Dear,
I am working on a soft-start system by micro-controller. In my design decision I don't want to go for thermistors or power resistors with relays bypass.
A Triac is a perfect device to use as switch and as soft-start, especially when used with a micro-controller.
Not only in my own designs I need a good soft-start system but I think in the DIY market there is also a demand for a good and clever soft-start system, and we are will offer soft-start boards for the DIY community as well once we really finalize it.
But to get there there are some unclear things for me.
The switching point. We will use a zero crossing detector. It is a bit unclear though where to switch the transformer on. In my understanding a coil (and therefor a transformer) has the current 90 degrees out of Phase with the voltage. So one would think we need to switch in on the peak of the mains voltage.
Also confirmed by this paper: http://relays.tycoelectronics.com/appnotes/app_pdfs/13c3206.pdf
However other people said that you should switch on the zero crossing point instead, because there is no current yet at switch on.
We also discovered that the switch off point is also important. The way a transformer is switched off defines the flux magnetization in the transformer. A magnetized transformer will have bigger inrush current with next time start up.
Of course the answer is easy to find by just building the circuit and measure myself, and I will, but before we have all the stuff it is nice to hear some input.
If there are any other suggestion for the soft-start with a Triac and a micro-controller, they are more then welcome.
The idea is to use the Triac as dimmer, and slowly rise the voltage as part of the soft-start.
With kind regards,
Bas
I am working on a soft-start system by micro-controller. In my design decision I don't want to go for thermistors or power resistors with relays bypass.
A Triac is a perfect device to use as switch and as soft-start, especially when used with a micro-controller.
Not only in my own designs I need a good soft-start system but I think in the DIY market there is also a demand for a good and clever soft-start system, and we are will offer soft-start boards for the DIY community as well once we really finalize it.
But to get there there are some unclear things for me.
The switching point. We will use a zero crossing detector. It is a bit unclear though where to switch the transformer on. In my understanding a coil (and therefor a transformer) has the current 90 degrees out of Phase with the voltage. So one would think we need to switch in on the peak of the mains voltage.
Also confirmed by this paper: http://relays.tycoelectronics.com/appnotes/app_pdfs/13c3206.pdf
However other people said that you should switch on the zero crossing point instead, because there is no current yet at switch on.
We also discovered that the switch off point is also important. The way a transformer is switched off defines the flux magnetization in the transformer. A magnetized transformer will have bigger inrush current with next time start up.
Of course the answer is easy to find by just building the circuit and measure myself, and I will, but before we have all the stuff it is nice to hear some input.
If there are any other suggestion for the soft-start with a Triac and a micro-controller, they are more then welcome.
The idea is to use the Triac as dimmer, and slowly rise the voltage as part of the soft-start.
With kind regards,
Bas
Many companies have used Triacs for soft start. peavey has done it for years. Also Biamp and Carver is the king of using triacs as a method for controlling a transformer. there early magnetic field amps use a method of Voltage regulation that is basically a big light dimmer!
You can also look at the Soundcraftsmen PCR series of amps for an alternate system using SCR's on the secondary side for what they call phase control regulation. same basic principals and lots to learn from.
You can also look at the Soundcraftsmen PCR series of amps for an alternate system using SCR's on the secondary side for what they call phase control regulation. same basic principals and lots to learn from.
Thank you Zero cool. I will google them. I am very intrigued by this.
With kind regards,
Bas
No one else?
I realize I place this topic in the wrong forum (chipamps) Is there any way I can place it in the right forum?
With kind regards,
Bas
I realize I place this topic in the wrong forum (chipamps) Is there any way I can place it in the right forum?
With kind regards,
Bas
Bas,
I have done some work on this and posted scope curves on the forum. I *think* it was in the now-closed thread about audible differences between cables, maybe it was the John Curl Blowtorch II thread...
Sorry don't exactly remember.
BTW A loaded transformer is not really an inductance (except for the stray inductance) but at the primary you see the (transformed) secondary load. Can be pure R.
jd
I have done some work on this and posted scope curves on the forum. I *think* it was in the now-closed thread about audible differences between cables, maybe it was the John Curl Blowtorch II thread...
Sorry don't exactly remember.
BTW A loaded transformer is not really an inductance (except for the stray inductance) but at the primary you see the (transformed) secondary load. Can be pure R.
jd
Thank you Jan for commenting.
I really try the search function in all possible ways on this forum, but wasn't able to find it. Sad that such a knowledge base is lost.
My main concern is, the first moment when you switch the transformer in, if the current is 90 degrees out of phase with the voltage, and where to switch in. The zero or the peak.
With kind regards,
Bas
Actually, it doesn't matter for a supply that is rectifier/capacitor loaded.
I'll try to find my results.
Edit: Try this:
http://www.diyaudio.com/forums/anal...rch-preamplifier-part-ii-431.html#post2176329
jd
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Thank you Janneman for the Effort. In that case the inrush current limit doesn't get reached by the switching point, but by the dimmer function of the triac, to slowy rise the voltage.
