Electronic – Correct EL Driver Rating for 8 Channel Sequencer

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I'm trying to use an 8 channel EL sequencer (found here) to control eight 10-foot segments of EL wire. I'm having trouble figuring out what the appropriate length rating is that I need on an inverter. I've read conflicting opinions. One states I need my inverter to handle the "total" length (in my case 80 feet), while another says it just needs to handle the longest strand (10 feet then). Which is correct?

I'm looking at this one currently (rated 5 to 15 feet). Is that enough? Or do I need to find one rated for at least 80 feet? Thanks.

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If I need an inverter that can handle 80 feet of EL wire then I'm worried that this will conflict with a sequencer. If the sequencer turns off say 3 channels (30 feet of the wire), then my load would only be 50 feet I would think. I've read that this is an issue because the inverter might break if I have less than the minimum load. Is this the case?

Best Answer

Lady Ada did a great write up

EL wire modeling

EL wire is not a resistive light (like an incandescent bulb) and it is not a diode light (like an LED), it acts like more like a capacitor! The stiff inner wire is one 'plate' of the capacitor, the corona wire is the other 'plate' and the phosphor coating being the insulator/dielectric (for more details on capacitors, see Wikipedia ). This means you cannot use dimming methods such as triac/chopping for resistive incandescents or PWM for LEDs.

In terms of thinking of how EL wire 'acts' you should model it as a capacitor that increases with the length of the wire. It is not a perfect capacitor, there is also some leakage which we will model as a resistor.

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Adding another meter, we duplicate the RC model in parallel alt text Of course, we can simplify by calculating the new capacitance and reistance. Remember that capacitance increases in parallel and resistance decreases

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The capacitance and resistance per meter depends on the 'thickness' of the EL wire, the brand and make, the voltage and frequency applied

Current draw

We can use this information to determine the power draw.

Assuming you have LyTec EL wire, 2.3mm diameter 'standard'...if have one meter, that is 6nF and 100KΩ in parallel. The capacitance has an impedance of 1/(2πfC) so at 2000 Hz, the impedence per meter is 12 KΩ, in parallel with 100 KΩ it is 11 KΩ total. For a 100V AC power source, the current draw is 100V/11KΩ = 9mA per meter. 100V * 9mA/meter = 0.9 Watts/meter!

If you are using our 'high brightness, long life' stuff, its about 1.5 Watts per meter.

Thus an inverter with a 100mA output capability can drive 10M or so of LyTec and 5M of 'high brightness' EL. The transformer and transistors used in an inverter are a big part of how much current an inverter can provide!

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