<div dir="ltr">It should work as far as I know. I would try it out as resistors aren't that expensive.<br>You could also use a potentiometer instead of the resistor.<br>Good luck!<br>Moshe<br><br><div class="gmail_quote">
On Tue, Jan 18, 2011 at 12:16 AM, Renato <span dir="ltr"><<a href="mailto:rennabh@gmail.com">rennabh@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin: 0pt 0pt 0pt 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">
<div class="im">On Mon, 17 Jan 2011 12:41:07 +0200<br>
Moshe Werner <<a href="mailto:moshwe@gmail.com">moshwe@gmail.com</a>> wrote:<br>
<br>
> Hi renato,<br>
><br>
> only saw the question now, so sorry for the late reply.<br>
> The input voltage of soundcards depends on the headroom of the AD of<br>
> your soundcard. (e.g. Dynamic Range).<br>
> Pro and semi-pro cards often have a line level of +4dBu (1.227V RMS,<br>
> 3.472V Peak) plus a bit headroom.<br>
> Hope this helps.<br>
> Moshe<br>
><br>
<br>
</div>Yes, thanks. I might just try sticking in my USB function generator and<br>
go up with voltage untill the sound card clips...<br>
<br>
Also, I realized that putting a zener diode in parallel will just clip<br>
the signal at the diode voltage, so it won't solve the problem.<br>
<br>
A simple resistor in series with the piezo (of appropriate value)<br>
should work, shouldn't it?<br>
<font color="#888888"><br>
renato<br>
</font></blockquote></div><br></div>