A transformer's inductive resistance is the result of the continuous re-magnetisation of the core. When a transformer is switched off, the core will always remain magnetized in one direction, except at one point that does however not coincide with the voltage's zero crossing.
If the incoming voltage at power on has the same polarity as the remanent magnetism in the core, the core will be driven into saturation. I. e. it cannot be re-magnetized any further and the inductive resistance vanishes. All you are left with is the copper resistance to limit the inrush current.
The options to deal with that depend on the transformer.
- Transformers with air-gap have a low remanence. It is usually low enough that the polarity does not matter and the best way to switch such a transformer on is at the voltage peak. This should be combined with power down at zero crossing, because then we get the lowest remanence we can determine. Zero remanence is somewhere shortly after the zero-crossing, but it is impossible to determine that exact point in relation to the voltage.
- Soft-start with phase angle control is limited to transformers with no more than 1,2 T induction (air-gap!), regulation of at least 5 % and they should not be started without load.
- Transformers with no air-gap, e. g. toroids have high remanence. Even when they are switched off at zero-crossing, the remanent magnetism will remain relatively high. A patented method magnetizes the core with DC pulses in a defined direction and then switches on at the zero-crossing where the half-wave with the opposite polarity starts. Transformer softstart Introduction : EMEKO, Michael Konstanzer
If you want to avoid a transformer's inrush current, you need to make sure that the voltage at power on has the opposite polarity of the remanent magnetism in the core. At the same time you need to make sure not to infringe Michael Konstanzer's patent, if you want to commercialize your design.
Switching off at the voltage zero-crossing is the correct choice, although not because the flux is zero.
If the incoming voltage at power on has the same polarity as the remanent magnetism in the core, the core will be driven into saturation. I. e. it cannot be re-magnetized any further and the inductive resistance vanishes. All you are left with is the copper resistance to limit the inrush current.
The options to deal with that depend on the transformer.
- Transformers with air-gap have a low remanence. It is usually low enough that the polarity does not matter and the best way to switch such a transformer on is at the voltage peak. This should be combined with power down at zero crossing, because then we get the lowest remanence we can determine. Zero remanence is somewhere shortly after the zero-crossing, but it is impossible to determine that exact point in relation to the voltage.
- Soft-start with phase angle control is limited to transformers with no more than 1,2 T induction (air-gap!), regulation of at least 5 % and they should not be started without load.
- Transformers with no air-gap, e. g. toroids have high remanence. Even when they are switched off at zero-crossing, the remanent magnetism will remain relatively high. A patented method magnetizes the core with DC pulses in a defined direction and then switches on at the zero-crossing where the half-wave with the opposite polarity starts. Transformer softstart Introduction : EMEKO, Michael Konstanzer
If you want to avoid a transformer's inrush current, you need to make sure that the voltage at power on has the opposite polarity of the remanent magnetism in the core. At the same time you need to make sure not to infringe Michael Konstanzer's patent, if you want to commercialize your design.
Switching off at the voltage zero-crossing is the correct choice, although not because the flux is zero.
is this supporting the idea that the effect is the same as if the primary coil were operating without an iron core?
Does the next cycle bring the core flux back to normal? i.e. symmetrical flux in either direction? Or does it take a few cycles to bring the saturated core back to symmetrical?
Can I take my first question a bit further?
If the explanation is that the new start up saturates the core and thus it cannot respond to the increasing current starting to flow, does the primary coil behave as if it were an air core inductor with a little bit of inductive reactance to add to the resistance of the primary?
I would start the triac by initially turning on almost at the end of every half-period, and then gradually move towards the beginning. With a micro-controller you can determine the frequency first, then detect zero-crossings from which to delay the turn on.
That way you would be pretty safe even with remanence in the core. But I would probably turn it off with the inverse pattern, leaving remanence small.
That way you would be pretty safe even with remanence in the core. But I would probably turn it off with the inverse pattern, leaving remanence small.
Yes, that was the plan. Inspired by how Triac based light dimmers works.