In the meanwhile. I have some good stuff to read as well for anyone who is interested in this: http://www.st.com/stonline/books/pdf/docs/1863.pdf
With kind regards,
Bas
In the meanwhile. I have some good stuff to read as well for anyone who is interested in this: http://www.st.com/stonline/books/pdf/docs/1863.pdf
With kind regards,
Bas
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Hi,
My view on this may not be right.
I have said this before and I don't recall anyone saying I was wrong.
Here goes.
When a iron cored inductor is sitting idle with no current flowing (OFF) there is no flux in the iron core and effectively the dormant inductor is an air cored inductor.
At the moment of start up the voltage is zero and the current is zero.
We have inductor (primary) resistance and Zero volts = Zero current.
As voltage rises, current flows through the primary resistance and starts to increase. This increase in current is resisted by the air cored inductor.
For the first few milliseconds after switch on the mains electrical flow is determined by the mains voltage, the mains resistance, the air cored inductor and the primary resistance of that air cored inductor.
This is the time during which the soft start is required to limit the peak current that will flow before the iron cored inductance has built up to react against the mains voltage.
Has anyone got ideas on the ratio of iron cored inductance of the primary of a transformer compared to that same coil with no iron core in it?
As the mains cycle continues, the flux in the iron core builds up and starts to follow the mains cycle. The transformer has now started. It now behaves as Janneman is telling you.
It's that very short period before the flux builds up that results in the high start up current.
Does this explanation hold water?
My view on this may not be right.
I have said this before and I don't recall anyone saying I was wrong.
Here goes.
When a iron cored inductor is sitting idle with no current flowing (OFF) there is no flux in the iron core and effectively the dormant inductor is an air cored inductor.
At the moment of start up the voltage is zero and the current is zero.
We have inductor (primary) resistance and Zero volts = Zero current.
As voltage rises, current flows through the primary resistance and starts to increase. This increase in current is resisted by the air cored inductor.
For the first few milliseconds after switch on the mains electrical flow is determined by the mains voltage, the mains resistance, the air cored inductor and the primary resistance of that air cored inductor.
This is the time during which the soft start is required to limit the peak current that will flow before the iron cored inductance has built up to react against the mains voltage.
Has anyone got ideas on the ratio of iron cored inductance of the primary of a transformer compared to that same coil with no iron core in it?
As the mains cycle continues, the flux in the iron core builds up and starts to follow the mains cycle. The transformer has now started. It now behaves as Janneman is telling you.
It's that very short period before the flux builds up that results in the high start up current.
Does this explanation hold water?
Andrew,
I don't understand the air core analogy. An unloaded xformer presents, at the primary, a very high inductance: it's an iron core inductor, you can disregard the secondary. So the unloaded current is very low.
That is also shown in my curves. In the curve where the switch-on point is at zero voltage, the current remains very low until the point where the rectifier diodes on the secondary side start to conduct. This load is transformed to the primary, of course, and you see the prim current rise sharply.
jd
I don't understand the air core analogy. An unloaded xformer presents, at the primary, a very high inductance: it's an iron core inductor, you can disregard the secondary. So the unloaded current is very low.
That is also shown in my curves. In the curve where the switch-on point is at zero voltage, the current remains very low until the point where the rectifier diodes on the secondary side start to conduct. This load is transformed to the primary, of course, and you see the prim current rise sharply.
jd
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Is there a reason why no-one uses a simple lm555 timer with resistor and relay for a slow start?
Dear Andrew,
Thanks for your (as always) detailed explanation. In theory I think you are right. But I remember some test (I will try to find them) on a dutch electro forum that there is flux remained in the core from the shut off point. If the transformer is shut off on the peak, there is no remaining flux, if the transformer is shut off on the zero crossing, there will be flux left, and the transformer is considered magnetized.
With kind regards,
Bas
If you shut off the xformer at 0-crossing, there is no load current on the sec side (the rectifier diodes are shut off, 'sperren') and no remaining flux. So you can then switch on at any time without heavy current.
If you shut of the xformer at the peak, that is where the load current (rectifiers charging the supply cap) is max, so then you have a lot of remaining flux.
A lot of confusion comes from the fact that people think that an amp supply load the xformer continuously, but that is incorrect. Again, look at my measurements.
Andrew is incorrect about the air core; read my post above and look at my posted measurements, they should be clear.
jd
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That doesn't sound quite right, Andrew.
If the coil is wound around an iron core, it is an iron core inductor. Even a very small, very brief current through the coil will generate a magnetic field through that iron. I would think that it would never behave as an air-core inductor, ...
