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11-01-2014 #1
I've known about zero switching for many years and was surprised myself.
here's 2 examples
Inrush current - Wikipedia, the free encyclopedia
Time-delay relay reduces inrush current | EDN
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11-01-2014 #2
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12-01-2014 #3
That is true for a pure resistance, however the issue is the transformer in this case is better modeled as an inductor. I followed Eddy's links but unfortunately they only give 'hand waving' explanations, so instead I've just calculated it and it transpires that if you model the transformer as such, using Faraday's law to find the voltage across the inductor as a function of magnetising flux, then solve this differential equation to find the magnetising flux as a function of applied voltage and switching angle, the result is sinosoidal with a DC-offset. This DC offset depends on the initial flux (i.e. residual flux) and the cosine of the switching angle, so clearly if the angle is pi/2 (i.e. a voltage peak), the cosine term is zero and you get the lowest inrush current. To make matters worse, the relationship between flux and current will be non linear since in normal operation the core operates adjacent to saturation, so when switched on the core is operating well into the saturation region. This means that although the flux implied by Faraday's law is only up to twice the rated value, the current is many times higher. Unfortunately things change a bit when you have capacitors connected to the output via a rectifier, as they essentially present a short circuit to the secondary.
To be honest my 600VA figure was just a rough estimate based on experience. Yes, you could consider the current capacity of the mains circuit you are connecting it to - e.g. if it's got a 30A RCD with no significant load connected then you're much less likely to have a problem, as the surge current trip is much greater than 30A...Last edited by Jonathan; 12-01-2014 at 12:33 AM. Reason: Forgot the A on VA!
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The Following User Says Thank You to Jonathan For This Useful Post:
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12-01-2014 #4
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The Following User Says Thank You to JAZZCNC For This Useful Post:
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11-01-2014 #5
I'm reading everywhere that zero crossing is the worst thing with regard to inrush.
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11-01-2014 #6
Not when starting something like a transformer. A mains powered transformer essentially has a zero current point 100 times a second, so if you can switch on at exactly that point, you won't see any surge on the input line, unlike if you were to switch it on when mains volatage is at it's peak you'll see a major surge as full peak voltage is essentially shorted out until the magnetic field builds up.
The issue with surges comes when you add things like capacitors, as these need time to charge. For smaller transformers, the transformer itself will usually limit current enough to avoid any major surge issues, however with larger transformers they can allow that much current to pass that you need to limit it until things get charged up.
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12-01-2014 #7
Following from Jonathan's post #99
Transformer Inrush - Open Electrical
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13-01-2014 #8
Good article, but they could have expressed the final equation more clearly to demonstrate their point. Also, it's a bit pointless to solve the differential equation using Laplace transforms when you can trivially separate the variables. This is how I did it:
From Faraday's law:
.
Seperate variables:
.
Use sin(A+B) identity to make it easy to integrate:
.
Integrate:
.
Use cos(A+B) identity to simplify:
. (1)
When the transformer is switched on, we have the initial condition relating to the residual flux:
.
So substitute this in to (1) to find constant, k:
.
So the solution is:
.
So now it's easier to see that the . term is a constant, which introduces a DC offset that disappears when the switching angle is . ... i.e at the peak voltage.Last edited by Jonathan; 13-01-2014 at 12:28 AM.
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13-01-2014 #9
I used to like Laplace transforms but that was 40 years ago, I've never needed to use them since
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13-01-2014 #10
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