For those who want to do it without programming a microcontroller, I think this little chip offers a nice solution. http://www.atmel.com/dyn/resources/prod_documents/doc4712.pdf
With kind regards,
Bas
At that moment it does not behave like a coil at all anymore, rather like a piece of wire.
No and yes. The absolute values of flux and remanence are symmetrical in either direction. They only change signs. The core is however magnetized further away from 0 than it would be during normal operation and it will take a few cycles until symmetry is restored. For the sizes we talk about in audio three to five cycles should be sufficient. Bigger transformers will take longer.
It starts up as a coil with iron core until it reaches saturation. At that moments the inductive resistance vanishes entirely. All that is left is a piece of wire with its copper resistance until the core leaves saturation and the primary becomes a coil with iron core again.
a few months ago a Member posted test results of the peak current into a transformer at start up.
Your description could fit his data. He found that peak current was about half that expected of resistance only. I think he surmised that the difference was due to inductance in the upstream cabling to the transformer.
Your explanation also fits a transformer blowing a low rated fuse even when there are no capacitors connected to the rectifier.
I am inclined to give your explanation more weight than mine. Thanks.
Anyone else with an hypothesis?
Timely, thread 'bas. I've been looking at triac for the past few days as the resistor+relay approaches in AN-1839 and P39 don't scale well to the amount of bypass capacitance in an amp I have planned. Another option I've been looking at besides the U2008B is the IES5528, specifically figure 8 in the datasheet. Bigger part but similar complexity and demo boards for its pin compatible predecessor (the OM5428) are available fairly cheaply. Haven't investigated enough to develop preferences yet.
I noticed you mention Bryston here. From a sampling of the schematics it seems only the PowerPac amps use triacs; is that correct?
I noticed you mention Bryston here. From a sampling of the schematics it seems only the PowerPac amps use triacs; is that correct?
Sounds like a good plan. But you can turn it off always at a zero crossing, since there is no secondary (and thus no prim) current then (assuming your use is for a rectifier-loaded audio supply). At around zero crossing of the mains (several mS before and after) the rectifiers are blocked, and the amp draws it's power from the supply caps, so there is no xformer current.
That way, you can guarantee zero remanence every time.
jd
Can anyone briefly explain the advantages this type of circuit has over a simpler timer/bypass method for a Power amp PSU?
Wouldn't the current at voltage zero crossing be whatever is determined by the inductance of the primary winding? And since it's a more or less "pure" inductor the current will be 90 degrees out of phase, i.e. at it's peak?
Yes, but unloaded, around the zero crossing, the primary is almost pure inductance, and very very high at that, and the current very very low with respect to the normal design current values. There's minimal flux at that point, although not zero.
Look at the curves I measured. The initial current pulses to charge the supply caps are about 28 Amps, gradually going down.
The current through the primary around the zero crossing is hardly visible on the graph.
The green pulses on the graphs are the mains zero-crossings (on the 2nd curve, one green is mains top, the other is zero crossing), the red curves are the primary current through the xformer.
You see that there is an initial pulse, that slowly goes down, to charge the supply caps (4*33.000uF at 36V here).
Even with switch-on at the mains top, you see that the current is almost the same. You also see that around the zero crossing there is no current visible through the primary. So, as long as you switch off at zero mains crossing, you can then switch on next time whenever you like, and there is no surge at all.
jd
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Agreed. Under normal conditions that works well. And under abnormal conditions you're probably more concerned with switching off as soon as possible, rather than trying to minimize remanence...
Possibly, yes, but trying to quickly switch of the mains to avoid disaster will be doubtfull with all that charge in the power supply caps. Anyway, if you use a triac, it will not switch off before zero crossing anyway.
If you were thinking about some kind of protection, you need to either disconnect the speakers or short the supply caps as soon as possible.
jd
Nice data; can you give a quick description of the circuiit measured? While the "SSR closes" event is obvious I'm not clear to me what the "relay closes" events indicate---is the SSR getting bypassed with a mechanical relay? I assume it's a toroidal trafo, full wave bridge, with the secondaries in series to form a center tapped output with 66,000uF per rail. But am curious what else, such as snubbers, might be present.