Except if the iron core saturates. Beyond that point the coil is basically air-core and the current will surge high, and fast. So how does this affect starting? It is due to the combination of a few factors at the moment of turn-on. Most transformers are built to operate fairly close to saturation since this means less iron and copper so less cost. The core can is usually left with a magnetic field in it depending on the direction and magnitude of the current through it at the point it was turned off. That is at a random point in the AC cycle. When you flip the switch to apply voltage again, the resulting current may add to this magnetic field (if it subtracts, that just means that it will add to that magnetic field on the next half-cycle). When you add the residual magnetic field to the magnetic field at turn-on, and add the extra magnetic field caused by the inrush current into empty filter capacitors on the secondary, the result is a saturated core, for only a brief time. It is during this brief time when the core is saturated that you see the very high surge of current into the transformer. After several AC cycles, the resisdual magnetic field is dissipated by the new alternating magnetic field (and the filter caps are also charged reducing current demand). So the core comes out of saturation.
(p.s. the contribution of the filter caps is real but not the main reason; a transformer, especially a toroid, will have a turn-on surge current even with open secondaries).
If the coil is wound around an iron core, it is an iron core inductor. Even a very small, very brief current through the coil will generate a magnetic field through that iron. I would think that it would never behave as an air-core inductor, ...
Except if the iron core saturates. Beyond that point the coil is basically air-core and the current will surge high, and fast. So how does this affect starting? It is due to the combination of a few factors at the moment of turn-on. Most transformers are built to operate fairly close to saturation since this means less iron and copper so less cost. The core can is usually left with a magnetic field in it depending on the direction and magnitude of the current through it at the point it was turned off. That is at a random point in the AC cycle. When you flip the switch to apply voltage again, the resulting current may add to this magnetic field (if it subtracts, that just means that it will add to that magnetic field on the next half-cycle). When you add the residual magnetic field to the magnetic field at turn-on, and add the extra magnetic field caused by the inrush current into empty filter capacitors on the secondary, the result is a saturated core, for only a brief time. It is during this brief time when the core is saturated that you see the very high surge of current into the transformer. After several AC cycles, the resisdual magnetic field is dissipated by the new alternating magnetic field (and the filter caps are also charged reducing current demand). So the core comes out of saturation.
(p.s. the contribution of the filter caps is real but not the main reason; a transformer, especially a toroid, will have a turn-on surge current even with open secondaries).
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Yes, but not if it was switched off at zero current; in that case there would be no remanent flux.
That was the case with my measurements, I always switched off at zero current to avoid the inrush current at next switch-on.
jd
I alluded to that with "The core is usually left with a magnetic field in it depending on the direction and magnitude of the current through it at the point it was turned off.".
so far the majority don't like my explanation.
To those two I ask.
Do the normal phase and current and voltage that apply to an iron cored inductor apply when a continuous sinewave stream of cycles are passed through the inductor?
The reactance XL = 2PiFL, Omega = 2 Pi F, L= Henries.
As far as I know this formula for reactance and the 90degree phase lag only applies with a continuous stream of sinewave cycles.
If we were to apply a half cycle, do any of the normal iron cored inductor laws hold true?
If we were to apply quarter wave cycle and cut off the voltage at the peak of the cycle, Then what applies in this non continuous pulse operation.
Now look at air cored inductors.
The back emf is directly proportional to the change in current. A zero current produces no back emf. A constant current produces no back emf.
For a saw tooth waveform where we would have an increasing or decreasing current at all times the back emf would change sign at each change in slope of the current flow. If we apply a single pulse that still holds true.
If we now put an iron core in that inductor, does a single pulse back emf equal the iron cored inductance times omega or does the iron core have a near zero effect for that single pulse and the initial reactance ignores the iron core and the inductor behaves as if there was an air core.
If the iron core has instant effect then the transformer will never blow fuses at switch on, with or without a load. The transformer could switch on and run with a tiny fuse that matches the transformer losses when the secondary is open circuit. But this is not the case.
Please offer another explanation that does not involve ignoring the iron core for single pulse operation.
To those two I ask.
Do the normal phase and current and voltage that apply to an iron cored inductor apply when a continuous sinewave stream of cycles are passed through the inductor?
The reactance XL = 2PiFL, Omega = 2 Pi F, L= Henries.
As far as I know this formula for reactance and the 90degree phase lag only applies with a continuous stream of sinewave cycles.
If we were to apply a half cycle, do any of the normal iron cored inductor laws hold true?
If we were to apply quarter wave cycle and cut off the voltage at the peak of the cycle, Then what applies in this non continuous pulse operation.
Now look at air cored inductors.
The back emf is directly proportional to the change in current. A zero current produces no back emf. A constant current produces no back emf.
For a saw tooth waveform where we would have an increasing or decreasing current at all times the back emf would change sign at each change in slope of the current flow. If we apply a single pulse that still holds true.
If we now put an iron core in that inductor, does a single pulse back emf equal the iron cored inductance times omega or does the iron core have a near zero effect for that single pulse and the initial reactance ignores the iron core and the inductor behaves as if there was an air core.
If the iron core has instant effect then the transformer will never blow fuses at switch on, with or without a load. The transformer could switch on and run with a tiny fuse that matches the transformer losses when the secondary is open circuit. But this is not the case.
Please offer another explanation that does not involve ignoring the iron core for single pulse operation.
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