Why do ultra-thin (40awg) secondaries burn out ?
Patrick, Sun Dec 30 2018, 09:57PMId like some input before i jump off into this SMPS you have all helped with.
Before making any permanent design commitments, why is it 40 awg wire in flybacks and iggies burn out ?
I assume :
First, there is burn out from over-voltage shorting. (ignore this case)
Second, there is burn out from current only.
Im winding my secondary out of double build 28 Awg (~0.004") vs 40 awg or finer (0.003") ive seen what i think is 45 or thinner when ive smashed open some coils.
So is it overall heating internally of the coil that becomes non-uniform then the copper picks that spot to scorch itself ? or is there a output voltage drop with infinite current through a spark gap that forces all the wattage across the coil, instead of a external wisely placed resistor ?
Re: Why do ultra-thin (40awg) secondaries burn out ?
Bjørn, Mon Dec 31 2018, 07:21PM
I don't know if it is relevant in this case but during the old experiments with lifters it was clear that very thin wire would cause the insulation to break down because of higher field strengths around very thin wire.
Bjørn, Mon Dec 31 2018, 07:21PM
I don't know if it is relevant in this case but during the old experiments with lifters it was clear that very thin wire would cause the insulation to break down because of higher field strengths around very thin wire.
Re: Why do ultra-thin (40awg) secondaries burn out ?
Patrick, Tue Jan 01 2019, 07:11PM
Im still reading PDFs but It may be due to proximity effect, which is different than skin effect.
Patrick, Tue Jan 01 2019, 07:11PM
Im still reading PDFs but It may be due to proximity effect, which is different than skin effect.
Re: Why do ultra-thin (40awg) secondaries burn out ?
Mr.Warwickshire, Tue Jan 29 2019, 06:36AM
Decreasing the diameter (AWG/size) of the wire will increase the voltage potential between winding layers. This is because one may fit much more turns in the same space, and with a constant value for volts per turn, more voltage is introduced in that same space. This leads me to another possible reason, which you mentioned. Because you are using thinner wire, it heats up more easily due to a multitude of factors. For one, thinner wire is again denser, and thereby heatsoaks more quickly. Thinner wire also has less ampacity and thermal mass, which is much more easily affected by joule heating. Field strengths may also play a factor, however I believe these issues are more significant.
Mr.Warwickshire, Tue Jan 29 2019, 06:36AM
Decreasing the diameter (AWG/size) of the wire will increase the voltage potential between winding layers. This is because one may fit much more turns in the same space, and with a constant value for volts per turn, more voltage is introduced in that same space. This leads me to another possible reason, which you mentioned. Because you are using thinner wire, it heats up more easily due to a multitude of factors. For one, thinner wire is again denser, and thereby heatsoaks more quickly. Thinner wire also has less ampacity and thermal mass, which is much more easily affected by joule heating. Field strengths may also play a factor, however I believe these issues are more significant.
Re: Why do ultra-thin (40awg) secondaries burn out ?
radiotech, Tue Jan 29 2019, 11:24PM
Quote:
So is it overall heating internally of the coil that becomes non-uniform then the copper picks that spot to scorch itself ? or is there a output voltage drop with infinite current through a spark gap that forces all the wattage across the coil, instead of a external wisely placed resistor ?
Your terminology about wattage is interesting.
The maximum power transfer theorem predicts that in a liven load.
when the conjugate of the source, equals the load, maximum power will
be transferred.
In a dynamic situation like an arc, heat may have an influence on
the source impedance, and it may tend to sustain the arc.
In real situations, the arc may, or may not blow itself out.
As to copper wire failure most insulation is reacted upon by copper.
It is why aluminum windings were chosen because of oxide on aluminum.
Also this, copper windings are most often 'glued' in place by dipping
but copper moves as it heats.
Eventually any copper winding will fail because of these factors. How long
it lasts under the control of the designer. i.e. why does a telephone
repeating coil last 100 years ?
radiotech, Tue Jan 29 2019, 11:24PM
Quote:
So is it overall heating internally of the coil that becomes non-uniform then the copper picks that spot to scorch itself ? or is there a output voltage drop with infinite current through a spark gap that forces all the wattage across the coil, instead of a external wisely placed resistor ?
Your terminology about wattage is interesting.
The maximum power transfer theorem predicts that in a liven load.
when the conjugate of the source, equals the load, maximum power will
be transferred.
In a dynamic situation like an arc, heat may have an influence on
the source impedance, and it may tend to sustain the arc.
In real situations, the arc may, or may not blow itself out.
As to copper wire failure most insulation is reacted upon by copper.
It is why aluminum windings were chosen because of oxide on aluminum.
Also this, copper windings are most often 'glued' in place by dipping
but copper moves as it heats.
Eventually any copper winding will fail because of these factors. How long
it lasts under the control of the designer. i.e. why does a telephone
repeating coil last 100 years ?
Re: Why do ultra-thin (40awg) secondaries burn out ?
Patrick, Wed Jan 30 2019, 05:55AM
The arc has nearly zero resistance right ? once struck. The voltage is high across a switch when open, then drops when closed.
So it seems the coil sees high voltage times the series arc current (V x I = W). or so Kirchhoff tells me.
But im thinking its mostly proximity effect. i need some calculations and the ability to post pics to pursue it further.
Patrick, Wed Jan 30 2019, 05:55AM
radiotech wrote ...
why does a telephone repeating coil last 100 years ?
It was well designed and not abused.why does a telephone repeating coil last 100 years ?
The arc has nearly zero resistance right ? once struck. The voltage is high across a switch when open, then drops when closed.
So it seems the coil sees high voltage times the series arc current (V x I = W). or so Kirchhoff tells me.
But im thinking its mostly proximity effect. i need some calculations and the ability to post pics to pursue it further.
Re: Why do ultra-thin (40awg) secondaries burn out ?
johnf, Wed Jan 30 2019, 06:12PM
Rubbish due to the inductance the voltage and current are out of step and there fore voltage can be many times higher. In inductors current lags so takes time to build up. Conversely breaking a connection the inductor wants to keep the current flowing and will raise the voltage in an attempt to make this happen (inductive kick). so drawing arcs causes dVdt voltage spikes that will overcome the insulation causing shorted turns
johnf, Wed Jan 30 2019, 06:12PM
Rubbish due to the inductance the voltage and current are out of step and there fore voltage can be many times higher. In inductors current lags so takes time to build up. Conversely breaking a connection the inductor wants to keep the current flowing and will raise the voltage in an attempt to make this happen (inductive kick). so drawing arcs causes dVdt voltage spikes that will overcome the insulation causing shorted turns
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