You can't ignore the iron core. You just can't. Unless the core saturates, because it is then that the iron begins to behave non-linearly and stange things happen. (even then you don't ignore it, you just adjust the permeability you use in the equations and carry on).
Yes, the equation for reactance, XL = 2*Pi*F*L is a simplification derived from the 'true' equation defining the inductor:
e = L*(di/dt)
where e is engineer speak for voltage (electromotive force), L is the inductance, and di/dt is the rate of change of the current through the inductor ("derivative of current with respect to time"). This equation and variations of it can be used to describe the resulting voltage for any current waveform or the current for any voltage waveform.
In addition to what I said in my earlier post above the residual flux left in the core from the turn-off, consider what happens when you switch on the AC when the voltage is zero. You already know that the current through the transformer (inductor) is zero because the circuit is/was open. But you also know that in steady-state AC operating conditions, the current through the inductor should lag the voltage by 90 degress. That means that at the voltage zero-crossing, the current "should" be at its peak. When the voltage peaks, the current should be crossing zero. So something is wrong already. Voltage and current are both zero. You know that for an inductor, whenever the voltage is positive, the current is increasing; when voltage is negative, the current is decreasing. The higher the voltage, the faster the current changes ( e = L*(di/dt) ).
So draw the graph of current in your mind. The current starts at zero. Voltage is at zero. For the first half AC cycle, voltage rises to its peak and drops back to zero. The current increases this entire time, reaching a peak. Next the voltage drops to the negative peak, and back up to zero. The current decreases this entire time. The decrease is equal to the increase, so the current basically goes back to zero. On the next AC cycle, this repeats. Note the current is always positive, in other words, it follows a sine wave, but with a DC offset. This is a problem, as iron is a non-linear magnetic material. Many stange things happen as a result of this DC current offset. It does settle down to pure AC eventually, and until then you hear a thud and hum from the transformer. The biggest problem of this DC offset is that it causes the magnetic flux to increase to twice the normal magnitude, and that can cause, amoung other things, saturation in the core, which can lead to a large increase in current (due to the lower inductance that results).
Now consider what happens if you flip the switch when voltage is at the positive peak (not crossing zero as above). The current starts at zero of course. So you flip the switch, and current begins to increase. The voltage is at peak, so the current immediately begins to increase from zero. In the first 1/4 AC cycle, the voltage decreases from peak to zero. The current has risen to it's peak. In the next 1/4 AC cycle, the voltage decreases to its negative peak, and current accordingly decreases. It will decrease the same amount as it increased in the first 1/4 cycle so it returns to zero. In the next 1/4 cycle the current continues to decrease, going to it's negative peak by the time voltage rises up to zero. In the final 1/4 cycle, the current returns up to zero as the voltage increases from zero to its positive peak. Note in this case, there is no DC offset in the current. It just follows that nice 90 degress phase shift that you expect to see. That's why you should switch on at the voltage peak, not at voltage zero-crossing.
Thanks from here as well to all of you including you Andrew. The theory here is a bit deep, but it helps me understand it better as well, and makes me able to design a better soft-start.
I would like to measure the effects myself, but sadly enough my otherwise wonderful Fluke PM3070 scope doesn't have a "storage" function 🙁
With kind regards,
Bas
Hi Macboy,
Read you post and understand the reasoning and theory. I just can't connect it to my measurements posted. Can you take a look at that and tell me what you think of it?
In the case of a rectifier-loaded transformer, where the rectifiers only conduct somewhere around the peak voltage, not around zero crossing, the transformer looks as just a huge inductance if you switch on at zero-crossing. That could mean that, although the current does rise to it's double value, it is only the 'unloaded' current (due to to the unloaded prim inductance). Could that explain that this current, even when twice the 'normal' current, is still very small?
jd
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