Need help on relay for 1200W Trafo soft start

[max7219]

Originally Posted by tatus

I missed the point that it was a 5-ch. Makes it (a bit :-) more reasonable for 5x75 watts rms output power (@4 ohms, would that be right?). I go the other way though and have good results. My (class D) stereo amp drives 250 w rms into each (8 ohm) speaker and is supplied with an SMPS capable of 100w continuous (600w short time peak). Plus large buffer capacitors. I don't manage to exceed 50°C at any heat sink no matter how loud I "listen".

Hi tatus,

could you please describe your SPMS for the class D amp?

Thanks,
max

[AndrewT]

not here.

[tatus]

Originally Posted by em2006

In my opinion, phase-control circuits used in such applications should be avoided, because they generate noise.
Spectrum generated is very large, these frequencies are radio noise.

Agree, on the other hand this will be the case during the first few seconds after turn on. After that period, it would be wise to short and also inhibit that circuit. Reverse procedure at turn off. This way I would expect that the relais has no heavy duty in either direction. Plus the circuit will be much smaller compared to resistive soft-start.

[tatus]

Originally Posted by max7219

Hi tatus,

could you please describe your SPMS for the class D amp?

Thanks,
max

See Problem with Hypex SMPS400/67 (OEM V4). Runs like a charm, except for that startup issue.

[tatus]

Originally Posted by AndrewT

To ALL,
this is the first time I have gone through this version of the extended discussion.
Can anyone see any errors that must be corrected, or even a better way of explaining why and how the soft start works?

Andrew, I agree especially to your safety message.

caxxxxxx, please judge by yourself: are you experienced enough in understanding electronics to work with dangerous mains voltage? Are you aware of the design principles, guidelines and regulations applicable for this type of work? This is not to sound arrogant, its just the necessary diligence that people like AndrewT must show in this kind of forum.


Regarding the electrical explanations, I agree to everything except the considerations at start-up. The transformer's primary is an inductance and it will behave as one. That means applying a constant voltage to it causes the current to ramp linearily and, provided the inductance is large, slowly. As the winding has a high quality factor (xl / r is large), the inductive term will always be dominant. In other words, just connecting the transformer to the mains, no matter at which phase, does not directly cause excessive inrush current.

What can actually happen at transformer startup (and what causes the fuse to blow) is core saturation.
(Remember: when you plot the flux curve of an inductor at steady-state operation at a sinusodial voltage, it lags the primary voltage by 90°. I am thinking of an unloaded transformer here.) The flux at the instance of connecting the transformer to mains is zero (neglecting remanence). Consider the mains voltage also starts with zero (zero phase). During normal operation, the instantaneous flux would have its lower extreme at this point of time. This means the flux has the wrong bias in our situation, and will reach a peak value of twice the peak value during steady-state. It's a bit difficulty for me to explain, I found this nice picture that shows it and also shows the process of moving into steady-state: http://www.mathworks.com/help/toolbo.../sess7swsn.gif . Normally transformers are constructed with enough core material to work with the steady-state peak flux, plus some margin. Going up twice as high will sature the core. In this situation, the core cannot magnetize to a higher level, which in turn renders it without effect. That means that the primary effectively becomes an air core, with much much less inductance, allowing that high current increase that blows the fuse. The funny point about this is: connecting the transfomer at mains voltage zero crossing is actually the worst situation. An interesting experiment could be using a 220V transformer at 110V. That one should never saturate and not require a startup circuit.

[AndrewT]

Originally Posted by tatus

Regarding the electrical explanations, I agree to everything except the considerations at start-up. The transformer's primary is an inductance and it will behave as one. That means applying a constant voltage to it causes the current to ramp linearily and, provided the inductance is large, slowly. As the winding has a high quality factor (xl / r is large), the inductive term will always be dominant. In other words, just connecting the transformer to the mains, no matter at which phase, does not directly cause excessive inrush current.

What can actually happen at transformer startup (and what causes the fuse to blow) is core saturation.
(Remember: when you plot the flux curve of an inductor at steady-state operation at a sinusodial voltage, it lags the primary voltage by 90°. I am thinking of an unloaded transformer here.) The flux at the instance of connecting the transformer to mains is zero (neglecting remanence). Consider the mains voltage also starts with zero (zero phase). During normal operation, the instantaneous flux would have its lower extreme at this point of time. This means the flux has the wrong bias in our situation, and will reach a peak value of twice the peak value during steady-state. It's a bit difficulty for me to explain, I found this nice picture that shows it and also shows the process of moving into steady-state: http://www.mathworks.com/help/toolbo.../sess7swsn.gif . Normally transformers are constructed with enough core material to work with the steady-state peak flux, plus some margin. Going up twice as high will sature the core. In this situation, the core cannot magnetize to a higher level, which in turn renders it without effect. That means that the primary effectively becomes an air core, with much much less inductance, allowing that high current increase that blows the fuse. The funny point about this is: connecting the transfomer at mains voltage zero crossing is actually the worst situation. An interesting experiment could be using a 220V transformer at 110V. That one should never saturate and not require a startup circuit.

I am not sure it happens this way.

When the transformer is off and you first apply voltage there is no flux in the core. It does not behave as an iron cored inductor on that first half cycle of mains power.

