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ESL wiresHi
Just a quick question about the proper wire to use in hooking secondary side of transformer to stators. I will be using at most 300W power amplifier for the panels, I think the peak to peak voltage is 160V? So then the current rating should be 300/160 amps? The transformer is involved which will change the amps output. What current and voltage rating should I consider? My bias will be no more then 4-5kV, I hope. Thanks Bryan |

Hi,
the power rating is the product of rms-values. 160Vp-p translates to 80Vp translates to ~57Vrms (80Vp/SQR2). So the ampere-rating is 300/57[A] ~5Arms. This translates to 7Ap, hence 14Ap-p. The transformer should be designed after the ´needs´ of the panel. By this I mean that You first have a look at how to match the impedance values. The impedance values are transformed by U˛ or 1/U˛ (U the transformation-factor). Lets say the panel measures around 1000pF of capacitance from stator to stator. At a high frequency of say 10kHz this calculates to X~16kOhm after the formula: X=(1/(2*pi*f*C). If Your amp is to see a minimum load of 1Ohms the transformation factor U must be SQR(16k/8) ~45:1. Now check if the voltage requirments are sufficient to drive the panel. The maximum bias voltage you can use is the maximum peak signal voltage. In case the signal voltage becomes higher than the bias voltage the panel changes from an audio transducer to a frequency doubler ;-). Lets assume that You use a stator-stator distance of 3mm, hence 2*1.5mm. After applying the bias voltage the membrane is deflected towards one of the stators. Depending on the mechanical tension the deflection maybe 0.5mm, thereby leaving 1mm and 2mm of free space left. Under normal conditions 1mm of air can withstand ~1.5kV-1.7kV of voltage (without any noise from discharging effects, etc.). Add some losses within the stator insulation and you may end up with a voltage of ~2.5kVdc at the output of the bias supply. 2.5kVdc of bias voltage equals 2.5kV of peak signal voltage. This translates to ~1.8kVrms or 5kVp-p signal voltage. 1.8kVrms signal voltage divided by the U of the transformer (45:1) results in 40Vrms. So the suggested 300W amp is sufficient to drive the panel and still got a bit of headroom. You can fudge a bit with the values (especially the lowest allowed impedance values at which frequency), but this set of values has worked well for me. You will see that working and designing this way everything fits perfectly togeteher in the end -the mechanical dimensions of the panel as well as the electrical values and the drive requirements. Keep in mind that flat segmented panels (wire stators) typically show much smaller capacitance values as unsegmented and curved panels (metal sheet stators). Therefore segmented panels can -and have to- work with much higher transformation factors. While this in praxis somehow doesn´t translate into better efficiency (!!) it easens the stress on the amplifier, because it offers a more constant impedance load. On the other hand it should be one of the primary goals to keep the U as small as possible! jauu Calvin |

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