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Registered Member #2662
Joined: Fri Jan 29 2010, 10:14AM
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Posts: 36
What happens to the impedance of a transformer’s primary when it is driven at the self resonant frequency of the secondary winding? Experience using ignition coils and audio amplifiers suggests that the impedance tends to zero (and destroys the amp), PSpice also shows a high primary current when driven at the secondary Fres… Could anyone enlighten me as to why? If the primary winding was also resonant that would make sense, but I don’t see how it can be given that it’s a few turns and no capacitance is added.
Registered Member #2463
Joined: Wed Nov 11 2009, 03:49AM
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Posts: 1546
If a capacitor is connected in series with a coil to achieve resonance, the impedance is the combination of the combined lossy parts of the coil (just the DC R and core losses with core driven toward saturation), and the ESR of the capacitor. Z=R
At this stage, the voltages across the coil will be high and equal to the capacitor voltage, but the two will cancel and the applied voltage will be low. The coil voltage will be I *6.28fl.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
The impedance of the secondary tends to zero at resonance, that means that it's the same as a shorted turn. So visualize your ignition coil as a transformer with a 1.7 ohm primary DCR and a shorted secondary. That's a horrible reactive load that would kill almost any cheap audio amp.
Registered Member #2662
Joined: Fri Jan 29 2010, 10:14AM
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Cheers... I get it now, I knew the secondary impedance approached zero, should have realised this represents a shorted turn! When such a transformer is driven by a square wave (say from a half bridge) then this issue isn't quite so bad, is this because the square wave has many harmonics above the resonant frequency, hence the transformer appear as a slightly higher impedance?
Registered Member #2463
Joined: Wed Nov 11 2009, 03:49AM
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This statement : So visualize your ignition coil as a transformer with a 1.7 ohm primary DCR and a shorted secondary
and this statement : ... I get it now, I knew the secondary impedance approached zero, should have realised this represents a shorted turn! When such a transformer is driven by a square wave
The first statement does imply the second.
I you drive the core to saturation with an open secondary it is the same as driving the core to saturation with no secondary winding at all. True, the secondary will have high voltage across it, however that voltage can in no way influence the magnetization because no secondary current is flowing.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
iJim wrote ...
When such a transformer is driven by a square wave (say from a half bridge) then this issue isn't quite so bad, is this because the square wave has many harmonics above the resonant frequency, hence the transformer appear as a slightly higher impedance?
No. (Well IMO :p ) It's because an inverter that generates a square wave is "non-dissipative". The energy stored in the inductor can just return to the DC bus through your MOSFETS' antiparallel diodes. But in an audio amp, the stored inductive energy mostly gets dissipated as heat in the output transistors.
So, an inverter driving a nasty inductive load will just draw very little real power from the DC bus, because it can recycle the reactive power. An audio amp will draw lots of real power, dissipate all of it in its own output devices, and burn itself out.
Better quality amps have protection circuits that limit the dissipation to a safe level, no matter how bad the load is. When this kind of protection kicks in on an inductive load, the leftover energy gets recycled to the DC bus, causing massive distortion of the waveform that some audio pundits tried to blame for "transistor sound".
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Impedance of transformer primary at resonance?