Could you try again at explaining it and maybe you can convince me of the true start up condition.

[caxxxxxx]

Lol, I am in no way an expert in anything, lol. But I think I use enough common sense to not have electrocuted myself yet with electronics in my limited use with it and mains wiring (keeps fingers crossed, knock on wood). I have thrown more than 1 breaker, but I trust myself enough that I lean or step away while powering it up first time. I'm safe, not crazy....fire extinguishers are close at hand, lol!!

Actually I have, in fact, rewired my first house and taken building trades course back in high school, which included installing and wiring an electrical panel with circuit breakers and ground wires and the wire(s) they hooked up to the meter, etc. I'm no expert, but common sense has served me well so far.

Anyway back on topic I think I see how this circuit works. Taking things step by step (give me time, I'm a slow learner but greatly appreciate the help), The Snubber's got me a little confused. Do I have to use a calculation or is there a standard that will work??? I think I can read about it for days and still get lost in the math. I'm gonna finish reading up on them, and I'm sure there will be a lot more questions to follow.

[tatus]

Originally Posted by AndrewT

When the transformer is off and you first apply voltage there is no flux in the core. It does not behave as an iron cored inductor on that first half cycle of mains power.

When you wind a conductor around a magnetizable core, you end up with an inductor with a specific inductance (ideally, neglecting all the dirt effects). Its inductance does not depend on the instantaneous magnetic flux in the core. It is there and will stay there - as long as you do not saturate the core. In saturation condition, the construct will behave like a wound conductor without the core, resulting in significant less inductivity provided the core's permeability is significant.

The primary winding of a transformer will behave in the same way when the transformer has no load at the secondary. In your mind, you can remove the secondary winding (it is open in this scenario) and what you get is a cored inductor. That will always behave like an inductor unless saturated. Here is a good excurse on saturation: How Transformers, Chokes and Inductors Work, and Properties of Magnetics : CWS Coil Winding Specialist, manufacturer of transformers, inductors, coils and chokes

Of course the situation is different when a (in the case of having a rectifier followed by a capacitor) initally short-circuiting load is present. In this case, excessive inrush current is caused by shorting the secondary (with the empty capacitor). As you say correctly, in this situation only the transformer's leakage inductance and ohmic/core losses limit that inrush current.

I hope I could convince you. If not - please explain me why an inductor with a core shall not behave like an inductor with a core when flux=0?

[tatus]

Originally Posted by caxxxxxx

Lol, I am in no way an expert in anything, lol. But I think I use enough common sense to not have electrocuted myself yet with electronics in my limited use with it and mains wiring (keeps fingers crossed, knock on wood). I have thrown more than 1 breaker, but I trust myself enough that I lean or step away while powering it up first time. I'm safe, not crazy....fire extinguishers are close at hand, lol!!

I believe you in that, but that was not my point. It was about knowing all the relevant design principles and regulations of working with mains voltage. Do you know how to handle PE, do you know the minimum creepage and clearance distances, do you know the difference between creepage and clearance, can you distinguish the difference between functional, single and reinforced insulationn, do you know the impact on the transformer and the way you have to mechanically construct the cabling inside your case? You should not feel uncomfortable with any of these questions before continuing. Imagine that you finish your project and start using it. But something gets loose, shorts L with your nice aluminium case, and your kid touches it. Or: the subcircuit that is continuously connected to mains voltage starts overheating at night because of a design mistake and causes fire in your house. Your insurance company will not pay, provided you survive it.

[tatus]

Originally Posted by caxxxxxx

The Snubber's got me a little confused. Do I have to use a calculation or is there a standard that will work???

The idea behind a snubber is connecting a capacitor across the contacts. This way the change of voltage cannot change rapidly. In other words, at the moment of time the relay contacts open, the capacitor will be empty and keep the voltage across the contacts zero (for a very short time, the more capacitance the longer). The capacitor will take the load current and get charged by it. That would be nice, if we would not have to close the contacts as well. The worst case here is that the mains voltage is just at its maximum, and due to this the capacitor is fully charged. If we now close the contacts the stored energy will instantaneously discharge into the contacts and can destroy them. For this reason, we also need a resistor in series with the capacitor that limits the current in this sitation to the maximum that the relay contacts can handle.

Summarizing:
1. capacitance as large as possible and reasonable, depending on load current.
2. resistance as small as possible, enough that the maximum switch current is not exceeded.

Condition 1 has no upper limit. Reasonable value is 100nF per 1A load current. For 10A load current you end up with a 1uF capacitor.

Condition 2: voltage = 250*1.41 = 350V, current = 10A (assuming 230V mains voltage, and that the relay is rated for 10A) ---> R=350/10 = 35 Ohms. Power rating depends on capacitor selection, as the resistor must be capable of dissipating the entire energy stored in the capacitor. The maximum stored energy in the capacitor is Q=1/2 * C * V^2 = 1/2 * 1uF * 350^2 = 0.06 joules. Assuming that the switch is not operated quicker than once per second, we get a rough picture of the required power rating of that resistor: P = Q/t = 0.06 / 1 = 0.06 W. So a standard 0.25 W resistor should be enough here.