of "minimum."

He and Ken seem to contradict the lesson plans where with 5.7 calculated they went with a 5 and with 1.2 calculated, they went with a 1. Ken says with 3.4 to go higher and use the 5.

I found in an earlier post that in series, the two 2Rs would be 4R, so that is answered (which is above 3.4 if that comes into play)

Getting this "minimum" definition clarified would tell me whether, with a 3.4 calculation I should use a 4 or 5 or whether I could use a 2. ( If going with Ken's low size amps where that 3.4 came from)

Can anybody help with that definition of "minimum" business? Thanks.

[LScamper]

ktmrfs, sorry for not using your complete response. I agree with you but was trying to add something else to think about.