Author Topic: Designing DC/AC sine inverter output able to trip mechanical circuit breaker  (Read 1071 times)

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Offline MiyukiTopic starter

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Hi folks.

I want to ask if you have any experience about designing something like this.
Or if you have some hints.

Required output: 230V AC up to 25A nominal
Ability to trip 25A class B circuit breaker (100ms 5 times In) at fault, without damage to power stage

Considered output topology NPC1 at moderate frequency 5-10kHz

I see two main problems to worry:
Output choke saturation
100ms is for transistors long time so inner two transistors must be rated to withstand 125Arms thermally for this time
 

Online Siwastaja

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Are you really sure this is the specification you want? It's hard to see it making any sense.

You have the full opportunity to have a soft shutdown at the inverter, which is way better and more accurate than the mechanical fuse, yet you want to massively oversize this inverter, and its protection level, to "force" the fault to be seen at the inferior mechanical device? Practically you would be building a hugely expensive, massive inverter designed to work in constant-current mode into a short for extended time, instead of hickuping or faulting like they normally do, which would be an increased fire risk as well, IMHO.

Of course, there is no magic in designing it. You just design it for current you want, which, in this case, would simply mean you are designing a ten times larger inverter than you actually need, so expect it to cost ten times more as well.
« Last Edit: May 31, 2019, 08:30:51 pm by Siwastaja »
 

Offline MagicSmoker

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Yeah, gotta agree with Siwastaja on this. It makes far more sense to reduce the switch duty cycle to actively limit fault current. That said, there needs to be an LC filter on the output for this approach to work reliably; "modified sine wave" inverters need not apply, then.

 

Online ejeffrey

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Yeah, mechanical generators can naturally supply huge surge currents for short periods of time, so it makes sense to use circuit breakers to stop overload conditions.  With inverters it makes more sense to use electronic control.  You want a fuse on the input to protect against an inverter fault like a shorted MOSFET.  Remember: the purpose of a circuit breaker is to protect the downstream wiring from exceeding its current rating.  If the inverter can't deliver more current than the wiring can take a circuit breaker doesn't do anything.  If for whatever reason you have a requirement to have a circuit breaker, go ahead and put it in, but I wouldn't oversize your inverter to allow it to deliver a dangerous current level only to ensure that the breaker can stop it.

Of course if you take a 100 amp inverter and split it between a bunch of 20 amp circuits then you want to have circuit breakers on each branch circuit.
 

Online jbb

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I'm going to make a guess here: do you intend to use this inverter to power multiple circuits of load (e.g. X and Y), and be able to trip a circuit breaker on circuit X without dumping the load on Y?

If so, and you really need to trip that breaker, then you'll need:
  • A larger DC link capacitor to push lots of energy towards the fault without too much bouncing around
  • Somewhat oversized transistors
  • Current sensors which provide OK measurement of nominal currents and don't saturate on high currents (i.e. high dynamic signal range)
  • Lots of inductor core material to handle extra current (combined with operating frequency, I suggest looking at nanocrystalline / amorphous iron, which has nice high peak flux density and OK losses below about 20kHz)
  • A control scheme which can handle strong overcurrent without going unstable and avoid big output voltage spikes/surges when the breaker opens.
  • A more sophisticated inverter protection scheme which can let enough energy through to open the breaker, but not so much that the inverter burns

It's certainly possible, but it's going to be difficult and cost a lot more than just equipping the inverter with a fast overcurrent trip.
 

Offline richard.cs

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As others have said this is a bit tricky to do. A lot of generators don't manage it either because once the voltage regulation runs out of control range you're left with the fairly high impedance of the winding and the voltage just collapses. Sometimes they don't even trip slowly on the thermal part because the generator looses excitation, the current drops to a few amps and never recovers.

Something you should probably do is design it with a current limit set to just beyond the maximum value needed to operate the magnetic part of the breaker, so for a 25 A type B you want to limit the current to not much more than the 125 A you need. Once they're on the magnetic part of the characteristic they don't get significantly faster with increasing current. How much energy you need to deliver in that time depends on the impedance of the fault, from say 100 J for a fault that collapses the voltage to 10 V or so to more like 3 kJ if the fault is a couple of Ohms and you have to deliver that 125 A into nearly full voltage. That's either a huge capacitor on the DC link or the ability to push the full peak power through from the source.
 

Offline NiHaoMike

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I would suggest designing the inverter to go into V/Hz scaleback on overcurrent, to ease starting of motor loads.
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