Perry, the inductor was shorted when I got the amp.
AndrewT, I'm not sure I understand your question. I have oscillation through the inductor, at idle with no audio input. Is that what you're asking?
AndrewT, I'm not sure I understand your question. I have oscillation through the inductor, at idle with no audio input. Is that what you're asking?
There is essentially no current flowing through the inductors at idle. The two inductors are in series in this amp. One is cool, the other heats up.
Then an inductor that is passing zero current cannot get hot.
I ask again:
Is it an inductor, or a transformer?
I ask again:
Is it an inductor, or a transformer?
It's an inductor. It's part of the output filter in a class D amp.
If the core material isn't right, it can get hot with the amp at idle.
If the core material isn't right, it can get hot with the amp at idle.
The ±80v carrier wave is reduced by half in each inductor. The good cores don't heat up. The cores that previously overheated have likely changed enough so that they're now being driven into saturation.
There is essentially no current at idle. Current isn't an issue at idle. The heating isn't from copper losses, it's due to core losses.
Is the core loss due to changing field in the core?
Is the changing field due to changing current in the winding/s?
Now tell us how the core can heat up if "There is essentially no current flowing through the inductors at idle"
Is the changing field due to changing current in the winding/s?
Now tell us how the core can heat up if "There is essentially no current flowing through the inductors at idle"
I'm not sure what the purpose of your question is. If you're saying that I'm wrong, explain why (I'm no magnetics guru). If you're trying to understand it better, say so.
It's both.
I need to ask myself simple questions to be able to analyse what I see/hear.
I present these questions hoping that answers come back that I can understand.
To it seems very simple. If there is zero current through the inductor then it can't heat up.
Conversely, if there is current through the inductor then there will be two sets of losses. The core loss and the winding loss.
I need to ask myself simple questions to be able to analyse what I see/hear.
I present these questions hoping that answers come back that I can understand.
To it seems very simple. If there is zero current through the inductor then it can't heat up.
Conversely, if there is current through the inductor then there will be two sets of losses. The core loss and the winding loss.
Ideally, a reactive component (inductor or capacitor) stores and releases 100% of the energy driven into it. Of course, in the real world, there are no ideal components.
Think of the inductor as the primary side of a transformer. With no load on the secondary of a transformer, there is 'essentially' no current through the primary. If you load the secondary, then you get a proportional current through the primary. When the inductor's core is lossy, it acts like a secondary winding to some degree. Across that winding there is essentially a resistance. The effective resistance would be very high for low loss cores and lower resistance for higher loss cores.
The current that it takes to heat the core can be very small. That's why I stated that there was essentially no current through the inductor. The current that it would take to heat the inductor due to resistive/copper losses would be MUCH greater.
Think of the inductor as the primary side of a transformer. With no load on the secondary of a transformer, there is 'essentially' no current through the primary. If you load the secondary, then you get a proportional current through the primary. When the inductor's core is lossy, it acts like a secondary winding to some degree. Across that winding there is essentially a resistance. The effective resistance would be very high for low loss cores and lower resistance for higher loss cores.
The current that it takes to heat the core can be very small. That's why I stated that there was essentially no current through the inductor. The current that it would take to heat the inductor due to resistive/copper losses would be MUCH greater.
If there is no current there is no heating.
That applies to both the core loss and the winding loss, unless my little understanding is way off.
That applies to both the core loss and the winding loss, unless my little understanding is way off.
There is a very small current. A coil of copper wire with no current applied to it, will create a magnetic field. Pass a magnet through that field and it will induce a very small current. Micro amps. But it's current, no the less.
Perry, I contacted Bytemark about these inductors. They were not helpful. He told me engineering could put something together for me, for about $500.00. I've tried to read up on figuring wire diameter and number of turns. But I'm just not getting it. Can you give me some idea on how to make these? Or could I pay you to make them?
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