How to wire lots of piezo tweeters?

Thanks guys!!! I did have a feeling it was some sort of hp filter.

"I wouldnt remove the cap, its there for a reason. What that reason is Im not sure but best to be on the safe side."

Total hogwash !!!

Throw it away and do it right.

If you don't understand something, ask.

Sorry for the late reply and thanks djk for the very helpful information. At the moment  we dont have time to wire the piezos but will do soon, one more question do we have to remove the cap that is already soldered on the piezos ?

Thanks for your time

The dog ate my homework (error message when I tried to post).

The figure #s I will reference in this post are here:

http://www.pulsardevelopments.com/products/detail/piezoan.html

The Motorola/CTS  KSN 1005 (orange squeezer) tweeters use the 0.13µF bender element, and are about 62R at 20Khz, increasing to about 250R in the 5Khz region (fig.7).

Unknown tweeters should be measured for capacitance for best results, most I have seen are about 0.1µF, or about 82R at 20Khz.

The LeSon (and a few others) have a matching transformer to make them more sensitive, and are outside the scope of this post.

The first thing to do is select a series resistor, I usually use one the same value as the capacitive reactance at 20Khz (50R~100R), 3W~10W capacity. For a 0.1µF bender element (the $1 generic piezo), 82R/5W is fine.

As Centauri suggests, hook them together in series pairs, and then hook the five pairs in parallel (fig.14). For a 0.1µF bender element this would give about a 0.25µF total load with about 32R8 in series, or about 64 ohms at 20Khz.

Crossovers (fig.10)

A passive crossover is possible with a few added parts. For our example of ten series/parallel connected tweeters we know we have about 0.25µF in series with 32R8 for the impedance. This is a far cry from a resistive load, so a normal crossover will not work. The easiest solution is to try and make the load look resistive. A 0.254mH inductor has about 32 ohms of reactance at 20Khz, about the same as our 0.25µF capacitor. If we wire a 33R/50W resistor in series with the 0.254mH inductor, and put that combination in parallel with the tweeter load of 0.25µF (in series with32R8), the the amplifier will have what looks like about a 33R resistive load to drive.

A 12dB/7Khz crossover for 32R might be 0.5µF and 1mH. Lower crossovers may be possible, but they will require a steeper slope, and a notch filter for the bending mode resonance (the impedance peak in the fig.7 graph)

Be sure and buy some extra units when using generics, and sort out the ones with no or low output.



Edited by _djk_ - 21 March 2010 at 9:35am

[QUOTE=2bitNum] Sorry my mistake with that picture, i want to wire ten units in one box

this what i wanted to post
[QUOTE=2bitNum]

The configuration above shows 2 sets of 4 series up which would be 32 ohms each and 1 set of 2 series up which would be 16 ohms. These are then paralleled down to give a final resistance of 6.4 ohms. The problem is that the set of 2 will receive twice as much power as the sets of 4 and at least one of them is likely to be damaged.

What I suggested before would share the power equally and running at 3.2 ohms per channel would be no problem to amps capable of running down to 2 ohms per channel, especially at high frequencies.

Didn't think piezo's loaded an amp, arn't they usually like 1000ohms or something?